CHAPTER 1 Section 1.1 Solutions -------------------------------------------------------------------------------1. Solve for x:
1 x = 2 360
2. Solve for x:
360 = 2 x, so that x = 180 .
3. Solve for x: −
360 = 4x, so that x = 90 .
1 x = 3 360
4. Solve for x: −
360 = −3x, so that x = −120 . (Note: The angle has a negative measure since it is a clockwise rotation.)
5. Solve for x:
1 x = 4 360
2 x = 3 360
720 = 2(360 ) = −3x, so that x = −240 . (Note: The angle has a negative measure since it is a clockwise rotation.)
5 x = 6 360
6. Solve for x:
7 x = 12 360
1800 = 5(360 ) = 6x, so that x = 300 .
2520 = 7(360 ) = 12x, so that x = 210 .
7. Solve for x: −
4 x = 5 360 1440 = 4(360 ) = −5x, so that
8. Solve for x: −
x = −288 . (Note: The angle has a negative measure since it is a clockwise rotation.)
x = −200 . (Note: The angle has a negative measure since it is a clockwise rotation.)
9.
10.
a) complement: 90 − 18 = 72
5 x = 9 360 1800 = 5(360 ) = −9x, so that
a) complement: 90 − 39 = 51
b) supplement: 180 − 18 = 162
b) supplement: 180 − 39 = 141
11.
12.
a) complement: 90 − 42 = 48
a) complement: 90 − 57 = 33
b) supplement: 180 − 42 = 138
b) supplement: 180 − 57 = 123
1
Chapter 1
13.
14.
a) complement: 90 − 89 = 1
a) complement: 90 − 75 = 15
b) supplement: 180 − 89 = 91
b) supplement: 180 − 75 = 105
15. Since the angles with measures ( 4x ) and ( 6x ) are assumed to be
complementary, we know that ( 4x ) + ( 6x ) = 90. Simplifying this yields
(10x ) = 90 , so that x = 9. So, the two angles have measures 36 and 54 .
16. Since the angles with measures ( 3x ) and (15x ) are assumed to be
supplementary, we know that ( 3x ) + (15x ) = 180. Simplifying this yields
(18x ) = 180 , so that x = 10. So, the two angles have measures 30 and 150 .
17. Since the angles with measures ( 8x ) and ( 4x ) are assumed to be
supplementary, we know that ( 8x ) + ( 4x ) = 180. Simplifying this yields
(12x ) = 180 , so that x = 15. So, the two angles have measures 60 and 120 .
18. Since the angles with measures ( 3x + 15 ) and (10x + 10 ) are assumed to be
complementary, we know that ( 3x + 15 ) + (10x + 10 ) = 90. Simplifying this yields
(13x + 25 ) = 90 , so that (13x ) = 65 and thus, x = 5. So, the two angles have measures 30 and 60 . 19. Since α + β + γ = 180 , we know that
20. Since α + β + γ = 180 , we know that
117 + 33 + γ = 180 and so, γ = 30 .
110 + 45 + γ = 180 and so, γ = 25 .
21. Since α + β + γ = 180 , we know that 4 β ) + β + ( β ) = 180 and so, β = 30. (
22. Since α + β + γ = 180 , we know that 3β ) + β + ( β ) = 180 and so, β = 36. (
= 150
= 155
= 6β
= 5β
Thus, α = 3β = 108 and γ = β = 36 .
Thus, α = 4 β = 120 and γ = β = 30 .
2
Section 1.1
23. α = 180 − ( 53.3 + 23.6 ) = 103.1
24. β = 180 − (105.6 + 13.2 ) = 61.2
25. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 4 2 + 32 = c 2 , which
simplifies to c 2 = 25, so we conclude that c = 5 . 26. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 32 + 32 = c 2 , which
simplifies to c 2 = 18, so we conclude that c = 18 = 3 2 . 27. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 6 2 + b 2 = 10 2 , which
simplifies to 36 + b 2 = 100 and then to, b2 = 64, so we conclude that b = 8 . 28. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes a 2 + 7 2 = 12 2 , which
simplifies to a 2 = 95, so we conclude that a = 95 . 29. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 82 + 52 = c 2 , which
simplifies to c 2 = 89, so we conclude that c = 89 . 30. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 6 2 + 52 = c 2 , which
simplifies to c 2 = 61, so we conclude that c = 61 . 31. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 7 2 + b 2 = 112 , which
simplifies to b 2 = 72, so we conclude that b = 72 = 6 2 . 32. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes a 2 + 52 = 92 , which
simplifies to a 2 = 56, so we conclude that a = 56 = 2 14 .
3
Chapter 1
33. Since this is a right triangle, we know from the Pythagorean Theorem that
a 2 + b2 = c 2 . Using the given information, this becomes a 2 +
( 7 ) = 5 , which 2
2
simplifies to a 2 = 18, so we conclude that a = 18 = 3 2 . 34. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 52 + b 2 = 10 2 , which
simplifies to b 2 = 75, so we conclude that b = 75 = 5 3 . 35. If x = 10 in., then the hypotenuse of this triangle has length
36. If x = 8 m, then the hypotenuse of
this triangle has length 8 2 ≈ 11.31 m .
10 2 ≈ 14.14 in.
37. Let x be the length of a leg in the given 45 − 45 − 90 triangle. If the hypotenuse of this triangle has length 2 2 cm, then 2 x = 2 2, so that x = 2.
Hence, the length of each of the two legs is 2 cm . 38. Let x be the length of a leg in the given 45 − 45 − 90 triangle. If the hypotenuse 10 10 of this triangle has length 10 ft., then 2 x = 10, so that x = = = 5. 2 2
Hence, the length of each of the two legs is
5 ft.
39. The hypotenuse has length 2 4 2 in. = 8 in.
40. Since
(
)
2 x = 6m x = 6 2 2 = 3 2m,
each leg has length 3 2 m.
41. Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and 3 x, the shorter leg must have length x. Hence, using the given information, we
know that x = 5 m. Thus, the two legs have lengths 5 m and 5 3 ≈ 8.66 m, and the hypotenuse has length 10 m. 42. Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and 3 x, the shorter leg must have length x. Hence, using the given information, we
know that x = 9 ft. Thus, the two legs have lengths 9 ft. and 9 3 ≈ 15.59 ft., and the hypotenuse has length 18 ft.
4
Section 1.1
43. The length of the longer leg of the given triangle is
3x = 12 yards. So,
12 12 3 = = 4 3. As such, the length of the shorter leg is 4 3 ≈ 6.93 yards, 3 3 and the hypotenuse has length 8 3 ≈ 13.9 yards.
x=
44. The length of the longer leg of the given triangle is x=
3x = n units. So,
n n 3 n 3 = units, and the . As such, the length of the shorter leg is 3 3 3
hypotenuse has length
2n 3 units. 3
45. The length of the hypotenuse is 2 x = 10 inches. So, x = 5. Thus, the length of the shorter leg is 5 inches, and the length of the longer leg is 5 3 ≈ 8.66 inches. 46. The length of the hypotenuse is 2 x = 8 cm. So, x = 4. Thus, the length of the shorter leg is 4 cm, and the length of the longer leg is 4 3 ≈ 6.93 cm. 47. For simplicity, we assume that the minute hand is on the 12. Let α = measure of the desired angle, as indicated in the diagram below. Since the measure of the angle formed using two rays emanating from the center 1 of the clock out toward consecutive hours is always 360 ) = 30 , it immediately ( 12
follows that α = 4 ⋅ ( −30 ) = −120 (Negative since measured clockwise.)
α
5
Chapter 1
48. For simplicity, we assume that the minute hand is on the 9. Let α = measure of the desired angle, as indicated in the diagram below. Since the measure of the angle formed using two rays emanating from the center 1 of the clock out toward consecutive hours is always 360 ) = 30 , it immediately ( 12
follows that α = 5 ⋅ ( −30 ) = −150 . (Negative since measured clockwise.)
α
49. The key to solving this problem is setting up the correct proportion. Let x = the measure of the desired angle. From the given information, we know that since 1 complete revolution corresponds to 360 , we obtain the following proportion: 360 x = 30 minutes 12 minutes
Solving for x then yields 360 x = 12 minutes = 144 . 30 minutes
(
)
50. The key to solving this problem is setting up the correct proportion. Let x = the measure of the desired angle. From the given information, we know that since 1 complete revolution corresponds to 360 , we obtain the following proportion: 360 x = 30 minutes 5 minutes
Solving for x then yields 360 x = 5 minutes = 60 . 30 minutes
(
)
6
Section 1.1
51. We know that 1 complete revolution corresponds to 360. Let x = time (in minutes) it takes to make 1 complete revolution about the circle. Then, we have the following proportion: 270 360 = 45 minutes x Solving for x then yields 270 x = 360 ( 45 minutes ) x=
360 ( 45 minutes )
= 60 minutes. 270 So, it takes one hour to make one complete revolution. 52. We know that 1 complete revolution corresponds to 360 . Let x = time (in minutes) it takes to make 1 complete revolution about the circle. Then, we have the following proportion: 72 360 = 9 minutes x Solving for x then yields 72 x = 360 ( 9 minutes ) x=
360 ( 9 minutes )
= 45 minutes. 72 So, it takes 45 minutes to make one complete revolution. 53. Let d = distance (in feet) the dog runs along the hypotenuse. Then, from the Pythagorean Theorem, we know that 30 2 + 80 2 = d 2 7,300 = d 2 85 ≈ 7,300 = d
So, d ≈ 85 feet . 54. Let d = distance (in feet) the dog runs along the hypotenuse. Then, from the Pythagorean Theorem, we know that 252 + 100 2 = d 2 10,625 = d 2 103 ≈ 10,625 = d
So, d ≈ 103 feet .
7
Chapter 1
55. Consider the following triangle T:
56. Consider the following triangle T:
Since T is a 30° − 60° − 90° triangle, the side opposite the 30° is of the form x, while the side opposite the 60° angle is of the form 3x. This gives us 400 3x = 400, so that x = ≈ 230.9 m. 3
Since T is a 30° − 60° − 90° triangle, the side opposite the 30° is of the form x, while the side opposite the 60° angle is of the form 3x. This gives us x = 400, so that
3x = 3 ( 400 ) ≈ 692.8 m.
57. Consider the following triangle T.
45
Since T is a 45 − 45 − 90 triangle, the two legs (i.e., the sides opposite the angles with measure 45 ) have the same length. Call this length x. Since the hypotenuse of such a triangle has measure
2 x, we have that 2 x = 100, so that
100 100 2 = = 50 2. 2 2 So, since lights are to be hung over both legs and the hypotenuse, the couple should buy 50 2 + 50 2 + 100 = 100 + 100 2 ≈ 241 feet of Christmas lights. x=
8
Section 1.1
58. Consider the following triangle T.
45
45
Since T is a 45 − 45 − 90 triangle, the two legs (i.e., the sides opposite the angles with measure 45 ) have the same length. Call this length x. Since the hypotenuse of such a triangle has measure
2 x, we have that
2x = 60, so that
60 60 2 = = 30 2. 2 2 So, since lights are to be hung over both legs and the hypotenuse, the couple should buy 30 2 + 30 2 + 60 = 60 + 60 2 ≈ 145 feet of Christmas lights. x=
59. Consider the following diagram:
represents the TREE and the vertices of the triangle The dashed line segment represent STAKES. Also, note that the two right triangles and are congruent (using the Side-Angle-Side Postulate from Euclidean geometry). Let x = distance between the base of the tree and one staked rope (measured in feet).
9
Chapter 1
For definiteness, consider the right triangle . Since it is a 30 − 60 − 90 triangle, the side opposite the 30 -angle (namely ) is the shorter leg, which has length x feet. Then, we know that the hypotenuse must have length 2x. Thus, by the Pythagorean Theorem, it follows that: x 2 + 17 2 = (2 x )2 x 2 + 289 = 4 x 2 289 = 3x 2 289 = x2 3 289 =x 3 So, the ropes should be staked approximately 9.8 feet from the base of the tree. 9.8 ≈
60. Using the computations from Problem 57, we observe that since the length of 289 , it follows that the length of each of the two the hypotenuse is 2x, and x = 3 289 ≈ 19.6299 feet. Thus, one should have 3 2 ×19.6299 ≈ 39.3 feet of rope in order to have such stakes support the tree.
ropes should be 2
61. Consider the following diagram:
The dashed line segment represents the TREE and the vertices of the triangle represent STAKES. Also, note that the two right triangles and are congruent (using the Side-Angle-Side Postulate from Euclidean geometry).
10
Section 1.1
Let x = distance between the base of the tree and one staked rope (measured in feet). For definiteness, consider the right triangle . Since it is a 30 − 60 − 90 triangle, the side opposite the 30 -angle (namely ) is the shorter leg, which has length 10 feet. Then, we know that the hypotenuse must have length 2(10) = 20 feet. Thus, by the Pythagorean Theorem, it follows that: x 2 + 10 2 = 20 2 x 2 + 100 = 400 x 2 = 300 x = 300 ≈ 17.3 feet So, the ropes should be staked approximately 17.3 feet from the base of the tree.
62. Using the computations from Problem 59, we observe that since the length of the hypotenuse is 20 feet, it follows that the length of each of rope tied from tree to the stake in this manner should be 20 feet in length. Hence, for four stakes, one should have 4 × 20 ≈ 80 feet of rope. 63. The following diagram is a view from one of the four sides of the tented area – note that the actual length of the side of the tent which we are viewing (be it 40 ft. or 20 ft.) does not affect the actual calculation since we simply need to determine the value of x, which is the amount beyond the length or width of the tent base that the ropes will need to extend in order to adhere the tent to the ground.
Now, solving this problem is very similar to solving Problem 57. The two right triangles labeled in the diagram are congruent. So, we can focus on the leftmost one, for definiteness. The side opposite the angle with measure 30 is the shorter leg, the length of which is x. So, the hypotenuse has length 2x. From the Pythagorean Theorem, it then follows that x 2 + 7 2 = (2 x )2 49 = 3x 2 4.0 ≈ 73 ≈
11
49 3
=x
Chapter 1
Hence, along any of the four edges of the tent, the staked rope on either side extends approximately 4 feet beyond the actual dimensions of the tent. As such, the actual footprint of the tent is approximately ( 40 + 2(4) ) ft. × ( 20 + 2(4) ) ft., which is 48 ft. × 28 ft. 64. The following diagram is a view from one of the four sides of the tented area – note that the actual length of the side of the tent which we are viewing (be it 80 ft. or 40 ft.) does not affect the actual calculation since we simply need to determine the value of x, which is the amount beyond the length or width of the tent base that the ropes will need to extend in order to adhere the tent to the ground.
Now, solving this problem is very similar to solving Problem 53. The two right triangles labeled in the diagram are congruent. So, we can focus on the leftmost one, for definiteness. Since this is a 45 − 45 − 90 triangle, the lengths of the two legs must be equal. So, x = 7. Hence, along any of the four edges of the tent, the staked rope on either side must extend 7 feet beyond the actual dimensions of the tent. As such, the actual footprint of the tent is approximately ( 40 + 2(7) ) ft. × (80 + 2(7) ) ft. , which is 54 ft. × 94 ft. . 65. The corner is not 90 because 102 + 152 ≠ 20 2. 66. x 2 + 82 = 17 2 x 2 = 225 x = 15 ft. 170 67. The speed is 1700 60 = 6 revolutions per second. Since each revolution
corresponds to 360 , the engine turns ( 170 6 ) ( 360 ) = 10,200 each second.
68. The speed is 300,000 = 20,000 per second. Since each revolution corresponds 15
to 360 , this amounts to 500 9 revolutions per second. Since 1 minute = 60 seconds, 10,000 the speed is 500 ≈ 3,333.3 RPMs. 9 × 60 = 3
12
Section 1.1
69. In a 30 − 60 − 90 triangle, the length opposite the 60 -angle has length 3 × ( shorter leg ) , not 2 × ( shorter leg ) . So, the side opposite the 60 -angle has
length 10 3 ≈ 17.3 inches. 70. The length of the hypotenuse must be positive. Hence, the length must be 5 2 cm. 71. False. Each of the three angles of an equilateral triangle has measure 60. But, in order to apply the Pythagorean theorem, one of the three angles must have measure 90. 72. False. Since the Pythagorean theorem doesn’t apply to equilateral triangles, and equilateral triangles are also isosceles (since at least two sides are congruent), we conclude that the given statement is false. 73. True. Since the angles of a right triangle are α , β , and 90 , and also we know that α + β + 90 = 180 , it follows that α + β = 90. 74. False. The length of the side opposite the 60 -angle is the side opposite the 30 -angle.
3 times the length of
75. True. The sum of the angles α , β ,90 must be 180 . Hence, α + β = 90 , so that α and β are complementary. 76. False. The legs have the same length x, but the hypotenuse has length
2x .
77. True. Angles swept out counterclockwise have a positive measure, while those swept out clockwise have negative measure. 78. True. Since the sum of the angles α , β ,90 must be 180 , α + β = 90. So, neither angle can be obtuse.
13
Chapter 1
79. First, note that at 12:00 exactly, both the minute and the hour hands are identically on the 12. Then, for each minute that passes, the minute hand moves 1 60 the way around the clock face (i.e., 6 ). Similarly, for each minute that passes,
the hour hand moves 601 the way between the 12 and the 1; since there are 1 12
(360 ) = 30 between consecutive integers on the clock face, such movement
corresponds to 601 (30 ) = 0.5. Now, when the time is 12:20, we know that the minute hand is on the 4, but the hour hand has moved 20 × 0.5 = 10 clockwise from the 12 towards the 1. The picture is as follows:
β
α1
α α3
The angle we seek is β + α1 + α 2 + α 3 . From the above discussion, we know that
α1 = α 2 = α 3 = 30 and β = 20. Thus, the angle at time 12:20 is 110 . 80. First, note that at 9:00 exactly, the minute is identically on the 12 and the hour hand is identically on the 9. Then, for each minute that passes, the minute hand moves 601 the way around the clock face (i.e., 6 ). Similarly, for each minute that passes, the hour hand moves 601 the way between the 9 and the 10; since there are 1 12
(360 ) = 30 between consecutive integers on the clock face, such movement
corresponds to 601 (30 ) = 0.5. Now, when the time is 9:10, we know that the minute hand is on the 2, but the hour hand has moved 10 × 0.5 = 5 clockwise from the 9 towards the 10, thereby leaving an angle of 25 between the hour hand and the 10. The picture is as follows:
14
Section 1.1
α α α α β
The angle we seek is β + α1 + α 2 + α 3 + α 4 . From the above discussion, we know that
α1 = α 2 = α 3 = α 4 = 30 and β = 25. Thus, the angle at time 9:10 is 145 . 81. Consider the following diagram:
58
Let x = length of DC and y = length of BD. Ultimately, we need to determine x. We proceed as follows. First, we find y. Using the Pythagorean Theorem on ΔABD yields 32 + y 2 = 52 , so that y = 4. Next, using the Pythagorean Theorem on ΔABC yields
15
Chapter 1
( y + x ) 2 + 32 =
( 58 )
2
(4 + x )2 + 9 = 58 (4 + x )2 = 49 4 + x = ±7 so that x = −11 or 3
So, the length of DC is 3. 82. Consider the following diagram:
41
Since we seek the length of DC (which we shall denote as DC), we first need only to apply the Pythagorean Theorem to ABD to find the length of BD. Indeed, observe that 4 2 + ( BD )2 = 52 , so that BD = 3. Next, we apply the Pythagorean Theorem to ABC to find DC: 4 2 + ( 3 + DC ) = 2
( 41)
2
16 + 9 + 6 ( DC ) + ( DC ) = 41 2
( DC ) + 6 ( DC ) − 16 = 0 2
( DC + 8)( DC − 2) = 0 DC = −8 , 2
So, the length of DC is 2 .
16
Section 1.1
83. Consider the following diagram:
Let x = length of AD and y = length of BD. Ultimately, we need to determine x. We proceed as follows. First, we find y. Using the Pythagorean Theorem on ABC yields 24 2 + ( y + 11)2 = 30 2 ( y + 11)2 = 324 324 y + 11 = ± =18
y = −11 ± 18 y = 7 or −29 so that y = 7. Next, using the Pythagorean Theorem on ABD yields 24 2 + y 2 = x 2
24 2 + 7 2 = x 2 625 = x 2 25 = x
So, the length of AD is 25. 84. Consider the following diagram:
17
Chapter 1
Let x = length of BD and y = length of AC . Ultimately, we need to determine y We proceed as follows. First, we find x. Using the Pythagorean Theorem on ABD yields x 2 + 60 2 = 612 x 2 = 121 x = 11 so that x = 11. Next, using the Pythagorean Theorem on ABC yields 602 + ( x + 36)2 = y2
602 + 472 = y2 5,809 = y2 76.2 ≈ y So, the length of AC is approximately 76.2. 85. Consider the following diagram:
Observe that BD = AC (in fact, by the Pythagorean Theorem, both = x 2 ). Furthermore, since the diagonals of a square bisect each other in a 90 − angle, the measure of each of the angles AEB, BEC, DEC, and AED is 90 . Further, x 2 . Hence, by the Side-Side-Side postulate from AE = EC = BE = ED = 2 Euclidean geometry, all four triangles in the above picture are congruent. Regarding their angle measures, since the diagonals bisect the angles at their endpoints, each of these triangles is a 45 − 45 − 90 triangle.
18
Section 1.1
86. Consider the following diagram:
Using the Pythagorean Theorem, we find that x 2 + (3x − 2)2 = (2 x + 2)2 x 2 + 9x 2 − 12 x + 4 = 4 x 2 + 8x + 4
6 x 2 − 20 x = 0 2 x(3x − 10) = 0 x = 0 or
10 3
So, x = 103 (since if x = 0, then there would be no triangle to speak of). 87. Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and 3 x, the shorter leg must have length x. Hence, using the given information, we know that x = 16.68ft. Thus, the two legs have lengths
16.68 ft. and 16.68 3 ≈ 28.89 ft., and the hypotenuse has length 33.36 ft. 88. Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and 3 x, the shorter leg must have length x. Hence, using the given information, we
know that x 3 = 134.75 cm. Thus, the two legs have lengths 134.75 cm and 134.75 ≈ 77.80 cm, and the hypotenuse has length 155.60 cm. 3
19
Section 1.2 Solutions --------------------------------------------------------------------------------1. B = 80 (vertical angles)
2. 80 + E = 180 (interior angles), so that E = 100
3. F = 80 (alternate interior angles)
4. G = 80 (corresponding angles)
5. H = 105° (supplementary angles)
6. 65° + D = 180° (interior angles), so that D = 105°.
7. B = 75 (corresponding angles) 8. Since F = 75 and D + F = 180 (interior angles), we know that D = 105. Hence, A = 105 (vertical angles). 9. Since G = 65 , it follows that F = 65 since vertical angles are congruent. Hence, B = 65 since corresponding angles are congruent. 10. Since G = 65 , it follows that F = 65 since vertical angles are congruent. Since A and B are supplementary angles, A = 115. 11. Since G = 65° we know that F = 65° (vertical angles). Then 65° + D = 180° (interior angles), and D =115°
12. C = 65° (corresponding angles)
13. 8x = 9x − 15 (vertical angles), so that x = 15. Thus, A = 8(15 ) = 120 = D. 14. 9x + 7 = 11x − 7 (corresponding angles), so that solving for x yields x = 7. Thus, B = (9(7) + 7) = 70 = F . 15. Since A + C = 180 and C = G, it follows that A + G = 180. As such, observe that (12 x + 14) + (9x − 2) = 180 21x + 12 = 180 21x = 168 x =8 Thus, A = (12(8) + 14 ) = 110 and G = ( 9(8) − 2 ) = 70.
20
Section 1.2
16. We know that (22x + 3) + (30x − 5) = 180 (interior angles). Solving for x yields (22 x + 3) + (30 x − 5) = 180 52 x − 2 = 180 52 x = 182 x = 182 52 = 3.5
Thus, C = ( 22(3.5) + 3 ) = 80 and E = ( 30(3.5) − 5 ) = 100. 17. b
18. c
19. a
20. f
21. d
23. By similarity,
a c 4 6 = , so that = . Solving for f yields f = 3 . 2 f d f
24. By similarity,
a b 12 9 = . Solving for d yields d = 4 . = , so that d 3 d e
25. By similarity,
b e 12 2 = , so that = . Solving for f yields f = 3. 18 f c f
26. By similarity,
9 e b e . Solving for e yields e = 1. = = , so that 7.2 0.8 c f
27. By similarity,
b a 7.5 a = , so that = . Solving for a yields a = 15 . 2.5 5 e d
22. e
=3
28. By similarity,
b 3.9 b c . Solving for b yields b = 2.1. = = , so that 1.4 2.6 e f
29. First, note that 26.25 km = 26,250 m. Now, observe that by similarity, a 1.1 m d a . Solving for a yields = , so that = 2.5 m 26,250 m f c (2.5 m)a = 28,875 m 2 2
m a = 28,875 = 11,550 m = 11.55 km 2.5 m
21
Chapter 1
30. First, note that 35 m = 3,500 cm. Now, observe that by similarity,
that
c b = , so f e
3,500 cm b . Solving for b yields = 14 cm 10 cm (14 cm)b = 35,000 cm 2 2
cm b = 35,000 = 2,500 cm = 25 m 14 cm
31. By similarity,
4 in. e b e = = , so that 5 . Solving for e yields 5 in. 3 in. c f 2 2
( 5 2 in.) e = ( 4 5 in.)( 3 2 in.) e= 32. By similarity,
( 4 5 in.)( 3 2 in.) = 12 25 in. ( 5 2 in.)
7 mm d a a = , so that 16 . Solving for a yields = 5 mm 1 mm e b 4 4
( 1 4 mm ) a = ( 7 16 mm )( 5 4 mm ) e=
( 716 mm )( 5 4 mm ) 35 = 16 mm ( 1 4 mm )
33. Note that 14 14 = 14.25. Now, consider the following two diagrams.
Let y = height of the tree (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain y 14.25 ft. = 4 ft. 1.5 ft. (1.5 ft.) y = 57 ft.2 y=
57 ft.2 = 38 ft. 1.5 ft.
So, the tree is 38 feet tall.
22
Section 1.2
34. Note that 34 = 0.75. Now, consider the following two diagrams.
Let y = height of the flag pole (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain y 15 ft. = 2 ft. 0.75 ft. (0.75 ft.) y = 30 ft.2 30 ft.2 y= = 40 ft. 0.75 ft.
So, the flag pole is 40 feet tall. 35. Consider the following two diagrams.
48 ft.
Let y = height of the lighthouse (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain 5 ft. 1.2 ft. = y 48 ft. (1.2 ft.) y = 240 ft.2 y=
240 ft.2 = 200 ft. 1.2 ft.
So, the lighthouse is 200 feet tall.
23
Chapter 1
36. Consider the following two diagrams.
Let y = length of the son’s shadow (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain 6 ft. 1 ft. 4 ft.2 2 so that (6 ft.) y = 4 ft.2 and hence, y = = = ft. 4 ft. 6 ft. 3 y So, the son’s shadow is 23 ft. long. Since 1 ft. = 12 in., this is equivalent to 8 in. 37. First, make certain to convert all quantities involved to a common unit. In order to avoid using decimals, use the smallest unit – in this particular problem, use cm. Now, consider the following two diagrams:
Let y = height of the lighthouse (in cm). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain 200 cm 5 cm = y 300 cm (5 cm) y = 60,000 cm2 60,000 cm2 y= = 12,000 cm = 120 m 5 cm So, the lighthouse is 120 m tall.
24
Section 1.2
38. Let H = the height of the pyramid (in meters). (Important! Don’t confuse this with the slant height.) Using the fact that the height is the length of the segment that extends from the apex of the pyramid down to the center of the square base, we have the following diagram. Also, we have
Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain H (115 + 16) m = 1m 0.9 m ( 0.9 m ) H = 131 m m H = 131 0.9 m = 145.56 m ≈ 146 m
So, the pyramid is approximately 146 m tall. 39. Let x = length of the planned island (in inches). (Note that 2′4′′ = 28 in.) Using similarity, we obtain x 28 in. = 11 1 4 in. 16 in. 11 28 ( 16 ) in.2 = ( 14 in.) x
x=
11 28 ( 16 ) in.2 1 4
in. So, the planned island is 6 ft. 5 in. long.
25
= 77 in. = 6 ft. 5 in.
Chapter 1
40. Let x = width of the pantry (in inches). Using similarity, we obtain 28 in. x = 7 1 4 in. 16 in.
28 ( 167 ) in.2 = ( 14 in.) x x=
28 ( 167 ) in.2 1 4
in.
= 49 in. = 4 ft. 1 in.
So, the pantry is 4 ft. 1 in. long. 1 in . The distance from Phoenix to Four Corners to 50 mi Albuquerque is 7 + 4 = 11 in. Thus, the total distance is 1 in 11 in x = 550 miles. = 50 mi x 41. First note that
1 in . The distance from Albuquerque to Denver to Salt Lake 50 mi City to Phoenix is 6 + 7.5 + 10.5 = 24 in. Thus, the total distance is 1 in 24 in x = 1200 miles. = 50 mi x
42. First note that
43. Consider the following two triangles:
The triangles are similar by AAA, so that 54 = 3.5 x . Solving for x yields x = 2.8 ft.
26
Section 1.2
44. Consider the following diagram:
60
Since this is a 30 − 60 − 90 triangle and x is the longer side (being opposite the larger of the two acute angles), we know that x = 15 3 ≈ 26 ft. 45. The two triangles are similar by AAA. So, we have 1.6 in. x ( 2.2 in.) x = (1.6 in.)(1.2 in.) x ≈ 0.87 in. = 2.2 in. 1.2 in. 46. If a right triangle is isosceles, then it must be a 45 − 45 − 90 triangle. Let x = length of a leg. Then, the hypotenuse is 2 x = 2 , so that x = 2 in. 47. The information provided is: 2 fd = 8 cm, dt = 2.5 cm, d s = 80 cm . We must determine the value of 2 fs . Using similar triangles, we have the following proportion: d s d s + dt 80 2.5 + 80 = . = 4 fs fd fs 320 Solving for fs yields: 82.5 fs = 320 fs = ≈ 3.87878 82.5 Thus, 2 fs ≈ 7.8 cm. 48. The information provided is: 2 fd = 4 cm, so that fd = 2 cm dt = 3.5 cm 2 fs = 3.8 cm, so that fs = 1.9 cm We must determine the value of d s . Using similar triangles, we have the following proportion: d s d s + dt d s d s + 3.5 = = . fs fd 1.9 2 Solving for d s yields: 2d s = 1.9d s + 6.65 0.1d s = 6.65 d s = 66.5 cm.
27
Chapter 1
49. The information provided is: 2 fd = 4 cm, so that fd = 2 cm d s = 60 cm 2 fs = 3.75 cm, so that fs = 1.875 cm We must determine the value of dt . Using similar triangles, we have the following proportion: d s d s + dt 60 + dt 60 = = . fs fd 1.875 2 Solving for dt yields: 120 = 112.5 + 1.875dt 7.5 = 1.875dt dt = 4 cm.
50. The information provided is: 2 fs = 4.15 cm, dt = 4.5 cm, d s = 60 cm. We must determine the value of 2 fd . Using similar triangles, we have the following proportion: d s d s + dt 60 60 + 4.5 = = . 2.075 fs fd fd 133.8375 Solving for fd yields: 60 fd = 133.8375 fd = ≈ 2.2306 60 Thus, 2 fd ≈ 4.5 cm. 51. The ratio is set up incorrectly. It should be
A B = . D E
52. In order to find F, one needs C instead of B. In such case,
A C = . D F
53. True. Two similar triangles must have equal corresponding angles, by definition. 54. True. If two triangles are congruent, then the ratio of corresponding sides equals 1, for all three pairs of sides, and all corresponding angles are equal. Hence, they must be similar. (Conversely, two similar triangles need not be congruent – see Problem 36.) 55. False. The third angle in such case would have measure 180 − ( 82 + 67 ) = 31.
28
Section 1.2
56. True. Consider two equilateral triangles, one whose sides have length x and a second whose sides have length y. Observe then that the ratio of corresponding sides is always xy (or xy , depending on which you refer to as Triangle 1 and
Triangle 2). Since all angles of an equilateral triangle have measure 60 , we conclude that they must be similar. However, if x = 1 and y = 2, then the triangles are NOT congruent. 57. False. Such angles are congruent, and hence unless they are both 90 , they need not be supplementary. 58. True. 59. False. You need the corresponding angles to be the same to ensure similarity of two isosceles triangles. 60. False. All we can say here is that corresponding sides of similar triangles are in proportion. 61. We label the given diagram as follows:
6
Observe that ACB = DCE (vertical angles). Since we are given that CBA = CDE , the third pair of angles must have the same measures. Hence, ΔABC is similar to ΔCDE . The corresponding sides between these two triangles are identified as follows: AB corresponds to ED, AC corresponds to EC, BC corresponds to DC AB AC 3 6 = , so that = , so that Using the ratio for similar triangles, we have ED EC x 4 x=2.
29
Chapter 1
62. Consider the following diagram:
Observe that ΔABE and ΔACD are similar triangles (since we are given that ABE = ACD and the triangles share A. Corresponding sides of these triangles are identified as follows: AB corresponds to AC, AE corresponds to AD, BE corresponds to CD Using the ratio for similar triangles, we know that
AB AE BE = = , so that AC AD CD
4 y 5 1 = = = (1). 4 + x y + 18 20 4 We now use different equalities from (1) to find x and y. 4 1 = so that 16 = 4 + x and hence, x = 12 . Find x: 4+x 4 y 1 Find y: = so that 4 y = y + 18 and hence, y = 6 . y + 18 4
63. Triangles 1 and 2 are similar because they both have a 90 angle, and the angles formed at the horizontal have the same measure since they are vertical angles. Similarly, Triangles 3 and 4 are similar. 64. a) Consider the following diagram:
D0 − f
H0 f
Hi
The two right triangles above are similar since they both have a 90 , and the measures of the angles opposite the sides with lengths H 0 and Hi are equal because H D −f . they are vertical angles. Hence, using the similar triangles ratio yields 0 = 0 Hi f 30
Section 1.2
b) Consider the following diagram:
H0
Di − f f
Again, from similarity if follows that
Hi
H0 f . = Hi Di − f
1 1 1 + = . D0 Di f Proof. Observe that from Problem 56, we obtain the following identities: H H f 0 + 1 = f 0 + f = D0 (from (a)) Hi Hi 65. Claim (Lens Law):
H H f i + 1 = f i + f = Di (from (b)) H0 H0
Hence, observe that H H f 0 + 1 + f i + 1 1 1 Di + D0 Hi H0 + = = D0 Di D0 Di H 0 Hi + 1 ⋅ f + 1 f Hi H 0 Hi H 0 H + H + 2 1 0 i 1 = = = , f f H 0 Hi + + 2 H H H H Hi H 0 f 2 0 ⋅ i + 0 + i + 1 Hi H 0 Hi H 0 =1 H H f i + 0 + 2 H 0 Hi
as desired.
31
Chapter 1
66. We begin with several observations: 1.) m ( ∠ABE ) = 90 since BF is parallel to CG, and ∠ABE ≈ ∠GCD since they
are corresponding angles. 2.) m ( ∠AEB ) = m ( ∠CGA) = m ( ∠FEG ) = 60
3.) m ( ∠CGD ) = m ( ∠CDG ) = 45 since ΔCDG is an isosceles right triangle. 4.) BC and FG have the same length. Hence, ΔABE ΔACG ΔGFE . 67. We begin with several observations: 1.) m ( ∠ABE ) = 90 since BF is parallel to CG, and ∠ABE ≈ ∠GCD since they
are corresponding angles. 2.) m ( ∠AEB ) = m ( ∠CGA) = m ( ∠FEG ) = 60
3.) m ( ∠CGD ) = m ( ∠CDG ) = 45 since ΔCDG is an isosceles right triangle. 4.) BC and FG have the same length. Hence, ΔABE ΔACG ΔGFE . Now, using this information, consider the following triangles:
Since ΔACG is a 30 − 60 − 90 triangle, we know that m( AC ) = 3 ⋅ m(CG ), and so m(CG ) = 5 3 3 . Hence, m( AG ) = 2 ⋅ m(CG ) = 103 3 . Next, since ΔACG ΔGFE , we know that 5 103 3 5 5 33 = y = 2 3 and = x = 3. 3 3 x y So, m( EF ) = 3 cm, m( EG ) = 2 3 cm. 68. Since m(CG ) = 5 3 3 cm and ΔDCG is 45 − 45 − 90 , we know that
( )
m( DG ) = 2 5 3 3 cm = 5 3 6 cm.
32
Section 1.3 Solutions --------------------------------------------------------------------------------1. sin θ =
8 4 = 10 5
2. cos θ =
6 3 = 10 5
3. csc θ =
1 1 5 = = 4 sin θ 4 5
4. sec θ =
1 1 5 = = 3 cos θ 3 5
5. tan θ =
8 4 = 6 3
6. cot θ =
1 1 3 = = 4 tan θ 4 3
7. Note that by the Pythagorean Theorem, the hypotenuse has length 5. So, cos θ =
1 5 . = 5 5
9. Note that by the Pythagorean Theorem, the hypotenuse has length 5. So, 1 1 sec θ = = = 5. 1 cos θ 5 11. tan θ =
2 = 2 1
13. sin θ =
5 5 34 = 34 34
8. Note that by the Pythagorean Theorem, the hypotenuse has length 5. So, sin θ =
2 2 5 . = 5 5
10. Note that by the Pythagorean Theorem, the hypotenuse has length 5. So, csc θ =
1 = sin θ
12. cot θ =
1 2 5
=
5 . 2
1 1 = tan θ 2
14. Using the Pythagorean Theorem, we know that the leg adjacent to angle θ
has length 3. So, cos θ =
33
3 3 34 . = 34 34
Chapter 1
15. Using the Pythagorean Theorem, we know that the leg adjacent to angle 5 θ has length 3. So, tan θ = . 3 17. Using the Pythagorean Theorem, we know that the leg adjacent to angle θ has length 3. So, sec θ =
1 = cos θ
34 . 3
19.
( )
sin ( 60 ) = cos ( 90 − 60 ) = cos 30
(
21. cos ( x ) = sin 90 − x
)
23.
( )
sin ( x + y ) = cos ( 90 − ( x + y ) )
(
= cos 90 − x − y
(
29.
(
)
cot ( 45 − x ) = tan 90 − ( 45 − x )
(
= tan 45 + x
20.
( )
sin ( 45 ) = cos ( 90 − 45 ) = cos 45
(
)
(
)
)
24. sec ( B ) = csc 90 − B
)
(
sin ( 60 − x ) = cos 90 − ( 60 − x )
(
= cos 30 + x
27. cos ( 20 + A) = sin 90 − ( 20 + A)
= sin 70 − A
34 5
18. Using the Pythagorean Theorem, we know that the leg adjacent to angle 1 3 = . θ has length 3. So, cot θ = tan θ 5
26.
(
1 = sin θ
22. cot ( A) = tan 90 − A
csc ( 30 ) = sec ( 90 − 30 ) = sec 60
25.
16. csc θ =
)
28.
)
cos ( A + B ) = sin ( 90 − ( A + B ) )
(
= sin 90 − A − B
)
30.
(
)
sec ( 30 − θ ) = csc 90 − ( 30 − θ )
(
= csc 60 + θ
34
)
)
)
Section 1.3
31. From the given information, we obtain the following triangle:
32. Consider the following triangle:
θ θ Hence, a round trip on the ATV corresponds to twice the length of the hypotenuse, which is 2 × ( 5 miles ) = 10 miles .
33. Consider the triangle:
opp y = = 1, so adj x y = x. Since we are given that 2 ( x + y ) = 200 (since a round trip is 200
Observe that tan θ =
yards), we see that x + y = 100 and hence, x = y = 50. As such, by the Pythagorean Theorem, the hypotenuse has length 50 2. So, the round trip of the ATV is 100 2 yards ≈ 141 yards . Applying the Pythagorean theorem yields 52 + 12 2 = z 2 z = 13. Thus, sin θ = 135 .
θ 34. The scenario can be described by the following two triangles:
θ
θ
35
Chapter 1
Applying the Pythagorean theorem yields z = 13 and w = 193. Hence, we have: 12 Bob: cos θ = 12 13 Neighbor: cos θ = 193 The value of cos θ is larger for Bob’s roof. The steeper the roof for the same horizontal distance, the longer the hypotenuse and hence denominator used in computing cos θ . 35.
Since we know the side opposite the 30° angle and we are looking for the hypotenuse, we can use the following equation: 200 cos30 = x 200 x= ≈ 230.9 ft cos30
36.
Since we know the side opposite the 30° angle and we are looking for the hypotenuse, we can use the following equation: 185 cos30 = x 185 x= ≈ 213.6 ft cos30
37. If tan θ = 3, we have the following diagram:
Since the hypotenuse has length 2 by the Pythagorean theorem, this is a 30 − 60 − 90 triangle. Hence, since θ is opposite the longer leg, it must be 60.
θ
36
Section 1.3
38. Since θ = 60 we have the following triangle:
Observe that 3 cos30 = 3Wft. W = cos30 ≈ 3.46 ft.
39. Consider the following triangle:
Observe that
40. We have the following triangle:
Now we have a similar triangle:
Observe that cos A = 35 , and the Pythagorean theorem implies that the height is 4, as shown.
As such,
1 41. Since sec θ = 1.75 cos θ = 1.75 , we have the following triangle:
θ
sin B = 35 .
3 4
= 8 xin. x = 6 in.
Note that 1.75 = 74 . Applying the Pythagorean theorem yields 2 1 + z 2 = ( 74 ) z = 433 . Hence, 33 33 sin θ = 4 = . 7 7 4
42. Note that tan θ = 1 when θ = 45. So, if tan θ = 1.2, then θ must be greater than 45. Hence, the hill is too steep.
37
Chapter 1
43. sin θ =
6 mm 1 = θ = 30 12 mm 2
45. a) tan θ =
44. sin θ =
2 mm 2 = 2 mm 2
θ = 45
10,000 2 1 5 = b) cot θ = = 25,000 5 tan θ 2
46. a) By the Pythagorean theorem, this equals
s 2 + p 2 . b) cos θ =
47. Opposite side has length 3, not 4.
48. tan x =
opp adj , not . adj opp
1 1 , not . cos x sin x
50. csc y =
1 1 , not . sin y cos y
49. sec x =
s s + p2 2
51. True. It follows from the co-function identity that sin ( 45 ) = cos ( 90 − 45 ) = cos ( 45 ) . 52. True. It follows from the co-function identity that sin ( 60 ) = cos ( 90 − 60 ) = cos ( 30 ) . 53. True. sec 60 = csc ( 90 − 60 ) . 54. True. cot ( 45 ) = tan ( 90 − 45 ) 56. Using Exercise 53, we have 3 sin 60 = cos 30 = 2 1 cos 60 = sin30 = 2
55. sin30 =
opp x 1 = = hyp 2 x 2
cos 30 =
adj 3x 3 = = hyp 2 x 2
38
Section 1.3
57. Using Exercises 53 and 54, we have: 3 1 sin30 1 3 sin 60 2 2 tan30 = = = = tan 60 and = = = 3. 3 1 cos 30 3 cos 60 3 2 2 (Alternatively, in order to avoid having to use Exercise 54 in the computation of tan 60 , you can instead use the co-function identity to obtain 1 1 tan 60 = cot 30 = = 1 = 3.) tan30 3 58. First, note that sin ( 45 ) = cos ( 45 ) =
we see that tan 45 =
sin ( 45 )
cos ( 45 )
59. First, note that sec 45 =
x 1 2 = = . Using these values, 2 2x 2
2
= 22 = 1. 2
1 1 1 = x = 1 = 2. Using the co-function cos 45 2x 2
identity, we see that csc 45 = sec 45 = 2. 60. Using Exercise 55, we see that tan 60 = 3 and cot 30 = tan 60 = 3. 61. −1 ≤ sin θ ≤ 1 and − 1 ≤ cos θ ≤ 1 62. Any real number 63. sec θ ≥ 1 and csc θ ≥ 1 64. sin θ increases from 0 to 1 as θ decreases from 0 to 90. 65. The cofunction of sin θ is cos θ . As θ increases from 0 to 90 , cos θ decreases from 1 to 0. 66. As θ increases from 0 to 90 , csc θ decreases to 1.
1 1 ≈ ≈ 2.92398 cos 70 0.342 b. The keystroke sequence 70, cos, 1/x yields 2.92380. 67. a. cos 70 ≈ 0.342 and so,
39
Chapter 1
1 1 ≈ ≈ 1.55521 sin 40 0.643 b. The keystroke sequence 40, sin, 1/x yields 1.55572. 68. a. sin 40 ≈ 0.643 and so,
1 1 ≈ ≈ 0.70274 tan54.9 1.423 b. The keystroke sequence 54.9, tan, 1/x yields 0.70281. 69. a. tan54.9 ≈ 1.423 and so,
1 1 ≈ ≈ 1.05485 cos18.6 0.948 b. The keystroke sequence 18.6, cos, 1/x yields 1.05511. 70. a. cos18.6 ≈ 0.948 and so,
40
Section 1.4 Solutions --------------------------------------------------------------------------------1. a
2. b
3. b
4. a
1 sin30 1 3 2 7. tan30 = = = = 3 cos30 3 3 2
8. tan 45 =
9. tan 60 =
3 sin 60 2 = = 1 cos 60 2
1 1 2 2 3 = = = 3 sin 60 3 3 2
15. Using Exercise 9, we see that 1 1 3 cot 60 = . = = tan 60 3 3
14. sec 60 =
1 1 = 1 = 2 cos 60 2
16. csc 45 =
1 1 = = 2 sin 45 2
17. sec 45 =
1 1 = 1 = cos 45 2
1 1 = 1 = 2 sin30 2
12. Using Exercise 7, we see that 1 1 cot 30 = = 3 = 3 . tan30 3
1 1 2 2 3 = = = 3 cos30 3 3 2
13. csc 60 =
6. c
2 sin 45 2 = =1 2 cos 45 2
10. csc 30 =
3
11.
sec 30 =
5. c
2
18. Using Exercise 8, we see that 1 1 cot 45 = = = 1. tan 45 1
2
19. 0.6018
20. 0.3057
21. 0.1392
22. 0.9278
23. 1.3764
24. 0.9391
25. 1.0098
26. 3.8637
27. 1.0002
28. 1.2868
29. 0.7002
30. 1.8040
41
Chapter 1
31.
32.
5 17′ 29′′
5 17′ 29′′
+ 63 28′ 35′′
+ 16 11′ 30′′
68 45′ 64′′ Remember that whenever the number of minutes or number of seconds is at least 60, you can further simplify the expression. In this case, since 64′′ = 1′ 4′′, the completely simplified
21 28′ 59′′ This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60.
answer is 68 46′ 4′′ . 33.
34.
63 28′ 35′′
63 28′ 35′′
+ 16 11′ 30′′
− 5 17′ 29′′
79 39′ 65′′ Remember that whenever the number of minutes or number of seconds is at least 60, you can further simplify the expression. In this case, since 65′′ = 1′ 5′′, the completely simplified
58 11′ 6′′ This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60.
answer is 79 40′ 5′′ . 35.
63 28′ 35′′ − 16 11′ 30′′
47 17′ 5′′ This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60.
42
Section 1.4
36. In order to compute
16 11′ 30′′ − 5 17′ 29′′
we borrow 1 from 16 and add it (in the form of 60′) to11′ in order to rewrite the problem in the equivalent form: 15 71′ 30′′ − 5 17′ 29′′ 10 54′ 1′′ This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60. 37. In order to compute 90 0′ 0′′
38. In order to compute 90 0′ 0′′
− 5 17′ 29′′
− 63 28′ 35′′
we borrow 1 from 90 and add it (in the form of 60′) to 0′ in order to rewrite the problem in the equivalent form: 89 60′ 0′′
we borrow 1 from 90 and add it (in the form of 60′) to 0′ in order to rewrite the problem in the equivalent form: 89 60′ 0′′ − 63 28′ 35′′ Next, we borrow 1′ from 60′ and add it (in the form of 60′) to 0′′ to rewrite this problem in the equivalent form: 89 59′ 60′′
− 5 17′ 29′′ Next, we borrow 1′ from 60′ and add it (in the form of 60′) to 0′′ to rewrite this problem in the equivalent form: 89 59′ 60′′
− 63 28′ 35′′
− 5 17′ 29′′
26 31′ 25′′
84 42′ 31′′ This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60.
43
This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60.
Chapter 1
20 39. First, note that 20′ = ≈ 0.33. 60
45 40. First, note that 45′ = ≈ 0.75. 60
So, 3320′ ≈ 33.33 .
So, 89 45′ = 89.75 .
27 41. First, note that 27′ = = 0.45. 60
13 42. First, note that 13′ = ≈ 0.22. 60
So, 5927′ = 59.45 .
So, 7213′ ≈ 72.22 .
43. First, note that
44. First, note that
15 15′′ = ≈ 0.004 3600
30 30′′ = ≈ 0.0083 3600
45 45′ = = 0.75. 60
5 5′ = ≈ 0.0833. 60
So, 27 45′15′′ = 27.754 .
So, 365′30′′ = 36.092 .
45. First, note that
46. First, note that
12 12′′ = ≈ 0.0033 3600
9 9′′ = = 0.0025 3600
10 10′ = ≈ 0.1667. 60
28 28′ = ≈ 0.4667. 60 So, 4228′12′′ = 42.470 .
So, 6310′9′′ = 63.169 .
47. Observe that 15.75 = 15 + 0.75
48. Observe that 15.50 = 15 + 0.50
60′ = 15 + ( 0.75 ) 1 = 15 + 45′
60′ = 15 + ( 0.50 ) 1 = 15 + 30′
So, 15.75 = 15 45′ .
So, 15.50 = 1530′ .
49. Observe that
60′ 22.35 = 22 + 0.35 = 22 + ( 0.35 ) = 22 + 21′ 1
So, 22.35 = 2221′ .
44
Section 1.4
50. Observe that
60′ 80.47 = 80 + 0.47 = 80 + ( 0.47 ) 1 = 80 + 28.2′ = 80 + 28′ + 0.2′ 60′′ = 80 + 28′ + ( 0.2′ ) 1′ = 80 + 28′ + 12′′
So, 80.47 = 8028′12′′ . 51. Observe that 30.175 = 30 + 0.175
52. Observe that 25.258 = 25 + 0.258
60′ = 30 + ( 0.175 ) 1 = 30 + 10.5′
60′ = 25 + ( 0.258 ) 1 = 25 + 15.48′
= 30 + 10′ + 0.5′
= 25 + 15′ + 0.48′
60′′ = 30 + 10′ + ( 0.5′ ) 1′ = 30 + 10′ + 30′′
60′′ = 25 + 15′ + ( 0.48′ ) 1′ = 25 + 15′ + 28.8′′
So, 30.175 = 3010′ 30′′ .
So, 25.258 ≈ 2515′ 29′′ .
53. Observe that 77.535 = 77 + 0.535
54. Observe that 5.995 = 5 + 0.995
60′ = 77 + ( 0.535 ) 1 = 77 + 32.1′
60′ = 5 + ( 0.995 ) 1 = 5 + 59.7′
= 77 + 32′ + 0.1′
= 5 + 59′ + 0.7′
60′′ = 77 + 32′ + ( 0.1′ ) 1′ = 77 + 32′ + 6′′
60′′ = 5 + 59′ + ( 0.7′ ) 1′ = 5 + 59′ + 42′′
So, 77.535 = 7732′ 6′′ .
So, 5.995 = 559′ 42′′ .
45
Chapter 1
55. Begin by converting 1025′ to DD:
56. Begin by converting 7513′ to DD:
25 Since 25′ = ≈ 0.4167 , 60 we see that 1025′ ≈ 10.4167. Then, observe that sin (10.4167 ) ≈ 0.1808 .
13 Since 13′ = ≈ 0.217 , 60 we see that 7513′ ≈ 75.217. Then, observe that cos ( 75.217 ) ≈ 0.2552 .
57. Begin by converting 2215′ to DD:
58. Begin by converting 6822′ to DD:
15 Since 15′ = = 0.25 , 60 we see that 2215′ = 22.25. Then, observe that tan ( 22.25 ) ≈ 0.4091.
22 Since 22′ = ≈ 0.3667 , 60 we see that 6822′ ≈ 68.3667. Then, observe that 1 sec ( 68.3667 ) = ≈ 2.7125 . cos ( 68.3667 )
59. Begin by converting 2825′35′′ to DD: Since
60. Begin by converting 5020′19′′ to DD: Since
25 25′ = ≈ 0.4167 60
20 20′ = ≈ 0.3333 60
35 35′′ = ≈ 0.0097 , 3600 we see that 28 25′35′′ ≈ 28 + 0.4167 + 0.0097
19 19′′ = ≈ 0.0053 , 3600 we see that 50 20′19′′ ≈ 50 + 0.3333 + 0.0053
≈ 28.426. Then, observe that 1 csc ( 28.426 ) = ≈ 2.1007 . sin ( 28.426 )
≈ 50.339. Then, observe that 1 sec ( 50.339 ) = ≈ 1.5668 . cos ( 50.339 )
46
Section 1.4
61. Using nλ = 2d sin θ , observe that 1(1.54) = 2d sin ( 45 )
62. Using nλ = 2d sin θ , observe that 4(1.67) = 2d sin ( 71.3 ) 6.68 = 2 sin ( 71.3 ) d
2 1.54 = 2d 2 1.54 ≈ 1.09 angstroms d= 2
d=
6.68 ≈ 3.53 angstroms 2 sin ( 71.3 )
63. Using ni sin (θi ) = nr sin (θ r ) ,
64. Using ni sin (θi ) = nr sin (θ r ) ,
observe that 1.00 sin ( 30 ) = nr sin (12 )
observe that 1.00 sin ( 30 ) = nr sin (18.5 )
nr =
1.00 sin ( 30 ) sin (12 )
≈ 2.405 .
nr =
1.00 sin ( 30 ) sin (18.5 )
≈ 1.576 .
65. Using ni sin (θi ) = nr sin (θ r ) ,
66. Using ni sin (θi ) = nr sin (θ r ) ,
observe that 1.00 sin ( 30 ) = nr sin ( 22 )
observe that 1.00 sin ( 30 ) = nr sin ( 20 )
nr =
1.00 sin ( 30 ) sin ( 22 )
≈ 1.335 .
nr =
1.00 sin ( 30 ) sin ( 20 )
67. 90 − 8528′ = 432′
68. 3619′ + 3619′ = 7238′
69. d = 15sin 60 + 4 3 = 15
15 70. d = csc65 + 4 ≈ 17.6 ft.
( )+4 3
≈ 1.462 .
3 2
= 232 3 ft. 71. 100 sin 45 = 100
( ) = 50 2 ft.
72. 100 sin 60 = 100
2 2
1 73. 40 sin30 = 40 = 20 ft 2
( )
3 2
2 74. 30 2 × sin 45 = 30 2 = 30 ft 2
(
)
1 75. 4018′27′′ = 40 + 18′ 601 ′ + 27′′ 3600 ′′ ≈ 40.3075
( ) = 50 3 ft.
47
Chapter 1
76. First, convert 3928′37′′ to DD: 1 3928′37′′ = 39 + 28′ 601 ′ + 37′′ 3600 ′′ ≈ 39.476944 .
So, h = 15tan ( 39.476944 ) ≈ 12.4 ft.
( )
(
)
78. We need to convert 15.8547° to DMS: 60 ' 15.8547° = 15° + .8547° 1° = 15° + 51.282′
77. 1 1 20°32′15′′ = 20° + 32′ + 15′′ 60 3600 ≈ 20.5375°
60′′ = 15° + 51′ + 0.282′ 1′ = 15°51′ 16.92′ = 15°51′ 17′′
1 3 79. cos ( 60 ) = , not 2 2
25 80. 25′ = ≈ 0.4167 ≈ 0.42 , not 60 0.25 . So, 3625′ ≈ 36.42 .
81. False.
82. False. Note that sin50 ≈ 0.7660 , which is not equal to 0.77.
sec ( 301 ) =
1 cos (
)
1 30
, not cos30
84. False. Since tan θ = cot ( 90 − θ ) ,
83. True.
we have tan ( 2050′ ) = cot ( 90 − 2050′ ) = cot ( 6910′ )
48
Section 1.4
85. Consider the following triangle:
86. Consider the following triangle:
a 2 + b2
a 2 + b2
b
sin ( 0 ) ≈
b
θ b
θ a
cos ( 0 ) ≈
≈ 0 since a is a lot
a a2 + b2
≈ 1 since a is a lot
a 2 + b2 bigger than b and so, the denominator is much larger than the numerator.
bigger than b and so, a ≈ a 2 + b 2 .
87. Using the co-function identity with Exercise 81, we see that cos ( 90 ) = sin ( 0 ) = 0 .
88. Using the co-function identity with Exercise 82, we see that sin ( 90 ) = cos ( 0 ) = 1 .
89. Since sec 45 =
1 2
2
3 3 , and cos 30 = , 3 2
1
= 2 , tan30 = 32 = 2
we have
(
)
2 3 3 2+ 3 2 + 33 3 2 + 3 2 sec 45 + tan30 2 6 +2 . = = ⋅ = = 3 cos30 3 9 3 3 2
90. Since 3 1 2 3 2 , csc 60 = 3 = , cot 30 = 2 = 3 , and sin 45 = 1 3 2 2 2
we have
( sin 45 )( cot 30 ) = ( ) ( 3 ) = ( csc 60 ) ( )
2 2
2
2 3 3
49
2
6 12
2 9
=
3 6 8
Chapter 1
91. 6− 2 1 1 − 2 cos 75 − csc 45 cos 30 = sin15 − cos 30 = 4 sin 45 2
= 92.
6− 2 6 − 6− 2 − = 4 2 4
cot 75 ( cos 45 + sin30 ) = tan15 ( cos 45 + sin30 )
(
= 2− 3
)( + ) = 2 − +1− 2 2
1 2
6 2
93. We begin by converting all to DD: 1 2510′15′′ = 25 + 10′ 601 ′ + 15′′ 3600 ′′ ≈ 25.17083
( ) ( ) 46 14′26′′ = 46 + 14′ ( ) + 26′′ ( ) ≈ 46.24056 23 17′23′′ = 23 + 17′ ( ) + 23′′ ( ) ≈ 23.28972
Then, we have
1 60′
1 3600′′
1 60′
1 3600′′
tan ( 2510′15′′ ) + sec ( 4614′26′′ ) csc ( 2317′23′′ )
≈ 0.7575 .
94. We begin by converting all to DD: 1 3338′1′′ = 33 + 38′ 601 ′ + 1′′ 3600 ′′ ≈ 33.6336
( ) ( ) 36 58′6′′ = 36 + 58′ ( ) + 6′′ ( ) ≈ 36.9683 50 33′2′′ = 50 + 33′ ( ) + 2′′ ( ) ≈ 50.5506
Then, we have
1 60′
1 3600′′
1 60′
1 3600′′
cos ( 3338′1′′ )
csc ( 50 33′2′′ )
− cot ( 3658′6′′ ) ≈ −0.6856 .
50
3 2
( ) 3 2
Section 1.4
96. a. Step 1: sin ( 40 ) ≈ 0.643
95. a. Step 1: cos ( 70 ) ≈ 0.342
Step 2: b.
1 ≈ 2.92398 cos ( 70 )
Step 2:
( cos (70 )) ≈ 2.92380
(
b. sin ( 40 )
−1
1 ≈ 1.55521 sin ( 40 )
) ≈ 1.55572 −1
The calculation in part b is more accurate since one does not round twice as in part a.
The calculation in part b is more accurate since one does not round twice as in part a.
97. This calculation sequence results in 3.240.
98. This calculation sequence results in 27 40′59′′ .
51
Section 1.5 Solutions --------------------------------------------------------------------------------1. three
2. four
3. two
4. one
5. 47
6. 10
7. 55
8. 28
9. 83
10. 74
11. Consider this triangle:
12. Consider this triangle:
Since cos ( 35 ) =
adj a = , we hyp 17 in.
have a = (17 in.) cos ( 35 ) ≈ 13.93 in. ≈ 14 in.
Since sin ( 35 ) =
opp b = , we have hyp 17 in.
b = (17 in.) sin ( 35 ) ≈ 9.751 in. ≈ 10 in.
52
Section 1.5
13. Consider this triangle:
14. Consider this triangle:
adj b opp a = , we have Since cos ( 55 ) = hyp = 22 ft. , we have hyp 22 ft. b = ( 22 ft.) cos ( 55 ) ≈ 12.619 ft. ≈ 13 ft. a = ( 22 ft.) sin ( 55 ) ≈ 18.02 ft. ≈ 18 ft.
Since sin ( 55 ) =
15. Consider this triangle:
16. Consider this triangle:
Since tan ( 20.5 ) = have
opp a = , we adj 14.7 mi.
a = (14.7 mi.) tan ( 20.5 )
Since tan ( 69.3 ) = have
≈ 5.496 mi. ≈ 5.50 mi.
opp b = , we adj 0.752 mi.
b = ( 0.752 mi.) tan ( 69.3 ) ≈ 1.990 mi.
≈ 1.99 mi. (since the number of significant digits is 3).
53
Chapter 1
17. Consider this triangle:
18. Consider this triangle:
Since cos ( 25 ) =
adj 11 km = , we hyp c
Since sin ( 75 ) =
have 11 km c= ≈ 12.137 km ≈ 12 km cos ( 25 ) (since the number of significant digits is 2).
c=
opp 26 km = , we have hyp c
26 km ≈ 26.917 km ≈ 27 km sin ( 75 )
(since the number of significant digits is 2). 20. Consider this triangle:
19. Consider this triangle:
opp 15.37 cm Since sin ( 48.25 ) = = , hyp c we have 15.37 cm c= ≈ 20.602 cm sin ( 48.25 )
≈ 20.60 cm (since the number of significant digits is 4).
Since cos ( 29.80 ) =
adj 16.79 cm = , hyp c
we have c=
16.79 cm ≈ 19.3485 cm cos ( 29.80 )
≈ 19.35 cm (since the number of significant digits is 4).
54
Section 1.5
21. Consider this triangle:
22. Consider this triangle:
α
β Since sin (α ) =
opp 29 mm = , hyp 38 mm
we have
adj 89 mm = , hyp 99 mm
Since cos ( β ) = we have
29 α = sin ≈ 49.74 ≈ 50 38 (since the number of significant digits is 2).
89 ≈ 25.974 ≈ 26 99 (since the number of significant digits is 2).
23. Consider this triangle:
24. Consider this triangle:
−1
β = cos −1
α
β Since cos (α ) =
adj 2.3 m = , hyp 4.9 m
we have
2.3 ≈ 62.005 ≈ 62 4.9 (since the number of significant digits is 2).
α = cos −1
Since sin ( β ) =
opp 7.8 m = , hyp 13 m
we have
7.8 ≈ 36.870 ≈ 37 13 (since the number of significant digits is 2).
β = sin −1
55
Chapter 1
25. Consider this triangle:
26. Consider this triangle:
α
β First, observe that
First, observe that 1 1 ≈ 21.2833 . α = 21 17′ = 21 + 17′ ′ ′ 27 21 27 21 = = + ≈ 27.35 . β ′ 60 60′ Since b opp , Since tan ( 27.35 ) = = a opp tan ( 21.2833 ) = , = adj 117.0 yd. adj 210.8 yd. we have we have b = (117.0 yd.) tan ( 27.35 ) a = ( 210.8 yd.) tan ( 21.2833 ) ≈ 60.52 yd. ≈ 82.117 yd.
(since the number of significant digits is 4).
≈ 82.12 yd. (since the number of significant digits is 4). 27. Consider this triangle:
Since cos (15.3333 ) =
adj 10.2 km = , we hyp c
have c=
10.2 km ≈ 10.58 km cos (15.3333 )
≈ 10.6 km (since the number of significant digits is 3).
β First, observe that 1 ≈ 15.3333 . ′ 60
β = 15 20′ = 15 + 20′
56
Section 1.5
28. Consider this triangle:
Since sin ( 65.5 ) =
opp 18.6 km = , hyp c
we have c=
18.6 km ≈ 20.440 km sin ( 65.5 )
≈ 20.4 km (since the number of significant digits is 3).
β First, observe that 1 ≈ 65.5 . β = 65 30′ = 65 + 30′ 60′
29. Consider this triangle:
30. Consider this triangle:
α
α
First, observe that α = 40 28′10′′
1 1 = 40 + 28′ + 10′′ 60′ 3600′′ ≈ 40.4694
First, observe that α = 28 32′ 50′′ 1 1 = 28 + 32′ + 50′′ 60′ 3600′′ ≈ 28.5472
Since sin ( 40.4694 ) =
opp 12,522 km , = hyp c
we have 12,522 km c= ≈ 19,293.05 km sin ( 40.4694 )
Since cos ( 28.5472 ) =
we have c=
17,986 km ≈ 20, 475.32 km cos ( 28.5472 )
≈ 20, 475 km
≈ 19,293 km (since the number of significant digits is 5).
adj 17,986 km = , hyp c
(since the number of significant digits is 5).
57
Chapter 1
31. Consider this triangle:
32. Consider this triangle:
β
β
First, observe that β = 180 − ( 90 + 32 ) = 58 .
First, observe that β = 180 − ( 90 + 65 ) = 25 .
Since sin ( 32 ) =
Since sin ( 65 ) =
a opp = , we have hyp 12 ft.
a = (12 ft.) ( sin32 ) ≈ 6.4 ft.
Similarly, since b adj cos ( 32 ) = = , we have hyp 12 ft.
b = (12 ft.) ( cos32 ) ≈ 10 ft.
33. Consider this triangle:
a opp = , we have hyp 37 ft.
a = ( 37 ft.) ( sin 65 ) ≈ 34 ft. Similarly, since cos ( 65 ) =
b adj = , hyp 37 ft.
we have b = ( 37 ft.) ( cos 65 ) ≈ 16 ft. 34. Consider this triangle: 12
44
β
β
First, observe that β = 180 − ( 44 + 90 ) = 46.
First, observe that β = 180 − (12 + 90 ) = 78.
Since cos ( 44 ) =
Since cos (12 ) =
adj 2.6 cm , = hyp c
we have
adj 10 m , = hyp c
we have 2.6 cm c= ≈ 3.6 cm cos ( 44 )
Similarly, since opp a = tan ( 44 ) = , adj 2.6 cm we have a = ( 2.6 cm ) tan ( 44 ) ≈ 2.5 cm.
c=
10 m ≈ 10.2 m cos (12 )
Similarly, since tan (12 ) =
opp a = , adj 10 m
we have a = (10 m ) tan (12 ) ≈ 2.13 m.
58
Section 1.5
35. Consider this triangle:
36. Consider this triangle:
60
β
β First, observe that β = 180 − ( 60 + 90 ) = 30 . Since cos ( 60 ) =
b adj = , hyp 5 in.
we have b = ( 5 in.) cos ( 60 ) ≈ 3 in. Since sin ( 60 ) =
a opp = , hyp 5 in.
we have a = ( 5 in.) ( sin 60 ) ≈ 4 in. 37. Consider this triangle:
First, observe that β = 180 − ( 9.67 + 90 ) = 80.33 . Since cos ( 9.67 ) =
b adj = , hyp 5.38 in.
we have b = ( 5.38 in.) cos ( 9.67 ) ≈ 5.30 in. Since sin ( 9.67 ) =
a opp = , hyp 5.38 in.
we have a = ( 5.38 in.) ( sin 9.67 ) ≈ 0.904 in. 38. Consider this triangle:
α
α
45
72
First, observe that α = 180 − ( 90 + 72 ) = 18 .
First, observe that α = 180 − ( 90 + 45 ) = 45 .
Since sin ( 72 ) =
Since sin ( 45 ) =
b opp = , hyp 9.7 mm
b opp = , hyp 7.8 mm
we have b = ( 9.7 mm ) ( sin 72 ) ≈ 9.2 mm .
we have b = ( 7.8 mm ) ( sin 45 ) ≈ 5.5 mm .
Similarly, since
Similarly, since
cos ( 72 ) =
a adj = , hyp 9.7 mm
we have a = ( 9.7 mm ) ( cos 72 ) ≈ 3.0 mm .
cos ( 45 ) =
a adj = , hyp 7.8 mm
we have a = ( 7.8 mm ) ( cos 45 ) ≈ 5.5 mm .
59
Chapter 1
39. Consider this triangle:
40. Consider this triangle:
α
54.2
β
47.2
First, observe that β = 180 − ( 90 + 54.2 ) = 35.8 . Since sin ( 54.2 ) =
opp 111 mi. = , hyp c
we have
c=
111 mi. ≈ 137 mi. sin ( 54.2 )
Similarly, since opp 111 mi. tan ( 54.2 ) = = , adj b we have 111 mi. b= ≈ 80.1 mi. tan ( 54.2 ) 41. Consider this triangle:
α
First, observe that α = 180 − ( 90 + 47.2 ) = 42.8 . Since cos ( 47.2 ) =
adj 9.75 mi. = , hyp c
we have
c=
9.75 mi. ≈ 14.4 mi. cos ( 47.2 )
Similarly, since b opp tan ( 47.2 ) = = , adj 9.75 mi. we have b = ( 9.75 mi.) tan ( 47.2 ) ≈ 10.5 mi. First, observe that α = 180 − ( 45 + 90 ) = 45 . Since the triangle is isosceles, we know that a = 10.2 km and c = 10.2 2 ≈ 14 km .
45
60
Section 1.5
42. Consider this triangle:
α
Since sin ( 85.5 ) =
opp 14.3 ft. = , hyp c
we have 14.3 ft. ≈ 14.344 ft. sin (85.5 ) Similarly, since opp 14.3 ft. tan ( 85.5 ) = = , adj a 14.3 ft. we have a = ≈ 1.13 ft. tan ( 85.5 )
c=
85.5
First, observe that α = 180 − ( 85.5 + 90 ) = 4.5 .
43. Consider this triangle:
44. Consider this triangle:
α
α
β
β First, observe that β = 180 − ( 90 + 28 23′ ) = 61 37′
1 = 61 + 37′ ≈ 61.62 ′ 60 1 α = 28 23′ = 28 + 23′ ≈ 28.38 ′ 60 adj 1,734 ft. Since cos ( 28.38 ) = = , hyp c we have 1,734 ft. c= ≈ 1,971 ft. cos ( 28.38 ) Similarly, since opp a tan ( 28.38 ) = = , adj 1,734 ft. we have a = (1,734 ft.) tan ( 28.38 ) ≈ 936.9 ft.
First, observe that β = 180 − ( 90 + 72 59′ ) = 17 1′ 1 ′ = 17 + 1 ≈ 17.02 60′ 1 α = 72 59′ = 72 + 59′ ≈ 72.98 ′ 60
Since tan ( 72.98 ) =
opp 2,175 ft. = , adj b
we have 2,175 ft. ≈ 665.7 ft. tan ( 72.98 ) Similarly, since opp 2,175 ft. sin ( 72.98 ) = = , hyp c we have 2,175 ft. c= ≈ 2,275 ft. sin ( 72.98 )
b=
61
Chapter 1
45. Consider this triangle:
46. Consider this triangle:
α
α
β
β 42.5 ft. , we obtain 28.7 ft. 42.5 α = tan −1 ≈ 56.0 . 28.7 So, β ≈ 180 − ( 90 + 56.0 ) ≈ 34.0 .
First, since tan α =
Next, using the Pythagorean Theorem, we obtain 2 2 c 2 = ( 28.7 ft.) + ( 42.5 ft.) = 2,629.94 ft.2 and so, c ≈ 51.3 ft.
19.8 ft. , we obtain 48.7 ft. 19.8 α = sin −1 ≈ 24.0 . 48.7 19.8 ft. Similarly, since cos β = , we obtain 48.7 ft. 19.8 β = cos −1 ≈ 66.0 . 48.7 Next, using the Pythagorean Theorem, we obtain
First, since sin α =
b 2 + (19.8 ft.) = ( 48.7 ft.) 2
2
b 2 = 1,979.65 ft.2 b ≈ 44.5 ft.
47. Consider this triangle:
α
β First, since sin α =
35,236 km , 42,766 km
we obtain 35,236 α = sin −1 ≈ 55.480 . 42,766 35,236 km Similarly, since cos β = , 42,766 km we obtain 35,236 β = cos −1 ≈ 34.520 . 42,766 Next, using the Pythagorean Theorem, we obtain b 2 + ( 35,236 km ) = ( 42,766 km ) 2
2
b 2 = 587,355,060 km 2 b ≈ 24,235 km
62
Section 1.5
48. Consider this triangle:
α
β First, since cos α =
0.1245 mm , 0.8763 mm
we obtain 0.1245 ≈ 81.83 . 0.8763 0.1245 mm Similarly, since sin β = , 0.8763 mm we obtain 0.1245 β = sin −1 ≈ 8.17 . 0.8763 Next, using the Pythagorean Theorem, we obtain
α = cos −1
a 2 + ( 0.1245 mm ) = ( 0.8763 mm ) 2
a 2 = 0.752401 mm 2 a ≈ 0.8674 mm
49. Consider this diagram:
50. Consider this diagram:
3
1
Observe that tan1 =
x=
5 ft. , so that x
2
Observe that tan3 =
5 ft. ≈ 286 ft. . tan1
x=
63
5 ft. , so that x
5 ft. ≈ 95 ft. . tan3
Chapter 1
51. Consider this triangle:
52. Consider this triangle:
36
36
100 ft. , so that H
Observe that sin36 =
Observe that sin36 =
So, the altitude between the two planes should be approximately 88 ft.
100 ft. ≈ 170 ft. sin36 So, the hose should be approximately 170 ft. long.
53. Consider this triangle:
54. Consider this triangle:
36
θ
a , so that 150 ft. a = ( sin36 ) (150 ft.) ≈ 88 ft.
Observe that sin3 = that
a , so 5,000 ft.
a = ( sin3 ) ( 5,000 ft.) ≈ 262 ft.
So, rounding to two significant digits, we see that the altitude should be approximately 260 ft.
H=
Let θ = glide slope angle. Observe that 450 ft. sin θ = , so that 5,200 ft. 450 ≈ 5 . . 5,200
θ = sin −1
Since an angle of 3 will allow the pilot to see both red and white lights while higher angles enable him to see only white lights, we conclude that in this case the pilot will see only white lights.
64
Section 1.5
55. Consider this triangle:
56. Consider this triangle:
θ
θ Let θ = glide slope angle. Observe that 3,000 ft. sin θ = , so that 15,500 ft. 3,000 ≈ 11.0 . 15,500 This angle is way too low (since the typical range is 18 − 20 ).
θ = sin −1
57. Consider the following diagram:
Let θ = glide slope angle. Observe that 2,500 ft. sin θ = , so that 7,800 ft. 2,500 ≈ 18.0 . 7,800 So, she is within safety specifications to land.
θ = sin −1
58. Consider the following diagram:
r , we have 150 ft. r = (150 ft.) ( tan15 ) ≈ 40 ft.
r , we have 500 ft. 4 = ( 500 ft.) ( tan15 ) ≈ 134 ft.
Since tan15 =
Since tan15 =
So, the diameter of the circle is 80 ft.
So, the diameter of the circle is 268 ft., which we round to 270 ft.
65
Chapter 1
59. Consider this triangle:
60. Consider this triangle:
θ
θ
First, note that θ = 1′′ ≈ 0.0002778. x Since tan ( 0.0002778 ) = , 35,000 km we have that x = ( 35,000 km ) tan ( 0.0002778 )
1 ′′ First, note that θ = ≈ 0.00013889. 2 x Since tan ( 0.00013889 ) = , 35,000 km we have that x = ( 35,000 km ) tan ( 0.00013889 )
≈ 0.1697 km ≈ 170 m
≈ 0.08484 km ≈ 85 m 61. Consider this triangle:
62. Consider this triangle:
θ
θ
First, note that 0.010 km = 10 m. 0.010 km , we have Since tan θ = 35,000 km that 0.010 θ = tan−1 ≈ 0.000016 . 35,000
First, note that 0.030 km = 30 m. 0.030 km Since tan θ = , we have that 35,000 km 0.030 ≈ 0.000049 . 35,000
θ = tan−1
66
Section 1.5
63. Consider this diagram:
64. Consider this diagram:
Observe that tan30 =
x=
15 ft. , so that x
Observe that cos55 =
15 ft. ≈ 26 ft. tan30
65. Consider this diagram:
x=
50 ft. , so that x
50 ft. ≈ 87 ft. cos55
66. Consider this diagram:
x , so that 92 x = 92 tan37° ≈ 69.33 feet
Observe that tan37° =
x , so that 2 x = 2 tan 72° ≈ 6.16 miles
Observe that tan 72° =
67
Chapter 1
67. Consider this diagram:
68. Consider this diagram:
x Observe that tan18° = , so that 150 x = 150 tan18° ≈ 48.7 feet
Observe that tan 75° =
69. Consider this diagram:
70. Consider this diagram:
Observe that tan87° =
x , so that 250 x = 250 tan87° ≈ 4770.28 feet
Observe that tan82° =
71. Consider this diagram:
Observe that sin 65 =
x , so that 150 x = 150 tan 75° ≈ 559.8 feet
x , so that 220 x = 220 tan 62° ≈ 1565.4 feet 4,000 ft. , x
so that 4,000 ft. x= ≈ 4, 414 ft. . sin 65
68
Section 1.5
72. Consider this diagram:
Observe that tan ( 33 + θ ) =
10 mi. , 6 mi.
so that
10 33 + θ = tan −1 6 10 θ = tan −1 − 33 ≈ 26 6
θ
So, the bearing would be N 26 E .
73. Consider this diagram:
Observe that tan ( 35 + θ ) =
8 mi. , 3 mi.
so that 8 35 + θ = tan −1 3 8 θ = tan −1 − 35 ≈ 34 3 So, the bearing would be N 34W .
θ
74. Consider this diagram:
Observe that tan ( 48 + θ ) =
17 mi. , 6 mi.
so that
θ
17 48 + θ = tan −1 6 17 θ = tan −1 − 48 ≈ 23 6 So, the bearing would be N 23 E
69
Chapter 1
75. We need to first consider a triangle that lies in the tennis court whose hypotenuse is the shadow of the path of the serve. This triangle has the following dimensions:
Here, S denotes the shadow of the serve in the court. Using the Pythagorean Theorem, we see that 1642 + 7202 = S 2 , so that S ≈ 738.44 in. ≈ 61.54 ft. Next, we consider the triangle that has as its hypotenuse the actual path of the serve, and whose base is the hypotenuse of the above triangle, and whose height is the height of the ball above ground the instant it is struck – as such, this triangle sticks up out of the court, perpendicular to it. This is illustrated below:
θ
We seek the measure of angle θ . Observe that tan θ =
9 ft. , so that 61.54 ft.
9 ft. ≈ 8.32 = 8° 61.54 ft.
θ = tan −1
70
Section 1.5
76. We need to first consider a triangle that lies in the tennis court whose hypotenuse is the shadow of the path of the serve. This triangle has the following dimensions:
Here, S denotes the shadow of the serve in the court. Using the Pythagorean Theorem, we see that 164 2 + 7202 = S 2 , so that S ≈ 738.44 in. ≈ 61.54 ft. Next, we consider the triangle that has as its hypotenuse the actual path of the serve, and whose base is the hypotenuse of the above triangle, and whose height is the height of the ball above ground the instant it is struck – as such, this triangle sticks up out of the court, perpendicular to it. This is illustrated below:
We seek the height h at which the ball must be struck so that it hits the ground at an h , so that h = 61.54 ft.(tan 44°) ≈ 59.4 ft. angle of 44°. Observe that tan 44° = 61.54 ft.
71
Chapter 1
77. Consider the following diagram:
78. Consider the following diagram:
θ α
θ α
First, observe that tan α =
2.219 , 2.354
First, observe that tan α =
2.097 , 2.354
so that
so that
2.219 ≈ 43.309 . 2.354 Since α + θ = 180 , we conclude that
2.097 ≈ 41.695 . 2.354 Since α + θ = 180 , we conclude that
θ ≈ 136.69 .
θ ≈ 138.30 .
79. Consider the following triangle:
80. Consider the following triangle:
α = tan−1
α = tan−1
Observe that h sin 75 = 100 h = 100 sin 75 ≈ 97 miles.
θ Observe that 25 tan θ = 150 θ ≈ 9.462. So, the heading is 80.5 east of north.
72
Section 1.5
81. The original canal has the following dimensions:
Observe that tan30 =
The diagram for the present day is:
w
2
5
=
w . 10
Solving for w yields
1 10 3 w = 10 tan30 = 10 = 3 ft. 3
Using similar triangles and solving for d yields d 5 = 4 10 3 3 20(3) 6 d= = ≈ 3.5 ft. 10 3 3
82. Consider the following diagram:
θ θ
Observe that 5 5 tan ( θ2 ) = 2 = θ2 = tan −1 ( 58 ) θ = 2 tan −1 ( 58 ) ≈ 1.117198 radians. 4 8 Converting to degrees (by multiplying by 180 π ) results in approximately 64. So, yes, erosion has widened the shape of the canal since the angle has increased by 4 degrees.
73
Chapter 1
83. Using the right-triangle suggested by the diagram (with the femur being the hypotenuse 18 inches in length and h denoting the height of elevation), we see that h sin15 = h = (18 in.) sin15 ≈ 4.7 in. 18 in. 84. Using the right-triangle suggested by the diagram (with the femur being the hypotenuse 18 inches in length, θ being the angle of elevation, and the height of elevation 6.2 in.), we see that 6.2 in. 6.2 sin θ = θ = sin −1 ≈ 20 18 in. 18 85. Should be β = tan −1 (80), not tan(80). 86. The calculation of b is correct. However, this value should not be then used to determine the value of a since doing so would introduce TWO computational a (as errors (due to round off). One should, instead, compute a using cos56 = 15 in Example 1). 87. True
88. True
89. False. Knowing the measures of all angles in a right triangle would not enable you to find any of the side lengths. In fact, you could produce infinitely many similar triangles with these same angles.
90. False. The angle opposite the hypotenuse in a right triangle is 90. If we only knew this and no other side length, we could not determine the measures of either of the two remaining angles.
91. False. There are only four significant digits here.
92. True
93. True
94. False. Use the Pythagorean theorem.
74
Section 1.5
95. Consider this diagram:
Ultimately, we wish to find x. We set up a system of two equations in x and y which we solve simultaneously: x x tan38 = (1) tan51 = (2) y + 1,900 y Equations (1) and (2) can be written equivalently as: x − ( tan38 ) y = 1,900 ( tan38 ) ( 3 ) x − ( tan51 ) y = 0
(4)
Subtracting (3) – (4) yields: − ( tan38 ) y + ( tan51 ) y = 1,900 ( tan38 ) so that y=
1,900 ( tan38 )
( tan51 ) − ( tan38 )
≈ 3,272.50 ft.
Substitute this into (4) to find that x is approximately 4,000 ft.
96. Consider this diagram:
97. Consider this diagram: 65
15
40
Let H = height of the shuttle (in miles) after 10 seconds of flight. H Observe that tan15 = , so that 8 mi. H = ( 8 mi.) ( tan15 ) ≈ 2.14 mi. Rounding to one significant digit, we see that the shuttle is approximately 2 miles above ground after 10 seconds of flight.
45
65
Observe that cos 45 = 12y y = 12 cos 45 cos 65 = 12x x = 12 cos 65 So, the width of the sidewalk is y − x = 12 cos 45 − cos 65 ≈ 3.4 ft.
75
Chapter 1
98. Consider the following diagram:
99. Consider the following diagram:
Note all angles shown – those not labeled are right angles. The angles were found using complementary and supplementary angles. Observe that Observe that yz y cos 48 = 10 yz = 10 cos 48 ≈ 6.691. tan35 = 5.33 y = 5.33tan35 . Also, we have Hence, xy +y cos 42 = 10 xy = 10 cos 42 ≈ 7.4314. tan 47 = x5.33 = x + 5.33tan35 , 5.33 Since we are given that xz = 10, we so that conclude that the perimeter of the x = 5.33 tan 47 − tan35 ≈ 1.98 . triangle A is approximately 24 units.
100. Consider the following diagram:
101. sin −1 ( sin 40 ) = 40
θ θ
102. cos −1 ( cos17 ) = 17 103. cos ( cos −1 0.8 ) = 0.8
Observe that −1 250 tan θ = 250 8 θ = tan ( 8 ) ≈ 88
104. sin ( sin −1 0.3 ) = 0.3
105. sin −1 ( sin θ ) = θ , for any
106. cos −1 ( cos θ ) = θ , for any
0 ≤ θ ≤ 90
0 ≤ θ ≤ 90
76
Chapter 1 Review Solutions ----------------------------------------------------------------------1. a) complement: 90 − 28 = 62
2. a) complement: 90 − 17 = 73
b) supplement: 180 − 28 = 152
b) supplement: 180 − 17 = 163
3.
4.
a) complement: 90 − 35 = 55
a) complement: 90 − 78 = 12
b) supplement: 180 − 35 = 145
b) supplement: 180 − 78 = 102
5. Since α + β + γ = 180 , we know that
6. Since α + β + γ = 180 , we know that
120 + 35 + γ = 180 and so, γ = 25 .
105 + 25 + γ = 180 and so, γ = 50 .
7. Since α + β + γ = 180 , we know that 7 β ) + β + ( β ) = 180 and so, β = 20 . (
8. Since α + β + γ = 180 , we know that
= 155
= 9β
= 130
6 β ) + β + ( β ) = 180 and so, β = 22.5 . ( = 8β
Thus, α = 7 β = 140 and γ = β = 20 .
Thus, α = 6 β = 135 and γ = β = 22.5 .
9. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 4 2 + b 2 = 12 2 , which simplifies to 16 + b 2 = 144 and then to, b 2 = 128 , so we conclude that b = 128 = 8 2 .
10. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes a 2 + 92 = 152 , which simplifies to a 2 + 81 = 225 and then to, a 2 = 144 , so we conclude that a = 12 .
11. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 7 2 + 4 2 = c 2 , which simplifies to c 2 = 65, so we conclude that c = 65 .
12. Since this is a right triangle, we know from the Pythagorean Theorem that a 2 + b2 = c 2 . Using the given information, this becomes 10 2 + 82 = c 2 , which simplifies to c 2 = 164, so we conclude that c = 164 = 2 41 . 77
Chapter 1
13. If the length of a leg in a 45 − 45 − 90 triangle is 12 yards, then the hypotenuse of this triangle has length 12 2 yd.
14. Let x be the length of a leg in a given 45 − 45 − 90 triangle. If the hypotenuse 8 = 4 = 2. Hence, of this triangle has length 8 ft., then 2 x = 8, so that x = 2 the length of each of the two legs is 2 ft. 15. Since the lengths of the two legs of the given30 − 60 − 90 triangle are x and 3 x, the shorter leg must have length x. Hence, using the given information, we know that x = 3 ft. Thus, the other leg has length 3 3 ft., and the hypotenuse has length 6 ft.
16. The length of the hypotenuse is 2 x = 12 km. So, x = 6. Thus, the length of the shorter leg is 6 km, and the length of the longer leg is 6 3 km. 17. Consider the following diagram:
Let α = measure of the desired angle between the hour hand and minute hand. Since the measure of the angle formed using two rays emanating from the center 1 of the clock out toward consecutive hours is always (360 ) = 30 , it immediately 12 follows that α = 5 ⋅ 30 = 150 .
78
Chapter 1 Review
18. Consider the following diagram:
Let α = measure of the desired angle between the hour hand and minute hand. Since the measure of the angle formed using two rays emanating from the center 1 of the clock out toward consecutive hours is always (360 ) = 30 , it immediately 12 follows that α = 3 ⋅ 30 = 90 .
19. G = 75 (vertical angles)
20. D = 105 (interior angles)
21. C = 75 (alternate interior angles)
22. E = 105 (supplementary angles)
23. B = 75 (corresponding angles)
24. A = 105 (since B = 75 , and A and B are supplementary )
25. By similarity,
A B 10 8 = , so that = . Solving for E yields E = 4 . 5 E D E
26. By similarity,
A B 15 12 = , so that = . Solving for D yields D = 5 . D 4 D E
27. First, convert all quantities involved to the common unit km. Observe that by A C 81 km C . Solving for C yields = , so that = similarity, 0.0045 km 0.0082 km D F 81 km C = ( 0.0082 km ) = 147.6 km 0.0045 km
79
Chapter 1
28. First, convert all quantities involved to the common unit m. Observe that by B C B 800 cm . Solving for B yields = , so that = similarity, 8 cm 14 cm E F 8 cm B = ( 800 cm ) ≈ 457 cm = 4.57 m . 14 cm 29. Consider the following two diagrams:
Let T = height of the tree (in meters). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain T 4m 4m so that T = ( 9.6 m ) = = 32 m. 9.6 m 1.2 m 1.2 m So, the tree is 32 m tall.
30. Note that 1 ft. 9 in. = 1.75 ft. Now, consider the following two diagrams.
80
Chapter 1 Review
Let M = height of the man (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain M 4 ft. 4 ft. so that M = (1.75 ft.) = = 7 ft. 1.75 ft. 1 ft. 1 ft. So, the man is 7 feet tall.
31. Let x = width of the built-in refrigerator (in inches). Using similarity, we obtain 4 in. 1 in. 3 4 3 ft. so that x = in. = = 4 ft. = 48 in. 3 ft. x 3 1 in. So, the built-in refrigerator is 48 inches wide. 32. Let x = width of the pantry (in inches). Using similarity, we obtain 5 in. 1 in. 4 5 3 ft. 15 15 = so that x = in. ft. = (12 in.) = 45 in. = 3 ft. x 4 1 in. 4 4 So, the pantry is 45 in. long. 33. Consider the following diagram:
34. Using the diagram and value of h in Exercise 33, we see that sin θ =
35. Using the diagram and information from Exercise 33, we see that
θ
sec θ =
From the Pythagorean Theorem, we see that h = 32 + 2 2 = 13. Thus, cos θ =
3 3 13 . = 13 13
1 = cos θ
13 . 2
36. Using Exercise 34, we see that
2 2 13 . = 13 13
csc θ =
81
1 = sin θ
13 . 3
Chapter 1
37. First, note that using the diagram and value of h in Exercise 33, we see that 3 sin θ = . 13 As such, we have 3 sin θ 3 = 13 = . tan θ = 2 cos θ 2 13
38. Using Exercise 37, we see that 1 2 cot θ = = . tan θ 3
41.
42. csc ( 60 ) = sec ( 90 − 60 ) = sec 30
( )
tan ( 45 ) = cot ( 90 − 45 ) = cot 45
43.
(
sin ( 30 − x ) = cos 90 − ( 30 − x )
= cos x + 60
)
(
(
45.
(
= sec 45 + x
47. b
48. a
)
)
49. b
1 sin30 1 3 2 = = = 53. tan30 = 3 cos30 3 3 2
57. csc 45 =
3 sin 60 2 = = 1 cos 60 2
1 1 = = 2 sin 45 2
)
( )
44.
(
cos ( 55 + A) = sin 90 − ( 55 + A)
(
46.
(
3
2
)
sec ( 60 − θ ) = csc 90 − ( 60 − θ )
(
= csc θ + 30
55. tan 60 =
(
40. cos ( A) = sin 90 − A
= sin 35 − A
csc ( 45 − x ) = sec 90 − ( 45 − x )
)
( )
39. cos ( 90 − 30 ) = cos 60
50. a 54. tan 45 =
51. c
)
)
)
52. c
2 sin 45 2 = =1 2 cos 45 2
56. csc 30 =
1 1 = 1 = 2 sin30 2
58. csc 60 =
1 1 2 2 3 = = = 3 sin 60 3 3 2
82
Chapter 1 Review
59. 1 1 2 2 3 sec 30 = = = = 3 cos30 3 3 2
60. sec 45 =
1 1 = 1 = cos 45 2
2
3
61. sec 60 =
1 1 = 1 = 2 cos 60 2
62. cot 30 =
1 1 = 3 = tan30 3
63. cot 45 =
1 1 = = 1 tan 45 1
64. cot 60 =
1 1 3 = = tan 60 3 3
65. 0.6691
66. 0.5446
67. 0.9548
68. 0.4706
69. 1.5399
70. 1.0446
71. 1.5477
72. 2.7837
17 73. First, note that 17′ = ≈ 0.28. 60
15 74. First, note that 15′ = = 0.25. 60
So, 3917′ ≈ 39.28 .
So, 68 15′ = 68.25 .
75. First, note that
76. First, note that
25 25′′ = ≈ 0.007 3600
15 15′′ = ≈ 0.004 3600
45 45′ = = 0.75. 60
30 30′ = = 0.50. 60 So, 2930′25′′ = 29.507 .
So, 25 45′15′′ = 25.754 .
77. Observe that 42.25 = 42 + 0.25
78. Observe that 60.45 = 60 + 0.45
60′ = 42 + ( 0.25 ) 1 = 42 + 15′
So, 42.25 = 42 15′ .
60′ = 60 + ( 0.45 ) 1 = 60 + 27′
So, 60.45 = 6027′ .
83
Chapter 1
79. Observe that 30.175 = 30 + 0.175
80. Observe that 25.258 = 25 + 0.258
60′ = 30 + ( 0.175 ) 1 = 30 + 10.5′
60′ = 25 + ( 0.258 ) 1 = 25 + 15.48′
= 30 + 10′ + 0.5′
= 25 + 15′ + 0.48′
60′′ = 30 + 10′ + ( 0.5′ ) 1′ = 30 + 10′ + 30′′
60′′ = 25 + 15′ + ( 0.48′ ) 1′ = 25 + 15′ + 28.8′′
So, 30.175 = 3010′ 30′′ .
So, 25.258 ≈ 2515′ 29′′ .
81. First, observe that
82. First, observe that
15 3715′ = 37 + = 37.25. 60 So, sin ( 37.25 ) ≈ 0.6053.
35 4235′ = 42 + = 42.583. 60 So, cos ( 42.583 ) ≈ 0.7363.
83. First, observe that
84. First, observe that
48 61 48′ = 61 + = 61.80. 60 So, cos ( 61.80 ) ≈ 0.4726.
17 20 17′ = 20 + = 20.283. 60 So, sin ( 20.283 ) ≈ 0.3467.
85. Using ni sin (θi ) = nr sin (θ r ) ,
86. Using ni sin (θi ) = nr sin (θ r ) , observe
observe that 1.00 sin ( 60 ) = nr sin (35.26 )
that 1.00 sin ( 60 ) = nr sin (36.09 )
nr =
nr =
1.00 sin ( 60 ) sin ( 35.26 )
≈ 1.50 .
84
1.00 sin ( 60 ) sin ( 36.09 )
≈ 1.47 .
Chapter 1 Review
87. Consider this triangle:
88. Consider this triangle:
Since cos ( 25 ) =
adj a = , we have hyp 15 in.
a = (15 in.) cos ( 25 ) ≈ 13.59 in. ≈ 14 in.
89. Consider this triangle:
Since sin ( 50 ) =
opp a = , we have hyp 27 ft.
a = ( 27 ft.) sin ( 50 ) ≈ 20.68 ft. ≈ 21 ft.
90. Consider this triangle:
Since tan ( 33.5 ) =
opp a , = adj 21.9 mi.
we have a = ( 21.9 mi.) tan ( 33.5 ) ≈ 14.5 mi.
Since sin ( 47.45 ) =
opp 19.22 cm , = hyp c
we have c=
85
19.22 cm ≈ 26.09 cm sin ( 47.45 )
Chapter 1
91. Consider this triangle:
92. Consider this triangle:
β First, observe that
First, observe that
1 ≈ 75.16667 . ′ 60
1 ≈ 37.75. β = 37 45′ = 37 + 45′ 60′ opp b , = Since tan ( 37.75 ) = adj 120.0 yd. we have b = (120.0 yd.) tan ( 37.75 ) ≈ 92.91 yd.
Since sin ( 75.16667 ) =
93. Consider this triangle:
94. Consider this triangle:
β = 75 10′ = 75 + 10′
opp 96.5 km , = hyp c 96.5 km ≈ 99.8 km we have c = sin ( 75.16667 )
α
65
β First, observe that β = 180 − ( 90 + 30 ) = 60. Since sin ( 30 ) =
opp a , = hyp 21 ft.
we have a = ( 21 ft.) ( sin30 ) ≈ 11 ft. Similarly, since adj b cos ( 30 ) = , = hyp 21 ft. we have b = ( 21 ft.) ( cos30 ) ≈ 18 ft.
First, observe that α = 180 − ( 90 + 65 ) = 25. Since sin ( 65 ) =
opp b , = hyp 8.5 mm
we have b = ( 8.5 mm ) ( sin 65 ) ≈ 7.7 mm. Similarly, since adj a cos ( 65 ) = , = hyp 8.5 mm we have a = ( 8.5 mm ) ( cos 65 ) ≈ 3.6 mm.
86
Chapter 1 Review
95. Consider this triangle:
96. Consider this triangle:
β
β
First, observe that β = 180 − ( 90 + 48.5 ) = 41.5 . Since sin ( 48.5 ) =
First, observe that β = 180 − ( 90 + 30 15′ ) = 59 45′
1 = 59 + 45′ = 59.75 60′ 1 ′ ′ = 30.25 α = 30 15 = 30 + 15 60′
opp 215 mi. = , hyp c
we have c=
215 mi. ≈ 287 mi. sin ( 48.5 )
Similarly, since opp 215 mi. tan ( 48.5 ) = = , adj b we have 215 mi. b= ≈ 190 mi. tan ( 48.5 )
Since cos ( 30.25 ) =
adj 2,154 ft. = , hyp c
we have c=
2,154 ft. ≈ 2, 494 ft. cos ( 30.25 )
Similarly, since tan ( 30.25 ) =
opp a = , adj 2,154 ft.
we have a = ( 2,154 ft.) tan ( 30.25 ) ≈ 1,256 ft.
87
Chapter 1
97. Consider this triangle:
98. Consider this triangle:
α
α
β
β 30.5 ft. First, since tan α = , we obtain 45.7 ft. 30.5 α = tan−1 ≈ 33.7 . 45.7 So, β ≈ 180 − ( 90 + 33.7 ) ≈ 56.3 . Next, using the Pythagorean Theorem, we obtain 2 2 c 2 = ( 30.5 ft.) + ( 45.7 ft.) = 3,018.74 ft.2 and so, c ≈ 54.9 ft.
First, since sin α =
11,798 km , we obtain 32,525 km
11,798 ≈ 21.268 . 32,525 11,798 km , Similarly, since cos β = 32,525 km we obtain 11,798 β = cos −1 ≈ 68.732 . 32,525 Next, using the Pythagorean Theorem, we obtain
α = sin −1
b 2 + (11,798 km ) = ( 32,525 km ) 2
2
b 2 = 918,682,821 km 2 b ≈ 30,310 km
99. Consider this triangle:
100. Consider this triangle:
100 ft. , so that H
Observe that sin30 =
Observe that sin30 =
So, the altitude should be about 75 ft.
100 ft. ≈ 200 ft. . sin30 So, the hose should be about 200 ft. long.
a , so that 150 ft. a = ( sin30 ) (150 ft.) ≈ 75 ft. .
H=
88
Chapter 1 Review
101. Consider this diagram:
102. Consider this diagram:
θ
θ
Observe that tan (15 + θ ) =
10 mi. , so 3 mi.
that
10 15 + θ = tan −1 3 10 θ = tan −1 − 15 ≈ 58 3 So, the bearing would be N 58 E
Observe that tan ( 20 + θ ) =
9 mi. , so 5 mi.
that
9 20 + θ = tan −1 5 9 θ = tan−1 − 20 ≈ 41 5 So, the bearing would be N 41W .
103. Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and 3 x , the shorter leg must have length x. Hence, using the given information, we know that x = 41.32 ft. Thus, the two legs have lengths 41.32 ft. and 41.32 3 ≈ 71.57 ft., and the hypotenuse has length 82.64 ft. 104. Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and 3 x, the shorter leg must have length x. Hence, using the given information, we know that x 3 = 87.65 cm Thus, the two legs have lengths 87.65 cm and 87.65 ≈ 50.60 cm, and the hypotenuse has length 101.20 cm. 3 1 1 ≈ ≈ 1.02041 sin 78.4 0.980 b. The keystroke sequence 78.4, sin, 1/x yields 1.02085.
105. a. sin78.4 ≈ 0.980 and so,
89
Chapter 1
1 1 ≈ ≈ 1.43885 tan34.8 0.695 b. The keystroke sequence 34.8, tan, 1/x yields 1.43881.
106. a. tan34.8 ≈ 0.695 and so,
107. 2.612
108. 0.125
109. −1
110. Error – dividing by zero.
90
Chapter 1 Practice Test Solutions ---------------------------------------------------------------1. Let x = measure of smallest angle in the triangle. Then, (1) implies that 5x = measure of the largest angle in the triangle, (2) implies that 3x = measure of the third angle in the triangle. So, x + 3x + 5x = 180 so that 9x = 180 and thus, x = 20. Thus, the three angles are 20 , 60 , 100 .
The angle opposite the 30 angle in a 30 − 60 − 90 triangle is the smallest of the three angles. So, the other leg must have length 5 3 cm and the hypotenuse must have length 10 cm.
2. Consider the following triangle:
60
30
3. Consider the following two diagrams:
θ θ
Let H = height of the Grand Canyon (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain 5 ft. 5 ft. H = = so that 600 ft. H ( ) 1 = 6,000 ft. 1 600 ft. 2 ft. 2 ft. So, the Grand Canyon is 6,000 ft. tall.
91
Chapter 1
4. Consider the following triangle:
θ
From the Pythagorean Theorem, we see that c = 32 + 12 = 10 .
a. sin θ =
1 10 = 3 cos θ 1 10 e. csc θ = = = 10 1 sin θ 1 f. cot θ = =3 tan θ
1 10 = 10 10
d. sec θ =
3 3 10 = 10 10 1 sin θ 10 = 1 c. tan θ = = 3 cos θ 3 10
b. cos θ =
5. a. sin ( 90 − θ ) = cos θ =
3 10 (by Exercise 4b) 10 b. sec ( 90 − θ ) = csc θ = 10 (by Exercise 4e) 1 c. cot ( 90 − θ ) = tan θ = (by Exercise 4c) 3
6. In the following computations, we use sin θ 1 1 1 tan θ = , cot θ = , sec θ = , csc θ = cos θ tan θ cos θ sin θ
θ 30 45 60
sin θ 1 2
2 2 3 2
cos θ
tan θ
3 2 2 2 1 2
3 3 1 3
92
cot θ 3
1 3 3
sec θ
2 3 3 2 2
csc θ 2
2
2 3 3
Chapter 1 Practice Test
7. sec ( 42.8 ) =
1 ≈ 1.3629 cos ( 42.8 )
9. First, note that
8. The first value ( cos θ = 23 ) is an exact value of the cosine function, while the second ( cos θ = 0.66667 ) would serve as an approximation to the first. 10. Consider the following triangle:
20 20′′ = ≈ 0.00556 3600
45 45′ = = 0.75 . 60 So, 33 45′20′′ ≈ 33.756 . Since tan ( 20 ) = b=
opp 10 cm = , we have adj b
10 cm ≈ 27 cm . tan ( 20 )
9.2 km , we obtain 23 km 9.2 β = cos −1 ≈ 66 . 23
11. Consider this triangle:
Since cos β =
β 12. Consider this triangle:
13. Consider this triangle:
α β
13.3 , we obtain Since sin β = 14.0
Since cos50 = 12 cmm , we see that
13.3 β = sin −1 ( 14.0 ) ≈ 71.8.
mm c = 12 ≈ 19 mm. cos50
93
Chapter 1
14. Consider this triangle:
15. Consider this triangle:
α
α
First, we convert α to DD: 3310′ = 33 + 10′ 601 ′ ≈ 33.17
Since tan α = 3.45 6.78 , we obtain
( )
Since sin ( 33.17 ) = 47am , we see that
α = tan−1 ( 3.45 6.78 ) ≈ 27.0 .
a = ( 47 m ) sin ( 33.17 ) ≈ 26 m.
16. Consider the following diagram:
Let α = measure of the desired angle between the hour hand and minute hand. Since the measure of the angle formed using two rays emanating from the center of 1 the clock out toward consecutive hours is always (360 ) = 30 , it immediately 12 follows that α = 30 .
17. Using ni sin (θi ) = nr sin (θ r ) , observe that
1.00 sin ( 25 ) = nr sin (16 ) nr =
1.00 sin ( 25 ) sin (16 )
94
≈ 1.53 .
Chapter 1 Practice Test
r , we have 150 ft. r = (150 ft.) ( tan 20 ) ≈ 54.6 ft.
18. Consider the following diagram:
Since tan 20 =
So, the diameter of the circle is approximately 109 ft.
19. 44.27 = 44 + ( 0.27 )
( ) = 44 + 16.20′ = 44 + 16′ + ( 0.20 )′ ( ) = 44 16′12′′
60′ 1
( )
(
60′′ 1′
)
1 20. 2210′23′′ = 22 + 10′ 601 ′ + 23′′ 3600 ′′ ≈ 22.1731
1 21. Since csc 75 = sin75 =
4 6+ 2
cos 45 3 = + 22. sin 60 + cot 30 2
2 3 1
2 2
, we have cos15 = sin 75 =
=
6+ 2 4
.
3 2 1 3 6 3 3+ 6 + ⋅ = + = 2 2 2 6 6 3
2
23. 12 40′ + 55 49′ = 6789′ = 67 + (60 + 29)′ = 67 + 60 ′ + 29′ = 68 29′. =1
24. 8227′ − 3539′ = 8187′ − 3539′ = ( 87 − 35 ) + ( 87 − 39 )′ = 46 48′
25. Consider the following triangle:
We have tan 72 = 200 hyards ,
so that h = ( 200 yards ) tan 72 ≈ 620 yards
95
CHAPTER 2 Section 2.1 Solutions -------------------------------------------------------------------------------1. QI
2. QII
3. QII
4. QII
5. QIV
6. QIV
7. y-axis
8. x-axis
9. Since −540 = −360 − 180, and −360 corresponds to one full revolution clockwise about the circle, the terminal side ends up on the x-axis.
10. Since −450 = −360 − 90, and −360 corresponds to one full revolution clockwise about the circle, the terminal side ends up on the y-axis.
11. QIII
13. QI
12. QIV
14. QII
15. Since 595 = 360 + 235, and 360 corresponds to one full revolution counterclockwise about the circle, the terminal side ends up in QIII.
16. Since 620 = 360 + 260, and 360 corresponds to one full revolution counterclockwise about the circle, the terminal side ends up in QIII.
17. Since 525 = 360 + 165, and 360 corresponds to one full revolution counterclockwise about the circle, the terminal side ends up in QII.
18. Since 1085 = 3 ( 360 ) + 5, and
19. Since −905 = 2 ( −360 ) − 185, and
20. Since −640 = −360 − 280, and −360 corresponds to one full revolution clockwise about the circle, the terminal side ends up in QI.
2 ( −360 ) corresponds to two full
revolutions clockwise about the circle, the terminal side ends up in QII.
3 ( 360 ) corresponds to three full
revolutions counterclockwise about the circle, the terminal side ends up in QI.
96
Section 2.1
21. Sketch of 135 :
22. Sketch of 225 :
23. Sketch of −405 :
24. Sketch of −450 :
25. Sketch of −225 :
26. Sketch of −330 :
97
Chapter 2
27. Sketch of 330 :
28. Sketch of −150 :
29. Sketch of 510 :
30. Sketch of −720 :
31. Sketch of 840 :
32. Sketch of −380 :
98
Section 2.1
33. Observe that −535 = −360 − 175. Since −175 + 185 = 360, the
coterminal angle corresponding to −535 is c . 34. Observe that −690 = −360 − 330. Since −330 + 30 = 360, the coterminal
angle corresponding to −690 is a . 35. Observe that 780 = 2 ( 360 ) + 60. So, 780 and 60 are coterminal angles
Thus, the answer is e . 36. Observe that 265 + −95 = 360 . So, they are coterminal angles Thus, the
answer is b . 37. Observe that −645 = −360 − 285. Since −285 + 75 = 360, we know
that −645 and 75 are coterminal angles Thus, the answer is f . 38. Observe that −560 = −360 − 200. Since −200 + 160 = 360, we know
that −560 and 160 are coterminal angles Thus, the answer is d . 39. Observe that −200 − 360 = −560, thus −200° and −560° are coterminal angles. So, the answer is d . 40. Observe that 750° − 2 ( 360° ) = 30°, thus 75 0° and 3 0° are coterminal angles.
So, the answer is a . 41. Observe that 545° − 360° =185°, thus 545 ° and 185° are coterminal angles. So, the answer is c . 42. Observe that 435° − 360° = 75°, thus 435 ° and 75° are coterminal angles. So, the answer is f . 43. Observe that −300° + 3 ( 360° ) = 780°, thus −300° and 780° are coterminal
angles. So, the answer is e . 44. Observe that −455° + 360° = −95°, thus −455° and −95° are coterminal angles. So, the answer is b .
99
Chapter 2
45. Observe that 412 = 360 + 52. So, the angle with the smallest positive
measure that is coterminal with 412 is 52 . 46. Observe that 379 = 360 + 19. So, the angle with the smallest positive
measure that is coterminal with 379 is 19 . 47. Observe that 268 = 360 − 92. So, the angle with the smallest positive
measure that is coterminal with −92 is 268 . 48. Observe that 173 = 360 − 187. So, the angle with the smallest positive
measure that is coterminal with −187 is 173 . 49. Observe that −390 = −360 − 30. Since 330 and −30 are coterminal, the
angle with the smallest positive measure that is coterminal with −390 is 330 . 50. Observe that 945 = 2 ( 360 ) + 225. So, the angle with the smallest positive
measure that is coterminal with 945 is 225 . 51. Observe that 510 = 360 + 150. So, the angle with the smallest positive
measure that is coterminal with 510 is 150 . 52. Observe that 1395 = 3 ( 360 ) + 315. So, the angle with the smallest positive
measure that is coterminal with 1395 is 315 . 53. Since the given angle is between 0 and 180 , the smallest angle coterminal with it is itself, namely 154. 54. 360 − 360 = 0 55. The terminal side of −1,050 coincides with that of −330. And, this is coterminal with 30. 56. The terminal side of 2,631 corresponds to that of 111. This is the angle with smallest positive measure with which it is coterminal.
100
Section 2.1
57. We seek the angle swept out by the second hand for a time duration of 3 min 20 sec. Note that 60 seconds corresponds to one complete revolution of the second-hand clockwise, and 20 seconds corresponds to 13 of one complete revolution clockwise. As such, we have 3 minutes corresponds to the angle 3 ( −360 ) = −1080 ,
20 seconds corresponds to the angle 13 ( −360 ) = −120.
Thus, we know that the result of such movement of the second-hand sweeps out the angle −1200 , regardless of its actual starting point on the clockface. 58. Note that the time span from Wednesday 3pm to Thursday 5pm constitutes 26 hours. For each hour, the hour hand moves 121 of a complete revolution 26 = 2 61 complete clockwise around the clockface. Hence, 26 hours corresponds to 12 revolutions around the clockface. Since one such complete revolution corresponds to −360 , the angle swept out during the given time period is 26 12
( −360 ) = −780 .
59. Alexandra’s serve corresponds to 3.5 × 360 , while Joe’s return is equivalent to
2 × ( −360 ) . The sum of these angles, namely 540 is the ball’s position.
60. Alexandra’s serve corresponds to 6 × 360 = 2,160 , while Joe’s return is
equivalent to 6 × ( −360 ) = −2,160.
61. Observe that Don: 900 = 2 ( 360 ) + 180 ,
Ron: −900 = 2 ( −360 ) − 180.
62. Observe that Dan: 3640 = 10 ( 360 ) + 40 ,
Stan: 1890 = 5 ( 360 ) + 90 .
Since 180 and −180 are coterminal, Don and Ron do end at the same place, namely 180 from their respective starting points.
Dan and Stan do not end at the same spot. Dan ends 40 counterclockwise from where he started, while Stan ends 90 counterclockwise from where he started.
101
Chapter 2
63. Note that each multiple of 5 cuts off a sector that corresponds to 18 of the entire circular face. Follow these instructions to obtain the total angle sum that corresponds to the total turning to unlock the lock with combination 35 – 5 – 20:
Step 1: 2 full clockwise turns:
2 ( 360 )
Step 2: Going clockwise, stop at 35:
Step 3: 1 full counterclockwise turn:
1 360 ) ( 8 (360 )
Step 4: From 35, go to 5 counterclockwise:
2 360 ) ( 8
5 360 ) ( 8 2 5 1 Now, the resulting total angle sum is 360 2 + +1+ + = 1, 440 . 8 8 8 Step 5: From 5, go to 20 clockwise:
64. Note that each multiple of 5 cuts off a sector that corresponds to 18 of the entire circular face. Follow these instructions to obtain the total angle sum that corresponds to the total turning to unlock the lock with combination 20 – 15 – 5:
Step 1: 2 full clockwise turns:
2 ( 360 )
Step 2: Going clockwise, stop at 20:
Step 3: 1 full counterclockwise turn:
4 360 ) ( 8 (360 )
Step 4: From 20, go to 15 counterclockwise:
7 360 ) ( 8
3 360 ) ( 8 4 7 3 Now, the resulting total angle sum is 360 2 + +1+ + = 1,665 . 8 8 8 Step 5: From 15, go to 5 clockwise:
102
Section 2.1
65. Since the Olympian makes 1.5 rotations counterclockwise, we know he turns 1.5 ( 360° ) = 540°.
66. Since the Olympian makes 1.5 rotations clockwise, we know she turns 1.5 ( −360° ) = −540°. 540° and −540° are both coterminal with 180° , thus they are coterminal with each other.
67. 1260° = 3.5 ( 360° ) , thus he made 3.5 68. He made two 1440° rotations for a total of 2880°. 2 880° = 8 ( 360° ) , thus he rotations. made 8 total rotations. 69. Consider the following triangle:
Since sin 60 = 100y ft. , we have y = (100 ft.) sin 60 ≈ 87 ft.
Similarly, since cos 60 = 100x ft. , we have x = (100 ft.) cos 60 ≈ 50 ft.
60
70. Consider the following triangle:
So, the horizontal distance is 50 ft. and the vertical distance is 87 ft.
Since sin 45 = 75yft. ,we have y = ( 75 ft.) sin 45 ≈ 53 ft.
Since the triangle is isosceles, x and y have the same length. So, the horizontal and vertical distances are both 53 ft. 45
71. QII
72. On the y-axis
103
Chapter 2
Note that 5 revolutions counterclockwise corresponds to 5 × 360 = 1,800 , and that 3 4 revolution
73. Consider the following diagram:
is equivalent to 3 4 ( 360 ) = 270. So, the angle made is the sum of these two, namely 2,070.
Note that
74. Consider the following diagram:
50 cos θ = 100 = 21 3 3 sin θ = −50 100 = − 2
Thus, θ = 53π , or equivalently 300. Hence, the resulting angle is 300.
θ
(
−
)
75. Coterminal angles are NOT supplementary angles. For instance, 300 and −60 are coterminal angles, but not supplementary (their sum does not add to 180 ). 76. All work except the last line is correct. Indeed, the terminal side of −150 is in QIII, not QII. 77. True, by definition.
104
Section 2.1
78. True. If α is an acute angle in standard position, then its terminal side is in QI if positive, and QIV if negative. However, an obtuse angle must have terminal side in QII if positive, and QIII if negative. 79. False. For example, when n = 30, the angles 30° and −30° are not coterminal. 80. Adding multiples of 360 yields coterminal angles. So, the desired angles are 30 + 360 k , where k is an integer. 81. All angles with positive measure coterminal with 30 are of the form 30 + n ( 360 ) , where n = 0, 1, 2, …
All angles with negative measure coterminal with 30 are of the form −330 − n ( 360 ) , where n = 0, 1, 2, …
82. Let x = measure of the desired angle. We are given that −360 < x < 0. Since the supplement of 130 is 50 , the
coterminal angle satisfying the given inequality is −310 . 83. 30 − 360 k, k = 0,1,2,3,... 84. These angles are given by −60 + 360 k , where k = 0, ± 1, ... , ± 5. So, there are 10 other angles coterminal with α . Including angle α itself, there are 11. 85. Consider the following diagram:
We are given ( x1 , y1 ) and ( x2 , y2 ) . Consider the new point ( x2 , y1 ) .
Observe that A2 + B 2 = d 2 by the Pythagorean theorem. This is the distance formula. d = ( x2 − x1 )2 + ( y2 − y1 )2
86. Since θ = 4.5 × 360 + 45 =
4 × 360 + ( 12 ×360 + 45 ) , the terminal side of
Back to positive x -axis
θ is in QIII. 105
= 225
Section 2.2 Solutions -------------------------------------------------------------------------------1. Consider the following diagram:
Using the Pythagorean Theorem, we obtain h = 32 + 4 2 = 5. opp 4 = sin θ = hyp 5 adj 3 = cos θ = hyp 5 opp 4 = tan θ = adj 3 1 3 cot θ = = tan θ 4 1 5 sec θ = = cos θ 3 1 5 csc θ = = sin θ 4
2. Consider the following diagram:
Using the Pythagorean Theorem, we obtain h = 12 2 + 52 = 13. opp 5 = sin θ = hyp 13 adj 12 = cos θ = hyp 13 opp 5 = tan θ = adj 12 1 12 cot θ = = tan θ 5 1 13 sec θ = = cos θ 12 1 13 csc θ = = sin θ 5
106
Section 2.2
3. Consider the following diagram:
sin θ =
opp 2 2 5 = = hyp 5 5
adj 1 5 = = hyp 5 5 opp 2 tan θ = = =2 adj 1 1 1 cot θ = = tan θ 2 1 1 sec θ = = 1 = 5 cos θ cos θ =
θ
5
csc θ =
Using the Pythagorean Theorem, we
1 1 5 = 2 = 2 sin θ 5
obtain h = 12 + 2 2 = 5. 4. Consider the following diagram:
sin θ =
opp 3 3 13 = = hyp 13 13
adj 2 2 13 = = hyp 13 13 opp 3 = tan θ = adj 2 1 2 cot θ = = tan θ 3 1 1 13 sec θ = = 2 = 2 cos θ cos θ =
θ
13
csc θ =
Using the Pythagorean Theorem, we
1 1 13 = 3 = 3 sin θ 13
obtain h = 32 + 2 2 = 13.
107
Chapter 2
5. Consider the following diagram:
opp 6 2 2 5 = = = hyp 3 5 5 5
sin θ =
adj 3 1 5 = = = hyp 3 5 5 5 opp 6 = =2 tan θ = adj 3 1 1 cot θ = = tan θ 2 1 1 sec θ = = 1 = 5 cos θ cos θ =
θ
5
csc θ =
1 1 5 = 2 = 2 sin θ 5
Using the Pythagorean Theorem, we obtain h = 32 + 6 2 = 45 = 3 5. 6. Consider the following diagram:
sin θ =
opp 4 1 5 = = = hyp 4 5 5 5
adj 8 2 2 5 = = = hyp 4 5 5 5 opp 4 1 = = tan θ = adj 8 2 1 cot θ = =2 tan θ 1 1 5 sec θ = = 2 = 2 cos θ cos θ =
θ
5
csc θ =
Using the Pythagorean Theorem, we
1 1 = 1 = 5 sin θ 5
obtain h = 82 + 4 2 = 80 = 4 5.
108
Section 2.2
7. Consider the following diagram:
h=
( ) +( ) =
1 4
cos θ =
1 1 10 adj = 241 = ⋅ hyp 10 2 41
5 5 41 = 41 41 2 opp 5 4 = = tan θ = adj 12 5
Using the Pythagorean Theorem, we obtain 2 2 5
2 opp 2 10 4 4 41 = 541 = ⋅ = = hyp 10 5 41 41 41
=
θ
1 2 2
sin θ =
+ 254 = 1041 .
8. Consider the following diagram:
cot θ =
1 5 = tan θ 4
sec θ =
1 1 41 = 5 = cos θ 5 41
csc θ =
1 1 41 = 4 = sin θ 4 41
sin θ =
2 opp 2 21 = 2 385 = ⋅ 3 2 85 hyp 21
7 7 85 = 85 85 4 4 21 adj cos θ = = 2 785 = ⋅ hyp 7 2 85 21 =
θ
6 6 85 = 85 85 2 opp 3 7 tan θ = = = adj 74 6 =
1 6 = tan θ 7 1 1 85 sec θ = = 6 = 6 cos θ
cot θ =
Using the Pythagorean Theorem, we obtain h=
( 74 ) + ( 23 ) = 2
2
16 49
2 85 + 94 = 340 21 = 21 .
85
csc θ =
1 1 85 = 7 = 7 sin θ 85
109
Chapter 2
9. Consider the following diagram:
sin θ =
opp 4 2 2 5 = = = hyp 2 5 5 5
adj −2 −1 5 = = =− hyp 2 5 5 5 opp 4 = = −2 tan θ = adj −2 1 1 cot θ = =− 2 tan θ 1 1 sec θ = = −1 = − 5 cos θ cos θ =
θ
5
csc θ =
1 1 5 = 2 = 2 sin θ 5
Using the Pythagorean Theorem, we obtain h = ( −2)2 + 42 = 20 = 2 5. 10. Consider the following diagram:
sin θ =
opp 3 3 10 = = hyp 10 10
adj 10 −1 = =− hyp 10 10 opp 3 tan θ = = = −3 adj −1 1 1 cot θ = =− tan θ 3 1 1 sec θ = = −1 = − 10 cos θ cos θ =
θ
10
csc θ =
1 1 10 = 3 = sin θ 3 10
Using the Pythagorean Theorem, we obtain h = ( −1)2 + 32 = 10.
110
Section 2.2
11. Consider the following diagram:
sin θ =
opp −7 7 65 = =− hyp 65 65
adj 4 65 −4 = =− hyp 65 65 opp −7 7 tan θ = = = adj −4 4 1 4 cot θ = = tan θ 7 1 1 65 sec θ = = −4 = − cos θ 4 cos θ =
θ
65
csc θ =
1 1 65 = −7 = − sin θ 7 65
Using the Pythagorean Theorem, we obtain h = ( −4)2 + (−7)2 = 65. 12. Consider the following diagram:
sin θ =
opp −5 5 106 = =− hyp 106 106
adj −9 9 106 = =− hyp 106 106 opp −5 5 = = tan θ = adj −9 9 1 9 cot θ = = tan θ 5 cos θ =
θ
sec θ =
1 1 106 = −9 = − 9 cos θ 106
csc θ =
1 1 106 = −5 = − sin θ 5 106
Using the Pythagorean Theorem, we obtain h = ( −5)2 + (−9)2 = 106.
111
Chapter 2
13. Consider the following diagram:
( − 2, 3 ) θ
sin θ =
opp 3 3 5 15 = = = hyp 5 5 5
cos θ =
adj − 2 − 2 5 10 = = =− hyp 5 5 5
tan θ =
opp 3 − 3 2 6 = = =− adj − 2 2 2
cot θ =
1 2 2 6 6 =− =− =− tan θ 6 3 6
sec θ =
1 1 5 10 = − 10 = − =− cos θ 2 10 5
csc θ =
1 1 5 5 15 15 = 15 = = = sin θ 15 3 15 5
sin θ =
opp 2 2 5 10 = = = hyp 5 5 5
cos θ =
adj − 3 − 3 5 15 = = =− hyp 5 5 5
tan θ =
opp 2 − 2 3 6 = = =− adj − 3 3 3
cot θ =
1 3 3 6 6 =− =− =− tan θ 6 2 6
sec θ =
1 1 5 15 = − 15 = − =− cos θ 3 15 5
csc θ =
1 1 5 5 10 10 = 10 = = = sin θ 10 2 10 5
Using the Pythagorean Theorem, we
( − 2 ) + ( 3 ) = 5. 2
obtain h =
2
14. Consider the following diagram:
( − 3, 2 ) θ −
Using the Pythagorean Theorem, we obtain h =
( − 3 ) + ( 2 ) = 5. 2
2
112
Section 2.2
15. Consider the following diagram:
−
θ
−
( − 5, − 3 )
sin θ =
opp − 3 − 3 2 6 = = =− hyp 2 2 4 4
cos θ =
adj − 5 − 5 2 10 = = =− hyp 2 2 4 4
tan θ =
opp − 3 3 5 15 = = = adj − 5 5 5
cot θ =
1 5 5 15 15 = = = 15 3 tan θ 15
sec θ =
1 1 4 2 10 = − 10 = − =− cos θ 5 10 4
csc θ =
1 1 4 = − 6 =− sin θ 6 4
Using the Pythagorean Theorem, we obtain h=
=−
( − 5 ) + ( − 3 ) = 8 = 2 2. 2
2
16. Consider the following diagram:
−
θ
−
( − 6, − 5 )
Using the Pythagorean Theorem, we obtain h =
sin θ =
opp − 5 − 5 11 55 = = =− hyp 11 11 11
cos θ =
adj − 6 − 6 11 66 = = =− hyp 11 11 11
tan θ =
opp − 5 5 6 30 = = = adj − 6 6 6
cot θ =
1 6 6 30 30 = = = 30 5 tan θ 30
sec θ =
1 1 11 66 = − 66 = − =− 6 cos θ 66 11
csc θ =
1 1 11 55 = − 55 = − =− sin θ 5 55 11
( − 5 ) + ( − 6 ) = 11. 2
4 6 2 6 =− 6 3
2
113
Chapter 2
17. Consider the following diagram:
sin θ =
4 opp 4 3 2 2 29 = 2 329 = ⋅ = = hyp 3 2 29 29 29 3
cos θ =
3 adj − 103 −10 = 2 29 = ⋅ hyp 3 2 29 3
−5 5 29 =− 29 29 4 2 opp tan θ = = 310 = − adj − 3 5 =
θ
1 5 =− tan θ 2 1 1 29 sec θ = = −5 = − cos θ 5
cot θ = Using the Pythagorean Theorem, we obtain
h=
( ) + ( ) = 1009 + 169 −10 2 3
4 2 3
29
csc θ =
2 29 = 116 3 = 3 .
1 1 29 = 2 = sin θ 2 29
18. Consider the following diagram:
sin θ =
opp − 13 −1 9 = 13 = ⋅ 3 13 hyp 9
−3 3 13 =− 13 13 2 adj − 9 −2 9 ⋅ cos θ = = 13 = hyp 9 13 9 =
θ
−2 2 13 =− 13 13 opp − 13 3 tan θ = = = adj − 29 2 =
Using the Pythagorean Theorem, we obtain h=
( −92 ) + ( −31 ) = 2
2
4 81
+ 91 = 913 .
cot θ =
1 2 = tan θ 3
sec θ =
1 1 13 = −2 = − cos θ 2 13
csc θ =
1 1 13 = −3 = − sin θ 3 13
114
Section 2.2
19. Consider the following diagram:
sin θ =
opp 2 2 5 = = hyp 5 5
adj 1 5 = = hyp 5 5 opp 2 tan θ = = =2 adj 1 1 1 cot θ = = tan θ 2 1 1 sec θ = = 1 = 5 cos θ cos θ =
θ
5
csc θ =
1 1 5 = 2 = 2 sin θ 5
Using the Pythagorean Theorem, we obtain h = 12 + 2 2 = 5. 20. Consider the following diagram:
sin θ =
opp 3 3 10 = = hyp 10 10
adj 1 10 = = hyp 10 10 opp 3 tan θ = = =3 adj 1 1 1 cot θ = = tan θ 3 1 1 sec θ = = 1 = 10 cos θ cos θ =
θ
10
csc θ =
1 1 10 = 3 = sin θ 3 10
Using the Pythagorean Theorem, we obtain h = 12 + 32 = 10.
115
Chapter 2
21. Consider the following diagram:
opp 1 5 = = hyp 5 5
sin θ =
adj 2 2 5 = = hyp 5 5 opp 1 tan θ = = adj 2 1 cot θ = =2 tan θ 1 1 5 sec θ = = 2 = 2 cos θ cos θ =
θ
5
Note that we can use ANY point on the line y = 12 x ( x ≥ 0) to form the triangle. In order to avoid the use of fractions, we use the point (2, 1). Using the Pythagorean Theorem, we
csc θ =
1 1 = 1 = 5 sin θ 5
obtain h = 12 + 2 2 = 5. 22. Consider the following diagram:
sin θ =
opp −1 5 = =− hyp 5 5
adj −2 2 5 = =− hyp 5 5 opp −1 1 tan θ = = = adj −2 2 1 =2 cot θ = tan θ 1 1 5 sec θ = = −2 = − 2 cos θ cos θ =
θ
5
Note that we can use ANY point on the line y = 12 x ( x ≤ 0) to form the triangle. In order to avoid the use of fractions, we use the point ( −2, −1). Using the Pythagorean Theorem yields
csc θ =
1 1 = −1 = − 5 sin θ 5
h = ( −1)2 + ( −2 )2 = 5. 116
Section 2.2
23. Consider the following diagram:
sin θ =
opp −1 10 = =− hyp 10 10
adj 3 3 10 = = hyp 10 10 opp 1 tan θ = =− adj 3 1 cot θ = = −3 tan θ cos θ =
θ
sec θ =
1 1 10 = 3 = cos θ 3 10
Note that we can use ANY point on the line y = − 13 x ( x ≥ 0) to form the triangle. In order to avoid the use of fractions, we use the point (3, −1). Using the Pythagorean Theorem
csc θ =
1 1 = −1 = − 10 sin θ 10
yields h = ( −1)2 + 32 = 10. 24. Consider the following diagram:
sin θ =
opp 1 10 = = hyp 10 10
adj 3 10 −3 = =− hyp 10 10 opp 1 tan θ = =− adj 3 1 cot θ = = −3 tan θ 1 1 10 sec θ = = −3 = − cos θ 3 cos θ =
θ
10
Note that we can use ANY point on the line y = − 13 x ( x ≤ 0) to form the triangle. In order to avoid the use of fractions, we use the point (−3,1). Using the Pythagorean Theorem, we
1 1 csc θ = = 1 = 10 sin θ 10
obtain h = 12 + ( −3)2 = 10. 117
Chapter 2
25. Consider the following diagram:
sin θ =
opp 2 2 13 = = hyp 13 13
adj −3 3 13 = =− hyp 13 13 opp 2 tan θ = =− adj 3 1 3 cot θ = =− 2 tan θ 1 1 13 sec θ = = −3 = − 3 cos θ cos θ =
θ
The equation of the line is y = − 23 x ( x ≤ 0 ). Note that we can use ANY point on the line to form the triangle. In order to avoid the use of fractions, we use the point ( −3,2). Using the Pythagorean Theorem, we obtain
13
csc θ =
1 1 13 = 2 = 2 sin θ 13
h = ( −3)2 + 2 2 = 13. 26. Consider the following diagram:
sin θ =
opp −2 2 13 = =− hyp 13 13
adj 3 3 13 = = hyp 13 13 opp 2 tan θ = =− adj 3 1 3 cot θ = =− 2 tan θ 1 1 13 sec θ = = 3 = 3 cos θ cos θ =
θ
The equation of the line is y = − 23 x (x ≥ 0). Note that we can use ANY point on the line to form the triangle. In order to avoid the use of fractions, we use the point (3, −2). Using the Pythagorean Theorem, we obtain
13
csc θ =
1 1 13 = −2 = − sin θ 2 13
h = 32 + ( −2)2 = 13. 118
Section 2.2
27. Since θ = 450 = 360 + 90 , the standard position of θ is 90. So,
28. Since θ = 540 = 360 + 180 , the standard position of θ is 180. So,
sin θ = 0 cos θ = −1 tan θ = 0
sin θ = 1 cos θ = 0 tan θ is undefined cot θ = 0 sec θ is undefined csc θ = 1
29. Since θ = 630 = 360 + 270 , the standard position of θ is 270. So,
cot θ is undefined sec θ = −1 csc θ is undefined 30. Since θ = 720 = 2(360 ), the standard position of θ is 0. So,
sin θ = 0 cos θ = 1 tan θ = 0
sin θ = −1 cos θ = 0 tan θ is undefined cot θ = 0 sec θ is undefined
cot θ is undefined sec θ = 1 csc θ is undefined
csc θ = −1
31. Since θ = −270 = −360 + 90 , the standard position of θ is 90. So,
32. Since θ = −180 = −360 + 180 , the standard position of θ is 180. So,
sin θ = 0 cos θ = −1 tan θ = 0
sin θ = 1 cos θ = 0 tan θ is undefined cot θ = 0
cot θ is undefined sec θ = −1 csc θ is undefined
sec θ is undefined csc θ = 1
119
Chapter 2
33. Since θ = −90 = −360 + 270 , the standard position of θ is 270. So,
34. Since the standard position of θ = −360 is 0 , we have
tan θ is undefined cot θ = 0 sec θ is undefined csc θ = −1
sin θ = 0 cos θ = 1 tan θ = 0 cot θ is undefined sec θ = 1 csc θ is undefined
35. Since θ = −450 = −2(360 ) + 270 , the standard position of θ is 270. So,
36. Since θ = −540 = −2(360 ) + 180 , the standard position of θ is 180. So,
sin θ = −1 cos θ = 0 tan θ is undefined
sin θ = 0 cos θ = −1 tan θ = 0 cot θ is undefined sec θ = −1 csc θ is undefined
sin θ = −1 cos θ = 0
cot θ = 0 sec θ is undefined csc θ = −1
37. Since θ = −630 = −2(360 ) + 90 , the standard position of θ is 90. So,
38. Since the standard position of θ = −720 = −2(360 ) is 0 , we have
sin θ = 0 cos θ = 1 tan θ = 0
sin θ = 1 cos θ = 0 tan θ is undefined
cot θ is undefined sec θ = 1 csc θ is undefined
cot θ = 0 sec θ is undefined csc θ = 1
120
Section 2.2
39. Since θ = 810 = 2(360 ) + 90 , the standard position of θ is 90. So,
40. Since θ = 900 = 2(360 ) + 180 , the standard position of θ is 180. So,
sin 900o = 0 cos 900o = −1 tan 900o = 0 csc 900o is undefined sec 900o = −1 cot 900o isundefined
sin 810o = 1 cos 810o = 0 tan 810o is undefined csc 810o = 1 sec 810o is undefined cot 810o = 0 41. Since θ = −810 = −2(360 ) − 90 , the standard position of θ is −90 , which is coterminal with 270. So,
42. Since θ = −900 = −2(360 ) − 180 , the standard position of θ is −180 , which is coterminal with 180. So,
sin (−810o) = −1 cos (−810o) = 0 tan (−810o) is undefined csc (−810o) = −1 sec (−810o) is undefined cot (−810o) = 0 43. Consider the following diagram:
sin (−900o) = 0 cos(−900o) = −1 tan (−900o) = 0 csc (−900o) = undefined sec (−900o) = −1 cot (−900o) = undefined First, find x using the Pythagorean Theorem: x 2 + 7 2 = 252 x = 576 = 24
θ
Hence, tan θ =
121
opp 7 . = adj 24
Chapter 2
44. Consider the following diagram:
First, find y using the Pythagorean Theorem: y2 = 84 2 + 132 y = 7,225 = 85
θ
45. Consider the following diagram:
θ
Hence, cos θ =
adj 13 = . hyp 85
We need to determine T. 11 22 Observe that cos θ = = . 61 122 Hence, using the Pythagorean Theorem yields: 2 ( 22 ft.) + T 2 = (122 ft.)2 T = 14, 400 ft.2 = 120 ft. So, the tree is 120 ft. high.
46. Consider the following diagram:
θ
We need to determine B, the distance between the base of the tree and point on the ground. 40 20 Observe that sin θ = = . 41 412 Hence, using the Pythagorean Theorem yields: 2 ( 20 ft.) + B 2 = ( 412 ft.)2 T=
122
81 4
ft. = 4.5 ft.
Section 2.2
47. Consider the following diagram:
We need to determine the height of the fuel plane (adj) when the hose (hyp) is 145 21 105 meters. Observe that cos θ = = . 29 145 Hence, using the Pythagorean Theorem yields: 2 2 2 (105 m ) + ( height ) = (145 m ) height = 10,000 m 2 = 100 m. So, the fuel plane is 100 m above the jet.
48. Consider the following diagram:
We need to determine the length of hose (hyp) when the fuel plane is 935 feet above 55 935 the jet. Observe that tan θ = = . 48 816 Hence, using the Pythagorean Theorem yields: 2 2 2 ( 935 ft ) + (816 ft ) = ( hose ) hose = 1,540,081ft 2 = 1241ft. So, the hose is 1,241 feet long.
49. Consider this triangle:
sin θ =
opp 4 4 17 = = hyp 17 17
adj 1 17 = = hyp 17 17 opp tan θ = =4 adj cos θ =
θ
123
Chapter 2
50. Consider this triangle: sin θ =
1 opp 5 = 52 = hyp 5 2
adj 1 2 5 =− 5 =− hyp 5 2 opp 1 tan θ = =− adj 2 cos θ =
θ
51. Since we are given that cos θ = 0.9, we have the triangle
The picture for our current situation is
θ θ
θ
19 By similar triangles, we see that 9.1 = 10x x ≈ 20.1 ft.
52. Since we are given that sec θ = 65 , and so cos θ = 65 , we have the triangle:
The picture for our current situation is
θ
θ θ
By similar triangles, we see that 8.911 = x6 x ≈ 16.1 ft. 124
Section 2.2
8 , we have the 10 following triangle (where the length of the remaining leg was computed using the Pythagorean theorem):
54. Using the Pythagorean theorem in the following triangle yields Revenue 2 + 32 = 42 Revenue = 7 .
53. Since sin θ =
θ θ Hence, Revenue 7 = < 1. Cost 3 Since tan θ < 1, this represents a loss. tan θ =
Hence, Revenue 8 = > 1. Cost 6 Since tan θ > 1, this represents a profit. tan θ =
4.8 , we have the 5.0 following triangle:
55. Since cos θ =
56. Since sec θ =
5.0 , we know that 4.9
4.9 and so, we have the 5.0 following triangle: cos θ =
θ Now, using the Pythagorean theorem yields 4.82 + d 2 = 5.02 d = 1.96 = 1.4 cm.
θ Now, using the Pythagorean theorem yields 4.92 + d 2 = 5.02 d = 0.99 ≈ 1.0 cm.
125
Chapter 2
57. Here, r = 12 + 2 2 = 5 , not 5.
1 does NOT equal 0, it is 0 undefined.
59. r represents the distance from the origin to the point ( 2, −1) , thus r is always positive. So we know 2 5 cos θ = . 5
60. Sine and cosine are cofunctions, not reciprocal identities. We can find 3 . cos 30° directly, thus cos30° = 2
61. False. Observe that 1 3 sin30 = , sin 60 = , sin 90 = 1 . 2 2 1 3 But, + ≠ 1. 2 2
62. False. Even though tan 0 = 0, thereby resulting in the statement tan 90 = tan 90 , this statement is not a truth since tan 90 is undefined.
63. True. The angles θ and θ + 360 n (where n is an integer) are coterminal, so that cos θ = cos (θ + 360 n ) .
64. True. This is true since sine is periodic with period 2π (which corresponds to 360 ).
58.
65. First, note that it is not possible for a = 0 since in such case there would be no triangle. Thus, there are only two cases to consider. Case 1: a > 0 Consider the following diagram: From the Pythagorean Theorem, it
follows that z = (4a )2 + (−3a )2 = 5a. So, cos θ =
θ
126
−3a 3 =− . 5a 5
Section 2.2
Case 2: < Consider the following diagram:
From the Pythagorean Theorem, it follows that z = (4a )2 + (−3a )2 = 5a. −3a 3 = − , which is the 5 5a same conclusion of Case 1.
So, cos θ =
θ
66. Referring to the diagram in Exercise 59, we observe that in both cases, 4a 4 sin θ = = . 5a 5 67. Case 1: m = 0 In such case, tan 0 = 0 = m. Case 2: m > 0
Consider the following diagram:
θ
Observe that tan θ =
m = m. 1
127
Chapter 2
Case 3: m < 0
Consider the following diagram:
θ
Similarly, tan θ =
sin θ − m = = m. cos θ −1
29 2 . , so that sin θ = − 2 29 Consider the following diagram:
68. First, note that csc θ = −
θ
θ
Using the Pythagorean Theorem, we see that x12 + ( −2)2 =
( 29 ) and x + ( −2) = ( 29 ) . 2
2 2
2
2
Hence, we have x1 = x2 = ±5. Thus, the two possible x values are ±5 .
128
Section 2.2
69. Consider the following diagram:
θ
Suppose the form of the equation of the line is given by y = mx + b. We need to determine m and b to find the equation of this line. First, substitute the point ( a, 0) into the equation: 0 = ma + b so that b = − ma. So, the equation of the line thus far is y = mx − ma = m( x − a ). Next, note that this line is parallel to the corresponding line translate to pass through the origin – the slope of that line is tan θ (by Exercise 61). Since parallel lines have the same slope, we know that m = tan θ . Thus, the equation of the line is y = (tan θ )( x − a ) . 70. First, note that we need to find a relationship between the slope the line and θ (the angle the line makes with the negative x-axis). This is done in cases, as in Exercise 61. Case 1: m = 0 In such case, tan 0 = 0 = m. Case 2: m > 0 Consider the following diagram:
θ
Observe that tan θ =
m = − m. −1
129
Chapter 2
Case 3: m < 0
Consider the following diagram:
θ
sin θ − m = = − m. cos θ 1 Now, to find the equation of a line with negative slope passing through the point ( a, 0) , we proceed as in Exercise 63. Suppose the form of the equation of the line is given by y = mx + b. We need to determine m and b to find the equation of this line. First, substitute the point ( a,0) into the equation: 0 = ma + b so that b = − ma. So, the equation of the line thus far is y = mx − ma = m( x − a ). Next, note that this line is parallel to the corresponding line translate to pass through the origin – the slope of that line is − tan θ (by the above discussion). Since parallel lines have the same slope, we know that m = − tan θ . Thus, the equation of the line is y = ( − tan θ )( x − a ) = (tan θ )( a − x ) . Similarly, tan θ =
71. sec x = cos1 x is undefined when cos x = 0. This occurs when x = 90° + (180k)° for any integer k. 72. If csc x = sin1 x is undefined, then sin x = 0 . 73. sin 270 = −1 75. tan 270 =
sin 270 is undefined. cos 270
74. cos 270 = 0 76. cot 270 =
cos 270 =0 sin 270
77. cos ( −270 ) = 0
78. sin ( −270 ) = 1
79. sec ( −270 ) is undefined.
80. sec ( 270 ) is undefined.
81. csc ( 270 ) = −1
82. csc ( −270 ) = 1
130
Section 2.3 Solutions -------------------------------------------------------------------------------1. QIV
2. QII
3. QII (Need cos θ < 0 and sin θ > 0.)
4. QIII (Need cos θ < 0 and sin θ < 0.)
5. QI (Need cos θ > 0 and sin θ > 0.)
6. QIII (Need cos θ < 0 and sin θ < 0.)
7. Consider the following diagram:
8. Consider the following diagram:
θ
θ
4 Observe that sin θ = − . 5
Observe that cos θ = −
9. Consider the following diagram:
10. Consider the following diagram:
θ
Using the Pythagorean Theorem, we obtain z = − 612 − 60 2 = −11. So, tan θ = −
60 . 11
12 . 13
θ
Using the Pythagorean Theorem, we obtain z = − 412 − 40 2 = −9. So, tan θ = −
131
9 . 40
Chapter 2
11. Consider the following diagram:
12. There is no solution here since there does not exist a value of θ for which cos θ > 1.
θ
Observe that sec θ =
1 4 4 7 . = = cos θ 7 7
13. Consider the following diagram:
14. Consider the following diagram:
θ
θ
Using the Pythagorean Theorem, we obtain z = ( −13)2 + (−84)2 = 85. So, 84 sin θ = − . 85
Using the Pythagorean Theorem, we obtain z = 252 − (−7)2 = 24. So, cos θ =
132
24 . 25
Section 2.3
15. Consider the following diagram:
16. Consider the following diagram:
θ
θ
Observe that cot θ = 17. Since sec θ =
1 1 = = 2. tan θ 12
1 = −2, we see cos θ
Observe that sin θ = −
−
95 . 10
18. Consider the following diagram:
1 that cos θ = − . Consider the 2 following diagram:
θ
θ Using the Pythagorean Theorem, we obtain z = 12 + 12 = 2. So, sin θ =
Using the Pythagorean Theorem, we obtain z = − 22 − ( −1)2 = − 3. So, tan θ =
− 3 = −1
3.
133
1 2 = . 2 2
Chapter 2
19. Consider the following diagram:
20. Consider the following diagram:
θ
θ
Using the Pythagorean Theorem, we obtain z = − 6 2 − tan θ =
( ) 11
2
= −5. So,
Using the Pythagorean Theorem, we obtain z = 12 + ( −2 ) = 5. So, 2
11 11 . = − 5 −5
cos θ =
−2 2 5 . = − 5 5
21. First, cos ( −270 ) = cos ( 90 ) = 0.
22. First, sin ( −270 ) = sin ( 90 ) = 1.
sin ( 450 ) = sin ( 90 ) = 1. Therefore,
cos ( 450 ) = cos ( 90 ) = 0. Therefore,
Also, since 450 = 360 + 90 , we have
Also, since 450 = 360 + 90 , we have
cos ( −270 ) + sin ( 450 ) = 1.
sin ( −270 ) + cos ( 450 ) = 1.
23. Since 630 = 360 + 270 , we have
24. Since −720 = 2 ( −360 ) , we have
sin ( 630 ) = sin ( 270 ) = −1. Further,
since −540 = −360 − 180 , we have tan ( −540 ) = tan ( −180 )
=
sin ( −180 )
cos ( −180 )
cos ( −720 ) = cos ( 0 ) = 1. Further,
since 720 = 2 ( 360 ) , we have tan ( 720 ) = tan ( 0 ) =
= 0.
sin ( 0 )
cos ( 0 )
cos ( −720 ) + tan ( 720 ) = 1.
So, sin ( 630 ) + tan ( −540 ) = −1.
134
= 0. So,
Section 2.3
25. Since 540 = 360 + 180 , we have
26. Since −450 = −360 − 90 , we have
since −540 = −360 − 180 , we have
Further, csc ( 270 ) =
sec ( −540 ) =
So, sin ( −450 ) + csc ( 270 ) = −1+ (−1) = −2 .
cos ( 540 ) = cos (180 ) = −1. Further,
cos ( −540 ) = cos ( −180 ) = −1, so that
=
1 cos ( −540 )
sin ( −450 ) = sin ( −90 ) = sin ( 270 ) = −1. 1 = −1. sin ( 270 )
1 = −1. cos ( −180 )
So, cos ( 540 ) − sec ( −540 ) = −1− (−1) = 0 . 27. Since −630 = −360 − 270 , we have csc ( −630 ) = csc ( −270 ) = csc ( 90 ) = 1.
Further, since 630 = 360 + 270 , we have cos ( 630 ) cos ( 270 ) = =0 cot ( 630 ) = sin ( 630 ) sin ( 270 ) So, csc ( −630 ) − cot ( 630 ) = 1 .
29. Since tan ( 720 ) = tan ( 0 ) = 0 and
sec ( 720 ) =
1 1 = 1, = cos ( 720 ) cos ( 0 )
28. Since −540 = −360 − 180 , we have
cos ( −540 ) = cos ( −180 ) = −1, so that
sec ( −540 ) = =
1 cos ( −540 ) 1 = −1. cos ( −180 )
Further, tan ( 540 ) = tan (180 ) = 0. So, sec ( −540 ) + tan ( 540 ) = −1.
30. Since 450 = 360 + 90 , we have
cos ( 450 ) = cos ( 90 ) = 0. Further,
we have tan ( 720 ) + sec ( 720 ) = 1 .
since −450 = −360 − 90. we have cos ( −450 ) = cos ( −90 ) = 0. So,
31. Possible since the range of sine is [ −1,1].
32. Not possible since the range of cosine is [ −1,1] .
cos ( 450 ) − cos ( −450 ) = 0 .
135
Chapter 2
33. Not possible since
2 6 > 1, and 3 hence out of the range of cosine (which is [ −1,1]).
34. Possible since −1 <
35. Possible since the range of tangent is (−∞,∞ ).
36. Possible since the range of cotangent is −∞ ∞
−4 < −1, and hence 7 in the range of secant (which is ( −∞, −1] ∪ [1,∞ ) ).
38. Possible since the range of cosecant is ( −∞, −1] ∪ [1,∞ ) .
39. Not possible since the 1 −1 < < 1, yet the range of cosecant is 2 ( −∞, −1] ∪ [1,∞ ) .
40. Possible since the range of cosecant is ( −∞, −1] ∪ [1,∞ ) .
41. Possible since the range of cotangent is ( −∞, ∞ ) .
42. Possible since the range of tangent is ( −∞, ∞ ) .
43.
44.
37. Possible since
2 < 1. 10
The given angle is shown in bold, while the reference angle is dotted. 1 cos ( 240 ) = − cos ( 60 ) = − 2
The given angle is shown in bold, while the reference angle is dotted. 1 cos (120 ) = − cos ( 60 ) = − 2
136
Section 2.3
45.
46.
The given angle is shown in bold, while the reference angle is dotted. sin ( 300 ) = − sin ( 60 ) = −
47. cot ( −150 ) =
cos ( −150 )
sin ( −150 )
49. cos ( −30 ) =
=
3 2
− 32 = 3 − 12
The given angle is shown in bold, while the reference angle is dotted. sin ( 315 ) = − sin ( 45 ) = −
48. sin ( −135 ) = −
2 2
2 2
50.
3 2
sec ( −135 ) =
51.
1 1 = 2 =− 2 cos ( −135 ) − 2
52.
137
Chapter 2
The given angle is shown in bold, while the reference angle is dotted. 1 sin30 2 tan ( 210 ) = tan ( 30 ) = = cos 30 3 2 =
The given angle is shown in bold, while the reference angle is dotted. −1 sec (135 ) = − sec ( 45 ) = cos ( 45 ) =−
2 = − 2 2
3 3
53.
54.
−
The given angle is shown in bold, while the reference angle is dotted. tan (135 ) = tan ( 45 ) = 1
−
The given angle is shown in bold, while the reference angle is dotted. 1 sec ( −330 ) = sec ( 30 ) = cos ( 30 ) =
138
2 2 3 = 3 3
Section 2.3
55.
56.
−
The given angle is shown in bold, while the reference angle is dotted. 1 csc ( −240 ) = csc ( 60 ) = sin ( 60 )
The given angle is shown in bold, while the reference angle is dotted. −1 = −2 csc ( 330 ) = − csc ( 30 ) = sin ( 30 )
=
57. cos θ =
3 when θ = 30 or 330 . 2 (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
58. sin θ =
1 when θ = 210 or 330 . 2 (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
60. cos θ = −
59. sin θ = −
2 2 3 = 3 3
3 when θ = 60 or 120 . 2 (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.) 1 when θ = 120 or 240 . 2 (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
139
Chapter 2
61. cos θ = 0 when θ = 90 or 270 . (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
62. sin θ = 0 when θ = 0 , 180 , or 360 .
63. θ = − when θ = (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
64. θ = − when θ = (Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
65. –0.8387
66. 0.7314
67. –11.4301
68. 7.1154
69. sec ( 421 ) =
1 ≈ 2.0627 cos ( 421 )
71. csc ( −111 ) = 73. cot (159 ) =
(Note: Since the right-side of the equation corresponds to standard angles, we refer the reader to the diagram in the text for a visual of the solutions to this equation.)
70. sec ( −222 ) =
1 ≈ −1.3456 cos ( −222 )
1 ≈ −1.0711 sin ( −111 )
72. csc ( 211 ) =
1 ≈ −1.9416 sin ( 211 )
1 ≈ −2.6051 tan (159 )
74. cot ( −82 ) =
1 ≈ −0.1405 tan ( −82 )
75. Since sin θ = 0.9397 , it follows that θ = sin −1 ( 0.9397 ) ≈ 70 . This angle has
76. Since cos θ = 0.7071 , it follows that θ = cos −1 ( 0.7071) ≈ 45 . This angle has
terminal side in QI. Since the desired angle has terminal side in QII, we have the following diagram. Note: The desired terminal side is the dashed segment.
terminal side in QI. Since the desired angle has terminal side in QIV, we have the following diagram. Note: The desired terminal side is the dashed segment.
140
Section 2.3
−
Hence, θ ≈ 70 + 40 = 110 .
Hence, θ ≈ 360 − 45 = 315 .
77. Since cos θ = −0.7986, it follows
78. Since sin θ = −0.1746, it follows that θ = sin −1 ( −0.1746 ) ≈ −10. This angle
that θ = cos −1 ( −0.7986 ) ≈ 143 . This angle has terminal side in QII. Since the desired angle has terminal side also in QII, no adjustment is necessary – this is pictured in the following diagram. Note: The desired terminal side is the dashed segment.
has terminal side in QIV. Since the desired angle has terminal side in QIII, we have the following diagram. Note: The desired terminal side is the dashed segment.
10
143
Hence, θ ≈ 180 + 10 = 190
141
−10
Chapter 2
79. Since tan θ = −0.7813, it follows that θ = tan −1 ( −0.7813 ) ≈ −38. This
angle has terminal side in QIV. Since the desired angle has terminal side also in QIV, we need only a corresponding angle with positive measure – this is pictured in the following diagram. Note: The desired terminal side is the dashed segment.
80. Since cos θ = −0.3420, it follows
that θ = cos −1 ( −0.3420 ) ≈ 110 . This angle has terminal side in QII. Since the desired angle has terminal side in QIII, we have the following diagram. Note: The desired terminal side is the dashed segment.
110 −110
322 −38
Hence, θ ≈ 360 − 110 = 250 . Hence, θ ≈ 360 − 38 = 322 . 81. Since tan θ = −0.8391, it follows that θ = tan −1 ( −0.8391) ≈ −40. This
82. Since tan θ = 11.4301, it follows that
angle has terminal side in QIV. Since the desired angle has terminal side also in QII, we have the following diagram. Note: The desired terminal side is the dashed segment.
terminal side in QI. Since the desired angle has terminal side also in QIII, we have the following diagram. Note: The desired terminal side is the dashed segment.
θ = tan −1 (11.4301) ≈ 85. This angle has
−
Hence, θ ≈ 180 − 40 = 140 .
Hence, θ ≈ 360 − 85 = 265 . 142
Section 2.3
83. Since sin θ = −0.3420, it follows that θ = sin −1 ( −0.3420 ) ≈ −20. This
84. Since sin θ = −0.4226, it follows that θ = sin −1 ( −0.4226 ) ≈ −25. This angle
angle has terminal side in QIV. Since the desired angle has terminal side also in QIV, we need only a corresponding angle with positive measure – this is pictured in the following diagram. Note: The desired terminal side is the dashed segment.
has terminal side in QIV. Since the desired angle has terminal side in QIII, we have the following diagram. Note: The desired terminal side is the dashed segment.
−
−
Hence, θ ≈ 360 − 20 = 340 .
Hence, θ ≈ 180 + 25 = 205 .
85. Use n1 sin (θ1 ) = n2 sin (θ 2 ) with
86. Use n1 sin (θ1 ) = n2 sin (θ 2 ) with
n1 = 1, θ1 = 30 , θ 2 = 22 to obtain
n1 = 1, θ1 = 30 , θ 2 = 18 to obtain
(1)sin ( 30 ) = n2 sin ( 22 ) n2 =
sin ( 30 )
sin ( 22 )
(1)sin ( 30 ) = n2 sin (18 )
≈ 1.3
n2 =
sin ( 30 ) sin (18 )
≈ 1.6
87. Use n1 sin (θ1 ) = n2 sin (θ 2 ) with
88. Use n1 sin (θ1 ) = n2 sin (θ 2 ) with
n1 = 1, n2 = 2.4, θ1 = 30 to obtain
n1 = 1, n2 = 1.9, θ1 = 30 to obtain
(1)sin ( 30 ) = (2.4)sin (θ 2 ) sin (θ 2 ) =
(1)sin ( 30 ) = (1.9)sin (θ 2 )
sin ( 30 )
sin (θ 2 ) =
2.4 sin ( 30 ) −1 ≈ 12 θ 2 = sin 2.4
sin ( 30 )
1.9 sin ( 30 ) −1 ≈ 15 θ 2 = sin 1.9
143
Chapter 2
89. Observe that θ = 30 + 3.25 ( 360 ) = 30 + 3 ( 360 ) + 0.25 ( 360 ) = 3 ( 360 ) +120.
Thus, sin θ = sin (120 ) = 23 .
90. Observe that θ = 120 + 32 ( −360 ) + 34 ( 360 ) = 150. Thus,
cos θ = cos (150 ) = − 23 .
91. Consider the following diagram:
She started at α = −90°. She went
2 of 3
2 ( 360° ) = 240°. So 3 β = −90° + 240° = 150°. Finally,
a rotation, which is
cos150° = −
3 . 2
92. Consider the following diagram:
From 87, she started at β = 150°. She 2 went of a rotation, which is 3 2 ( 360° ) = 240°. So 3 γ = 150° + 240° = 390°. Then, 390° is coterminal with 30°. Finally, 1 sin30° = . 2
93. Since tan α = 0 α = 0 or 180 , the minute hand is on the 3 or the 9. 94. First, note that sin α = 12 α = 30 or 150. Since the angle between
consecutive hours on the face of the clock is 30 , the minute hand is on either the 2 or the 10. 95. The measure of the reference angle is 180 − 165 = 15. Physically, this means that the lower leg is bent at the knee in a backward direction at an angle of 15 degrees.
144
Section 2.3
96. The measure of the reference angle is 180 − 160 = 20. Physically, an angle whose measure is greater than 180 degrees represents a hyper-extended knee. 97. The diagram is as follows:
α=
98. Observe that cos (104.5 ) = − cos ( 75.5 ) ≈ −0.2504 99. The reference angle is measured between the terminal side and the x-axis, not the y-axis. So, in this case, the reference angle is 60. Since 1 1 cos 60 = and cos120 = − , we see 2 2 1 that sec120 = = −2. 1 − 2
θ=
The reference angle is 180 − 104.5 = 75.5 The cosine of angles with terminal sides in QII and QIII is negative. The calculator correctly produces the angle 104 , but since we want the terminal side to lie in QIII, we need to use the correct reference angle. In this case, we angle would be 360 − 104 = 256.
100. Consider the following diagram:
−
101. θ = 26° is not located in quadrant IV. Instead, take θ = −26° and find a coterminal angle in quadrant IV. θ = −26° + 360° = 334°.
102. Tangent is only positive in quadrants I and III so the values of θ are 60° and 240°.
145
Chapter 2
103. True. For any 0 < θ < 90 , sin θ and cos θ are both positive. Since all of the trigonometric functions are defined as quotients of sine, cosine, and 1, they are also positive for such values of θ . 104. False. This requires both sin θ and cos θ to be negative. But, in such case, tan θ and cot θ are both positive. 105. False. For instance, 2 cos ( −45 ) = . 2
106. False. For instance, 1 cos (120 ) = − . 2
107. Assume that a > 0, b > 0. Consider the following diagram:
108. Assume that a > 0, b > 0. Consider the following diagram:
θ
θ
+
+
Observe that sin θ = −
a a + b2 2
.
109. If cos θ = − 23 , then
θ = 150 or 210. The reference angle
Observe that cos θ = −
b a + b2 2
.
110. tan ( 45 + 180 k ) = tan ( 45 ) = 1,
for any integer k.
for both is 30 . 111. If the reference angle for β is 60 , then β could be any of 60 , 120 , 240 , 300. The cosine of these angles is either − 12 or 12 .
146
Section 2.3
112. Observe that csc ( 3(−210 ) + 30 ) − 12 = csc ( −600 ) − 12 = csc ( −240 ) − 12
=
1 1 2 3 1 4 3 −3 − = − = 3 2 6 sin ( −240 ) 2
113. Observe that sin80 ≈ 0.9848 ≈ sin100. Also, sin −1 (0.9848) ≈ 80. So, the calculator is assuming that the terminal side of the angle lies in QI.
114. Observe that sin 260 ≈ −0.9848 ≈ sin 280. Also, sin −1 ( −0.9848) ≈ −80 , which corresponds to the angle 280. So, the calculator is assuming that the terminal side of the angle lies in QIV.
115. Observe that cos 75 ≈ 0.2588 ≈ cos( −75 ). Also, cos −1 (0.2588) ≈ 75. So, the calculator is assuming that the terminal side of the angle lies in QI.
116. Observe that cos105 ≈ −0.2588 ≈ cos(255 ). Also, cos −1 ( −0.2588) ≈ 105. So, the calculator is assuming that the terminal side of the angle lies in QII.
117. Observe that tan 65 ≈ 2.1445 ≈ tan 245. Also, tan −1 (2.1445) ≈ 65. So, the calculator is assuming that the terminal side of the angle lies in QI.
118. Using a calculator yields: tan (136 ) ≈ −0.9657 ≈ tan ( 316 ) tan −1 ( −0.9657 ) ≈ −44 So, the calculator assumes the terminal side is in QIV when calculating inverse tangent of a negative quantity.
147
Section 2.4 Solutions -------------------------------------------------------------------------------1. sec θ =
1 8 = cos θ 7
2. csc θ =
1 7 = − sin θ 3
3. csc θ =
1 1 1 5 = = 3= − sin θ −0.6 − 5 3
4. sec θ =
1 1 1 5 = = 4 = cos θ 0.8 5 4
5. cot θ =
1 1 1 = = − 5 tan θ −5
6. cot θ =
1 1 1 = =1= 2 tan θ 0.5 2
7. Since
5 1 = csc θ = , we see that sin θ 2
8. Since
11 1 = sec θ = , we see 2 cos θ
that cos θ =
2 2 11 = . 11 11
1 1 5 5 7 = = − = − cot θ − 57 7 7
10. tan θ =
1 1 1 2 = =7 = cot θ 3.5 2 7
11. tan θ =
sin θ − 12 1 3 = 3 =− = − 3 cos θ 3 2
12. cot θ =
3 cos θ = 21 = − 3 sin θ − 2
13. tan θ =
sin θ 4 = − cos θ 3
14. cot θ =
cos θ 3 = − sin θ 4
15. cot θ =
cos θ −0.8 − 54 4 = = = sin θ −0.6 − 35 3
16. tan θ =
sin θ −0.6 − 53 3 = = = cos θ −0.8 − 54 4
sin θ =
2 2 5 = . 5 5
9. tan θ =
18. Using Exercise 17, observe that
17. 11 sin θ 6 = − 11 ⋅ − 6 = tan θ = = 5 6 5 cos θ − 6
−
11 5
148
cot θ =
1 1 5 5 11 = = = tan θ 11 11 11 5
Section 2.4
5 5 19. sin θ = ( sin θ ) = = 64 8
20. cos3 θ = ( cos θ ) = ( 0.1) = 0.001
21. csc3 θ = ( csc θ ) = ( −2 ) = −8
22. tan2 θ = ( tan θ ) = ( −5 ) = 25
2
2
3
2
3
3
2
3
2
23. Solving the equation sin2 θ + cos 2 θ = 1 for cos θ yields cos θ = ± 1− sin2 θ .
Since the terminal side of θ is in QIII, we use cos θ = − 1 − sin2 θ . Hence, cos θ = − 1 − ( − 12 ) = − 2
3 4
= − 23 .
24. Solving the equation sin2 θ + cos 2 θ = 1 for cos θ yields cos θ = ± 1− sin2 θ .
Since the terminal side of θ is in QIII, we use cos θ = − 1 − sin2 θ . Hence, cos θ = − 1 − ( − 53 ) = − 2
16 25
= − 54 .
25. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QIV, we use sin θ = − 1 − cos 2 θ . Hence, sin θ = − 1 − ( 25 ) = − 2
21 25
= − 521 .
26. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QIV, we use sin θ = − 1 − cos 2 θ . Hence, sin θ = − 1 − ( 72 ) = − 2
45 49
= − 3 75 .
27. Solving the equation 1 + tan2 θ = sec 2 θ for sec θ yields sec θ = ± 1+ tan2 θ . Since the terminal side of θ is in QIII, cos θ < 0 and so sec θ < 0. So, we use sec θ = − 1 + tan2 θ . Hence, sec θ = − 1+ tan2 θ = − 1+ (4)2 = − 17 .
28. Solving the equation 1 + tan2 θ = sec 2 θ for sec θ yields sec θ = ± 1+ tan2 θ . Since the terminal side of θ is in QII, cos θ < 0 and so sec θ < 0 . So, we use sec θ = − 1+ tan2 θ . Hence, sec θ = − 1+ tan2 θ = − 1+ (−5)2 = − 26 .
149
Chapter 2
29. Solving the equation 1 + cot 2 θ = csc 2 θ for csc θ yields csc θ = ± 1+ cot 2 θ . Since the terminal side of θ is in QIII, sin θ < 0 and so csc θ < 0. So, we use csc θ = − 1 + cot 2 θ . Hence, csc θ = − 1+ cot 2 θ = − 1+ (2)2 = − 5 .
30. Solving the equation 1 + cot 2 θ = csc 2 θ for csc θ yields csc θ = ± 1+ cot 2 θ . Since the terminal side of θ is in QII, sin θ > 0 and so csc θ > 0. So, we use csc θ = 1 + cot 2 θ . Hence, csc θ = 1 + cot 2 θ = 1+ (−3)2 =
10 .
31. Solving the equation sin2 θ + cos 2 θ = 1 for cos θ yields cos θ = ± 1− sin2 θ .
Since the terminal side of θ is in QII, we use cos θ = − 1 − sin2 θ . Hence, cos θ = − 1− ( 158 ) = − 2
161 225
161 = − 15 .
8 sin θ 8 −8 161 . Thus, tan θ = = 15 = − = cos θ 161 161 161 − 15
32. Solving the equation sin2 θ + cos 2 θ = 1 for cos θ yields cos θ = ± 1− sin2 θ .
Since the terminal side of θ is in QIII, we use cos θ = − 1 − sin2 θ . Hence, cos θ = − 1− ( − 157 ) = − 2
sin θ Thus, tan θ = = cos θ
176 225
176 = − 15 .
7 15 = 7 = 7 = 7 11 . 44 176 176 4 11 − 15
−
33. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QIII, we use sin θ = − 1 − cos 2 θ . Hence, sin θ = − 1− ( − 157 ) = − 2
Thus, csc θ =
1 − 15 −15 −15 11 = = = . 44 sin θ 176 4 11
150
176 225
176 = − 15 .
Section 2.4
34. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QII, we use sin θ = 1 − cos 2 θ . Hence, sin θ = 1 − ( − 158 ) = 2
Thus, csc θ =
161 225
161 = 15 .
1 15 15 161 = = . 161 sin θ 161
35. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QIV, we use sin θ = − 1− cos 2 θ . From the given 1 3 information, observe that cos θ = = . Hence, we obtain sec θ 13 sin θ = − 1 −
( ) =− 3 13
2
4 13
= − 213 = − 2 1313 .
36. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QII, we use sin θ = 1 − cos 2 θ . From the given 1 5 information, observe that cos θ = =− . Hence, we obtain sec θ 61
(
sin θ = 1 − − 561
) = 2
36 61
= 661 = 6 6161 .
4 and the terminal 3 side of θ lies in QII, we conclude that 4 4 = sin θ = 25 5 3 3 =− . cos θ = − 5 25
37. Consider the following diagram:
Since tan θ = −
25
θ
151
Chapter 2
3 and the terminal 4 side of θ lies in QII, we conclude that 3 4 sin θ = , cos θ = − . 5 5
38. Consider the following diagram:
Since tan θ = −
5
θ
Since the terminal side of θ lies in QIV, we conclude that 1 26 =− sin θ = − 26 26
39. Consider the following diagram:
cos θ =
θ
5 5 26 = 26 26
26
Since the terminal side of θ lies in QIV, we conclude that 2 2 13 =− sin θ = − 13 13
40. Consider the following diagram:
cos θ =
θ 13
152
3 3 13 = 13 13
Section 2.4
Since tan θ = 2 and the terminal side of θ lies in QIII, we conclude that 2 2 5 =− sin θ = − 5 5
41. Consider the following diagram:
1 5 =− . 5 5
cos θ = −
θ 5
Since tan θ = 5 and the terminal side of θ lies in QIII, we conclude that 5 5 26 =− sin θ = − 26 26
42. Consider the following diagram:
1 26 =− . 26 26
cos θ = −
θ 26
−3 and the −5 terminal side of θ lies in QIII, we conclude that 3 3 34 =− sin θ = − 34 34
43. Consider the following diagram:
Since tan θ = 0.6 =
θ
cos θ = −
34
153
5 5 34 =− 34 34
Chapter 2
−4 and the −5 terminal side of θ lies in QIII, we conclude that 4 4 41 =− sin θ = − 41 41
44. Consider the following diagram:
Since tan θ = 0.8 =
θ
cos θ = −
41
45. sec θ cot θ =
1 cos θ 1 ⋅ = sin θ cos θ sin θ
46. csc θ tan θ =
1 sin θ 1 ⋅ = cos θ sin θ cos θ
sin2 θ 1 sin2 θ − 1 − cos 2 θ − = = = −1 cos 2 θ cos 2 θ cos 2 θ cos 2 θ Equivalently, you can compute as follows: tan2 θ − (1 + tan2 θ ) = tan2 θ − 1 − tan2 θ = −1
47. tan2 θ − sec 2 θ =
cos 2 θ 1 cos 2 θ − 1 − sin2 θ − = = = −1 sin2 θ sin2 θ sin2 θ sin2 θ Equivalently, you can compute as follows: cot 2 θ − (1 + cot 2 θ ) = cot 2 θ − 1 − cot 2 θ = −1
48. cot 2 θ − csc 2 θ =
49. csc θ − sin θ =
1 1− sin2 θ cos 2 θ − sin θ = = sin θ sin θ sin θ
50. sec θ − cos θ =
1 1− cos 2 θ sin2 θ − cos θ = = cos θ cos θ cos θ
154
5 5 41 =− 41 41
Section 2.4
1 1 sec θ cos θ 51. = = sin θ sin θ tan θ cos θ 53.
1 1 csc θ sin θ 52. = = cos θ cot θ cos θ sin θ
( sin θ + cos θ ) = sin2 θ + 2 sin θ cos θ + cos 2 θ 2
= ( sin 2 θ + cos 2 θ ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ
54.
( sin θ − cos θ ) = sin2 θ − 2 sin θ cos θ + cos 2 θ 2
= ( sin2 θ + cos 2 θ ) − 2 sin θ cos θ = 1 − 2 sin θ cos θ
55. Consider the following diagram:
56. Consider the following diagram:
θ
θ
101
409
Observe that tan θ = 101 = 0.1 . So, 10%.
Observe that 409 sec θ = 20 , so that cos θ = 20 . 409 Hence, tan θ = 203 = 0.15 . So, 15%.
155
Chapter 2
57. Consider the following triangle:
58. Consider the following triangle:
θ
θ
Observe that tan θ = 125 .
Observe that sec θ = 257 , so that cos θ = 257 . Hence, tan θ = 247 .
59. Observe that r = 8sin θ cos 2 θ = 8sin θ (1− sin2 θ ) = 8sin θ − 8sin3 θ .
θ 30
60 90
r
r = 8sin ( 30 ) − 8sin ( 30 ) = 8 ( 12 ) − 8 ( 12 ) = 4 −1 = 3
3
r = 8 sin ( 60 ) − 8sin3 ( 60 ) = 8
3
( ) − 8( ) = 4 3 − 3 3 = 3 3 2
3 2
3
r = 8sin ( 90 ) − 8sin3 ( 90 ) = 8 (1) − 8 (1) = 8 − 8 = 0 3
60. Observe that r = 8sin θ cos 2 θ = 8sin θ (1− sin2 θ ) = 8sin θ − 8sin3 θ .
θ −30
−60 −90
r
r = 8sin ( −30 ) − 8sin3 ( −30 ) = 8 ( − 12 ) − 8 ( − 12 ) = −4 +1 = −3 3
( ) ( ) = −4 3 + 3 3 = − 3
r = 8sin ( −60 ) − 8sin3 ( −60 ) = 8 − 23 − 8 − 23
3
r = 8sin ( −90 ) − 8sin3 ( −90 ) = 8 ( −1) − 8 ( −1) = −8 + 8 = 0 3
156
Section 2.4
sin θ 4 5 4 = = ≈ 1.33. This means that each dollar spent on cos θ 3 5 3 costs produces $1.33 in revenue.
61. Note that tan θ =
5 sin θ 1 = 2 55 = = 0.5. This means that each dollar spent on cos θ 2 5 costs produces $0.50 in revenue.
62. Note that tan θ =
63. Cosine is negative in QIII, so that cos θ = −
2 2 . 3
64. Must use the Pythagorean identity here, not the ratio identities. Further, note that −4 is not a possible value for cos θ (since the only tenable values are in the interval [ −1,1] ). In this case, the correct values would be: sin θ =
1 17 = , 17 17
cos θ = −
4 4 17 =− 17 17
65. The substitution was incorrect. To simplify a trigonometric function, we should be using identities to substitute. Using the Pythagorean Identity sin2 θ + cos 2 θ = 1 1 − sin 2 θ = cos 2 θ , we simplify to get: (1− sin θ ) (1 + sin θ ) = 1− sin2 θ = cos 2θ 66. In quadrant III, tan θ is positive so the answer should be tan θ = 15. 67. False. Since csc θ =
1 , they have the same sign for a given value of θ . sin θ
68. False. Since sec θ =
1 , they have the same sign for a given value of θ . cos θ
69. cot θ =
cos θ b = , so that a ≠ 0. sin θ a
70. tan θ =
sin θ a = , so that b ≠ 0. cos θ b
157
Chapter 2
1 , so that sin2 θ = 1. Hence, we seek those sin θ values of θ for which either sin θ = 1 or sin θ = −1 . The only value of θ such that 0 < θ ≤ 180 that satisfies either one of these equations is θ = 90. (Note: The equation sin θ = −1 has no solution when θ is restricted to this interval.) 71. Observe that sin θ = csc θ =
1 , so that cos 2 θ = 1. Hence, we seek those cos θ values of θ for which either cos θ = 1 or cos θ = −1. The only value of θ such that 0 < θ ≤ 180 that satisfies either one of these equations is θ = 180 . (Note: Even though cos ( 0 ) = 1, 0 is not included in the interval and so, the equation 72. Observe that cos θ = sec θ =
cos θ = 1 has no solution when θ is restricted to this interval.)
cos θ 1− sin2 θ =± , where the sign is chosen appropriately depending sin θ sin θ on the angle θ .
73. cot θ =
1 1 , where the sign is chosen appropriately depending = sin θ ± 1 − cos 2 θ on the angle θ .
74. csc θ =
75.
64 − ( 8sin θ ) = 64 (1 − sin2 θ ) = 64 cos 2 θ = 8 cos θ
76.
36 − ( 6 cos θ ) = 36 (1 − cos 2 θ ) = 36 sin2 θ = 6 sin θ
2
2
Observe that cos θ = x1 .
77. Consider the following diagram:
So, tan θ = −
θ − x2 − 1
158
x2 −1 1
=−
x 2 −1.
Section 2.4
78. Observe that cos θ cos θ cos 2 θ 1 − sin2 θ = = = tan θ (1− sin θ ) sin θ 1 − sin θ ( ) sin θ (1 − sin θ ) sinθ (1 − sinθ ) cos θ
=
79. a. sin ( 22 ) ≈ 0.3746
(1+ sin θ ) (1 − sin θ ) 1 + sin θ = sin θ sin θ (1− sin θ ) b. cos ( 22 ) ≈ 0.9272 c.
sin ( 22 )
cos ( 22 )
≈ 0.4040
d. tan ( 22 ) ≈ 0.4040 Yes, the values in parts c. and d. are the same. 80. a. sin (160 ) ≈ 0.3420 b. cos (16 ) ≈ 0.9613 c.
sin (160 ) cos (16 )
≈ 0.3558
d. tan (10 ) ≈ 0.1763 No, the values in parts c. and d. are not the same. 81. a. sin ( 35 ) ≈ 0.5736 b. 1− cos ( 35 ) ≈ 0.1808 c.
1− cos 2 ( 35 ) ≈ 0.5736
The values in parts a and c are the same. 82. a. tan ( 68 ) ≈ 2.4751 b. 1+ tan ( 68 ) ≈ 3.4751 c.
None of the values are the same.
159
1+ tan2 ( 68 ) ≈ 2.6695
Chapter 2 Review Solutions ----------------------------------------------------------------------1. QIV
2. QIV
3. QII
4. x-axis ( x < 0)
5. Sketch of 120 :
6. Sketch of −30 :
7. Sketch of 450 :
8. Sketch of −185 :
9. Observe that 1,000 = 2 ( 360 ) + 280. So, angle with
10. Observe that −800 = 2 ( −360 ) − 80. Since −80 and
smallest positive measure that is coterminal with 1,000 is 280 .
280 are coterminal, the angle with smallest positive measure that is coterminal with −800 is 280 .
160
Chapter 2 Review
11. Observe that 480 = 360 + 120. So, the angle with smallest positive measure that is coterminal with 480 is 120 .
12. Observe that −10 = −360 + 350. So, angle with smallest positive measure
that is coterminal with −10 is 350 .
13. QII 14. We are looking for the angle swept out by the minute hand for a time duration of 2 hours and 40 minutes. Note that 60 minutes corresponds to one complete 2 revolution of the minute-hand clockwise, and 40 minutes corresponds to of one 3 complete revolution clockwise. As such, we have: 2 hours corresponds to the angle 2 ( −360° ) = −720°,
2 ( −360° ) = −240°. 3 Thus, we know that the result of such movement of the minute-hand sweeps out the angle −960°, regardless of its actual starting point on the clockface. 40 minutes corresponds to the angle
opp −8 −4 = = hyp 10 5 adj 6 3 cos θ = = = hyp 10 5 opp −8 −4 = = tan θ = adj 6 3 1 −3 cot θ = = 4 tan θ 1 5 = sec θ = cos θ 3 1 −5 = csc θ = sin θ 4
15. Consider the following diagram:
sin θ =
θ
Using the Pythagorean Theorem, we obtain h = 6 2 + (−8)2 = 10.
161
Chapter 2
opp −7 = hyp 25 adj −24 = cos θ = hyp 25 opp −7 7 = = tan θ = adj −24 24 1 24 = cot θ = tan θ 7 1 −25 sec θ = = cos θ 24 1 −25 csc θ = = sin θ 7
16. Consider the following diagram:
sin θ =
θ
Using the Pythagorean Theorem, we obtain h = ( −7)2 + (−24)2 = 25. 17. Consider the following diagram:
sin θ =
opp 2 10 = = hyp 2 10 10
adj −6 −3 10 = = hyp 2 10 10 opp 2 −1 = = tan θ = adj −6 3 1 = −3 cot θ = tan θ cos θ =
θ
1 − 10 = 3 cos θ 1 = 10 csc θ = sin θ
sec θ =
Using the Pythagorean Theorem, we obtain h = ( −6)2 + 2 2 = 40 = 2 10.
162
Chapter 2 Review
18. Consider the following diagram:
opp 9 = hyp 41 adj −40 = cos θ = hyp 41 opp 9 −9 = = tan θ = adj −40 40 1 −40 cot θ = = 9 tan θ 1 −41 sec θ = = cos θ 40 1 41 csc θ = = sin θ 9 sin θ =
θ
Using the Pythagorean Theorem, we obtain h = ( −40)2 + 92 = 41. 19. Consider the following diagram:
sin θ =
opp 1 = hyp 2
cos θ =
adj 3 = hyp 2
opp 1 3 = = adj 3 3 1 = 3 cot θ = tan θ
tan θ =
θ
1 2 2 3 = = 3 cos θ 3 1 =2 csc θ = sin θ
sec θ =
Using the Pythagorean Theorem, we obtain h =
( 3 ) + 1 = 2. 2
2
163
Chapter 2
20. Consider the following diagram:
sin θ =
opp −9 −1 − 2 = = = hyp 9 2 2 2
adj −9 −1 − 2 = = = hyp 9 2 2 2 opp −9 tan θ = = =1 adj −9 1 cot θ = =1 tan θ 1 sec θ = =− 2 cos θ 1 csc θ = =− 2 sin θ cos θ =
θ
Using the Pythagorean Theorem, we obtain
h = ( −9)2 + (−9)2 = 162 = 9 2. 21. Consider the following diagram:
sin θ =
opp −1 − 2 = = hyp 2 2
adj −1 − 2 = = hyp 2 2 opp −1 tan θ = = =1 adj −1 1 cot θ = =1 tan θ 1 sec θ = =− 2 cos θ 1 csc θ = =− 2 sin θ cos θ =
θ
Note that we can use ANY point on the line y = x ( x ≤ 0) to form the triangle. For convenience, we use ( −1, −1). Using the Pythagorean Theorem, we obtain
h = ( −1)2 + (−1)2 = 2.
164
Chapter 2 Review
22. Consider the following diagram:
sin θ =
opp 2 2 5 = = hyp 5 5
adj −1 − 5 = = hyp 5 5 opp 2 = = −2 tan θ = adj −1 −1 1 = cot θ = tan θ 2 1 =− 5 sec θ = cos θ cos θ =
θ
Note that we can use ANY point on the line y = −2x ( x ≤ 0) to form the triangle. For convenience, we use the point (−1,2). Using the Pythagorean Theorem, we obtain
csc θ =
1 5 = sin θ 2
h = ( −1)2 + 22 = 5. opp −3 = hyp 5 adj 4 cos θ = = hyp 5 opp −3 tan θ = = adj 4 1 −4 = cot θ = 3 tan θ 1 5 sec θ = = cos θ 4 1 −5 csc θ = = 3 sin θ
23. Consider the following diagram:
sin θ =
θ
Note that we can use ANY point on the line y = − 34 x ( x ≥ 0) to form the triangle. In order to avoid the use of fractions, we use (4, −3). Using the Pythagorean Theorem, we obtain
h = 42 + ( −3)2 = 5.
165
Chapter 2
24. Consider the following diagram:
opp 5 = hyp 13 adj 12 cos θ = = hyp 13 opp 5 tan θ = = adj 12 1 12 cot θ = = tan θ 5 1 13 sec θ = = cos θ 12 1 13 csc θ = = sin θ 5 sin θ =
θ
Note that we can use ANY point on the line y = 125 x ( x ≥ 0) to form the triangle. In order to avoid the use of fractions, we use the point (12,5). Using the Pythagorean Theorem, we obtain h = 122 + 52 = 13. 25. For θ = 270 , we have: sin θ = −1 cos θ = 0 tan θ is undefined
26. Since θ = 1080 = 3(360 ) , the standard position of θ is 0 . So, sin θ = 0
cos θ = 1 tan θ = 0
cot θ = 0
cot θ is undefined sec θ = 1 csc θ is undefined
sec θ is undefined csc θ = −1
166
Chapter 2 Review
27. Since θ = −1080 = 3( −360 ) + 0 , we know that −1080 is coterminal with 0. So, sin θ = 0
28. Let n be an integer. Observe that since θ = ( −360 n ), the standard position of θ is 0. So, sin θ = 0
cos θ = 1
cos θ = 1
tan θ = 0
tan θ = 0
cot θ is undefined sec θ = 1
cot θ is undefined sec θ = 1 csc θ is undefined
csc θ is undefined 29. Consider the following diagram:
Using the Pythagorean Theorem, we 2
5 3 obtain h = 1 + = . 4 4 3 opp 3 sin θ = = 4= , hyp 5 5 4 adj 1 4 cos θ = = = , hyp 5 5 4 3 opp 3 tan θ = = 4= , adj 1 4 2
7 30. Since we are given that cos θ = , 8 we have the following triangle.
By similar triangles, we see that
where the missing side is given by 82 − 7 2 = 15 The picture for our current situation is:
15 8 = x ≈ 14.87 meters 7.2 x
167
Chapter 2
31. Recall that tan θ =
sin θ . So, if cos θ sin θ < 0 and tan θ > 0, it follows that cos θ < 0. Hence, the terminal side of θ must be in QIII.
1 > 0, it follows sin θ that sin θ > 0. Since also we have cos θ < 0, the terminal side of θ is in QII.
1 > 0, it follows cos θ that cos θ > 0. Since also we have sin θ < 0, it follows that tan θ = cos θ sin θ < 0, so that the terminal side of θ is in QIV.
sin θ . So, if cos θ sin θ < 0 and tan θ < 0, it follows that cos θ > 0. Hence, the terminal side of θ must be in QIV.
35. Consider the following diagram:
36. Consider the following diagram:
32. Since csc θ =
33. Since sec θ =
34. Recall that tan θ =
θ
θ
Using the Pythagorean Theorem, we
Using the Pythagorean Theorem, we obtain h = ( −3)2 + 4 2 = 5. So, we have 4 that sin θ = . 5
obtain h = 17 2 − (−8)2 = 15. So, we have that cos θ =
168
15 . 17
Chapter 2 Review
37. Consider the following diagram:
38. Consider the following diagram:
−
θ
Using the Pythagorean Theorem, we
Using the Pythagorean Theorem, we
obtain h = − 3 − 1 = −2 2. So, we 2
have that tan θ =
θ
2
obtain h =
−2 2 = −2 2 . 1
( − 3 ) + ( −1) = 2. So, we 2
2
−1 . 2
have that sin θ =
39. First, cos ( 270 ) = 0. Also,
40. Since 450 = 360 + 90 , we have
cos ( 270 ) + sin ( −270 ) = 1.
tan ( 450 ) is undefined and hence, one
cos ( 450 ) = cos ( 90 ) = 0. Therefore,
sin ( −270 ) = sin ( 90 ) = 1. Therefore,
cannot compute tan ( 450 ) − cot ( 540 ) .
41. First, since −720 = 2 ( −360 ) , we
42. Since θ = 1080 = 3(360 ), we have
have cos ( −720 ) = cos ( 0 ) = 1, so that
tan (1080 ) =
1 sec ( −720 ) = = 1. Also, since 1 −450 = −360 − 90 , we have
=
sin ( −450 ) = sin ( −90 ) = sin ( 270 ) = −1,
so that csc ( −450 ) =
1 = −1. sin ( 270 )
sin (1080 )
cos (1080 ) sin ( 0 )
cos ( 0 )
=0
Further, sin ( −1080 ) = − sin (1080 ) = 0
So, tan (1080 ) − sin ( −1080 ) = 0 .
So, sec ( −720 ) + csc ( −450 ) = 1 − 1 = 0
169
Chapter 2
43. Not possible because the equation sin θ = − 2 has no solution since the range of sine is [ −1,1] .
44. Possible since −1 ≤ 0.0004 ≤ 1.
45. Possible since the tangent function can attain any real value for the appropriate choice of θ .
46. Observe that solutions to the 1 15 = need to equation csc θ = sin θ 17 17 satisfy sin θ = . However, the latter 15 equation has no solutions since the 17 > 1. range of sine is [ −1,1] , and 15
47.
48.
330
60 −30 −300
The given angle is shown in bold, while the reference angle is dotted. 1 sin ( 330 ) = − sin ( 30 ) = − 2
The given angle is shown in bold, while the reference angle is dotted. 1 cos ( −300 ) = cos ( 60 ) = 2
170
Chapter 2 Review
49.
50.
150 30 −45
315
The given angle is shown in bold, while the reference angle is dotted. sin (150 ) sin30 tan (150 ) = = cos (150 ) − cos 30 =
The given angle is shown in bold, while the reference angle is dotted. cos ( 315 ) cos 45 cot ( 315 ) = = = −1 sin ( 315 ) − sin 45
12 3 = − 3 − 3 2
51.
52.
−
The given angle is shown in bold, while the reference angle is dotted. 1 1 sec ( −150 ) = = cos ( −150 ) − cos ( 30 ) =
The given angle is shown in bold, while the reference angle is dotted. 1 −1 csc ( 210 ) = = = −2 sin ( 210 ) sin ( 30 )
−2 −2 3 = 3 3 171
Chapter 2
53. –0.2419
54. 0.0872
55. 1.0355
56. cot ( 500 ) =
1 ≈ −1.1918 tan ( 500 )
58. csc ( 215 ) =
1 ≈ −1.7434 sin ( 215 )
60. csc ( −59 ) =
1 ≈ −1.1666 sin ( −59 )
57. sec (111 ) =
1 ≈ −2.7904 cos (111 )
59. cot ( −57 ) =
1 ≈ −0.6494 tan ( −57 )
3 α = 30° or 3 210°. Since the angle between consecutive numbers on the face of the clock is 30°, the minute hand is on the 2 or the 8. 61. Since tan α =
63. csc θ =
1 1 11 = 7 = − sin θ − 11 7
62. Observe that 1 1 θ = 75° + ( 360° ) + ( −360° ) = 135°. 2 3 2 . Thus, sin θ = sin (135° ) = 2
64. Since cot θ =
that tan θ =
65. tan θ =
sin θ = cos θ
8 17 −15 17
= −
8 15 2
2 3 4 67. sin θ = ( sin θ ) = = 27 9 2
2
69. tan3 θ = ( tan θ ) = 3
66. cot θ =
5 3 5
1 3 5 = , we see tan θ 5
=
5 5 5 . = 15 3
−5 cos θ 5 = −1312 = sin θ 12 13
68. 2 2 cos 2 θ = ( cos θ ) = ( −0.45 ) = 0.2025
( 3) = 3 3 3
172
Chapter 2 Review
70. Solving the equation sin2 θ + cos 2 θ = 1 for cos θ yields cos θ = ± 1− sin2 θ .
Since the terminal side of θ is in QII, we use cos θ = − 1 − sin2 θ . Hence, cos θ = − 1 − ( 11 61 ) = − 2
3600 612
= − 60 61 .
11 sin θ −11 . Thus, tan θ = = 61 = 60 cos θ − 60 61
71. Solving the equation sin2 θ + cos 2 θ = 1 for sin θ yields sin θ = ± 1− cos 2 θ .
Since the terminal side of θ is in QIV, we use sin θ = − 1− cos 2 θ . Hence, sin θ = − 1 − ( 257 ) = − 2
576 252
= − 24 25 .
7 cos θ −7 . Thus, cot θ = = 25 = sin θ − 24 24 25
72. Solving the equation 1 + tan2 θ = sec 2 θ for sec θ yields sec θ = ± 1+ tan2 θ . Since the terminal side of θ is in QIV, cos θ > 0 and so sec θ > 0. So, we use sec θ = 1 + tan2 θ . Hence, sec θ = 1 + tan2 θ = 1+ (−1)2 =
2.
73. Solving the equation 1 + cot 2 θ = csc 2 θ for cot θ yields cot θ = ± csc 2 θ −1. Since the terminal side of θ is in QII, sin θ > 0 and cos θ < 0, so cot θ < 0. So, we 2
5 13 use cot θ = − csc θ −1. Hence, cot θ = − csc θ −1 = − −1 = − . 12 12 2
2
173
Chapter 2
74. Consider the following diagram:
θ
75. Consider the following diagram:
24 and the terminal side 7 of θ lies in QII, we conclude that 24 sin θ = 25 7 cos θ = − 25
Since tan θ = −
(
h = ( −3 ) 2 + − 3
) = 12 = 2 3. 2
3 and the terminal side 3 of θ lies in QIII, we conclude that 3 3 3 3 sin θ = − =− =− 2 6 2 3 Since cot θ =
−
θ
cos θ = −
3 3 1 =− =− 6 2 2 3
Using the Pythagorean Theorem, we see that 76. Consider the following diagram:
−
θ
Since cot θ = 3 and the terminal side of θ lies in QIII, we conclude that 1 sin θ = − 2 3 cos θ = − 2
174
Chapter 2 Review
cos θ 78. sin θ cot θ = sin θ = cos θ sin θ
77. sec θ cot 2 θ =
1 cos θ cos θ ⋅ = 2 sin2 θ cos θ sin θ 2
sin θ tan θ cos θ 79. = = sin θ 1 sec θ cos θ
80.
1 ( cos θ + sec θ ) = cos θ + cos θ
2
2
( cos θ +1) = 2
2
cos 2 θ cos 4 θ + 2 cos 2 θ + 1 = cos 2 θ 1 = cos 2 θ + 2 + cos 2 θ 81.
cot θ ( sec θ + sin θ ) = =
cos θ 1 cos θ 1 cos θ sin θ + sin θ = + sin θ cos θ sin θ cos θ sin θ
(
)
1 + cos θ sin θ
sin2 θ sin2 θ 82. = = sin3 θ . 1 csc θ sin θ
83. sec180 = −1
84. csc180 is undefined.
85. Using a calculator yields: sin ( 218 ) ≈ −0.6157
86. Using a calculator yields: cos ( 28 ) ≈ 0.8829
sin −1 ( −0.6157 ) ≈ −38 So, the calculator assumes the terminal side is in QIV.
cos −1 ( 0.8829 ) ≈ 152 So, the calculator assumes the terminal side is in QII.
87. a. cos ( 73 ) ≈ 0.2924 b. 1− sin ( 73 ) ≈ 0.4370 c.
1− sin2 ( 73 ) ≈ 0.2924
The values in parts a and c are the same. 88. a. cot ( 28 ) ≈ 1.8807 b. 1+ cot ( 28 ) ≈ 2.8807 c.
None of the values are the same. 175
1+ cot 2 ( 28 ) ≈ 2.1301
Chapter 2 Practice Test Solutions ---------------------------------------------------------------1. sin θ cos θ
0 0 1
QI + +
90 1 0
1 > 0, it follows that cos θ cos θ > 0. Since also we have cos θ < 0, it follows that cot θ = sin θ sin θ < 0, so that the terminal side of θ is in QIV.
2. Since sec θ =
180 0 −1
QII + −
QIII − −
270 −1 0
QIV − +
360 0 1
sin θ is undefined cos θ when cos θ = 0, which occurs when 3. Note that tan θ =
θ = 90 , 270 .
4. Observe that 890 = 2 ( 360 ) + 170. This angle is not coterminal with −170.
The angle with the smallest positive measure that is coterminal with 890 is 170 . 5. Observe that −500 = −360 − 140. So, the angle with the smallest positive
measure that is coterminal with −500 is 220 . 6. Consider the following diagram:
Using the Pythagorean Theorem, we obtain h = ( −5)2 + (−2)2 = 29. So, we have that cos θ =
θ
176
−2 −2 29 . = 29 29
Chapter 2 Practice Test
7. Consider the following diagram:
θ
θ
Case 1: x ≥ 0 Consider the point (1,1) 1 2 . on y = x. Then, sin θ = = 2 2 Case 2: x < 0 Consider the point (−1,1) 1 2 . on y = x. Then, sin θ = = 2 2 2 if 2 the terminal side of θ lies along the graph of y = x . Thus, we conclude that sin θ =
8. Consider the following diagram:
The given angle is shown in bold, while the reference angle is dotted. 1 sin ( 210 ) = − sin ( 30 ) = − 2
9.
2 2
10. −
177
3 3
Chapter 2
11.
12.
−
The given angle is shown in bold, while the reference angle is dotted. sin ( −315 ) sin 45 tan ( −315 ) = = = 1 cos ( −315 ) cos 45
The given angle is shown in bold, while the reference angle is dotted. 1 1 sec (135 ) = = cos (135 ) − cos ( 45 ) =
13. cot ( 222 ) =
1 ≈ 1.1106 tan ( 222 )
−2 = − 2 2
14. sec ( −165 ) =
15. First, we convert the angle to DD: 11011′ = 110 + 11′ 601 ′ ≈ 110.1833
Hence, cos (11011′ ) ≈ −0.3450 .
( )
178
1 ≈ −1.0353 cos ( −165 )
Chapter 2 Practice Test
16. Consider the following diagram:
θ
17. Consider the following diagram:
θ
−
4 and the terminal side 3 of θ lies in QIII, we conclude that −4 sin θ = 5 −3 cos θ = 5
Since tan θ =
1 and the terminal side 6 of θ lies in QIV, we see that − 35 . So, sin θ = 6 − 35 sin θ tan θ = = 6 = − 35 . 1 cos θ 6
Since cos θ =
= sin θ cos 2 θ 1 − cos θ ) (1 + cos θ ) 1 − cos 2 θ ( 2 2 sin 18. = = θ = cos θ 2 sin2 θ tan2 θ sin θ 2 cos θ 2
19. a. Not possible since the range of cosine is [ −1,1]. b. Possible since the range of tangent is . 20. cot θ =
cos θ = sin θ
15 4 1
= 15
4
179
Chapter 2
21. cot 2 θ + 1 = csc 2 θ cot θ = ± csc 2 θ −1 = ±
4 3
−1 = ± 33
22. cos 2 θ + sin2 θ = 1 cos θ = ± 1− sin2 θ = ± 1− ( 73 ) = ± 7 = ± 2
40
2 10 7
23. Since tan θ = 52 , we have the triangle, and the subsequent similar one:
θ
θ
Since sides are in proportion, we have 52 = 100hyds. h = 250 yds. 24. cos θ = 0.1125 θ = cos −1 ( 0.1125 ) ≈ 83.54. Since the terminal side is in
QIV, we use θ = 360 − 83.54 ≈ 276. 25. sec ( 870 ) = sec ( 720 + 150 ) =
1 1 1 2 3 = = 3 =− 3 cos ( 720 + 150 ) cos (150 ) − 2
26. a. undefined since dividing by zero. b. defined and equals 0 c. undefined since dividing by zero.
180
Chapter 2 Cumulative Review Solutions ------------------------------------------------------1. a. 90 − 37 = 53
b. 180 − 37 = 143
2. Using the Pythagorean theorem, we have 24 2 + b 2 = 252 b = 7. 3. The central angle of the sector formed between consecutive minutes on the face of a clock is 360 12 = 30 . Since there are 7 such sectors used in forming 35 minutes,
the corresponding central angle would be 7 × 30 = 210. 4. If x = 15 ft. is a leg of a 45 − 45 − 90 triangle, then the hypotenuse is 15 2 ft. ≈ 21.21 ft. 5. Consider the following two triangles:
θ
θ
Using similar triangles, we obtain h 6 ft. = h = 40 ft. 46 ft. 2.3 ft. 3 6. Note that angles A and B are supplementary, so that A = 95. Since angles A and E are corresponding angles, E = 95. 7. tan 60 = cot ( 90 − 60 ) = cot ( 30 )
(
)
8. cos ( 30 − x ) = sin 90 − ( 30 − x ) = sin ( 60 + x ) 9.
3 2
10. 0.4695
181
( )
11. 17.525 = 17 + 0.525 601 ′ = 17 + 31.5′ = 17 + 31′ + 0.5′ ( 601′′′ ) = 1731′ 30′′ 12. Consider the following triangle:
Observe that cos ( 58.32 ) =
11.6 mi. , and so c 11.6 mi. c= ≈ 22.1 mi. cos ( 58.32 )
58.32
13. Consider the following triangle:
α
14. Consider the following triangle:
Observe that −1 39 sin α = 39 48 α = sin ( 48 ) ≈ 54 .
Observe that ft. sin 63 = 4,200d ft. d = 4,200 ≈ 4,713 ft. sin63
63 15. QIII
16. 360 + ( −170 ) = 190
17. QIV
182
Chapter 2 Practice Test
18. Consider the following triangle:
sin θ =
opp 2 29 = hyp 29
adj 5 29 =− hyp 29 opp 2 tan θ = =− adj 5 1 5 cot θ = =− tan θ 2 1 29 sec θ = =− cos θ 5 1 29 csc θ = = sin θ 2 cos θ =
θ
19. Since θ = −1170 = −3(360 ) − 90 , the standard position of θ is 270. So, sin θ = −1 cos θ = 0 tan θ is undefined cot θ = 0 sec θ is undefined csc θ = −1 20. Consider the following triangle:
sin θ =
opp 3 10 = hyp 10
adj 10 =− hyp 10 opp tan θ = = −3 adj
θ
cos θ =
183
21. Consider the following triangle:
θ
sec θ =
1 1 25 = = 24 cos θ 24 25
22. Not possible since −1 ≤ sin θ ≤ 1 and so, either sin1θ ≥ 1 or sin1θ ≤ −1. Since π2 is between −1 and 1, there is no value of θ for which sin1θ = π2 .
3 23. − 2 25. tan θ ( csc θ + cos θ ) =
24. (0.2)3 = 0.008
1 sin θ 1 + sin θ + cos θ = cos θ sin θ cos θ
184
CHAPTER 3 Section 3.1 Solutions -------------------------------------------------------------------------------1. θ =
s 2 cm 1 = = = 0.2 r 10 cm 5
2. θ =
s 2 cm 1 = = = 0.1 r 20 cm 10
3. θ =
s 4 in. 2 = = ≈ 0.18 r 22 in. 11
4. θ =
s 1 in. 1 = = ≈ 0.17 r 6 in. 6
5.
6.
s 20 mm 2 cm 1 θ= = = = = 0.02 r 100 cm 100 cm 50 7. θ =
s 321 in. 1 = = = 0.125 r 14 in. 8
s r
θ= = 8. θ =
9.
2 cm 2 cm 1 = = = 0.02 1 m 100 cm 50
s 143 cm 4 = = = 0.29 r 34 cm 14
10.
s 5 mm 5 mm 1 θ= = = = = 0.2 r 2.5 cm 25 mm 5 11. 30 = 30 ⋅
13. 45 = 45 ⋅
π 180
π 180
15.
315 = 315 ⋅
π
=
π
14. 90 = 90 ⋅
4
π 180
π 180
16.
π 180
= 7 ( 45 ) ⋅
0.2 mm 0.2 mm 1 = = = 0.0125 1.6 cm 16 mm 80
12. 60 = 60 ⋅
6
=
s r
θ= =
270 = 270 ⋅
π
4 ( 45 )
=
7π 4
=
=
3
π 2
π
180
= 3 ( 90 ) ⋅
185
π
π
2 ( 90 )
=
3π 2
Chapter 3
17.
75 = 75 ⋅
18.
π 180
= 5 (15 ) ⋅ 19.
170 = 170 ⋅
180
= 17 (10 ) ⋅
780 = 780 ⋅
π
12 (15 )
π
21.
5π 12
340 = 340 ⋅
π
18 (10 )
=
π
180
= 5 ( 20 ) ⋅
17π 18
= 13 ( 60 ) ⋅
540 = 540 ⋅
π
3 ( 60 )
π 180
= −7 ( 30 ) ⋅
=
13π 3
−320 = −320 ⋅
6 ( 30 )
= −
π
9 ( 20 )
7π 6
5π 9
=
17π 9
π
180
= 3 (180 ) ⋅
24.
π
9 ( 20 )
=
180
= 17 ( 20 ) ⋅
π
π
22.
180
−210 = −210 ⋅
=
20.
π
23.
100 = 100 ⋅
π 180
= 3π
π
180
= −16 ( 20 ) ⋅
π
9 ( 20 )
= −
16π 9
π 180 6 ( 30 ) = ⋅ = = 30 25. 6 6 π 6
π 180 4 ( 45 ) = ⋅ = = 45 26. 4 4 π 4
3π 3 π 180 3 4 ( 45 ) 27. = ⋅ = = 135 4 4 4 π
7π 7 π 180 7 6 ( 30 ) 28. = ⋅ = = 210 6 6 6 π
π
29.
3π 3 π 180 = ⋅ = 67.5 8 8 π
π
30.
11π 11 π 180 119 ( 20 ) = ⋅ = = 220 9 9 π 9
186
Section 3.1
5π 5 π 180 512 (15 ) = ⋅ = = 75 31. 12 12 π 4
33. 9π = 9 π ⋅
180
π
= 1620
19π 19 π 180 19 20 ( 9 ) = ⋅ = = 171 20 20 4 π
37.
715 (12 ) 7π 7 π 180 − =− ⋅ =− = −84 15 15 π 15
39. 4 = 4 ⋅
π
180
π
153 = ≈ 48.70 π
43.
−2.7989 = −2.7989 ⋅
180
π
π 180
47. 112 = 112 ⋅ 49.
=
π 180
π
38.
8 9 ( 20 ) 8π 8 π 180 − =− ⋅ =− = −160 9 9 9 π
40. 3 = 3 ⋅
180
540 = ≈ 171.89 π
π
42.
3.27 = 3.27 ⋅
180
π
588.6 = ≈ 187.36 π
44.
−5.9841 = −5.9841⋅
180
π
−1,077.138 = ≈ −342.86 π 46. 65 = 65 ⋅
112π ≈ 1.95 180
48. 172 = 172 ⋅
56.5π 56.5 = 56.5 ⋅ ≈ 0.986 = 180 180
= −1080
47π ≈ 0.820 180
=
π
13π 13 π 180 13 36 ( 5 ) = ⋅ = = 65 36 36 36 π
−503.802 = ≈ −160.37 π 45. 47 = 47 ⋅
180
36.
720 = ≈ 229.18 π
41.
0.85 = 0.85 ⋅
34. −6π = −6 π ⋅
35.
180
7π 7 π 180 7 3 ( 60 ) = ⋅ = = 420 32. 3 3 π 3
π 180
50.
298.7 = 298.7 ⋅
187
=
π
=
172π ≈ 3.00 180
=
298.7π ≈ 5.21 180
180
π 180
65π ≈ 1.13 180
Chapter 3
51. Consider the following diagram: 2π 3
52. Consider the following diagram:
3π 4
π
3
π
4
The reference angle is
π 3
.
In degrees:
The reference angle is
π 180 3 ( 60 ) = ⋅ = = 60 3 3 π 3
π
53. Consider the following diagram:
π 4
.
In degrees:
π 180 4 ( 45 ) = ⋅ = = 45 4 4 π 4
π
54. Consider the following diagram:
π
π
4
4
5π 4 7π 4
The reference angle is In degrees:
π 4
=
π 4
The reference angle is
.
In degrees:
4
.
π 180 4 ( 45 ) = ⋅ = = 45 4 4 π 4
π
4 ( 45 ) π 180 ⋅ = = 45 4 π 4
π
188
Section 3.1
55. Consider the following diagram:
56. Consider the following diagram: 7π 12
5π 12
The given angle is its own reference angle. In degrees: 5π 5 π 180 512 (15 ) = ⋅ = = 75 12 12 π 4
57. Consider the following diagram:
The reference angle is
5π 12
5π . 12
In degrees:
5π 5 π 180 512 (15 ) = ⋅ = = 75 12 12 π 4
58. Consider the following diagram:
π
π
9π 4, 4
3
4π 3
The reference angle is In degrees:
π 3
.
The reference angle is
π 180 3 ( 60 ) = ⋅ = = 60 3 3 π 3
π
In degrees:
π 4
189
=
π 4
.
π 180 4 ( 45 ) ⋅ = = 45 4 π 4
Chapter 3
2 π 59. sin = sin ( 45 ) = 2 4
3 π 60. cos = cos ( 30 ) = 2 6
61.
62.
2 7π π sin = − sin = − sin ( 45 ) = − 2 4 4
1 2π π cos = − cos = − cos ( 60 ) = − 2 3 3
2 3π 63. sin − = − 2 4
3 7π 64. cos − =− 2 6
3 4π 65. sin =− 2 3
3 11π 66. cos = 6 2
1 π 67. sin − = − 2 6
2 π 68. cos − = 4 2
5π 1 69. cos − = 3 2
5π 1 70. sin = 6 2
71.
72.
11π π sin − sin 11π 6 = 6 tan = 6 cos 11π cos π 6 6 − sin ( 30 )
− 12
−1 = = 3 = cos ( 30 ) 3 2 = −
5π π sin − sin 5π 3 = 3 tan = 3 cos 5π cos π 3 3
=
− sin ( 60 ) cos ( 60 )
= − 3
3 3
190
=
− 23 1 2
Section 3.1
73.
74.
π sin sin 30 ( ) π 6 = tan = 6 cos π cos ( 30 ) 6 =
1 2 3 2
5π π sin sin 5π 6 = 6 tan = 6 cos 5π − cos π 6 6
1 3 = = 3 3
= =
75.
77.
1
2 = − cos ( 30 ) − 23
1 − 3
= −
3 3
76.
5π sin − 1 5π 6 = −2 tan − = 6 cos − 5π − 23 6 =
sin ( 30 )
3π sin − 2 3π 4 =− 2 =1 tan − = 4 cos − 3π − 22 4
1 3 = 3 3 78.
3π cos 3π 2 =0= 0 cot = 2 sin 3π 1 2
1 1 π csc − = = = −1 2 sin − π −1 2
80.
79. 1 1 1 = = sec ( 5π ) = = −1 cos ( 5π ) cos (π ) −1
81.
3π cot − = 2
3π cos − 2 =0= 0 3π 1 sin − 2
82.
( )
(
)
( )
sin 134π = sin 2π + 54π = sin 54π = −
2 2
( )
(
)
( ) 12
cos 113π = cos 2π + 53π = cos 53π =
191
Chapter 3
83.
(
cos −
17 π 6
84.
) = cos ( −2π − ) (
)
= cos − 56π = −
= sin ( − 23π ) = −
3 2
85. Convert 270 to radians:
270 = 270 ⋅
π
180
= 3 ( 90 ) ⋅
110 = 110 ⋅
π
2 ( 90 )
=
π 180
=
(36 ) ⋅
3 2
86. Convert 110 to radians:
3π 2
π
180
= 11 (10 ) ⋅
87. Convert 36 to radians: 36 = 36 ⋅
sin ( − 83π ) = sin ( −2π − 23π )
5π 6
π
18 (10 )
=
11π 18
88. Convert 72 to radians:
π
5 ( 36 )
=
π 5
72 = 72 ⋅
π 180
( ) ⋅
= 2 36
π
5 ( 36 )
=
2π 5
89. The second hand moves 360 (or 2π radians) for every minute. Hence, in 2.5 2π + 2π + 12 ( 2π ) = 5π . minutes, it must move 1 minute
1 minute
1 minute 2
90. The second hand moves 360 (or 2π radians) for every minute. Hence, in 3.25 13π minutes, it must move 2π + 2π + 2π + 14 ( 2π ) = . 2 1 minute 1 minute 1 minute 1 4
minute
91. Assuming that the 32 spikes are evenly spaced along the circle, they divide the 2π π . = circle into 32 congruent sectors, each with a central angle of 32 16 92. Since one revolution takes 45 minutes, we know that the restaurant turns by an angle measuring 2π radians in 45 minutes. As such, in 25 minutes, the 25 min 10π restaurant must rotate by an angle measuring radians . ( 2π ) = 9 45 min 93. Since the Olympian makes 1.5 rotations counterclockwise, we know he turns 1.5 ( 2π ) = 3π
192
Section 3.1
94. Since the Olympian makes 1.5 rotations clockwise, we know she turns 1.5 ( −2π ) = −3π . 3π and −3π are both coterminal with π , thus they are coterminal with each other. 95. 1260°⋅
π 180°
= 7π = 3.5 ( 2π ) , thus he made 3.5 rotations.
96. He made two 1440° rotations for a total of 2880°.
2 880°⋅
π
180°
= 16π = 8 ( 2π ) , thus he made 8 total rotations.
97. Using the formula s = rθ , we have 15 = 20θ , so that θ = 0.75. 98. Using the formula s = rθ , we have 35 = 15θ , so that θ ≈ 2.3. 99. First, note that 1500 revolutions per minute is equivalent to 1500 60 = 25 revolutions per second. Since 1 revolution corresponds to 2π radians, the engine turns 25 × 2π ≈ 157.1 radians each second. 100. Converting 15,000 to radians yields 15,000 × 101. The reference angle is π −
3π π = or 45. 4 4
102. The reference angle is π −
2π π = or 60. 3 3
103. Convert 105 to radians: 105 = 105 ⋅
π
180
104. Convert 120 to radians: 120 = 120 ⋅
180
193
12 (15 )
=
105. Convert 134.3 to radians: 134.3 = 134.3 ⋅
π
180
7 (15 ) ⋅ π
=
π
( ) ≈ 261.8 radians.
2 ( 60 ) ⋅ π
π 180
3 ( 60 )
=
=
7π 12
=
2π 3
1,343π ≈ 2.3440 1,800
Chapter 3
106. Convert 109.5 to radians: 109.5 = 109.5 ⋅
π 180
=
1,095π ≈ 1.9111 1,800
107. The radius and the arc length must be converted to the same units prior to 6 cm 6 cm = = 0.03. using the radian formula. So, θ = 2 m 200 cm 108. The radius and the arc length must be converted to the same units prior to 2 in. 2 in. 1 = = . using the radian formula. So, θ = 1 ft. 12 in. 6 109. Note that the ‘45’ in the expression tan 45 is in radians, not degrees. Hence, we have tan 45 ≈ 1.61977 rather than tan 45 = 1. 110. Note that all inputs of the functions involved are in radians, not degrees. The value of each of the three expressions was obtained using the calculator set in degree mode rather than radian mode. The correct calculations (in radians) would be: cos 42 ≈ −0.40, tan 65 ≈ −1.47, sin12 ≈ −0.54 111. False. Note that 4 radians = 4 ×
( ) = 229 , the terminal side of which is in 180
π
QIII. 112. False. The measure of an angle can be any real number. 113. False. This is only possible when θ = 0 , which corresponds to 0 radians. 114. True. The sum of the three angle measures of a triangle is 180 , which corresponds to π radians. 115. If α and β are complementary, then α + β = 90 , which corresponds to π2
radians. 116. One revolution corresponds to 2π radians. So, 92 radians corresponds to 92 ≈ 14.64 revolutions. So, 14 complete revolutions are made. 2π
194
Section 3.1
117. Consider this diagram:
118. Consider this diagram:
Observe that s 900 mi. 9 θ= = = = 0.225 radians r 4000 mi. 40
Observe that θ =
s so that r s = θ r = ( 0.22 ) ( 6400 km ) = 1408 km,
which we round to 1400 km. 119. Consider the following diagram:
θ2
θ1
The sector formed between each consecutive hour on the clock face corresponds 2π π = . Note that had the hour hand NOT moved from the to a central angle of 12 6 8, there would be four such sectors to form angle θ1 , yielding a radian measure of
π 2π . However, at this time, the hour hand has moved 13 the distance from 4 = 3 6 the 8 to 9 (in 20 minutes), thereby creating a slightly larger angle whose measure is 1π π the sum of θ1 and θ 2 = = . So, the radian measure of θ is 3 6 18 1 angle of one sector 3
θ1 + θ 2 =
2π π 13π . + = 3 18 18
195
Chapter 3
120. Consider the following diagram:
θ1 θ2
The desired angle is 2π −
θ1 − θ 2 ) (
. To begin, we first find θ1 and θ 2 .
smaller angle at 9:05
Note that if the hands were exactly on the 9 and the 1, then θ1 , the central angle of this sector, constitutes 13 of one revolution around the clockface and so, has π 2π . Now, at 9:05, the hour hand has moved 121 the distance measure θ1 = 4 = 6 3 1 π π from the 9 to the 10, corresponding to the angle θ 2 = = . As such, we 12 6 72 2π π 47π − = . So, the larger angle at 9:05 is given by have θ1 − θ 2 = 3 72 72 47π 144π − 47π 97π 2π − = = . 72 72 72 121.
π π π 2π 3π 5 cos 3 + − 2 sin 2 + 5 = 5 cos − 2 sin +5 2 3 3 2 3 3 = −2 + 5 = − 3 + 5 2
122.
π −π 8π π −2 cos 3( −π ) + − 2 sin + 5 = −2 cos − − 2 sin − + 5 3 6 3 6 1 1 = −2 − − 2 − + 5 = 7 2 2
196
Section 3.1
123. In radian mode we get sin 42 = −0.917, while in degree mode we get
180 180 . sin 42 ⋅ = −0.917. This makes sense because 42 radians = 42 ⋅ π π
124. In radian mode we get cos5 = 0.284, while in degree mode we get
180 180 cos 5 ⋅ = 0.284. This makes sense because 5 radians = 5 ⋅ . π π
197
Section 3.2 Solutions -------------------------------------------------------------------------------1. Using the formula s = θ r r, we obtain s = 3 ( 4 mm ) = 12 mm .
2. Using the formula s = θ r r, we obtain
3. Using the formula s = θ r r, we obtain s = 12π ( 8 ft.) = 23π ft.
4. Using the formula s = θ r r, we obtain
5. Using the formula s = θ r r, we obtain s = 27π ( 3.5 m ) = π m .
6. Using the formula s = θ r r, we obtain
π 7. Using the formula s = θ d r, 180 we obtain π 11π s = 22 (18 μ m ) = 5 μ m . 180
π 8. Using the formula s = θ d r, we 180 π 7π obtain s = 14 (15 μ m ) = 6 μ m . 180
π 9. Using the formula s = θ d r, 180 we obtain π 200π s = 8 (1500 km ) = 3 km . 180
π 10. Using the formula s = θ d r, 180 we obtain π s = 3 (1800 km ) = 30π km . 180
π 11. Using the formula s = θ d r, 180 we obtain π 32π s = 48 ( 24 cm ) = 5 cm . 180
π 12. Using the formula s = θ d r, 180 we obtain π s = 30 (120 cm ) = 20π cm . 180
s = 4 ( 5 cm ) = 20 cm .
s = π8 ( 6 yd.) = 34π yd.
s = π4 (10 in.) = 52π in.
198
Section 3.2
13. Using the formula r =
s
θr
14. Using the formula r =
,
s
θr
16. Using the formula r =
,
s
θr
18. Using the formula r =
,
we obtain 12π yd. 12π 5 r= 5 = ⋅ yd. = 3 yd. 4π 5 4π 5 19. Using the formula r =
s
π θd 180
, we obtain
s
θr
, we obtain
5π km 5π 180 r= 9 = ⋅ km = 100 km . π 9 π 180
we obtain 24π in. 24π 5 r= 5 = ⋅ in. = 8 in. 3π 5 3π 5 17. Using the formula r =
θr
5π m 5π 12 r= 6 = ⋅ m = 10 m . π 6 π 12
we obtain 5π ft. 5π 10 r= 2 = ⋅ ft. = 25 ft. π 2 π 10 15. Using the formula r =
s
r=
,
we obtain 4π yd. 4π 9 = ⋅ yd. = 4 yd. r= 9 π 9 π 20 180
s
θr
, we obtain
4π in. 2 8 = 4π ⋅ in. = in . 3π 3π 3 2
20. Using the formula r =
s
π θd 180
,
we obtain 11π cm 11π 12 = ⋅ cm = 22 cm . r= 6 π 6 π 15 180
199
Chapter 3
21. Using the formula r =
s
π θd 180
,
we obtain 8π mi. 8π 9 r= 3 = ⋅ mi. = 12 mi. π 3 2π 40 180 23. Using the formula r =
s
π θd 180
,
22. Using the formula r =
s
π θd 180
,
we obtain
π
r=
μm π 6 4 = ⋅ μ m = 1.5 μ m . π 4 π 30 180
24. Using the formula r =
s
π θd 180
2π km 8 km = we obtain r = 11 11 π 45 180
3π ft. 27 ft. = we obtain r = 16 28 π 35 180
25. Using the formula s = θ r r, we
26. Using the formula s = θ r r, we
obtain s = ( 3.3 )( 0.4 mm ) ≈ 1.3 mm .
obtain s = ( 2.4 ) ( 5.5 cm ) ≈ 13cm .
27. Using the formula s = θ r r, we
28. Using the formula s = θ r r, we
obtain s = ( 15π ) ( 8 yd.) = 815π yd. ≈ 1.7 yd.
obtain s = ( 10π ) ( 6 ft.) = 35π ft. ≈ 1.9 ft.
29. Using the formula s = θ r r, we
30. Using the formula s = θ r r, we
,
obtain s = ( 4.95 ) ( 30 mi.) = 150 mi.
obtain s = ( 78π ) (17 mm ) ≈ 47 mm .
π 31. Using the formula s = θ d r, 180 we obtain π s = 79.5 (1.55 μ m ) ≈ 2.2 μ m . 180
π 32. Using the formula s = θ d r, 180 we obtain π s = 19.7 ( 0.63 μ m ) ≈ 0.22 μ m . 180
200
Section 3.2
π 33. Using the formula s = θ d r, 180 we obtain π s = 29 ( 2,500 km ) ≈ 1,300 km . 180
π 34. Using the formula s = θ d r, 180 we obtain π s = 11 ( 2,200 km ) ≈ 420 km . 180
π 35. Using the formula s = θ d r, 180 we obtain π s = 57 ( 22 ft.) ≈ 22 ft. 180
π 36. Using the formula s = θ d r, 180 we obtain π s = 127 ( 58 in.) ≈ 130 in. 180
1 37. Using the formula A = r 2θ r , 2 we obtain 1 49π 2 π ft.2 ≈ 12.8 ft.2 A = ( 7 ft.) = 2 6 12
1 38. Using the formula A = r 2θ r , 2 we obtain 1 9π 2 π A = ( 3 in.) = in.2 ≈ 2.83 in.2 2 5 10
1 39. Using the formula A = r 2θ r , 2 we obtain 1 2 3π A = ( 2.2 km ) ≈ 2.85 km 2 2 8
1 40. Using the formula A = r 2θ r , 2 we obtain 1 845π 2 5π A = (13 mi.) mi.2 = 2 6 12 ≈ 221 mi.2
1 41. Using the formula A = r 2θ r , 2 we obtain 1 2 3π A = (10 cm ) ≈ 42.8 cm 2 2 11
1 42. Using the formula A = r 2θ r , 2 we obtain 1 2 2π 2 A = ( 33 m ) ≈ 1,140 m 2 3
201
Chapter 3
43. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = ( 4.2 cm ) ( 56 ) 2 180
44. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = ( 2.5 mm ) ( 27 ) 2 180
≈ 8.62 cm 2
45. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = (1.5 ft.) (1.2 ) 2 180
≈ 1.47 mm 2
46. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π 2 A = ( 3.0 ft.) (14 ) ≈ 1.10 ft. 2 180
≈ 0.0236 ft.2
47. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = ( 2.6 mi.) ( 22.8 ) 2 180
48. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = (15 km ) ( 60 ) 2 180
≈ 1.35 mi.2
≈ 118 km 2
π 49. Using the formula s = θ d r, we obtain 180 π s = 45 400 km + 300 km 6, = 1,675π km ≈ 5,262 km. 180 Earth's radius Orbit above surface So, the satellite covers approximately 5,262 km during the ground station track. π 50. Using the formula s = θ d r, we obtain 180 6,650π π 6, 400 km 250 km km ≈ 3,482 km. s = 30 + = 6 180 Earth's radius Orbit above surface So, the satellite covers approximately 3,482 km during the ground station track.
202
Section 3.2
51. Consider the following diagram:
θ
Let θ be the central angle of a sector of the clock face corresponding to 25 minutes. Since the radian measure of one of the twelve congruent sectors (formed between consecutive hours) is 2π , we see that 12 2π 5π θ = 5 = . 12 6 Hence, 35π 5π ft. s = θr r = (14 ft.) = 3 6 ≈ 36.65 ft. = 37 ft.
52. Consider the following diagram:
θ
Let θ be the central angle of a sector of the clock face corresponding to 35 minutes. Since the radian measure of one of the twelve congruent sectors (formed between consecutive hours) is 2π , we see that 12 2π 7π θ = 7 = . 12 6 Hence, 49π 7π ft. s = θr r = (14 ft.) = 3 6 ≈ 51.31 ft. = 51 ft.
203
Chapter 3
53. Consider the following diagram:
θ =π
Since the diameter is 400 ft., the radius is 200 ft. Since you load the passenger car on the ground, one would make ½ of a complete revolution in order to reach the highest point the first time. This corresponds to an angle measure of π radians. The distance traveled during such a trip is s = θ r r = (π )( 200 ft.) ≈ 628 ft.
54. Since the 32 cars are equally positioned along the circle, the 6th stop from once you enter the corresponding sector formed between your car and the 6th one after 3π 6 has radian measure ( 2π ) = . Since the radius of the eye is 200 ft., the 32 8 distance traveled during such a trip is 600π 3π s = θ r r = ( 200 ft.) = ft. = 75π ft. ≈ 236 ft. 8 8 55. The arc length of the smaller gear is given by: 10π π π s = θd cm. r = 120 ( 5 cm ) = 3 180 180 Since the arc length of the smaller and larger gears are the same, we can find the 10π angle by which the larger gear has rotated using s = cm and r = 12.1 cm. 3 Indeed, we have (in radians) 10π cm s 3 θ larger = = ≈ 0.86545252. r 12.1 cm Converting to degrees subsequently yields 180 ( 0.86545252 ) ⋅ ≈ 50 . π
204
Section 3.2
56. The arc length of the smaller gear is given by: π π s = θd r = 420 ( 3 in.) = 7π in. 180 180 Since the arc length of the smaller and larger gears are the same, we can find the angle by which the larger gear has rotated using s = 7π in. and r = 15 in. Indeed, we have (in radians) s 7π in. 7π = . θ larger = = r 15 in. 15 Converting to degrees subsequently yields 7π 180 = 84 . ⋅ 15 π 57. The front crank (chain and pedal) turns one full rotation, which corresponds to an arc length of 2 ( 2.2π ) in. = 4.4π in. Hence, the chain moves this distance.
4.4π . The 3 wheel which corresponds to the back crank then turns at the same angle. Since its radius is 13 in., the resulting arc length on the wheel is approximately 60 in. = 5 ft. As such, the back crank (whose radius is 3 in.) must turn an angle of
58. The front crank (chain and pedal) turns one full rotation, which corresponds to an arc length of 2 ( 4π ) in. = 8π in. Hence, the chain moves this distance. As
8π . The wheel 1 which corresponds to the back crank then turns at the same angle. Since its radius is 13 in., the resulting arc length on the wheel is 8π ×13 in. ≈ 326.7 in. = 27.2 ft. such, the back crank (whose radius is 1 in.) must turn an angle of
205
Chapter 3
59. First, we determine the central angle corresponding to the arc length of 1000 mi. with a radius of 15.1 in. (We use the fact that 1 mi. = ( 5,280 × 12 ) in. = 63,360 in.) s 1000 mi. 63,360,000 in. = ≈ 4,196,026.49 15.1 in. r 15.1 in. Next, we calculate the arc length of the smaller tire using this angle: 30 in. s = rθ r = ( 4,196,026.49 ) = 62,940,397.35 in. 2 1 mi. = ( 62,940,397.35 in.) ⋅ ≈ 993 mi. 63,360 in.
θ= =
60. Since for every 1,000 miles with the larger tires, the onboard computer indicates that approximately 993.3774834 miles have been driven, we can set up a proportion to determine the corresponding number of miles it will display once we have traveled 50,000 miles on the larger tires. Indeed, observe that 993.3774834 mi. x so that x = 49,668.87 mi. ≈ 49,669 mi. = 1,000 mi. 50,000 mi. 61. Here, we are given that r = 7 in and θ d = 19°. So, using the formula
π π s = rθ d ≈ 2.3 in. , we see that the length the bug crawled was s = 7 ⋅19° 180° 180° 62. Here, we are given that r = 6 ft and θ d = 53°. So, using the formula
π π s = rθ d ≈ 5.6 ft. , we see that the length she crawled was s = 6 ⋅ 53° 180° 180° 63. Here, we are given that r = 20 ft. and θ d = 45 . So, using the formula
1 π A = r 2θ d , we see that the area of the region that the sprinkler covers is 2 180 1 2 π 2 2 A = ( 20 ft.) ( 45 ) = 50π ft. ≈ 157 ft. 2 180
206
Section 3.2
64. Here, we are given that r = 22 ft. and θ d = 60. So, using the formula
1 π A = r 2θ d , we see that the area of the region that the sprinkler covers is 2 180 1 2 π 242π A = ( 22 ft.) ( 60 ) ft.2 ≈ 253 ft.2 = 2 3 180 65. We seek the area of the following region:
70
1 2 π 1 π 2 Area of the BIG sector = A = r 2θ d = (12 in.) ( 70 ) = 28π in. 2 180 2 180 Area of the SMALL sector = 1 2 π 1 π 28π in.2 A = r 2θ d = ( 4 in.) ( 70 ) = 2 9 180 2 180 The area of the desired region = Area of the BIG sector – Area of the SMALL sector 28π 2 252π − 28π 2 224π 2 2 = 28π − in. = in. = in. ≈ 78 in. 9 9 9
207
Chapter 3
66. We seek the area of the following region:
65
1 2 π 1 π 1,573π Area of the BIG sector = A = r 2θ d in.2 = (11 in.) ( 65 ) = 2 180 2 180 72 Area of the SMALL sector = 1 2 π 1 π 26π in.2 A = r 2θ d = ( 4 in.) ( 65 ) = 2 180 2 180 9 The area of the desired region = Area of the BIG sector – Area of the SMALL sector 1,573π 26π 2 252π − 208π 2 1,365π 2 2 = − in. = in. = in. ≈ 60 in. 9 72 72 72 67. Using a simple proportion, we see that 45 x = x = 27,000. 0.05 sec. 30 sec.
Since 1 revolution corresponds to 360 , the wheel turns 27,000 = 75 revolutions in 360 30 seconds. 68. Using a simple proportion, we see that 2π rad. x 3 = x = 837.758 rad. 0.075 sec. 30 sec. ≈ 133 revolutions Since 1 revolution corresponds to 2π rad., the wheel turns 837.758 2π in 30 seconds.
208
Section 3.2
69. We know that the radius is 13 in. and the circumference is 26π in.. Hence, a simple proportion yields 20 in. 26π in. = x = 0.41841 sec. x 0.10 sec. So, there is 1 revolution in 0.41841 sec. Hence, there are 2.4 revolutions per second. 70. Note that “4 revolutions per second” is equivalent to “ 8π radians per second.” Hence, it turns 4π radians in 0.5 seconds, which is approximately equal to 13 radians. 71. For Lisa: We are given that r = 9 in and θ d = 68°. So, using the formula
π π s = rθ d ≈ 10.7 in. , we see that the length of the crust is s = 9 ⋅ 68° 180° 180° For Mark: We are given that r = 7 in and θ d = 42°. So, using the formula π π s = rθ d ≈ 5.13 in. , we see that the length of the crust is s = 7 ⋅ 42° 180° 180° Since he had two slices, the total length is 2 ⋅ 5.13 in ≈ 10.3 in 72. For Lisa: We are given that r = 9 in and θ d = 68°. So, using the formula
1 π A = r 2θ d , we see that the area of the pizza she ate was 2 180° 1 π 2 s = ⋅ 92 ⋅ 68° ≈ 48.1 in . 80° 2 1 For Mark: We are given that r = 7 in and θ d = 42°. So, using the formula π s = rθ d , we see that the area of the pizza he ate was 180° 1 π 2 s = ⋅ 72 ⋅ 42° ≈ 18.0 in . Since he had two slices, the total area of pizza 2 180° 2 is 36 in .
1 π 73. Using the formula A = r 2θ d yields 2 180 1 2 π 1 π 189π mi.2 ≈ 74 mi.2 A = r 2θ d = ( 9 mi.) (105 ) = 2 8 180 2 180
209
Chapter 3
π 74. First, using the formula s = r θ d yields 180 π π 21π s = r θd mi. = ( 9 mi.) (105 ) = 4 180 180 The two radial segments each measure 9 miles. So, the perimeter is given by 21π 9 + 9 + mi. ≈ 34 mi. 4 75. First, note that 4.5 nm = ( 4.5×10 −9 ) m. Since there are 10 twists for the entire
circle, the interior angle corresponding to each twist is
2π π = . So, the arc length 10 5
between consecutive twists of DNA is given by 4.5×10 −9 π −9 rθ = m ≈ (1.4 ×10 ) m = 1.4 nm. 2 5 76. First, note that 2.0 nm = ( 2.0 ×10 −9 ) m. Since there are 40 twists for the entire
circle, the interior angle corresponding to each twist is
2π π = . So, the arc length 40 20
between consecutive twists of DNA is given by 2.0 ×10 −9 π −9 rθ = m ≈ ( 0.157 ×10 ) m ≈ 0.2 nm. 2 20 77. The arc length formula for radian measure was used when the corresponding formula for degree measure should have been used. The correct computation would be π s = ( 5 cm ) ( 65 ) . 180 78. The area of a sector formula for radian measure was used when the corresponding formula for degree measure should have been used. The correct computation would be 1 2 π A = ( 2.2 cm ) ( 25 ) . 2 180 79. False. Must give the length in radians, not degrees. It should be π 4 radians.
210
Section 3.2
80. True. Since r = 1, using the formula s = θ r r yields the result. 81. True. Let θ be a fixed central angle, and let sr be the arc length corresponding to an arc with central angle θ and radius r. Then, s2 r is the arc length corresponding to an arc with central angle θ and radius 2r. Observe that sr = θ r and s2r = θ ( 2r ) = 2 (θ r ) = 2 sr . 82. False. The area, in fact, quadruples (rather than doubles). To see this, let θ be the central angle (in radians), and let Ar be the area of a sector corresponding to an arc with central angle θ and radius r. Then, A2r is the area of a sector corresponding to an arc with central angle θ and radius 2r. Observe that 1 1 2 1 A2 r = ( 2r ) θ = 4r 2θ = 4 r 2θ = 4 Ar . 2 2 2
π 83. First, note that the arc length of the smaller gear is ssmaller = θ1 r1 (1) 180 Let the arc length of the larger gear be denoted by slarger . Recall from the text
discussion that ssmaller = slarger . Hence, using the value in (1) as slarger with the radius r2 , we calculate the measure of θ 2 to be:
θ2 =
slarger r2
=
π r1 180 = θ π r1 = 1 r2 180 r2
θ1
Radian measure of θ2
84. Consider the following diagram:
s1
θ
r r 2
θ1 1 Degree measure of θ2
Let θ be a fixed central angle (in radians). Observe that 1 1 1 2 2 s A = ( r1 ) θ = ( r1 ) 1 = r1s1 , r1 2 2 2 Since s1 =θ r1
r1
which is the desired formula.
211
Chapter 3
85.
1 π Using the formula A = r 2θ d yields 2 180 1 2 π 2 A = ( 95 ft.) (150 ) ≈ 11,800 ft. 2 180 86. We need the area of the triangle. Observe that the height is 90 ft., and the base has length 90 ft. Hence, the area of the triangle is 1 Area = ( 90 ft.) ( 90 ft.) = 4,050 ft.2 2
So, the area of the infield is approximately 11,800 ft.2 − 4,050 ft.2 = 7,750 ft.2 1 π 87. Using the formula A = r 2θ d , we see that the area in front of home 2 180 1 2 π 2 plate “in the dirt” is A = (13 ft.) ( 90 ) ≈ 133 ft. 2 180 88. From Exercise 85, the home plate circle area is approximately 133 ft.2 1 2 π 2 The pitcher mound circle area is approximately ( 9 ft.) (180 ) ≈ 127 ft. 2 180 Now, the first and third basemen each cover ½ of the remaining area, which is 2 2 2 1 ≈ 1,895 ft.2 4,050 ft. − 133 ft. + 127 ft. 2 Blue Triangle Home Base Pitcher Mound
212
Section 3.3 Solutions -------------------------------------------------------------------------------1. Using the formula v =
v=
2. Using the formula v =
2m 2 m = . sec. 5 sec. 5
3. Using the formula v =
v=
s yields t
s yields t
v=
5. Using the formula v =
v=
s yields t
v=
s yields t
s yields t
8. Using the formula v =
(
s = 2.8 m
) 3.5 sec.) = 9.8 m .
10. Using the formula v =
(
v=
)
s yields t
12.2 mm ≈ 3.59 mm . min. 3.4 min.
12. Using the formula s = vt yields
(
sec. (
13. (Note that 20 min. = 13 hr.) Using the formula s = vt yields 1 s = 4.5 mi. hr.) = 1.5 mi. hr. ( 3
s yields t
2 cm 1 cm . v= 5 = hr. 8 hr. 20
3 m 3 m 10 v= . = sec. 5.2 sec. 52
11. Using the formula s = vt yields
7,524 mi. . = 627 mi. day 12 days
s yields t 3.6 μ m v= = 0.4 μ m . ns 9 ns
1 in. 1 in. v = 16 . = min. 4 min. 64
9. Using the formula v =
s yields t
6. Using the formula v =
1.75 nm = 7 nm . ms 0.25 ms
7. Using the formula v =
12 ft. = 4 ft. . min. 3 min.
4. Using the formula v =
68,000 km = 272 km . hr. 250 hr.
s yields t
s = 6.2 km
) 4.5 hr.) = 27.9 km .
hr. (
14. (Note that 2 min. = 120 sec.) Using the formula s = vt yields s = 5.6 ft. 120 sec.) = 672 ft. sec. (
(
213
)
Chapter 3
15. (Note that 15 min. = 14 hr.) Using the formula s = vt yields 1 s = 60 mi. hr. = 15 mi. hr. 4
)
16. (Note that 10 min. = 61 hr.) Using the formula s = vt yields 1 s = 72 km hr. = 12 km . hr. 6
17. (Note that 4 days = 4 ( 24 × 60 ) min. = 5,760 min.) Using the formula s = vt yields s = 750 km 5,760 min.) min. (
18. (Note that 27 min. = 27(60) sec. = 1,620 sec.) Using the formula s = vt yields s = 120 ft. 1,620 sec.) = 194, 400 ft. sec. (
(
(
)
(
)
(
)
= 4,320,000 km . 19. (Note that 3 min. = 3(60) sec. = 180 sec.) Using the formula s = vt yields s = 23 ft. 180 sec.) = 4,140 ft. sec. (
(
)
21. Using the formula ω =
θ t
(θ
measured in radians) yields 25π rad. 5π rad. ω= = sec. 10 sec. 2
23. Using the formula ω =
θ t
(θ
measured in radians) yields 100π rad. ω= = 20π rad. min. 5 min.
25. Using the formula ω =
θ t
(θ
measured in radians) yields 7π rad. 7π rad. ω= 2 = 24 hr. 12 hr.
20. (Note that 20 min. = 20/60 hr. = 1/3 hr.) Using the formula s = vt yields
(
s = 46 km
)
hr. ( 3 1
hr.) ≈ 15 km
22. Using the formula ω =
θ t
(θ
measured in radians) yields 3π rad. 9π rad. ω= 4 = sec. 1 2 sec. 6 24. Using the formula ω =
θ t
(θ
measured in radians) yields
π
rad. 2 = 5π rad. ω= min. 1 min. 10
26. Using the formula ω =
θ
(θ t measured in radians) yields 18.3 rad. ω= = 0.601 rad. hr. 30.45 hr.
214
Section 3.3
27. First, note that
π 10π . = 9 180
28. First, note that
θ = 200
Using the formula ω =
θ t
( θ measured
π π = . 180 3
θ = 60
Using the formula ω =
in radians) yields 10π rad. 2π rad. 9 ω= = sec. 5 sec. 9
in radians) yields
29. First, note that π 13π θ = 780 . = 3 180
30. First, note that
Using the formula ω =
Using the formula ω =
θ t
( θ measured
θ t
( θ measured
π
rad. 5π rad. 3 ω= = sec. 0.2 sec. 3
π 7π . = 180 3
θ = 420
θ t
( θ measured
in radians) yields 13π rad. 13π rad. ω= 3 = min. 3 min. 9
in radians) yields 7π rad. 7π rad. ω= 3 = min. 6 min. 18
31. First, note that
32. First, note that π 35π θ = 350 . = 180 18
π = 5π . 180
θ = 900
Using the formula ω =
θ t
( θ measured
Using the formula ω =
θ t
( θ measured
in radians) yields 5π rad. 10π rad. ω= = sec. 3.5 sec. 7
in radians) yields 35π rad. 98π rad. ω = 18 = sec. 5.6 sec. 9
33. Using the formula v = rω yields 2π in. v = ( 9 in.) = 6π sec. 3 sec.
34. Using the formula v = rω yields 3π cm v = ( 8 cm ) = 6π sec. 4 sec.
35. Using the formula v = rω yields
36. Using the formula v = rω yields
π π mm v = ( 5 mm ) = sec. 20 sec. 4
5π 15π ft. v = ( 24 ft.) = sec. 2 16 sec. 215
Chapter 3
37. Using the formula v = rω yields
38. Using the formula v = rω yields
4π 2π in. v = ( 2.5 in.) = sec. 3 15 sec.
8π 12π cm v = ( 4.5 cm ) = sec. 5 15 sec.
39. Using the formula v = rω yields
40. Using the formula v = rω yields
7 16π 112π yd. v = yd. = sec. 9 3 3 sec.
π 51π in. v = (10.2 in.) = min. 40 8 min.
41. Using the formula v = rω yields 10π cm v = ( 40 cm ) = 400π sec. 1 sec.
42. Using the formula v = rω yields 27.3 mm v = ( 22.6 mm ) ≈ 617 sec. 1 sec.
43. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Using this formula yields π s = ( 5 cm ) (10 sec.) 6 sec. 25π cm ≈ 26.2 cm = 3
44. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Using this formula yields 6π s = ( 2 mm ) (11 sec.) 1sec.
45. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Further, note that 10 min. = 10(60) sec. = 600 sec. Using this with the above formula yields π s = ( 5.2 in.) ( 600 sec.) 15 sec.
46. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Further, note that 3 min. = 3(60) sec. = 180 sec. Using this with the above formula yields π s = ( 3.2 ft.) (180 sec.) 4 sec.
= 132π mm ≈ 415 mm
= 208π in. ≈ 653 in. or 54.5 ft.
216
= 144π ft. ≈ 452 ft.
Section 3.3
47. Note that using the two equations s v = and v = rω together generates t the formula: s = r ω t . Hence, 3π s = (12 m ) (100 sec.) ≈ 5,650 m 2 sec.
48. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Hence, 2π 60 sec. s = (6.5 cm) (50.5 min.) 15 sec. 1 min. ≈ 8,250 cm
49. Note that using the two equations s v = and v = rω together generates t the formula: s = r ω t . Hence,
50. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Hence,
π s = ( 30 cm ) ( 25 sec.) ≈ 236 cm 10 sec.
5π 60 sec. s = ( 5 cm ) ( 9 min.) 3 sec. 1 min.
≈ 14,100 cm 51. Note that using the two equations s v = and v = rω together generates t the formula: s = r ω t . Further, note that (i) 15 min. = 15(60) sec. = 900 sec. ) radians radians (ii) ω = 5 rotations = 5(2π sec. = 10π sec. sec. Now, using these observations with the above formula yields 10π s = (15 in.) ( 900 sec.) 1 sec.
52. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Further, note that (i) 10 min. = 10(60) sec. = 600 sec. ) radians radians (ii) ω = 6 rotations = 6(2π sec. = 12π sec. sec. Now, using these observations with the above formula yields 12π s = (17 in.) ( 600 sec.) 1 sec.
= 135,000π in. Since 1 mi. = 5,280 ft. = 63,360 in., this simplifies to 1 mi. s = (135,000π in.) ⋅ 63,360 in.
= 122, 400π in. Since 1 mi. = 5,280 ft. = 63,360 in., this simplifies to 1 mi. s = (122, 400π in.) ⋅ 63,360 in.
≈ 6.69 mi.
≈ 6.07 mi.
217
Chapter 3
53. First, since 1 mi. = 63,360 in., we see that mi. 65 63,360 ) in. v 65 hr. = ( hr. ≈ 338,962 rad. ω= = hr. r 12.15 in. 12.15 in. Next, using the formula v = r ω with r = 13.05 in. and ω ≈ 338,962 rad.
(
v = (13.05 in.) 338,962 rad.
hr.
)
= 4, 423, 454 in.
hr.
=
4, 423, 454 in. 63,360 in.
hr.
yields
hr. ≈ 69.81 mi.
mi.
So, she is actually traveling at a speed of approximately 69.81 mph. 54. First, since 1 mi. = 63,360 in., we see that mi. 70 63,360 ) in. v 70 hr. = ( hr. ≈ 357,677 rad. ω= = hr. r 12.4 in. 12.4 in. Next, using the formula v = r ω with r = 13.5 in. and ω ≈ 357,677 rad.
(
v = (13.5 in.) 357,677 rad.
=
4,828,639.5 in.
hr.
) = 4,828,639.5 in. hr.
hr.
yields
hr. ≈ 76.21 mi.
hr. mi. So, she is actually traveling at a speed of approximately 76.21 mph. 63,360 in.
7,926 55. The radius of the earth is r = mi. = 3,963 mi. 2 Calculation 1: (Assuming that one rotation takes 24 hours) 2π rad. π rad. . Under this assumption, the In such case, we see that ω = = hr. 24 hr. 12 earth is spinning at the following velocity: π rad. v = r ω = ( 3,963 mi.) = 330.25π mph ≈ 1,037.51 mph hr. 12 Calculation 2: (Assuming that one rotation takes 23 hr. 56 min. 4 sec.) 4 Since 56 min. = 56 60 hr. and 4 sec. = 3,600 hr., we see that 4 23 hr. 56 min. 4 sec. = ( 23 + 56 60 + 3,600 ) hr. = 23.934 hr.
218
hr.
Section 3.3
2π rad. . Under this hr. 23.934 assumption, the earth is spinning at the following velocity: 2π rad. ≈ 331.15π mph ≈ 1,040.35 mph v = r ω = ( 3,963 mi.) hr. 23.934
So, one rotation in this amount of time yields ω =
56. One rotation in 9.9 hours yields ω =
2π rad. hr. 9.9
88,846 Also, since the diameter is 88,846 mi., we know that r = mi. = 44,423 mi. 2 2π rad. So, v = r ω = ( 44, 423 mi.) ≈ 28,193.73 mph . hr. 9.9
5(2π ) rad. = 10π rad. min. 1 min. Since the diameter of the carousel is 60 ft., its radius is 30 ft. So, v = r ω = ( 30 ft.) 10π rad. = 300π ft. min. min. . 1 min. = 300π ft. = 5π ft. ≈ 15.71 ft. min. 60 sec. sec. sec. 57. Observe that ω = 5 revolutions per min. =
(
(
)
)(
)
30(2π ) rad. = 60π rad. min. 1 min. Since the diameter of the carousel is 6 ft., its radius is 3 ft. So, v = r ω = ( 3 ft.) 60π rad. = 180π ft. min. min. . 1 min. = 180π ft. = 3π ft. ≈ 9.42 ft. min. 60 sec. sec. sec.
58. Observe that ω = 30 revolutions per min. =
(
(
)(
59.
ω = 33 13 revolutions per minute =
)
)
33 13 ( 2π ) rad. = 66 23 π rad. = 2003 π rad. min. min. min.
60. From Exercise 59, we know that ω = 2003 π rad.
. min. Since the diameter of the record is 12 in., its radius is 6 in. 200π rad. in. ≈ 1,256.64 in. So, v = r ω = ( 6 in.) = 400π min. min. min. 3
219
Chapter 3
61. We are given that ω = 5π rad.
sec.
. Further, since the diameter is 27 in., the
(
radius is 13.5 in. Hence, v = r ω = (13.5 in.) 5π rad.
sec.
) = 67.5π in. sec. .
Converting the mph yields: 1 mi. 67.5π in. 3,600 sec. = sec. 1 sec. 1 hr. 5,280(12) in. 243,000π mi. = ≈ 12.05 mph 63,360 hr.
v = 67.5π in.
62. We are given that ω = 5π rad.
sec.
. Further, since the diameter is 22 in., the
(
radius is 11 in. Hence, v = r ω = (11 in.) 5π rad.
sec.
) = 55π in. sec. .
Converting the mph yields: 1 mi. 55π in. 3,600 sec. = sec. 1 sec. 1 hr. 5,280(12) in. 198,000π mi. = ≈ 9.82 mph 63,360 hr.
v = 55π in.
63. If the smaller pulley has radius 1 in. and runs at 1,600 revolutions per minute, then the larger pulley (with radius 2.5 in.) to which it is connected must run at 1 in. (1,600 ) = 640 revolutions per minute . 2.5 in. 64. If the smaller pulley has radius 1.25 in. and runs at 1,800 revolutions per minute, then the larger pulley (with radius 2 in.) to which it is connected must run 1.25 in. at (1,800 ) = 1,125 revolutions per minute . 2 in. 65. We are given that r = 29 ft. and v = 200 mph . We seek ω . Use the formula v ω = . Observe that r 200 mph ω= ≈ 36, 413.79 rad. hr. 29 mi. 5,280
220
Section 3.3
Converting to radians per second, we see that 36, 413.79 rad. 1 hr. rad. ω ≈ = 1.6 rotations per second . ⋅ = 10.11 sec. hr. 3,600 sec. 66. We are given that r = 29 ft. and ω = 2π rad.
v = ω r . Observe that
(
v = 2π rad.
sec.
. We seek v. Use the formula
) 29 ft.) = 58π ft/sec. ≈ 182 ft/sec.
sec. (
67. We are given that r = 7m. Also, the fact that it rotates 24 times per minute implies that ω = 24 ( 2π ) rad. . We seek v. Use the formula v = ω r. Observe min. that 336π m 1 min. m m 7m ) = v = 48π rad. ( = 5.6π sec. ≈ 17.59 sec. min. min. 60 sec.
(
)
68. We are given that r = 7m = 0.007 km. Also, the fact that it rotates 30 times . We seek v. Use the formula v = ω r. per minute implies that ω = 30 ( 2π ) rad. min. Observe that 0.42π km 1 min. v = 60π rad. 0.007 km ) = ( min. min. 601 hr.
(
= 25.2π km
)
hr.
≈ 79.17 km
hr.
2π π cm 69. Using v = rω yields v = (10 cm ) = sec. Converting to meters 60 sec. 3 π 1 m m per second, we obtain v = × sec. ≈ 0.01 sec. 3 100
70. Since v = rω , it follows that ω = vr = 10 cmsec . = 0.10 rad. sec. 0.01 m
71. First, note that 2 rotations gives us θ = 2 ( 2π ) = 4π . Using the formula
ω=
θ t
yields ω =
4π rad ≈ 1.4 rad/s. 9 sec
221
Chapter 3
72. First, note that 27 rotations gives us θ = 27 ( 2π ) = 54π . Using the formula v = rω yields v = (15 ft ) ( 54π ) ≈ 2,544.7 ft/min.
73. First, note that 3 rotations gives us θ = 3 ( 2π ) = 6π . Using the formula v = rω yields v = 9 ( 6π ) = 54π m/min. Then changing the seconds gives us
54π m 1 min ⋅ ≈ 2.8 m/s. 1 min 60 sec 1.65π m 60 s ⋅ = 99π m/min. Using the formula 1 min 1s v 99π m/min ω = yields ω = ≈ 34.6 revolutions per min. 9m r
74. First, note that v =
75. Angular velocity must be expressed in radians (not degrees) per second. Here, in place of 180 one should use π rad. sec. sec. 76. Angular velocity must be expressed in radians (not degrees) per second. Here, in place of 180 one should use π rad. sec. sec.
1 77. False. Since ω = v, we see that angular velocity and linear velocity are r directly proportional. 1 78. True. Observe that ω = v. r 79. Standard Tires: v1 = r1ω Upgraded Tires: v2 = r2ω Note that the angular velocity ω is the same in both. v r Hence, v2 = r2ω = r2 1 = 2 v1. r1 r1 80. Actual mileage: s2 = v2t
Odometer mileage: s1 = v1t (1)
r r s r Observe that s2 = v2t = 2 v1 t = 2 1 t (by (1) ) = 2 s1. r1 r1 t r1
222
Section 3.3
81. The arc lengths of gears corresponding to a given angle are the same. So, we have the following information for the small gear: 1 rev. 2π rad. ωsmall = rsmall = 1 cm = sec. sec. 2π rad. 2π cm So, vsmall = rsmall ωsmall = (1 cm ) . ≈ 6.3 cm = sec. sec. sec. 82. The arc lengths of gears corresponding to a given angle are the same. So, we have the following information for the small gear: 1.5 rev. (1.5)(2π ) rad. 3π rad. ωsmall = rsmall = 1 cm = = sec. sec. sec. 3 π r ad. 3 π cm So, vsmall = rsmall ωsmall = (1 cm ) ≈ 9.4 cm . = sec. sec. sec. 83. Note that vred = 10ω versus vblue = 5ω = 12 × vred . Hence, the linear speed for the red ball is twice that of the blue ball. 84. Since the radius is 50 ft., using v = rω yields v = rω = ( 50 ft.) ( 235π sec.) ≈ 8.976 ft. sec. , So, the linear speed of a point halfway in
between is 4.49 ft/sec (since the radius is exactly half of the above).
223
Section 3.4 Solutions -------------------------------------------------------------------------------3 5π 1. sin = − 2 3
5π 1 2. cos = 3 2
3 7π 3. cos = − 2 6
1 7π 4. sin = − 2 6
2 3π 5. sin = 2 4
2 3π 6. cos = − 2 4
7π 2 sin − 4 7π 7. tan = 2 = −1 = 7 π 4 2 cos 4 2
8. Using Exercise 7, we see that 1 1 7π cot = −1 = = 4 tan 7π −1 4
1 1 2 3 5π 9. sec = = − = 3 3 6 cos 5π − 6 2
1 1 11π 10. csc = = −2 = 6 sin 11π − 1 2 6
11.
12.
1 sec ( 225 ) = = cos ( 225 ) =−
csc ( 300 ) =
1 −
2 2
2 = − 2 2
1 = sin ( 300 )
=−
13.
1 −
3 2
2 2 3 = − 3 3
14.
3 = 2 = tan ( 240 ) = 1 cos ( 240 ) − 2
sin ( 240 )
3 = 2 = − 3 cot ( 330 ) = sin ( 330 ) − 1 2
−
15.
3
cos ( 330 )
16. 3 2π 2π sin − = − sin = − 2 3 3
2 2 5π 5π sin − = = − sin = − − 2 4 4 2
224
Section 3.4
3 π π 17. sin − = − sin = − 2 3 3
18.
2 3π 3π 19. cos − = cos = − 2 4 4
5π 5π 1 20. cos − = cos = 3 3 2
3 5π 5π 21. cos − = cos = − 2 6 6
2 7π 7π 22. cos − = cos = 2 4 4
23.
24.
7π 7π 1 1 sin − = − sin = −− = 6 6 2 2
sin ( −225 ) = − sin ( 225 )
sin ( −180 ) = − sin (180 ) = 0
2 2 = − − = 2 2 25. sin ( −270 ) = − sin ( 270 ) = − ( −1) = 1
26. sin ( −60 ) = − sin ( 60 ) = −
27. cos ( −45 ) = cos ( 45 ) =
28. cos ( −135 ) = cos (135 ) = −
2 2
30. cos ( −210 ) = cos ( 210 ) = −
3 2
2 2
29. cos ( −90 ) = cos ( 90 ) = 0
31. θ = 33. θ = 35. θ = 37. θ =
π 11π 6
,
32. θ =
6
4π 5π , 3 3
34. θ =
π 5π ,
36. θ =
3π 5π , 4 4
38. θ =
3 3
225
5π 7π , 6 6
π 2π 3
,
3
5π 11π , 6 6
π 3π , 4 4
3 2
Chapter 3
39. θ = 0, π , 2π , 3π , 4π
40. θ =
41. θ = π , 3π
42. θ =
3π 7π , 2 2
π 3π 5π 7π 2
,
2
,
2
,
2
43. We seek angles θ for which sin θ = cos θ and sin θ , cos θ have
44. We seek angles θ for which sin θ = cos θ and sin θ , cos θ have the
opposite signs. These two conditions 3π 7π occur when θ = , . 4 4
same sign. These two conditions occur π 5π . when θ = , 4 4
45. We seek angles θ for which 1 = − 2 , which is equivalent to cos θ 1 2 cos θ = − =− . 2 2
46. We seek angles θ for which 1 = 2 , which is equivalent to sin θ 1 2 sin θ = = . 2 2
This occurs when θ =
3π 5π , . 4 4
This occurs when θ =
π 3π 4
,
4
.
47. We seek angles θ for which 1 sin θ = 0 (since then csc θ = is sin θ undefined). This occurs when θ = 0, π , 2π .
48. We seek angles θ for which 1 cos θ = 0 (since then sec θ = is cos θ undefined). This occurs when π 3π θ= , . 2 2
49. We seek angles θ for which sin θ cos θ = 0 (since then tan θ = is cos θ undefined). This occurs when π 3π θ= , . 2 2
50. We seek angles θ for which cos θ sin θ = 0 (since then cot θ = is sin θ undefined). This occurs when θ = 0, π , 2π .
226
Section 3.4
51. −0.4154
52. 0.9848
53. 1.3764
54. −1.2540
55. −0.7568
56. 0.7539
57. −0.7470
58. 1.1884
59. Note that February 15 corresponds to x = 46. Observe that 2π (46 − 31) T ( 46 ) = 50 − 28cos 365
60. Note that August 15 corresponds to x = 227. Observe that 2π (227 − 31) T ( 227 ) = 50 − 28cos 365
≈ 22.9 F
≈ 77.2 F
61. Note that 6 am corresponds to x = 6. Observe that π T ( 6 ) = 99.1 − 0.5sin 6 + ≈ 99.1 F 12
62. Note that 9 pm corresponds to x = 21. Observe that π T ( 21) = 99.1 − 0.5sin 21+ 12 ≈ 98.8 F
63. Note that 3 pm corresponds to x = 15. Observe that π h (15 ) = 5 + 4.8sin (15 + 4) ≈ 2.6 ft. 6
64. Note that 5 am corresponds to x = 5. Observe that π h (15 ) = 5 + 4.8sin (5 + 4 ) ≈ 0.2 ft. 6
65. Since x = 6 corresponds to June, we see that π w(6) = 145 + 10 cos (6) 6 = 145 + 10 cos (π )
66. Since x = 12 corresponds to December, we see that π w(12) = 145 + 10 cos (12) 6 = 145 + 10 cos ( 2π )
= 135 lbs.
= 155 lbs.
227
Chapter 3
67. Consider the following diagram:
68. Consider the following diagram:
We need to determine the length of the ladder (hyp) when the height of the building (opp) is 50 feet. Observe that π 50 sin = . 3 x 3 50 100 Then = x= ≈ 58 feet. 2 x 3
We need to determine the length of the ladder (hyp) when the horizontal distance from the building (adj) is 24 π 24 . meters. Observe that tan = 4 x 24 x = 24 meters. Then 1 = x Note: Since θ =
π
, we know this is an 4 isosceles triangle. This is confirmed by the fact that both legs of the triangle are 24 m.
69. Consider the following diagram:
70. Consider the following diagram:
We need to determine the horizontal distance (adj) from the point to the base of the tree when the height of the tree π 17 (opp) is 17 feet. Observe that tan = . 4 x 17 Then 1 = x = 17 feet. x
We need to determine the height of the tree (opp) when the horizontal distance (adj) from the point to the base of the tree is 12 3 feet. Observe that π x tan = . 6 12 3
Note: Since θ =
Then
π
, we know this is an
4 isosceles triangle. This is confirmed by the fact that both legs of the triangle are 17 feet.
228
3 x = x = 12 feet. 3 12 3
Section 3.4
71. Consider the following diagram:
72. Consider the following diagram:
We need to determine the length of the cables (hyp) when the vertical distance π 400 (opp) 400 feet. Observe that sin = . x 4 2 400 Then x ≈ 566 feet. = x 2
We need to determine the length of the road (hyp) when the vertical distance π 1 (opp) 1 km. Observe that sin = . 6 x 1 1 Then = x = 2 km. 2 x
73. Since x = 2 corresponds to February, we see that π n(2) = 30,000 + 20,000 sin (2 + 1) 2
74. Since x = 12 corresponds to December, we see that π n(2) = 30,000 + 20,000 sin (12 + 1) 2 13π 2
= 10,000 guests = 50,000 guests
75. T (6) = 60 − 20 cos π = 80 F
76. T (10) = 65 − 25 cos ( 53π ) = 52.5 F
77. y(2.5) = 3sin 25 ≈ −0.40 cm
78. y(10) = 5sin36 ≈ − 5 cm
79. y(3.5) = −15sin ( 72 (3.5) + 72π ) ≈ 14 in.
80. y(5) = −15sin ( 4.6(5) − π2 ) ≈ −8.0 in.
π 81. C (0) = 15.4 − 4.7 sin = 15.4 − 4.7 = 10.7 μ g/μ L 2
π π 82. C (6) = 15.4 − 4.7 sin (6) + = 15.4 − 4.7 sin(2π ) = 15.4 μ g/μ L 2 4
229
Chapter 3
83. The input t = 0 corresponds to Midnight. So,
3 π π T (0) = 36.3 − 1.4 cos (0 − 2) = 36.3 −1.4 cos − = 36.3 − 1.4 ≈ 35 C 12 6 2 84. The input t = 14.75 corresponds to 2:45 pm. (because 45 minutes = 75% of an hour). So, π T (14.75) = 36.3 − 1.4 cos (14.75 − 2) ≈ 38C . 12 3 5π 85. Should have used cos =− 2 6 5π 1 and sin = . 6 2
86.
87. Sine is the reciprocal of cosecant.
88.
3 11π Should have used cos . = 6 2
7π 7π 6 = 3 tan = 6 cos 7π 3 6 sin
89. True. The angles ( 2 nπ + θ ) and θ
90. True. The angles ( 2 nπ + θ ) and θ
are coterminal for any integer n. Hence, they have the same cosine and sine values.
are coterminal for any integer n. Hence, they have the same cosine and sine values.
3π 91. False. For instance, sin = −1 2 (which corresponds to n = 1).
92. False. For instance, cos (π ) = −1 (which corresponds to n = 1).
93. Cosecant is an odd function because 1 1 1 csc ( −θ ) = = =− = − csc (θ ) . sin ( −θ ) − sin θ sin θ 94. Tangent is an odd function because sin ( −θ ) − sin θ sin θ = =− = − tan (θ ) . tan ( −θ ) = cos ( −θ ) cos θ cos θ
230
Section 3.4
95. When considering the restriction 0 ≤ θ ≤ 2π , the equation sin θ = cos θ is π 5π . satisfied when θ = , 4 4 96. If one removes the restriction in Exercise 71, then the equation sin θ = cos θ is π 5π satisfied not only when θ = , , but also at any angle coterminal with either of 4 4 these. As such, the solutions are given by π 5π + 2 nπ , where n is an integer. θ = + 2nπ , 4 4
This can be written more succinctly as θ =
π
4
+ nπ , where n is an integer.
97. The period of y = cos ( 2π t ) is 1, and in any given period, cos(2π t ) = 1 twice,
namely when 2π t = 0, π . Since this curve undergoes 12 periods in the interval [0,12], we conclude that the given expression holds 25 times.
98. The period of y = sin ( π2 t ) is 4, and in any given period, sin ( π2 t ) = 1 twice,
namely when π2 t = π2 , 32π . Since this curve undergoes 2.5 periods in the interval [0,10], we conclude that the given expression holds 5 times. 99. This is expression is true at
π 3π 5π 7π
, , , . 4 4 4 4
1 = cos(t ) , or 1 = cos 2 t. cos(t ) And this is further equivalent to cos(t) = ±1, which occurs when t = 0, π ,2π .
100. Note that sec ( t ) = cos ( t ) is equivalent to
231
Chapter 3
101. Consider the following diagram:
We want those central angles x for which the terminal side falls on the dashed portion of the circle. This occurs when 5π 5π <x< . 3 6
102. Consider the following diagram:
Note that tan x < 1 when either 0 < sin x < cos x or when exactly one of sin x, cos x is negative. This occurs for any π2 < x < 2π . The additional restriction that sec x < 0 is equivalent to cos x < 0 , which requires that the terminal side of x be in QII or QIII. Hence, we conclude that π2 < x < 32π .
103. sin ( 423 ) ≈ 0.891
104. cos ( 227 ) ≈ −0.682
π 105. cos = 0.5 3
π 106. sin ≈ 0.866 3
sin ( −423 ) = − sin ( 423 ) ≈ −0.891
cos ( −227 ) = cos ( 227 ) ≈ −0.682
232
Chapter 3 Review Solutions ----------------------------------------------------------------------1.
135 = 135 ⋅
π 180
3.
4 ( 45 )
180
= 11 ( 30 ) ⋅
216 = 216 ⋅
180
= 14 ( 36 ) ⋅
−150 = −150 ⋅
6 ( 30 )
=
π
3 ( 60 )
108 = 108 ⋅
5 ( 36 )
=
6π 5
600 = 600 ⋅
5 ( 36 )
=
180
=π
π
180
π
= 3 ( 36 ) ⋅
π
π
4π 3
=
11π 6 6.
π
14π 5
5 ( 36 )
=
3π 5
=
10π 3
π
180
= 10 ( 60 ) ⋅
π
3 ( 60 )
10.
π
−15 = −15 ⋅
180
= −5 ( 30 ) ⋅
π
6 ( 30 )
= −
π 180
= −
π 12
5π 6
π 180 3 ( 60 ) = ⋅ = = 60 11. 3 3 π 3 π
= 4 ( 60 ) ⋅
8.
π
9.
180
180 = 180 ⋅
π
180
= 6 ( 36 ) ⋅
504 = 504 ⋅
3π 4
π
7.
=
π
4.
π
5.
240 = 240 ⋅
π
= 3 ( 45 ) ⋅
330 = 330 ⋅
2.
11π 11 π 180 116 ( 30 ) = ⋅ = = 330 12. 6 6 6 π
233
Chapter 3
5π 5 π 180 5 4 ( 45 ) = ⋅ = = 225 13. 4 4 4 π
2π 2 π 180 2 3 ( 60 ) = ⋅ = = 120 14. 3 3 3 π
5π 5 π 180 5 9 ( 20 ) = ⋅ = = 100 15. 9 9 9 π
16.
17.
18.
17π 17 π 180 1710 (18 ) = ⋅ = = 306 10 10 10 π
13π 13 π 180 13 4 ( 45 ) = ⋅ = = 585 4 4 4 π
19.
−
5π 5 π 180 =− ⋅ 18 18 π =−
18
20.
518 (10 )
− = −50
13π 13 π 180 =− ⋅ 36 36 π =−
7π π = 4 4
22. π −
5π π = 6 6
7π π −π = 6 6
24. π −
2π π = 3 3
21. 2π −
23.
11π 11 π 180 113 ( 60 ) = ⋅ = = 660 3 3 3 π
13 36 ( 5 ) 36
= −65
obtain s = π3 ( 5 cm ) = 53π cm ≈ 5.24 cm .
26. Using the formula s = θ r r , we obtain s = 56π (10 in.) = 253π in. ≈ 26.18 in.
π 27. Using the formula s = θ d r , 180 we obtain π 25π s = 100 ( 5 in.) = 9 in. ≈ 8.73 in. 180
π 28. Using the formula s = θ d r , 180 we obtain π 12π s = 36 (12 ft.) = 5 ft. ≈ 7.54 ft. 180
25. Using the formula s = θ r r , we
234
Chapter 3 Review
29. θ =
s 6 in. = = 0.5 r 12 in.
31. θ =
s 4π ft. 2π = = r 6 ft. 3
30. Note that 10 ft. = 10(12) in. = 120 in. s 27 in. So, θ = = = 0.225 . r 120 in. 32. θ =
33. Note that 5 ft. = 5(12) in. = 60 in. s 10 in. 1 = . So, θ = = r 60 in. 6 35. Using the formula r =
s
θr
34. Note that 4 km = 4,000 m. So, s 4m θ= = = 0.001 . r 4,000 m 36. Using the formula r =
, we
obtain
r=
s 2π m π = = r 8m 4
r=
8 =
s
π θd 180
,
, we
3π km 3 = 3π ⋅ km 2π 2π 3 9 km = 4.5 km 2
38. Using the formula r =
s
π θd 180
we obtain
we obtain 14π m 180 r= = 14π ⋅ m π 150 π ( ) 150 180
14π in. 180 r= = 14π ⋅ in. π 63 π ( ) 63 180
=
θr
obtain
π in. 8 8 = π ⋅ in. = in. = 1.6 in. 5π 5 5π
37. Using the formula r =
s
= 40 in.
84 m = 16.8 m 5
235
,
Chapter 3
39. Using the formula r =
s
π θd 180
,
we obtain 5π yd. 5π 18 r = 18 = ⋅ yd. = 5 yd. π 18 π 10 180
40. Using the formula r =
s
π θd 180
,
we obtain 5π ft. 5π 180 3 ft. r= = ⋅ π 3 π ( 80 ) 80 180 = 3.75 ft.
41. Using the formula A =
1 2 r θ r , we 2
obtain
42. Using the formula A =
1 2 r θ r , we 2
obtain
A=
1 2 π ( 24 mi.) = 96π mi.2 2 3
A=
≈ 302 mi.2
≈ 53 in.2
43. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = ( 60 m ) ( 60 ) 2 180
44. Using the formula 1 π A = r 2θ d , we obtain 2 180 1 2 π A = ( 36 cm ) ( 81 ) 2 180 = 291.6π cm 2 ≈ 916 cm 2
= 600π m 2 ≈ 1,885 m 2 45. Using the formula v =
v=
v=
s yields t
46. Using the formula v =
3 ft. 1 ft. = . sec. 9 sec. 3
47. Using the formula v =
1 405π 2 5π in.2 ( 9 in.) = 2 24 12
s yields t
v=
5,280 ft. = 1,320 ft. . min. 4 min.
48. Using the formula v =
15 mi. . = 5 mi. min. 3 min.
v=
236
s yields t
s yields t
12 cm . = 48 cm sec. 0.25 sec.
Chapter 3 Review
49. (Note that 1 day = 24 hr.) Using the formula s = vt yields s = 15 mi. 24 hr.) = 360 mi. hr. (
50. (Note that 1 min. = 60 sec.) Using the formula s = vt yields s = 16 ft. 60 sec.) = 960 ft. sec. (
51. (Note that 15 min. = 14 hr.) Using the formula s = vt yields 1 s = 80 mi. hr. = 20 mi. hr. 4
1 52. (Note that 6 sec. = 600 hr.) Using the formula s = vt yields 1 s = 1.5 cm hr. = 0.0025 cm hr. 600
(
)
(
)
53. Using the formula ω =
θ t
(θ
(
(
)
)
54. Using the formula ω =
θ t
(θ
measured in radians) yields 6π rad. 2π rad. ω= = sec. 9 sec. 3
measured in radians) yields π rad. ω= = 20π rad. sec. 0.05 sec.
π 180 55. Using the formula ω = t ( θ measured in degrees) yields π 225 180 = 225π rad. ω= sec. 20 sec. 3,600
π 180 56. Using the formula ω = t ( θ measured in degrees) yields π 330 180 = π rad. ω= sec. 22 sec. 12
θd
=
π rad. 16
θd
sec.
57. Using the formula v = rω yields 5π m v = (12 m ) = 10π sec. 6 sec.
58. Using the formula v = rω yields
π 3π in. v = ( 30 in.) = sec. 2 20 sec.
237
Chapter 3
59. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Using this with the above formula yields π s = (10 ft.) ( 30 sec.) = 75π ft. 4 sec.
60. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Using this with the above formula yields 3π s = ( 6 in.) ( 6 sec.) = 27π in. 4 sec.
61. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Using this with the above formula yields 2π s = (12 yd.) ( 30 sec.) = 240π yd. 3 sec.
62. Note that using the two equations s v = and v = rω together generates the t formula: s = r ω t . Further, note that 3 min. = 3(60) sec. = 180 sec. Using this with the above formula yields π s = (100 in.) (180 sec.) 18 sec. = 1,000π in.
63. First, note that the radius r = 2 in. (half the diameter). Next, observe that 60(2π ) rad. = 120π rad. So, ω = 60 revolutions per min. = min. 1 min. ≈ 754 in. v = r ω = ( 2 in.) 120π rad. = 240π ft. . min. min. min. Thus, the lady bug is traveling at approximately 754 in. min.
(
64. We are given that ω = 10π rad.
)
sec.
. Further, since the diameter is 30 in., the
(
radius is 15 in. Hence, v = r ω = (15 in.) 10π rad.
sec.
) = 150π in. sec. .
Converting the mph yields: 1 mi. 150π in. 3,600 sec. v = 150π in. = ≈ 27 mph . sec. 1 sec. 1 hr. 5,280(12) in.
238
Chapter 3 Review
65.
66.
5π 1 sin 6 5π = 2 tan = π 5 6 3 cos − 6 2
=
3 5π cos = − 2 6
67. 1 11π sin = − 2 6
−1 − 3 = 3 3
68.
69.
1 1 2 2 3 11π sec = = = = 3 3 3 6 cos 11π 6 2
70.
5π csc = 4
1 −2 1 = − 2 = = 5π 2 2 sin − 4 2
1 1 5π cot = =1 = 4 tan 5π 1 4
3π 71. sin = −1 2
3π 72. cos = 0 2
73. cos (π ) = −1
74.
75.
tan ( 315 ) =
sin ( 315 )
cos ( 315 )
=
−
2 2 = −1 2 2
cos ( 60 ) =
76. sin ( 330 ) = −
1 2
77. 1 5π 5π 1 sin − = − sin = − = − 2 6 6 2
1 2
78.
79.
2 5π 5π cos − = cos = − 2 4 4
cos ( −240 ) = cos ( 240 ) = −
239
1 2
Chapter 3
80. sin ( −135 ) = − sin (135 ) = − 82. θ =
81. θ =
2 2
2π 4π , 3 3
7π 11π , 6 6
83. We seek angles θ in the interval 0 ≤ θ ≤ 4π for which sin θ sin θ = 0 (since then tan θ = is 0). cos θ This occurs when θ = 0, π , 2π , 3π , 4π .
3π 7π , (Remember, any angle 2 2 3π is a that is coterminal with 2 solution.)
so that θ r ≈ 1.7589 rad ≈ 100.78 . Upon using the command ANGLE/DMS on the TI-calculator, we see that this is equivalent to 100 46′48′′.
86. Since s = rθ r , we have 139.2 = 56.9θ r ,
87. cos ( 1312π ) ≈ −0.9659
85. Since s = rθ r , we have 19.7 = 11.2θ r ,
84. θ =
so that θ r ≈ 2.4464 rad ≈ 140.17. Upon using the command ANGLE/DMS on the TI-calculator, we see that this is equivalent to 14010′12′′.
88. sin ( 56π ) ≈ 0.5000
240
Chapter 3 Practice Test Solutions ---------------------------------------------------------------1. Note that 20 cm = 200 mm. So, s 4 mm θ= = = 0.02 . r 200 mm
13π 13 π 180 13 4 ( 45 ) = ⋅ = = 585 2. 4 4 π 4
3.
4. 217 = 217 ⋅
260° = 260° ⋅
π 180°
= 13 ( 20° ) ⋅
π
9 ( 20° )
=
13π 9
5. Consider the following diagram:
π
180
≈ 3.79
6. Using the formula v = st yields 10 cm 0.5 cm v = 20 min = 1 min . Then, since v = rω r = ωv , we have 0.5 r= ≈ 4.77 . 2π
60
So, the radius is approximately 4.77 cm.
( ) , we obtain s = (0.05 mi.) (120 ) ( ) ≈ 1.0 mi. 7. Using the formula s = r θ d
The reference angle (the dashed arc) 7π 5π of the given angle . (in bold) is 12 12 8. Using the formula s = θ r r , we obtain s = 15π ( 8 yd.) = 815π yd. .
π
180
π
180
1 π 9. Using the formula A = r 2θ d , 2 180 we obtain 1 2 π 625π A = ( 25 ft.) ( 30 ) ft.2 = 2 180 12 ≈ 164 ft.2
241
Chapter 3
10. First, note that the radius r = 15 in. Next, observe that 5(2π ) rad. = 10π rad. So, ω = 5 revolutions per sec. = sec. 1 sec. . v = r ω = (15 in.) 10π rad. = 150π in. sec. sec. Converting the miles per minute yields: 150π mi. 1 mi. 150π in. 60 sec. = = . v = 150π in. sec. 1 sec. 1 min. 5,280(12) in. 1,056 min. So, the distance (in mile) traveled in 15 minutes is given by 150π mi. s = vt = (15 min.) ≈ 6.7 miles . min. 1,056
(
)
11. The arc length of the smaller gear is given by: 3π π π s = θd cm . r = 135 ( 2 cm ) = 2 180 180 Since the arc length of the smaller and larger gears are the same, we can find the 3π angle by which the larger gear has rotated using s = cm and r = 5.2 cm . 2 3π cm s 2 ≈ 0.9062286501 . Converting to Indeed, we have (in radians) θ larger = = r 5.2 cm 180 degrees subsequently yields ( 0.9062286501) ⋅ ≈ 52 . π 12. We know that the radius if 5 ft. So, using the formula s = θ r r , we see that 55 = 5θ θ = 11 rad.
π 13. We know that the radius is 6 ft. So, using the formula s = θ d r , we 180 obtain π s = 40 (6 ft.) ≈ 4.2 ft. 180
242
Chapter 3 Practice Test
14. Area of the big sector is 21 ( 4.5 in.) ( π4 ) units2. 2
Area of the small sector is 21 (1 in.) ( π4 ) units2. 2
Hence, the area in between is the difference of these two, namely approx. 8 inches2. 2π rad. 4π rad. = . Hence, using the formula v = rω , we 1.5 sec. 3 sec. 4π rad. in. obtain y = (9 in.) = 12π sec. . 3 sec.
15. Observe that ω =
7π 7π 1 1 sin − = − sin = −− = 6 2 2 6
7π 2 sin − 4 7π = 2 = −1 17. tan = 7 π 4 2 cos 4 2
18.
19.
16.
1 1 3π = csc − = 4 sin − 3π − sin 3π 4 4 1 2 =− =− = − 2 2 2 2
3π cos − 3π 2 =0= 0 cot − = 2 sin − 3π 1 2
1 7π 7π is undefined. 20. Since cos − it follows that 0 , = sec − = 2 2 cos − 7π 2
243
Chapter 3
21. Consider the following diagram:
Suppose the central angle, θ , of a sector was formed by having the hands of the clock lie identically on the 10 and the 2. Since the radian measure of one of the twelve congruent sectors (formed 2π between consecutive hours) is , we see 12 that 2π 2π θ = 4 = . 12 3 However, at the time 10:10, once the minute hand is on the 2, the hour hand has moved 61 the distance from the 10 to the 11 – this slight movement corresponds to a central angle of 5 , or
π π radians. Hence, the 5 = 180 36 2π π 23π − = . desired angle is 36 3 36 22.
4π 5π , 3 3
3 occurs when 3 cos θ = 23 , sin θ = 21 OR π 7π . These are satisfied when θ = , 6 6
23. Note that tan θ =
cos θ = − 23 , sin θ = − 12 .
24. s(3) = 500 − 125 cos ( π2 ) = 500 dollars.
( )
25. p(17) = 50 − 12 cos ( 1712π − π4 ) = 50 − 12 cos ( 76π ) = 50 − 12 − 23 ≈ 60 units
244
Chapter 3 Cumulative Review Solutions ------------------------------------------------------1. γ = 180 − (α + β ) = 40 2. Let x = length of the shortest side. Hypotenuse = 2x = 24 m, so that x = 12 m. As such, the larger side has length x 3 m = 12 3 m . 3. Angles D and H are equal (since they are corresponding angles) and H + G = 180 (since they are supplementary). Hence, D + G = 180 ( 9x + 6 ) + ( 7 x − 2 ) = 180
16 x + 4 = 180 x = 11 So, D = 105 , G = 75 . 4. We have the following two similar triangles:
Since the sides are in proportion, we have 54 6 12 54 = 6 s = ( 12 ) ( 1216 ) s = 61 ( 1254 ) ( 1216 ) = 1ft. 16 s 12 5. cos θ = 159 = 53
6. csc 30 = sec ( 90 − 30 ) = sec 60
7. Observe that 74 13′ 29′′
8. We first convert the angle to DD: 78 25′ = 78 + 25′ 601 ′ ≈ 78.41667
− 9 24′ 15′′
64 49′ 15′′
( ) 1 sec ( 78.41667 ) = cos ( 78.41667 ) Now,
≈ 4.9803.
245
Chapter 3
9. First, β = 180 − ( 90 + 37.4 ) = 52.6 . 132 mi. Next, we have tan37.4 = 132bmi. b = tan37.4 ≈ 173 mi.
Finally, the Pythagorean theorem yields c = 132 2 + 1732 ≈ 217 mi. 10. QII
11. positive y-axis
12. Consider the following diagram:
sin θ =
opp 3 3 13 = = hyp 13 13
adj 2 2 13 =− =− hyp 13 13 opp 3 tan θ = =− adj 2 1 2 =− cot θ = tan θ 3 1 1 13 sec θ = = 2 =− cos θ 2 cos θ =
θ
13
13
csc θ =
1 1 13 = 3 = sin θ 3 13
13. Since θ = −900 = −2(360 ) − 180 , the standard position of θ is −180 , which is coterminal with 180 . So, sin (−900°) = 0
14. Consider the following diagram:
cos (−900°) = −1 tan (−900°) = 0 csc (−900°) = undefined sec (−900°) = −1
θ
cot (−900°) = undefined
csc θ = sin1θ =
246
1 41 =− 40 − 41 40
Chapter 3 Cumulative Review
15. Since 540 = 360 + 180 and −540 = −360 −180 , both have standard 1 = 0 − (−1) = 1. positions as 180 . So, sin540 − sec ( −540 ) = sin180 − cos180 16. tan θ = 1.4285 θ = tan −1 1.4285 ≈ 55 , which is in QI. Since the terminal side of the given angle is in QIII, we use 55 + 180 = 235. 1 3 5 17. tan θ = cot1 θ = − 5 = − 5 3
18. Consider the following triangle:
θ
Observe that sin θ = − 635 .
19. Consider the following triangle:
θ
− 35
37
Observe that sin θ = − 6 3737 , cos θ = − 3737 .
20. First, note that 1.6 cm = 16 mm. Now, using the formula s = rθ r , we obtain 4 mm = (16 mm )θ r θ r = 0.25 rad. 21. Note that 1 minute corresponds to 2π rad. and 45 seconds = 34 minute
corresponds to 34 ( 2π rad.) = 32π rad. Hence, in 1 minute 45 seconds, the second
hand moves a total of ( 2π + 32π ) rad. = 72π rad.
22. Using the formula s = rθ r , we obtain 97π m = ( 29 ) r r = 27π m. m s 23. Using the formula v = st , we obtain 2.6 sec. = 3.3 sec. s = (2.6)(3.3 m ) = 8.58 m.
247
Chapter 3 rad. π in. 24. Using the formula v = rω , we obtain v = (12 in.) ( 5πsec. . Now, ) = 60sec.
60 × 60 converting to miles per hour, we multiply this quantity by 5,280 × 12 (since there are
60 seconds in one minute and 60 minutes in one hour, and 5,280 feet in one mile and 12 inches in one foot). This yields v ≈ 10.71 mph. 25. tan θ = 1 sin θ = cos θ θ = π4 , 54π
248
CHAPTER 4 Section 4.1 Solutions -------------------------------------------------------------------------------1. c (Observe that this graph is the graph of y = sin x reflected over the x-axis.)
2. d
3. a
4. j (Observe that this graph is the graph of y = cos x reflected over the x-axis.)
5. h (Observe that the amplitude is 2, and the graph has the same intercepts and shape as y = sin x.)
6. i (Observe that the amplitude is 2, and the graph has the same intercepts and shape as y = cos x.)
7. b (Observe that the graph has the same shape as y = sin x, but is horizontally stretched by a factor of 2.)
8. f (Observe that the graph has the same shape as y = cos x, but is horizontally stretched by a factor of 2.)
9. e (Observe that this graph has the 10. g (Observe that this graph has the 1 same shape as y = cos ( 2 x ) , but it has an same shape as y = sin ( 12 x ) , but it has an
amplitude of 2, is stretched by a factor of 2, and reflected over the x-axis.)
amplitude of 2, is stretched by a factor of 2, and reflected over the x-axis.)
11. Since A =
3 and B = 3, we 2 conclude that the amplitude is 3/2 and 2π the period is p = . 3
2 and B = 4, we conclude 3 that the amplitude is 2/3 and the period 2π π is p = = . 4 2
13. Since A = −1 and B = 5, we conclude that the amplitude is 1 and 2π the period is p = . 5
14. Since A = −1 and B = 7, we conclude that the amplitude is 1 and the 2π . period is p = 7
12. Since A =
249
Chapter 4 15. Since A =
2 3 and B = , we 3 2 conclude that the amplitude is 2/3 and 2π 4π the period is p = = . 3 3 2
3 2 and B = , we 2 3 conclude that the amplitude is 3/2 and 2π the period is p = = 3π . 2 3
17. Since A = −3 and B = π , we conclude that the amplitude is 3 and 2π = 2. the period is p =
18. Since A = −2 and B = π , we conclude that the amplitude is 2 and the 2π = 2. period is p =
π
19. Since A = 5 and B =
π
π
20. Since A = 4 and B =
π
3 conclude that the amplitude is 5 and 2π the period is p = = 6. π 3
, we 4 conclude that the amplitude is 4 and the 2π period is p = =8. π 4
21. Since A = − 54 and B = 12 , we conclude that the amplitude is 54 and 2π = 4π . the period is p =
22. Since A = 3 and B = π1 , we conclude that the amplitude is 3 and the period is 2π p= = 2π 2 .
1
, we
16. Since A =
23. Since A = − 13 and B = 14 , we conclude that the amplitude is 13 and 2π = 8π . the period is p = 1
1
π
2
24. Since A = 1 and B = π3 , we conclude that the amplitude is 1 and the period is 2π p= = 6. π
4
250
3
Section 4.1
25. We have the following information about the graph of y = 8cos x on [ 0,2π ] :
Amplitude: 8
Period:
2π = 2π 1
x-intercepts occur when cos x = 0. This happens when x =
π 3π
, . So, the 2 2
π 3π points , 0 , , 0 are on the graph. 2 2 The maximum value of the graph is 8 since the maximum of cos x is 1, and the amplitude is 8. This occurs when x = 0, 2π . So, the points (0,8), (2π ,8) are on the graph. The minimum value of the graph is –8 since the minimum of cos x is –1, and the amplitude is 8. This occurs when x = π . So, the point (π , −8) is on the graph. Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Using this information, we find that the graph of y = 8cos x on [ 0,2π ] is given by:
251
Chapter 4 26. We have the following information about the graph of y = 7 sin x on [ 0,2π ] :
2π = 2π 1 x-intercepts occur when sin x = 0 . This happens when x = 0, π , 2π . So, the points ( 0, 0 ) , (π , 0 ) , ( 2π , 0 ) are on the graph. Amplitude: 7
Period:
The maximum value of the graph is 7 since the maximum of sin x is 1, and π π the amplitude is 7. This occurs when x = . So, the point , 7 is on 2 2 the graph. The minimum value of the graph is –7 since the minimum of sin x is –1, 3π and the amplitude is 7. This occurs when x = . So, the point 2 3π , − 7 is on the graph. 2 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Using this information, we find that the graph of y = 7 sin x on [ 0,2π ] is given by:
252
Section 4.1
π 27. We have the following information about the graph of y = sin 4x on 0, : 2 Amplitude: 1 2π π Period: = 4 2 sin 4x = 0 x-intercepts occur when . This happens when 4 x = nπ , and so nπ π π x= , where n is an integer. So, the points ( 0, 0 ) , , 0 , , 0 are on 4 4 2 the graph. The maximum value of the graph is 1 since the maximum of sin 4 x is 1, and the amplitude is 1. This occurs when 4x =
π
2
, so that x =
π
8
. So, the
π point ,1 is on the graph. 8 The minimum value of the graph is –1 since the minimum of sin 4 x is –1, 3π 3π and the amplitude is 1. This occurs when 4x = , so that x = . So, 2 8 3π the point , −1 is on the graph. 8 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. π Using this information, we find that the graph of y = sin 4 x on 0, is given by: 2
253
Chapter 4 2π 28. We have the following information about the graph of y = cos3x on 0, : 3 Amplitude: 1 2π Period: 3 cos 3x = 0 x-intercepts occur when . This happens when (2n + 1)π (2n + 1)π , where n is an integer. So, the points 3x = , and so x = 2 6 π π , 0 , , 0 are on the graph. 6 2 The maximum value of the graph is 1 since the maximum of cos 3x is 1, 2π and the amplitude is 1. This occurs when 3x = 0, 2π , so that x = 0, . 3 2π So, the points ( 0,1) , ,1 are on the graph. 3 The minimum value of the graph is –1 since the minimum of cos 3x is –1, and the amplitude is 1. This occurs when 3x = π , so that x =
π
3
. So, the
π point , −1 is on the graph. 3 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis.
2π Using this information, we find that the graph of y = cos3x on 0, is given by: 3
254
Section 4.1
3 4π 29. We have the following information about the graph of y = sin x on 0, : 2 3 Amplitude: 1 2π 4π Period: = 3 3 2 3 3 x-intercepts occur when sin x = 0. This happens when x = nπ , and so 2 2 2 2π 4π x = nπ , where n is an integer. So, the points ,0 , ,0 are on 3 3 3 the graph. 3 The maximum value of the graph 1 since the maximum of sin x is 1, and 2 3 π π the amplitude is 1. This occurs when x = , so that occurs at x = . 2 2 3 π So the point ,1 is on the graph. 2 3 The minimum value of the graph is –1 since the minimum of sin x is –1, 2 3π 3 and the amplitude is 1. This occurs when x = , so that occurs at 2 2 x = π . So, the point (π , −1) is on the graph. Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. 3 4π Using this information, we find the graph of y = sin x on 0, is given by: 2 3
255
Chapter 4 1 30. We have the following information about the graph of y = cos x on [ 0, 6π ] : 3 Amplitude: 1 2π Period: = 6π 1 3 ( 2n + 1) π , 1 1 x-intercepts occur when cos x = 0. This happens when x = 3 2 3 3 ( 2 n + 1) π , where n is an integer. So, the points and so x = 2 3π 9π ,0 , ,0 are on the graph. 2 3 1 The maximum value of the graph 1 since the maximum of cos x is 1, and 3 1 the amplitude is 1. This occurs when x = 0, 2π , so that occurs at 3 x = 0, 6π . So the points ( 0,1) and ( 6π ,1) are on the graph. 1 The minimum value of the graph is –1 since the minimum of cos x is –1, 3 and the amplitude is 1. This occurs when , so that occurs at . So, the point is on the graph. Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. 1 Using this information, we find the graph of y = cos x on [0,6π] is given by: 3
256
Section 4.1
31. We have the following information about the graph of y = −2 cos x on [0,2π ]:
Period: 2π
Amplitude: 2
x-intercepts occur when cos x = 0. This happens when x =
(2n + 1)π , 2
π 3π where n is an integer. So, the points , 0 , , 0 are on the graph. 2 2 The maximum value of the graph is 2 since the maximum of − cos x is 1, and the amplitude is 2. This occurs when x = π . So, the point (π ,2 ) is on the graph. The minimum value of the graph is –2 since the minimum of − cos x is –1, and the amplitude is 2. This occurs when x = 0,2π . So, the points ( 0, −2 ) , ( 2π , −2 ) are on the graph. Since the coefficient used to compute the amplitude is negative, there is a reflection over the x-axis. Using this information, we find that the graph of y = −2 cos x on [ 0,2π ] is given by:
257
Chapter 4 32. We have the following information about the graph of y = − 12 sin 2 x on
[0,π ] :
Period: π Amplitude: 12 x-intercepts occur when sin 2 x = 0. This happens when 2x = nπ , where n is π an integer. So, the points ( 0,0 ) , , 0 , (π , 0 ) are on the graph. 2 The maximum value of the graph is 12 since the maximum of − sin 2x is 1
and the amplitude is 12 . This occurs when 2 x = 32π . So, the point ( 34π , 12 ) is on the graph. The minimum value of the graph is − 12 since the minimum of − sin 2 x is −1 and the amplitude is 12 . This occurs when 2 x = π2 . So, the point
( π4 , − 21 ) is on the graph.
Since the coefficient used to compute the amplitude is negative, there is a reflection over the x-axis. Using this information, we find that the graph of y = − 12 sin 2 x on [ 0, π ] is given by:
258
Section 4.1
33. We have the following information about the graph of y = −4 cos ( 2 x ) on [ 0, π ] :
Amplitude: −4 = 4
Period:
2π =π 2
x-intercepts occur when cos ( 2x ) = 0. This happens when 2x =
π 3π
π 3π
, , and 2 2
, . So, the points ( π4 , 0 ) , ( 34π , 0 ) are on the graph. 4 4 The maximum value of the graph is 4 since the maximum of − cos ( 2 x ) is
so x =
1, and the amplitude is 4. This occurs when 2x = π , and so x = π2 . So, the
points ( π2 , 4 ) are on the graph.
The minimum value of the graph is –4 since the minimum of − cos ( 2 x ) is −1, and the amplitude is 4. This occurs when 2x = 0,2π , and so x = 0, π . So, the points ( 0, −4 ) , (π , −4 ) are on the graph. Since the coefficient used to compute the amplitude is negative, there is a reflection over the x-axis.
Using this information, we find that the graph of y = −4 cos ( 2 x ) on [ 0, π ] is given by:
259
Chapter 4 34. We have the following information about the graph of y = sin 0.5x on [ 0, 4π ]:
2π = 4π 0.5 x-intercepts occur when sin 0.5x = 0. This happens when 0.5x = nπ , and so x = 2nπ , where n is an integer. So, the points ( 0, 0 ) , ( 2π , 0 ) , ( 4π , 0 ) are on the graph. The maximum value of the graph is 1 since the maximum of sin 0.5x is 1,
Amplitude: 1
Period:
and the amplitude is 1. This occurs when 0.5x = point (π ,1) is on the graph.
π
2
, so that x = π . So, the
The minimum value of the graph is –1 since the minimum of sin 0.5x is –1, 3π and the amplitude is 1. This occurs when 0.5x = , so that x = 3π . So, 2 the point ( 3π , −1) is on the graph. Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Using this information, we find that the graph of y = sin 0.5x on [ 0, 4π ] is given by:
260
Section 4.1
1 35. We have the following information about the graph of y = −3cos x on 2 [0, 4π ]: Amplitude: −3 = 3
Period:
2π = 4π 1 2
1 π 3π 1 x-intercepts occur when cos x = 0. This happens when x = , , 2 2 2 2 and so x = π , 3π . So, the points (π , 0 ) , ( 3π , 0 ) are on the graph. 1 The maximum value of the graph is 3 since the maximum of cos x is 1, 2 1 and the amplitude is 3. This occurs when x = 0, 2π , and so x = 0, 4π . 2 So, the points (0,3), (4π ,3) are on the graph. 1 The minimum value of the graph is –3 since the minimum of cos x is 2 1 –1, and the amplitude is 3. This occurs when x = π , and so x = 2π . So, 2 the point (2π , −3) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 3cos x over the x-axis. 2 1 Using this information, we find that the graph of y = −3cos x on [ 0, 4π ] is given 2 by:
261
Chapter 4 1 36. We have the following information about the graph of y = −2 sin x on 4 [0,8π ]: Amplitude: −2 = 2
Period:
2π = 8π 1 4
1 1 x-intercepts occur when sin x = 0. This happens when x = 0, π , 2π , and 4 4 so x = 0, 4π , 8π . So, the points ( 0, 0 ) , ( 4π , 0 ) , ( 8π , 0 ) are on the graph. 1 The maximum value of the graph is 2 since the maximum of sin x is 1, 4 1 π and the amplitude is 2. This occurs when x = , and so x = 2π . So, the 4 2 point (2π ,2) is on the graph. 1 The minimum value of the graph is –2 since the minimum of sin x is 4 1 3π –1, and the amplitude is 2. This occurs when x = , and so x = 6π . 4 2 So, the point (6π , −2) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 2 sin x over the x-axis. 4 1 Using this information, we find that the graph of y = −2 sin x on [ 0,8π ] is given by: 4
262
Section 4.1
37. We have the following information about the graph of y = −3sin (π x ) on
[0,2] :
Amplitude: −3 = 3
Period:
2π
π
=2
x-intercepts occur when sin (π x ) = 0. This happens when π x = 0, π , 2π , and so x = 0, 1, 2. So, the points ( 0, 0 ) , (1, 0 ) , ( 2, 0 ) are on the graph.
The maximum value of the graph is 3 since the maximum of sin (π x ) is 1, and the amplitude is 3. This occurs when π x =
π
1 , and so x = . So, the 2 2
1 point , 3 is on the graph. 2 The minimum value of the graph is –3 since the minimum of sin (π x ) is –1, and the amplitude is 3. This occurs when π x =
3π 3 , and so x = . So, 2 2
3 the point , − 3 is on the graph. 2 Since the coefficient used to compute the amplitude is negative, we reflect the graph of y = 3sin (π x ) over the x-axis. Using this information, we find that the graph of y = −3sin (π x ) on [ 0,2 ] is given by:
263
Chapter 4 38. We have the following information about the graph of y = −2 cos (π x ) on
[0,2]:
Amplitude: −2 = 2
Period:
2π
π
=2
x-intercepts occur when cos (π x ) = 0. This happens when π x =
π 3π
, , 2 2
1 3 1 3 and so x = , . So, the points , 0 , , 0 are on the graph. 2 2 2 2 The maximum value of the graph is 2 since the maximum of cos (π x ) is 1, and the amplitude is 2. This occurs when π x = 0, 2π , and so x = 0, 2. So, the points (0,2), (2,2) are on the graph. The minimum value of the graph is –2 since the minimum of cos (π x ) is –1, and the amplitude is 2. This occurs when π x = π , and so x = 1. So, the point (1, −2) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect the graph of y = 2 cos (π x ) over the x-axis. Using this information, we find that the graph of y = −2 cos (π x ) on [ 0,2 ] is given by:
264
Section 4.1
39. We have the following information about the graph of y = 5 cos 2π x on [ 0,1]:
Amplitude: 5
Period:
2π =1 2π
(2n + 1)π 2 (2 n + 1) 1 3 and so, x = , where n is an integer. So, the points , 0 , , 0 4 4 4 are on the graph. The maximum value of the graph is 5 since the maximum of cos 2π x is 1, and the amplitude is 5. This occurs when 2π x = 0, 2π , so that x = 0, 1 . So, the points ( 0,5 ) , (1,5 ) are on the graph.
x-intercepts occur when cos 2π x = 0. This happens when 2π x =
The minimum value of the graph is –5 since the minimum of cos 2π x is 1 –1, and the amplitude is 5. This occurs when 2π x = π , so that x = . So, 2 1 the point , −5 is on the graph. 2 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Using this information, we find that the graph of y = 5 cos 2π x on [ 0,1] is given by:
265
Chapter 4 40. We have the following information about the graph of y = 4 sin 2π x on [ 0,1] :
2π =1 2π x-intercepts occur when sin 2π x = 0. This happens when 2π x = nπ , and so n 1 x = , where n is an integer. So, the points ( 0, 0 ) , , 0 , (1, 0 ) are on 2 2 the graph. The maximum value of the graph is 4 since the maximum of sin 2π x is 1, π 1 and the amplitude is 4. This occurs when 2π x = , so that x = . So, the 2 4 1 point ,4 is on the graph. 4 The minimum value of the graph is –4 since the minimum of sin 2π x is 3π 3 –1, and the amplitude is 4. This occurs when 2π x = , so that x = . 2 4 3 So, the point , −4 is on the graph. 4 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Amplitude: 4
Period:
Using this information, we find that the graph of y = 4 sin 2π x on [ 0,1] is given by:
266
Section 4.1
π 41. We have the following information about the graph of y = −3sin x on [ 0,8] : 4 2π Amplitude: −3 = 3 Period: =8 π 4 π π x-intercepts occur when sin x = 0. This happens when x = 0, π , 2π , 4 4 and so x = 0, 4, 8. So, the points ( 0, 0 ) , ( 4, 0 ) , ( 8, 0 ) are on the graph. π The maximum value of the graph is 3 since the maximum of sin x is 1, 4 and the amplitude is 3. This occurs when point ( 2, 3 ) is on the graph.
π
4
x=
π
2
, and so x = 2. So, the
π The minimum value of the graph is –3 since the minimum of sin x is 4 π 3π –1, and the amplitude is 3. This occurs when x = , and so x = 6. So, 4 2 the point ( 6, − 3 ) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect π the graph of y = 3sin x over the x-axis. 4 π Using this information, we find that the graph of y = −3sin x on [ 0,8] is given by: 4
267
Chapter 4 π 42. We have the following information about the graph of y = −4 sin x on [ 0, 4 ] : 2 2π Amplitude: −4 = 4 Period: =4 π 2 π π x-intercepts occur when sin x = 0. This happens when x = 0, π , 2π , 2 2 and so x = 0, 2, 4. So, the points ( 0, 0 ) , ( 2, 0 ) , ( 4, 0 ) are on the graph. π The maximum value of the graph is 4 since the maximum of sin x is 1, 2 and the amplitude is 4. This occurs when point (1, 4 ) is on the graph.
π
2
x=
π
2
, and so x = 1. So, the
π The minimum value of the graph is –4 since the minimum of sin x is 2 π 3π –1, and the amplitude is 4. This occurs when x = , and so x = 3. So, 2 2 the point ( 3, − 4 ) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect π the graph of y = 4 sin x over the x-axis. 2 π Using this information, we find that the graph of y = −4 sin x on [ 0, 4 ] is given by: 2
268
Section 4.1
43. We have the following information about the graph of y = 2 cos ( π2 x ) on [ −8,8] :
Amplitude: 2
Period:
2π π
=4
2
x-intercepts occur when cos ( π2 x ) = 0. This happens when and so, x = (2 n + 1), where n is an integer. So, the points ( 2n + 1, 0 ) , n = −4, −3,...,2,3 are on the graph.
π 2
x=
(2n + 1)π 2
The maximum value of the graph is 2 since the maximum of cos ( π2 x ) is 1, and the amplitude is 2. This occurs when π2 x = 2 nπ , where n is an integer. So, x = −8, −4,0, 4,8. So, the points ( −8,2 ) , ( −4,2 ) , ( 0,2 ) , ( 4,2 ) , ( 8,2 ) are on the graph. The minimum value of the graph is –2 since the minimum of cos ( π2 x ) is –1, and the amplitude is 2. This occurs when π2 x = (2 n + 1)π , where n is an integer. So, , where n is an intger. So, the points are on the graph. Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Using this information, we find that the graph of on is given by:
269
Chapter 4 π 44. We have the following information about the graph of y = −3sin x on [ −8,8] : 2 2π Amplitude: −3 = 3 Period: =4 π 2 π π x-intercepts occur when sin x = 0. This happens when x = 0, π , 2π , 2 2 and so x = 0, 2, 4. So, the points ( 0, 0 ) , ( 2, 0 ) , ( 4, 0 ) are on the graph. π The maximum value of the graph is 3 since the maximum of sin x is 1, 2
π
π
+ 2nπ , where n is an 2 2 integer. So, x = 1 + 4 n, where n is an integer. So, the points ( −7,3) , ( −3,3 ) , (1,3 ) , (5,3) are on the graph. and the amplitude is 3. This occurs when
x=
π The minimum value of the graph is –3 since the minimum of sin x is 2 π 3π –1, and the amplitude is 3. This occurs when x = + 2 nπ , where n is 2 2 an integer. So, x = 3 + 4 n, where n is an integer. So, the points ( −5, −3 ) , ( −1, −3 ) , (3, −3) , ( 7, −3) are on the graph. Since the coefficient used to compute the amplitude is negative, we reflect π the graph of y = 3sin x over the x-axis. 2 π Using this information, we find that the graph of y = −3sin x on [ −8,8] is given by 2
270
Section 4.1
45. We have the following information about the graph of y = 6 sin3x on − 43π , 43π :
2π 3 x-intercepts occur when sin3x = 0. This happens when 3x = nπ , and so nπ x= , where n is an integer. 3 The maximum value of the graph is 6 since the maximum of sin3x is 1,
Amplitude: 6
Period:
and the amplitude is 6. This occurs when 3x =
2
+ 2nπ , so that
2nπ π = (1 + 4n ) , where n is an integer. 6 3 6 The minimum value of the graph is –6 since the minimum of sin3x is –1, 3π and the amplitude is 6. This occurs when 3x = + 2nπ , so that 2 π 2nπ π x= + = ( 3 + 4 n ) , where n is an integer. 2 3 6 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis.
x=
π
π
+
Using this information, we find that the graph of y = 6 sin3x on − 43π , 43π is given by:
271
Chapter 4 46. We have the following information about the graph of y = −2 cos ( π3 x ) on [ −12,12 ] :
Amplitude: −2 = 2
Period:
2π π
=6
3
x-intercepts occur when cos ( π3 x ) = 0. This happens when
π 3
x=
π 2
+ nπ ,
3(1 + 2 n ) , where n is an integer. 2 The maximum value of the graph is 2 since the maximum of cos ( π3 x ) is 1, so that x =
and the amplitude is 2. This occurs when π3 x = 2 nπ , and so x = 6n, where n is an integer. The minimum value of the graph is –2 since the minimum of cos ( π3 x ) is –1, and the amplitude is 2. This occurs when π3 x = (2 n + 1)π , and so x = 3(2n + 1), where n is an integer. Since the coefficient used to compute the amplitude is negative, there is a reflection over the x-axis. Using this information, we find that the graph of y = −2 cos ( π3 x ) on [ −12,12 ] is given by:
272
Section 4.1
1 47. In order to construct the graph of y = −4 cos x on [ −8π , 8π ] , we first 2 gather the information essential to forming the graph for one period, namely on the interval [ 0, 4π ]. Then, we extend this picture to cover the entire given interval. Indeed, we have the following: Amplitude: −4 = 4
Period:
2π = 4π 1 2
1 π 3π 1 x-intercepts occur when cos x = 0. This happens when x = , , 2 2 2 2 and so x = π , 3π . So, the points (π , 0 ) , ( 3π , 0 ) are on the graph. 1 The maximum value of the graph is 4 since the maximum of cos x is 1, 2 1 and the amplitude is 4. This occurs when x = 0, 2π , and so x = 0, 4π . 2 So, the points (0, 4), (4π , 4) are on the graph. 1 The minimum value of the graph is –4 since the minimum of cos x is – 2 1 1, and the amplitude is 4. This occurs when x = π , and so x = 2π . So, 2 the point (2π , −4) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 4 cos x over the x-axis. 2 1 Now, using this information, we can form the graph of y = −4 cos x on 2 [0, 4π ] , and then extend it to the interval [ −8π , 8π ], as shown below:
273
Chapter 4 1 48. In order to construct the graph of y = −5sin x on [ −8π , 8π ] , we first 2 gather the information essential to forming the graph for one period, namely on the interval [ 0, 4π ]. Then, we extend this picture to cover the entire given interval. Indeed, we have the following: Amplitude: −5 = 5
Period:
2π = 4π 1 2
1 1 x-intercepts occur when sin x = 0. This happens when x = 0, π , 2π , 2 2 and so x = 0, 2π , 4π . So, the points ( 0, 0 ) , ( 2π , 0 ) , ( 4π , 0 ) are on the graph. 1 The maximum value of the graph is 5 since the maximum of sin x is 1, 2 1 π and the amplitude is 5. This occurs when x = , and so x = π . So, the 2 2 point (π , 5) is on the graph. 1 The minimum value of the graph is –5 since the minimum of sin x is 2 2π –1, and the amplitude is 5. This occurs when = 2π , and so x = 3π . So, B the point (3π , −5) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 5sin x over the x-axis. 2 1 Now, using this information, we can form the graph of y = −5sin x on 2 [0, 4π ] , and then extend it to the interval [ −8π , 8π ], as shown below:
274
Section 4.1
2π 2π 49. In order to construct the graph of y = − sin ( 6 x ) on − , , we first 3 3 gather the information essential to forming the graph for one period, namely on π the interval 0, . Then, extend this picture to cover the entire interval. Indeed, 3 we have the following: 2π π Amplitude: −1 = 1 Period: = 6 3 x-intercepts occur when sin 6x = 0 . This happens when 6 x = nπ , and so nπ π π , where n is an integer. So, the points ( 0, 0 ) , , 0 , , 0 are on x= 6 6 3 the graph. The maximum value of the graph is 1 since the maximum of sin 6 x is 1, and the amplitude is 1. This occurs when 6x =
π
2
, so that x =
π
12
. So, the
π point ,1 is on the graph. 12 The minimum value of the graph is –1 since the minimum of sin 6 x is –1, 3π π and the amplitude is 1. This occurs when 6x = , so that x = . So, the 2 4 π point , −1 is on the graph. 4 Since the coefficient used to compute the amplitude is negative, we reflect the graph of y = sin ( 6 x ) over the x-axis. π Now, using this information, we can form the graph of y = − sin ( 6 x ) on 0, , 3 2π 2π and then extend it to the interval − , , as shown below: 3 3
275
Chapter 4 50. In order to construct the graph of y = − cos ( 4 x ) on [ −π , π ] , we first gather
the information essential to forming the graph for one period, namely on the π interval 0, . Then, extend this picture to cover the entire interval. Indeed, we 2 have the following: 2π π Amplitude: −1 = 1 Period: = 4 2 x-intercepts occur when cos 4x = 0 . This happens when (2n + 1)π (2n + 1)π 4x = , and so x = , where n is an integer. So, the points 2 8 π 3π , 0 , , 0 are on the graph. 8 8 The maximum value of the graph is 1 since the maximum of cos 4 x is 1, and the amplitude is 1. This occurs when 4x = 0, 2π , so that x = 0,
π
2
. So,
π the points ( 0,1) , ,1 are on the graph. 2 The minimum value of the graph is –1 since the minimum of cos 4 x is –1, and the amplitude is 1. This occurs when 4x = π , so that x =
π
4
. So, the
π point , −1 is on the graph. 4 Since the coefficient used to compute the amplitude is negative, we reflect the graph of y = cos ( 4 x ) over the x-axis. π Now, using this information, we can form the graph of y = − cos 4x on 0, , 2 and then extend it to the interval [ −π , π ] , as shown below:
276
Section 4.1
π 51. In order to construct the graph of y = 3cos x on [ −16,16 ] , we first gather 4 the information essential to forming the graph for one period, namely on the interval [ 0,8] . Then, extend this picture to cover the entire interval. Indeed, we have the following: Amplitude: 3
2π
Period:
π
=8
4
π (2n + 1)π π x-intercepts occur when cos x = 0. This happens when x = 4 2 4 and so, x = 2(2n + 1), where n is an integer. So, the points ( 2, 0 ) , ( 6, 0 ) are on the graph. π The maximum value of the graph is 3 since the maximum of cos x is 1, 4 and the amplitude is 3. This occurs when
π
4 So, the points ( 0,3 ) , ( 8,3 ) are on the graph.
x = 0, 2π , so that x = 0, 8.
π The minimum value of the graph is –3 since the minimum of cos x is 4 –1, and the amplitude is 3. This occurs when the point ( 4, −3 ) is on the graph.
π
4
x = π , so that x = 4. So,
Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. π Now, using this information, we can form the graph of y = 3cos x on [ 0,8] , 4 and then extend it to the interval [ −16,16 ] , as shown below:
277
Chapter 4 π 52. In order to construct the graph of y = 4 sin x on [ −16,16 ] , we first gather 4 the information essential to forming the graph for one period, namely on the interval [ 0,8] . Then, extend this picture to cover the entire interval. Indeed, we have the following: Amplitude: 4
Period:
2π
π
=8
4
π π x-intercepts occur when sin x = 0 . This happens when x = nπ and 4 4 so, x = 4 n , where n is an integer. So, the points ( 0, 0 ) , ( 4, 0 ) , ( 8, 0 ) are on the graph. π The maximum value of the graph is 4 since the maximum of sin x is 1, 4 and the amplitude is 4. This occurs when point ( 2, 4 ) is on the graph.
π
4
x=
π
2
, so that x = 2 . So, the
π The minimum value of the graph is –4 since the minimum of sin x is 4 π 3π –1, and the amplitude is 4. This occurs when x = , so that x = 6 . So, 4 2 the point ( 6, −4 ) is on the graph. Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. π Now, using this information, we can form the graph of y = 4 sin x on [ 0,8] , 4 and then extend it to the interval [ −16,16 ] , as shown below:
278
Section 4.1
53. In order to construct the graph of y = sin ( 4π x ) on [ −1,1] , we first gather the
information essential to forming the graph for one period, namely on the interval 1 0, 2 . Then, extend this picture to cover the entire interval. Indeed, we have the following: Amplitude: 1 2π 1 Period: = 4π 2 x-intercepts occur when sin 4π x = 0 . This happens when 4π x = nπ , and so n 1 1 x = , where n is an integer. So, the points ( 0, 0 ) , , 0 , , 0 are on 4 4 2 the graph. The maximum value of the graph is 1 since the maximum of sin 4π x is 1, π 1 and the amplitude is 1. This occurs when 4π x = , so that x = . So, the 2 8 1 point ,1 is on the graph. 8 The minimum value of the graph is –1 since the minimum of sin 4π x is –1, 3π 3 and the amplitude is 1. This occurs when 4π x = , so that x = . So, 2 8 3 the point , −1 is on the graph. 8 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. 1 Now, using this information, we can form the graph of y = sin ( 4π x ) on 0, , 2 and then extend it to the interval [ −1,1] , as shown below:
279
Chapter 4 2 2 54. In order to construct the graph of y = cos ( 6π x ) on − , , we first gather 3 3 the information essential to forming the graph for one period, namely on the 1 interval 0, . Then, extend this picture to cover the entire interval. Indeed, we 3 have the following: Amplitude: 1 2π 1 Period: = 6π 3 (2n + 1)π x-intercepts occur when cos 6π x = 0. This happens when 6π x = 2 (2 n + 1) 1 1 and so, x = , where n is an integer. So, the points , 0 , , 0 12 12 4 are on the graph. The maximum value of the graph is 1 since the maximum of cos 6π x is 1, 1 and the amplitude is 1. This occurs when 6π x = 0, 2π , so that x = 0, . 3 1 So, the points ( 0,1) , ,1 are on the graph. 3 The minimum value of the graph is –1 since the minimum of cos 6π x is –1, 1 and the amplitude is 1. This occurs when 6π x = π , so that x = . So, the 6 1 point , −1 is on the graph. 6 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. Now, using this information, we can form the graph of y = cos ( 6π x ) on, and then 2 2 extend it to the interval − , , as shown below: 3 3
280
Section 4.1
55. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx ). y-intercept is (0,0). Since one full period occurs in the interval [0, π ] , the period is π and so, 2π = π . Doing so yields B = 2 . we can determine the value of B by solving B Since the maximum y-value is 1 and the minimum y-value is –1, we know maximum y-value − minimum y-value = 1. that the amplitude A = 2 The original graph of y = sin x is reflected over the x-axis to result in this graph; so we know that the correct choice of A is –1. Thus, the equation is y = − sin 2 x . 56. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx ). y-intercept is (0, −2). Since one full period occurs in the interval [0,2π ] , the period is 2π and so, 2π we can determine the value of B by solving = 2π . Doing so yields B B = 1. Since the maximum y-value is 2 and the minimum y-value is –2, we know maximum y-value − minimum y-value that the amplitude A = = 2. 2 The original graph of y = cos x is reflected over the x-axis to result in this graph; so we know that the correct choice of A is –2. Thus, the equation is y = −2 cos x .
281
Chapter 4 57. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely, y = A sin Bx or y = A cos Bx ). The y-intercept is ( 0, 0 ) .
2π 2π Since one full period occurs in the interval 0, , the period is and 3 3 2π 2π so, we can determine the value of B by solving = . Doing so yields B 3 B = 3. 1 1 Since the maximum y-value is and the minimum y-value is − , we 2 2 maximum y value − minimum y value 1 = . know that the amplitude A = 2 2 The original graph of y = sin x is reflected over the x-axis to result in this 1 graph; so we know that the correct choice of A is − – . 2 1 Thus, the equation is y = − sin ( 3x ) . 2 58. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). The y-intercept is ( 0, 0 ) .
Since one full period occurs in the interval [ 0, 4π ] , the period is 4π and so,
2π 1 = 4π . Doing so yields B = . 2 B 1 1 Since the maximum y-value is and the minimum y-value is − , we 2 2 maximum y value − minimum y value 1 = . know that the amplitude A = 2 2 The shape of the graph coincides with the graph of y = cos x and hence, it is 1 not reflected over the x-axis; so we know that the correct choice of A is . 2 1 1 Thus, the equation is y = cos x . 2 2 we can determine the value of B by solving
282
Section 4.1
59. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). y-intercept is (0, 1). Since one full period occurs in the interval [0,2], the period is 2 and so, we 2π can determine the value of B by solving = 2. Doing so yields B = π . B Since the maximum y-value is 1 and the minimum y-value is –1, we know maximum y-value − minimum y-value = 1. that the amplitude A = 2 The shape of the graph coincides with the graph of y = cos x and hence, it is not reflected over the x-axis; so we know that the correct choice of A is 1. Thus, the equation is y = cos π x . 60. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). y-intercept is (0, 0). Since one full period occurs in the interval [0,2], the period is 2 and so, we 2π can determine the value of B by solving = 2. Doing so yields B = π . B Since the maximum y-value is 1 and the minimum y-value is –1, we know maximum y-value − minimum y-value = 1. that the amplitude A = 2 The shape of the graph coincides with the graph of y = sin x and hence, it is not reflected over the x-axis; so we know that the correct choice of A is 1. Thus, the equation is y = sin π x .
283
Chapter 4 61. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). y-intercept is (0,0). Since one full period occurs in the interval [0, 4], the period is 4 and so, we π 2π can determine the value of B by solving = 4. Doing so yields B = . B 2 Since the maximum y-value is 2 and the minimum y-value is –2, we know maximum y-value − minimum y-value = 2. that the amplitude A = 2 The original graph of y = sin x is reflected over the x-axis to result in this graph; so we know that the correct choice of A is –2.
π Thus, the equation is y = −2 sin x . 2 62. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). y-intercept is (0,−3). Since one full period occurs in the interval [0, 4], the period is 4 and so, we 2π π = 4. Doing so yields B = . can determine the value of B by solving B 2 Since the maximum y-value is 3 and the minimum y-value is –3, we know maximum y-value − minimum y-value that the amplitude A = = 3. 2 The original graph of y = cos x is reflected over the x-axis to result in this graph; so we know that the correct choice of A is –3.
π Thus, the equation is y = −3cos x . 2
284
Section 4.1
63. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). y-intercept is (0, 0). 1 1 and so, Since one full period occurs in the interval 0, , the period is 4 4 2π 1 we can determine the value of B by solving = . Doing so yields B 4 B = 8π . Since the maximum y-value is 1 and the minimum y-value is –1, we know maximum y-value − minimum y-value that the amplitude A = =1. 2 The shape of the graph coincides with the graph of y = sin x and hence, it is not reflected over the x-axis; so we know that the correct choice of A is 1. Thus, the equation is y = sin(8πx). 64. In order to determine the equation, we make the following key observations that enable us to determine the values of A and B in the appropriate choice of standard form (namely y = A sin Bx or y = A cos Bx). y-intercept is (0, 1). 1 1 and so, Since one full period occurs in the interval 0, , the period is 2 2 2π 1 we can determine the value of B by solving = . Doing so yields B 2 B = 4π . Since the maximum y-value is 1 and the minimum y-value is –1, we know maximum y-value − minimum y-value that the amplitude A = = 1. 2 The shape of the graph coincides with the graph of y = cos x and hence, it is not reflected over the x-axis; so we know that the correct choice of A is 1. Thus, the equation is y = cos 4π x . 65. The maximum is 10 and the minimum is 3. So, the amplitude is 102−3 = 3.5 (thousands of widgets). Thus, we conclude that the amplitude is 3500 widgets. 66. Consecutive maximum values of d occur at t = 4 and t = 16. So, the period is 16 – 4 = 12 months.
285
Chapter 4 67. We plot the data graphically, as follows:
The amplitude is 8−26 = 1 mg/L . The period is 8 weeks.
68. We plot the data graphically, as follows:
The amplitude is 9−25 = 2 mg/L . The period is 8 weeks.
t k k 69. Observe that y = 4 cos = 4 cos t . So, the amplitude is 4 cm and 2 4 the mass m = 4 g. k 70. Observe that y = 3cos 3t k = 3cos t . So, the amplitude is 3 cm and 1 9 the mass m = 1 g. 9
(
)
t 71. First, since the period of y = 3cos is 2π 1 = 4π , the frequency of 2 2 cycles per second. oscillation is given by 1 4π 72. First, since the period of y = 3.5 cos ( 3t ) is 2π , the frequency of oscillation 3 3 is given by cycles per second. 2π
286
Section 4.1
73. From the equation y = 0.005sin ( 2π (256t ) ) we see that the amplitude is 0.005
cm and the frequency is 256 hertz. 74. From the equation y = 0.005sin ( 2π (288t ) ) we see that the amplitude is 0.005 cm and the frequency is 288 hertz. 75. Observe that y = 0.008sin ( 750π t ) = 0.008sin ( 2π (375t ) ) . So, the amplitude is 0.008 cm and the frequency is 375 hertz. 76. Observe that y = 0.006 cos (1,000π t ) = 0.006 cos ( 2π (500t ) ) . So, the amplitude
is 0.006 cm and the frequency is 500 hertz. 330 m
sec. = 1 with Mach number M = 2, we see that V M the speed of the plane V = 2 330 m = 660 m sec. sec.
77. Using the equation
(
)
330 m
sec. = 1 with speed V = 990 m , we see that sec. V M the Mach number M satisfies the following: 330 m sec. = 1 so that M = 3 . M 990 m sec.
78. Using the equation
m θ 330 sec. 79. Use the equation sin = with θ = 60 to find V. V 2 60 1 Observe that sin = sin (30 ) = . Substituting this into the above equation 2 2 yields
m 1 330 sec. = so that V = 660 m . sec. 2 V
287
Chapter 4 m θ 330 sec. 80. Use the equation sin = with θ = 30 to find V. V 2 30 Observe that sin = sin (15 ) . Substituting this into the above equation yields 2 V=
330 m
sec. ≈ 1,275 m sec. sin (15 )
81. The correct graph is reflected over the x-axis. 82. The correct graph is reflected over the x-axis. Also, since the period is π 2π = π , one should form the table in steps of so that points other than the 2 4 x-intercepts (such as the maximum and minimum points) are obtained. 83. True. In general, the graphs of y = f (x) and y = − f ( x) are reflections of each other over the x-axis. 84. True. Observe that since sine is odd, y = A sin( −Bx) = −Asin(Bx), which is a reflection of y = A sin(Bx) over the x-axis. 85. False. Observe that since cosine is even, y = −A cos(−Bx) = −Acos(Bx), which is a reflection of y = A cos(Bx) over the x-axis. 86. True. Observe that since sine is odd, y = −A sin(− Bx ) = − ( − A sin( Bx ) ) = A sin( Bx ). 87. Since A cos ( B ⋅ 0 ) = A(1) = A, the y-intercept of y = A cos(Bx) is (0, A). 88. Since A sin ( B ⋅ 0 ) = A(0) = 0, the y-intercept of y = A sin(Bx) is (0, 0). 89. Assuming that A ≠ 0, the x-intercepts of y = A sin(Bx) occur when
sin(Bx) = 0, which happens when Bx = nπ , where n is an integer. So, x = where n is an integer.
288
nπ , B
Section 4.1
90. Assuming that A ≠ 0, the x-intercepts of y = A cos(Bx) occur when (2n +1)π cos(Bx) = 0, which happens when Bx = , where n is an integer. So, 2 (2n +1)π x= , where n is an integer. 2B NOTE: Regarding problems 91–94, the convention for the three graphs displayed in each case is as follows. The dotted graph corresponds to the first term of the sum (including the negative if it is present); the dashed graph corresponds to the second term (no negative, even if it is present), and the solid graph is the sum (or difference if the second term has a negative in front of it) of the two functions. 91.
92.
93.
94.
289
Chapter 4 95. No. Observe that the graphs of y2 = sin5x and y1 = 5sin x are different, as seen below:
96. No. Observe that the graphs of y2 = cos 3x and y1 = 3cos x are different, as seen below:
97. The graphs of y1 = sin x and
98. The graphs of y1 = cos x and
π y2 = cos x − are the same, as seen 2 below:
π y2 = sin x + are the same, as seen 2 below:
290
Section 4.1
π 99. a. The graph of y2 = cos x + is 3 the graph of y1 = cos x (dotted graph) shifted to the left
π 3
units, as seen
π b. The graph of y2 = cos x − is the 3 graph of y1 = cos x (dotted graph) shifted to the right
π 3
units, as seen
below:
below:
π 100. a. The graph of y2 = sin x + 3 is the graph of y1 = sin x (dotted graph)
π b. The graph of y2 = sin x − is the 3 graph of y1 = sin x (dotted graph)
shifted to the left below:
π 3
units, as seen
shifted to the right below:
291
π 3
units, as seen
Chapter 4 101. Consider the graphs of the following functions on the interval [0,2π ]: y1 = e −t , y2 = sint, y3 = e −t sint
Observation: As t increases, the graph of y1 approaches the t-axis from above, the graph of y2 oscillates, keeping its same form, the graph of y3 oscillates, but dampens so that it approaches the t-axis from below and above.
102. Consider the graphs of the following functions: y1 = e −t sin t
y2 = e −2t sin t y3 = e −4t sin t
Observation: As the value of k increases, the effect of the damping increases, causing the graph to approach the t-axis more quickly.
292
Section 4.2 Solutions -------------------------------------------------------------------------------1. c This is the graph of y = sin x
shifted right
π 2
units.
3. a This is the graph of y = cos x
shifted left
π
units, and then reflected
2. b This is the graph of y = cos x
shifted left
π 2
units.
4. h This is the graph of y = sin x
shifted right
π
units, and then reflected
2 over the x-axis.
2 over the x-axis.
5. e This is the graph of y = cos x translated vertically up 1 unit.
6. g This is the graph of y = sin x reflected over the x-axis, then translated vertically up 1 unit.
7. f This is the graph of y = sin x translated vertically down 1 unit.
8. d This is the graph of y = cos x reflected over the x-axis, then translated vertically up 1 unit.
9. In order to graph y = −3 + sin x on one period, follows these steps:
Step 1: Graph y = sin x on [ 0,2π ] :
Step 2: Translate the graph in Step 1 vertically down 3 units to obtain the graph of the desired equation y = −3 + sin x on [ 0,2π ] :
293
Chapter 4 10. In order to graph y = 3 − cos x on one period, follow these steps:
Step 1: Graph y = cos x on [ 0,2π ] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of y = − cos x on [ 0,2π ] :
Step 3: Translate the graph in Step 2 vertically up 3 units to obtain the graph of the desired equation y = 3 − cos x on [ 0,2π ] :
294
Section 4.2
11. In order to graph y = 2 + cos x over one period, follow these steps: Step 1: Graph y = cos x on [ 0, 2π ] :
Step 2: Translate the graph in Step 1 vertically up 2 units to obtain the graph of the desired equation y = 2 + cos x on [0, 2π ] :
12. In order to graph y = −1− cos x over one period, follow these steps: Step 1: Graph y = cos x on [ 0, 2π ] :
Step 2: Reflect the graph in Step 1 over the x -axis to obtain the graph of y = − cos x on [ 0, 2π ] :
Step 3: Translate the graph in Step 2 vertically down 1 unit to obtain the graph of the desired equation y = −1 − cos x on [ 0, 2π ] :
295
Chapter 4 13. In order to graph π y = cos x + over one period, 3 follow these steps: Step 1: Graph y = cos x on [ 0,2π ] :
14. In order to graph π y = sin x − over one period, 6 follow these steps: Step 1: Graph y = sin x on [ 0, 2π ] :
Step 2: Translate the graph in Step 1 horizontally left
π
units to obtain the 3 graph of the desired equation π π 5π y = cos x + on − , : 3 3 3
Step 2: Translate the graph in Step 1 horizontally right
π
units to obtain the 6 graph of the desired equation π π 13π y = sin x − on , : 6 6 6
296
Section 4.2
15. In order to graph y = sin ( x + π2 ) on one period, follow these steps:
Step 1: Graph y = sin x on [ 0,2π ] :
Step 2: Shift the graph in Step 1 to the left π2 units to obtain the graph of y = sin ( x + π2 ) on − π2 , 32π :
16. In order to graph y = cos ( x − π4 ) on one period, follow these steps:
Step 1: Graph y = cos x on [ 0,2π ] :
Step 2: Shift the graph in Step 1 to the right π4 units to obtain the graph of y = cos ( x − π4 ) on π4 , 94π :
297
Chapter 4 17. In order to graph y = 1 − sin( x − π ) on one period, follow these steps:
Step 1: Graph y = sin x on [ 0,2π ] :
Step 2: Shift the graph in Step 1 to the right π units to obtain the graph of y = sin( x − π ) on [π ,3π ] :
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of y = − sin( x − π ) on [π ,3π ] :
Step 4: Translate the graph in Step 3 vertically up 1 unit to obtain the graph of the desired equation y = 1 − sin( x − π ) on [π ,3π ] :
298
Section 4.2
18. In order to graph y = −1 + cos( x + π ) on one period, follow these steps:
Step 1: Graph y = cos x on [ 0,2π ] :
Step 2: Shift the graph in Step 1 to the left π units to obtain the graph of y = cos( x + π ) on [ −π , π ] :
Step 3: Translate the graph in Step 2 vertically down 1 unit to obtain the graph of y = −1 + cos( x + π ) on [ −π , π ] :
299
Chapter 4 19. In order to graph y = 2 + cos 2x on one period, follow these steps:
Step 1: Note that the period of y = cos 2x is π . So, following the approach of Section 4.1, we graph y = cos 2x on [ 0, π ] :
Step 2: Translate the graph in Step 1 up 2 units to obtain the graph of y = 2 + cos 2 x on [ 0, π ] :
20. In order to graph y = 3 + sin π x on one period, follow these steps:
Step 1: Note that the period of y = sin π x is 2. So, following the approach of Section 4.1, we graph y = sin π x on [ 0,2 ] :
Step 2: Translate the graph in Step 1 up 3 units to obtain the graph of y = 3 + sin π x on [ 0,2 ] :
300
Section 4.2
21. In order to graph y = 2 − sin 4x on one period, follow these steps:
Step 1: Note that the period of y = sin 4 x is
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of π y = − sin 4x on 0, : 2
π
. So, following the 2 approach of Section 4.1, we π graph y = sin 4 x on 0, : 2
Step 3: Translate the graph in Step 2 vertically up 2 units to obtain the graph π of y = 2 − sin 4 x on 0, : 2
301
Chapter 4 22. In order to graph y = 1 − cos 2x on one period, follow these steps:
Step 1: Note that the period of y = cos 2x is π . So, following the approach of Section 4.1, we graph y = cos 2x on [ 0, π ] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of y = − cos 2x on [ 0, π ] :
Step 3: Translate the graph in Step 2 vertically up 1 unit to obtain the graph of y = 1 − cos 2 x on [ 0, π ] :
302
Section 4.2
1 23. In order to graph y = 3 − 2 cos x on one period, follow these steps: 2
Step 1: Note that the period of Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of 1 y = 2 cos x is 4π and the amplitude 1 2 y = −2 cos x on [ 0, 4π ] : 2 is 2. So, following the approach of 1 Section 4.1, we graph y = 2 cos x on 2 [0, 4π ] :
Step 3: Translate the graph in Step 2 vertically up 3 units to obtain the graph 1 of y = 3 − 2 cos x on [ 0, 4π ] : 2
303
Chapter 4 1 24. In order to graph y = 2 − 3sin x on one period, follow these steps: 2
Step 1: Note that the period of 1 y = 3sin x is 4π and the amplitude 2 is 3. So, following the approach of 1 Section 4.1, we graph y = 3sin x on 2 [0, 4π ] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of 1 y = −3sin x on [ 0, 4π ] : 2
Step 3: Translate the graph in Step 2 vertically up 2 units to obtain the graph 1 of y = 2 − 3sin x on [ 0, 4π ] : 2
304
Section 4.2
π 25. In order to graph y = −1 + 2 sin x on one period, follow these steps: 2
Step 1: Note that the period of Step 2: Translate the graph in Step 2 vertically down 1 unit to obtain the π y = 2 sin x is 4 and the amplitude is π 2 graph of y = −1 + 2 sin x on [ 0, 4 ] : 2 2. So, following the approach of Section π 4.1, we graph y = 2 sin x on [ 0, 4 ] : 2
π 26. In order to graph y = −2 + cos x on one period, follow these steps: 2
Step 1: Note that the period of π y = cos x is 4 and the amplitude is 1. 2 So, following the approach of Section π 4.1, we graph y = cos x on [ 0, 4 ] : 2
Step 2: Translate the graph in Step 2 vertically down 2 units to obtain the π graph of y = −2 + cos x on [ 0, 4 ] : 2
305
Chapter 4
π 27. In order to graph y = 2 + sin x + on one period, follow these steps: 4 Step 1: Graph y = sin x on [ 0,2π ] :
Step 2: Shift the graph in Step 1 to the left π4 units to obtain the graph of
π y = sin x + on − π4 , 74π : 4
Step 3: Translate the graph in Step 2 vertically up 2 units to obtain the graph π of y = 2 + sin x + on − π4 , 74π : 4
306
Section 4.2
28. In order to graph y = −1 + cos( x − π ) on one period, follow these steps:
Step 1: Graph y = cos x on [ 0,2π ] :
Step 2: Shift the graph in Step 1 to the right π units to obtain the graph of y = cos ( x − π ) on [π ,3π ] :
Step 3: Translate the graph in Step 2 vertically down 1 unit to obtain the graph of y = −1+ cos( x − π ) on [π ,3π ] :
307
Chapter 4
29. Amplitude
1 , Period 2π , 3
Phase Shift
30. Amplitude 10, Period 2π , 31. Amplitude 2, Period
2π
π
= 2,
33. Amplitude 5, Period
2π , 3
38. Amplitude
π π 1
= −π
2 3
,
Phase Shift
3 4
2π = 2, −π
Phase Shift
2π = −2 −π
34. Amplitude 7, Period C = −
37. Amplitude 3, Period
1
Phase Shift −
Phase Shift −
π
3π 4
Phase Shift
2π = 2π , 1
36. Amplitude 3, Period
2
Phase Shift −
32. Amplitude 4, Period
35. Amplitude 6, Period
π
2π
−π
2
= 4,
Phase Shift
2
2π = π, 2
1 , Period 2, 2
Phase Shift
Phase Shift
1 2
39. Amplitude 2,
Period
2π , 3
Phase Shift −
40. Amplitude 1,
Period
2π , 3
Phase Shift
308
π 3
π 3
3π 8
−π −π
2 =1 2
Section 4.2
1 3 + cos(2x + π ) : 2 2 3 2π −π 1 Amplitude , Period , Vertical Shift unit up = π , Phase Shift 2 2 2 2 1 3 3π 3π In order to graph y = + cos(2 x + π ) on − , , follow these steps: 2 2 2 2 41. First, we have the following information for the equation y =
Step 2: Extend the graph in Step 1 to 3π 3π the interval − , using 2 2 periodicity:
3 Step 1: Graph y = cos(2x + π ) on 2 π π π π − 2 , − 2 + π = − 2 , 2 :
Step 3: Translate the graph in Step 2 1 vertically up unit to obtain the graph 2 1 3 3π 3π of y = + cos(2x + π ) on − , : 2 2 2 2
309
Chapter 4 1 2 42. First, we have the following information for the equation y = + sin(2x − π ) : 3 3 2 2π π 1 Amplitude , Period = π , Phase Shift , Vertical Shift unit up 2 2 3 3 1 2 3π 3π In order to graph y = + sin(2 x − π ) on − , , follow these steps: 3 3 2 2 Step 2: Extend the graph in Step 1 to 3π 3π the interval − , using 2 2 periodicity:
2 Step 1: Graph y = sin(2x − π ) on 3 π π π 3π 2 , 2 + π = 2 , 2 :
Step 3: Translate the graph in Step 2 1 vertically up unit to obtain the graph 3 1 2 3π 3π of y = + sin(2x − π ) on − , : 3 3 2 2
310
Section 4.2
43. First, we have the following information for the equation y =
Amplitude −
1 1 1 π − sin x − : 2 2 2 4
1 1 = (negative in front indicates reflection over x-axis), 2 2
π
π 1 2π Period = 4π , Phase Shift 4 = , Vertical Shift unit up 1 1 2 2 2 2 1 1 1 π 7π 9π , In order to graph y = − sin x − on − , follow these steps: 2 2 2 4 2 2 1 1 π Step 1: Graph y = sin x − on 2 2 4 π π π 9π π , + 4 2 2 = 2 , 2 :
Step 2: Extend the graph in Step 1 to 7π 9π the interval − , using 2 2 periodicity:
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of 1 1 π 7π 9π y = − sin x − on − , : 2 2 4 2 2
Step 4: Translate the graph in Step 3 1 vertically up unit to obtain the graph of 2 1 1 1 π 7π 9π y = − sin x − on − , : 2 2 2 4 2 2
311
Chapter 4 1 1 π 1 44. First, we have the following information for the equation y = − + cos x + : 2 2 4 2
π
1 1 π 2π , Period = 4π , Phase Shift − 4 = − , Vertical Shift unit 1 1 2 2 2 2 2 down 1 1 π 1 9π 7π In order to graph y = − + cos x + on − , , follow these steps: 2 2 4 2 2 2 Amplitude
1 π 1 Step 1: Graph y = cos x + on 2 4 2 π π π 7π − , − + 4 π 2 2 = − 2 , 2 :
Step 2: Extend the graph in Step 1 to 9π 7π the interval − , using 2 2 periodicity:
Step 3: Translate the graph in Step 2 1 vertically down unit to obtain the 2 1 1 π 1 graph of y = − + cos x + on 2 2 4 2 9π 7π − 2 , 2 :
312
Section 4.2
45. First, we have the following information for the equation y = −3 + 4 sin (π ( x − 2 ) ) :
Amplitude 4 , Period
2π
π
= 2 , Phase Shift
2π
π
= 2 , Vertical Shift 3 units down
In order to graph y = −3 + 4 sin (π ( x − 2 ) ) on [ 0, 4 ] , follow these steps: Step 1: Graph y = 4 sin (π ( x − 2 ) ) on
[2,2 + 2] = [2, 4 ] :
Step 2: Extend the graph in Step 1 to the interval [ 0, 4 ] using periodicity:
Step 3: Translate the graph in Step 2 vertically down 3 units to obtain the graph of y = −3 + 4 sin (π ( x − 2 ) ) on
[0, 4] :
313
Chapter 4 46. First, we have the following information for the equation y = 4 − 3cos (π ( x +1) ) :
Amplitude −3 = 3 , Period
2π
π
= 2 , Phase Shift −
π = −1, Vertical Shift 4 units up π
In order to graph y = 4 − 3cos (π ( x +1) ) on [ −1,3] , follow these steps: Step 1: Graph y = 3cos (π ( x + 1) ) on
Step 2: Extend the graph in Step 1 to the interval [ −1,3] using periodicity:
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of y = −3cos (π ( x +1) ) on [ −1,3] :
Step 4: Translate the graph in Step 3 vertically up 4 units to obtain the graph of y = 4 − 3cos (π ( x +1) ) on [ −1,3] :
[ −1, −1+ 2] = [ −1,1] :
314
Section 4.2
π 47. First, we have the following information for the equation y = 2 − 3cos 3x − : 2 π
2π π , Phase Shift 2 = , Vertical Shift 2 units up 3 3 6 π π 5 π In order to graph y = 2 − 3cos 3x − on − , , follow these steps: 2 2 6 Amplitude −3 = 3 , Period
π Step 1: Graph y = 3cos 3x − on 2 π π 2π π 5π 6 , 6 + 3 = 6 , 6 :
Step 2: Extend the graph in Step 1 to π 5π the interval − , using 2 6 periodicity:
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of π π 5π y = −3cos 3x − on − , : 2 2 6
Step 4: Translate the graph in Step 3 vertically up 2 units to obtain the graph π π 5π of y = 2 − 3cos 3x − on − , : 2 2 6
315
Chapter 4 48. First, we have the following information for the equation π y = 1 − 2 sin 3x + : 2
π
π 2π Amplitude −2 = 2 , Period , Phase Shift − 2 = − , Vertical Shift 1 unit up 3 6 3 π 5π π In order to graph y = 1 − 2 sin 3x + on − , , follow these steps: 2 6 2
π Step 1: Graph y = 2 sin 3x + on 2 π 2π π π π − 6 , − 6 + 3 = − 6 , 2 :
Step 2: Extend the graph in Step 1 to 5π π the interval − , using 6 2 periodicity:
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of π 5π π y = −2 sin 3x + on − , : 2 6 2
Step 4: Translate the graph in Step 3 vertically up 1 unit to obtain the graph π 5π π of y = 1 − 2 sin 3x + on − , : 2 6 2
316
Section 4.2
π 1 π 1 49. First, since sine is odd, we have y = 2 − sin − x − = 2 + sin x − . 2 2 2 2 Further, we have the following information for this equation: 2π
π
4 = 1 , Vertical Shift 2 units up π π 2 2 2 π 1 7 9 In order to graph y = 2 + sin x − on − , , follow these steps: 2 2 2 2 Amplitude 1, Period
= 4 , Phase Shift
π 1 Step 1: Graph y = sin x − on 2 2 1 1 1 9 2 , 2 + 4 = 2 , 2 :
Step 2: Extend the graph in Step 1 to 7 9 the interval − , using periodicity: 2 2
Step 3: Translate the graph in Step 2 vertically up 2 units to obtain the graph π 1 7 9 of y = 2 + sin x − on − , : 2 2 2 2
317
Chapter 4 π 1 π 1 50. First, since cosine is even, we have y = 1 − cos − x + = 1 − cos x + . 2 2 2 2 Further, we have the following information for this equation:
π
2π
1 = 4 , Phase Shift − 4 = − , Vertical Shift 1 unit up π π 2 2 2 π 1 9 7 In order to graph y = 1 − cos x + on − , , follow these steps: 2 2 2 2
Amplitude −1 = 1 , Period
π 1 Step 1: Graph y = cos x + on 2 2 1 1 1 7 − 2 , − 2 + 4 = − 2 , 2 :
Step 2: Extend the graph in Step 1 to 9 7 the interval − , using periodicity: 2 2
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of π 1 9 7 y = − cos x + on − , : 2 2 2 2
Step 4: Translate the graph in Step 3 vertically up 1 unit to obtain the graph π 1 9 7 of y = 1 − cos x + on − , : 2 2 2 2
318
Section 4.2
51. Assume that the equation is either of the form y = K + Asin(Bx +C ) or y = K + A cos( Bx + C ) . We make the following preliminary observations: 2π Since the period is 2, we know that = 2 , so that B = π . B Since the maximum y-value is 2 and the minimum y-value is 0, we have: 2+0 =1 • Vertical shift K = 2 2−0 • Amplitude A = =1 2 Possibility 1: The form of the equation is y = K + A sin( Bx + C ). In this case, since the general shape of the graph is that of a reflected sine curve, we know that A = 1. So, the equation thus far is y = 1 + sin(π x + C ). To find C, use the fact that the point (1,1) is on the graph: 1 = 1 + sin(π (1) +C ), so that sin(π + C ) = 0. This equation has infinitely many solutions given by π + C = nπ , where n is an integer, so that C = ( n − 1)π , where n is an integer. The simplest choice is to use C = 0. Doing so results in the equation y = 1 + sin(π x ) .
Possibility 2: The form of the equation is y = K + A cos( Bx + C ). In this case, since the general shape of the graph is that of a shifted, non-reflected cosine curve, we know that again A = 1. So, the equation thus far is y = 1 + cos(π x + C ). To find C, use the fact that the point (1,1) is on the graph: 1 = 1 + cos(π (1) +C ), so that cos(π + C ) = 0. (2 n +1)π This equation has infinitely many solutions given by π + C = , where n is 2 (2n −1)π , where n is an integer. The simplest choice is to an integer, so that C = 2
π
use C = − . Doing so results in the equation 2
π 1 y = 1 + cos π x − = 1 + cos π x − . 2 2
319
Chapter 4 52. Assume that the equation is either of the form y = K + Asin(Bx +C ) or y = K + A cos( Bx + C ). We make the following preliminary observations: 2π Since the period is 2, we know that = 2, so that B = π . B Since the maximum y-value is 2 and the minimum y-value is 0, we have: 2+0 =1 • Vertical shift K = 2 2−0 • Amplitude A = =1 2 Possibility 1: The form of the equation is y = K + A sin( Bx + C ). In this case, since the general shape of the graph is that of a translated, reflected sine curve, we know that A = −1. So, the equation thus far is y = 1 − sin(π x + C ). To find C, use the fact that the point (1,2 ) is on the graph: 2 = 1 − sin(π (1) +C ), so that sin(π + C ) = −1. Though this equation has infinitely many solutions, the simplest one is given by 3π π π + C = , so that C = . Doing so results in the equation 2 2
π 1 y = 1 − sin π x + = 1 − sin π x + . 2 2 Possibility 2: The form of the equation is y = K + Acos(Bx +C ). In this case, since the general shape of the graph is that of a reflected cosine curve, we know that again A = −1. So, the equation thus far is y = 1 − cos(π x + C ). To find C, use the fact that the point (1,2 ) is on the graph: 2 = 1 − cos(π (1) + C ), so that cos(π + C ) = −1. Though this equation has infinitely many solutions, the simplest one is given by π + C = π , so that C = 0. Doing so results in the equation y = 1 − cos (π x ) .
320
Section 4.2
53. Assume that the equation is either of the form y = K + Asin(Bx +C ) or y = K + A cos( Bx + C ) . We make the following preliminary observations: 2π Since the period is 2, we know that = 2, so that B = π . B Since the maximum y-value is 1 and the minimum y-value is −1, we have: 1 + ( −1) • Vertical shift K = = 0 – so there is no vertical shift 2 1 − ( −1) • Amplitude A = =1 2 Possibility 1: The form of the equation is y = K + Asin(Bx +C ). In this case, since the general shape of the graph is that of a translated, nonreflected sine curve, we know that A = 1 . So, the equation thus far is y = sin(π x + C ). To find C, use the fact that the point (1,1) is on the graph:
1 = sin(π (1) + C ) Though this equation has infinitely many solutions, the simplest one is given by
π +C =
π
π
, so that C = − . Doing so results in the equation 2 2
π 1 y = sin π x − = sin π x − . 2 2 Possibility 2: The form of the equation is y = K + A cos( Bx + C ). In this case, since the general shape of the graph is that of a translated, nonreflected cosine curve, we know that again A = 1 . So, the equation thus far is y = cos(π x + C ). To find C, use the fact that the point (1,1) is on the graph:
1 = cos(π (1) + C ) Though this equation has infinitely many solutions, the simplest one is given by π + C = 2π , so that C = π . Doing so results in the equation y = cos (π x + π ) = cos (π ( x + 1) ) .
321
Chapter 4 54. Assume that the equation is either of the form y = K + Asin(Bx +C ) or y = K + A cos( Bx + C ). We make the following preliminary observations: 2π Since the period is 2, we know that = 2, so that B = π . B Since the maximum y-value is 1 and the minimum y-value is −1 , we have: 1 + ( −1) • Vertical shift K = = 0 – so there is no vertical shift 2 1 − ( −1) • Amplitude A = =1 2 Possibility 1: The form of the equation is y = K + A sin( Bx + C ). In this case, since the general shape of the graph is that of a translated, nonreflected sine curve, we know that A = 1. So, the equation thus far is y = sin(π x + C ). To find C, use the fact that the point (1,0 ) is on the graph:
0 = sin(π (1) + C ) Though this equation has infinitely many solutions, the simplest one is given by π + C = 0, so that C = −π . Doing so results in the equation y = sin (π x − π ) = sin (π ( x −1) ) .
Possibility 2: The form of the equation is y = K + Acos(Bx +C ). In this case, since the general shape of the graph is that of a translated, nonreflected cosine curve, we know that again A = 1. So, the equation thus far is y = cos(π x + C ). To find C, use the fact that the point (1,0 ) is on the graph:
0 = cos(π (1) + C ) Though this equation has infinitely many solutions, the simplest one is given by
π +C =
π
π
, so that C = − . Doing so results in the equation 2 2
π 1 y = cos π x − = cos π x − . 2 2 NOTE: Regarding problems 55–66, the convention for the three graphs displayed in each case is as follows. The dotted graph corresponds to the first term of the sum (including the negative if it is present); the dashed graph corresponds to the second term (no negative, even if it is present), and the solid graph is the sum (or difference if the second term has a negative in front of it) of the two functions.
322
Section 4.2
55.
56.
57.
58.
59.
60.
323
Chapter 4 61.
62.
63.
64.
65.
66.
324
Section 4.2
67. November 20 corresponds to t = 365 – (31 + 10) = 324. So, f (324) = 3.1cos(324) + 7.4 ≈ 4.56 mg/L. 68. June 30 corresponds to t = 181. So, f (181) = 3.1cos(181) + 7.4 ≈ 8.49 mg/L. 69. The amplitude is 0.7 = $70. This means the quarterly values will vary by $70 in either direction starting from the baseline of $210. 2π 70. The period is 1.02 ≈ 6 quarters. This means that quarterly values fluctuate between a high of $280 to a low of $140 every 6 quarters.
71. The vertical shift is $210. This is the baseline around which the quarterly sales fluctuate. 4 72. The phase shift is 1.02 ≈ 4. This indicates that quarterly sales will begin their upward trend in the fourth quarter.
73. The amplitude is 0.87 thousand = $870. This means the quarterly sales will vary by $870 in either direction starting from the baseline of $4210. 2π 74. The period is 1.54 ≈ 4 quarters. This means that quarterly sales fluctuate between a high of $5080 to a low of $3340 every 4 quarters (every year).
75. The vertical shift is $4210. This is the baseline around which the quarterly sales fluctuate. 3.24 76. The phase shift is 1.54 ≈ 2. This indicates the quarterly sales will begin their upward trend in the second quarter (summer).
1 77. Consider the function I (t ) = 220 sin 20π t − , t ≥ 0. We make the 100 following observations: 1 Maximum current occurs when sin 20π t − = 1, and is 220 amps. 100 1 Minimum current occurs when sin 20π t − = −1, and is −220 amps. 100 2π 1 Period of I = = = 0.1 second 20π 10 1 Phase Shift of I = = 0.01 second 100
325
Chapter 4 1 78. Consider the function I (t ) = 120 sin 30π t − , t ≥ 0. We make the 200 following observations: 1 Maximum current occurs when sin 30π t − = 1, and is 120 amps. 200 1 Minimum current occurs when sin 30π t − = −1, and is −120 amps. 200 2π 1 Period of I = = second 30π 15 1 Phase Shift of I = = 0.005 second 200 79. Assume that the equation is of the form y = K + Asin(Bx +C ), where x represents the month (here, 1 corresponds to January, and 12 corresponds to December). We make the following preliminary observations: 2π π Since the period is 12, we know that = 12, so that B = . B 6 Since the maximum y-value is 79.3 and the minimum y-value is 39.3, we have: 79.3 + 39.3 = 59.3 • Vertical shift K = 2 79.3 − 39.3 • Amplitude A = = 20 2 Since the general shape of the graph is that of a non-reflected sine curve, we know π that A = 20. So, the equation thus far is y = 59.3 + 20 sin x + C . To find C, we 6 could use any of the 12 data points. For definiteness, we use the first point (1,39.3) :
π 39.3 = 59.3 + 20 sin (1) +C 6 π −20 = 20 sin +C 6 π −1 = sin +C 6
326
Section 4.2
Though this equation has infinitely many solutions, the simplest one is given by π 3π 3π π 4π +C = , so that C = − = . 6 2 2 6 3 Thus, the Charlotte monthly temperature (in degrees Fahrenheit) is given by the 4π π equation y = 59.3 + 20 sin x + . 3 6 80. Assume that the equation is of the form y = K + A sin(Bx +C ), where x represents the month (here, 1 corresponds to January, and 12 corresponds to December). We make the following preliminary observations: 2π π Since the period is 12, we know that = 12, so that B = . B 6 Since the maximum y-value is 82.6 and the minimum y-value is 50.4, we have: 82.6 + 50.4 • Vertical shift K = = 66.5 2 82.6 − 50.4 • Amplitude A = = 16.1 2 Since the general shape of the graph is that of a non-reflected sine curve, we know π that A = 16.1. So, the equation thus far is y = 66.5 + 16.1sin x + C . To find C, 6 we could use any of the 12 data points. For definiteness, we use the first point (1,50.4 ) :
π 50.4 = 66.5 + 16.1sin (1) +C 6 π −16.1 = 16.1sin +C 6 π −1 = sin +C 6 Though this equation has infinitely many solutions, the simplest one is given by π 3π 3π π 4π +C = , so that C = − = . 6 2 2 6 3 Thus, the Houston monthly temperature (in degrees Fahrenheit) is given by the 4π π equation y = 66.5 + 16.1sin x + . 3 6 327
Chapter 4 π 81. Consider the equation y = 20 sin x + 30, where x is measured in feet. 400 π The maximum of y occurs when sin x = 1, and so has value 50. 400 Thus, the roller coaster reaches a maximum height of 50 feet. 2π The period of y is = 800. So, in order to complete 3 cycles, the π 400 coaster will have traveled 3(800) = 2400 feet. π 82. Consider the equation y = 25sin x + 30, where x is measured in feet. 500 π The maximum of y occurs when sin x = 1, and so has value 55. 500 Thus, the roller coaster reaches a maximum height of 55 feet. 2π The period of y is = 1000. So, in order to complete 4 cycles, the π 500 coaster will have traveled 4(1000) = 4000 feet. π 83. Consider the equation y = 100 sin t + 500, where t is measured in years. 4 π The maximum of y occurs when sin t = 1, and so has value 600. Thus, 4 the maximum number of deer is 600. π The minimum of y occurs when sin t = −1, and so has value 400. 4 Thus, the minimum number of deer is 400. 2π The period of y is = 8 years. So, the difference between two years π 4 during which the numbers are the highest is 8 years.
328
Section 4.2
π 84. Consider the equation y = 500 sin t + 1000, where t is measured in years. 2 π The maximum of y occurs when sin t = 1, and so has value 1500. 2 Thus, the maximum number of deer is 1500. π The minimum of y occurs when sin t = −1, and so has value 500. 2 Thus, the minimum number of deer is 500. 2π The period of y is = 4 years. So, the difference between two years π 2 during which the numbers are the highest is 4 years. 85. Assume that the equation is of the form y = K + A sin(Bt), where t is measured in hours. We make the following preliminary observations: 2π π Since the period is 22 hours, we know that = 22, so that B = . 11 B Since the maximum y-value is 12 and the minimum y-value is 6, we have: 12 + 6 12 − 6 Vertical shift K = = 9 Amplitude A = =3 • 2 2
π So, the equation is y = 3sin t + 9 . 11 2 π 86. a. y(46) = 3sin (46) + 9 ≈ 10.62 micromoles of carbon per m sec. 11 b. Although you can calculate this immediately with a calculator, it is perhaps more instructive to see how periodicity enters in. Indeed, observe that π π π y(46) = 3sin (68) + 9 = 3sin (46 + 22) + 9 = 3sin (46) + 2π + 9 11 11 11
2 π = 3sin (46) + 9 ≈ 10.62 micromoles per m sec. 11 Notice that the answers in parts a and b are the same (due to periodicity of this particular Circadian rhythm).
329
Chapter 4 87. Assume that the equation is of the form y = K + A sin(Bt +C ) , where t is measured in seconds. We make the following preliminary observations: 2π π Since the period is 4 seconds, we know that = 4, so that B = . B 2 Since the maximum y-value is 50 and the minimum y-value is 0, we have: 50 + 0 50 − 0 = 25 Amplitude A = = 25 • Vertical shift K = 2 2 π So, the equation thus far is y = 25 + 25sin t + C . To find C, we use the fact 2 that the point ( 0,0 ) is on the graph:
π 0 = 25 + 25sin (0) +C − 1 = sin (C ) 2 Though this equation has infinitely many solutions, the simplest one is given by C =−
π π π . So, the equation is y = 25 + 25sin t − = 25 + 25sin (t −1) . This 2 2 2 2
π
is equivalent to y = 25 − 25 cos ( π2t ) . Also, observe that the intensity at 4 minutes is 0 candelas per m2 because 4 minutes = 4(60) seconds, and the period of the cycle is 4 seconds. π 88. The graph of 25 + 25sin (t − 1) for ½ minute occurs on the interval [0, 30], 2 since t is measured in seconds. The graph is given by:
330
Section 4.2
89. By definition, the phase shift is − C
shift is π
2
B
. Since C = −π and B = 2, the phase
units, not π units.
90. The period is 2π
2
= π , not 2π as the graphs indicate.
91. False. The period is 2π
decreases as ω increases.
ω , which
92. False. The phase shift is φ .
ω
93. True. 94. False. Take ω = 1 and t = 0, for instance. Observe that
π π A sin ωt − ω = A sin − = − A, 2 2 π while A cos (ωt ) = A cos(0) = A. Hence, A sin ωt − ω ≠ A cos (ωt ) , in general. 2 95. Solving 1 + cos (ω x ) = 0 yields: cos (ω x ) = −1
96. Solving sin (ω x − π2 ) = 0 yields:
ω x − π2 = nπ ω x = π2 + nπ = ( n + 12 ) π
ω x = (2n + 1)π x = (2 nω+1)π , n is an integer
x = 22nω+1 π , n is an integer
97. Consider y = K + A cos (ω t − φ ) . The period is 2π
ω and the phase shift is
φ . So, one full period of the graph would be covered in the interval ω φ φ 2π ω , ω + ω . 98. Consider y = k + A cos (ωt + φ ) . The period is 2π
ω and the phase shift is
− φ . So, one full period of the graph would be covered in the interval
ω φ φ 2π − ω , − ω + ω .
331
Chapter 4 99. Using the identity cos θ = sin (θ + π2 ) , we see that the given equation is
equivalent to y = sin ( 2 x + π6 ) .
100. Solving the equation sin ( 4 x + π2 ) = 1 yields: 4 x + π2 = π2 + 2 nπ 4 x = 2 nπ ⇔ x = n2π , n is an integer.
101. The graphs of the following functions are provided below. Notice they are very close on [0, 0.2]. x3 y1 = x − , y2 = sin x 6
102. The graphs of the following functions are provided below. Notice they are very close on [0, 0.2]. x2 y1 = 1 − , y2 = cos x 2
103. The following is the graph of 4π π y = 59.3 + 20 sin x + : 3 6
104. The following is the graph of 4π π y = 66.5 + 16.1sin x + : 3 6
332
Section 4.3 Solutions -------------------------------------------------------------------------------1. b This graph is the reflection of y = tan x over the x-axis, and has vertical asymptotes x = ±
π 2
.
2. f This graph is the reflection of y = csc x over the x-axis, and has vertical asymptotes x = 0, x = ±π .
3. h This graph is y = sec x horizontally compressed by a factor of 2, and it has vertical asymptotes π 3π x=± , x=± . 4 4
4. a This graph is y = csc x horizontally compressed by a factor of 2, and it has vertical asymptotes
5. c This graph is y = cot x horizontally compressed by a factor of π , and it has vertical asymptotes x = 0, x = ±1.
6. e This is the graph of Exercise 5 reflected over the x-axis.
7. d This graph has the same shape and vertical asymptotes as y = sec x, but with vertices at ±3 rather than ±1.
8. g This graph has the same shape and vertical asymptotes as y = csc x, but with vertices at ±3 rather than ±1.
9. Period = π
Phase Shift = π
10. Period = π3
Phase Shift = none
11. Period = π
Phase Shift = π4
12. Period = 4π
Phase Shift = − π2
13. Period = 23π
Phase Shift = π6
14. Period = 8π
Phase Shift = − 32
15. Period = 2π
Phase Shift = −4π
16. Period = 20π
Phase Shift = −π
17. Period = 1
Phase Shift = 12
x = 0, x = ±
π
2
, x = ±π .
18. Period = π2
333
Phase Shift = −1
Chapter 4
19. In order to graph y = tan ( 4x ) on [ −π , π ] we first note that A = 1, B = 4.
π
=
π
•
The period
•
Find two vertical asymptotes we use Bx = − Bx =
π
B
x=
4
.
π
π 2
x=−
π 8
and
•
. 8 2 Find the x -intercept between the two asymptotes Bx = 0 x = 0.
•
Divide the period of x
− −
π 8
π 16 0
π 16
π 8 •
π
4
into four equal parts in steps of
π
16
.
y = tan ( 4 x )
( x, y )
undefined
vertical asymptote, x = −
π 4
π − , −1 16 ( 0,0 ) x -intercept
−1 0
1
π ,1 16
undefined
vertical asymptote, x = −
π 4
Graph the points from the table and connect with a smooth curve. Repeat until reaching the interval endpoints.
334
Section 4.3
20. In order to graph y = cot ( 4 x ) on [ −π , π ] we first note that A = 1, B = 4.
π
The period
•
Find two vertical asymptotes we use Bx = 0 x = 0 and Bx = π x =
•
Find the x -intercept between the two asymptotes Bx =
•
Divide the period of x 0
π 16
π 8 3π 16
π 4 •
B
=
π
•
4
.
π 4
into four equal parts in steps of
π 2
x=
π
16
π 8
undefined
vertical asymptote, x = 0 π , −1 16 π , 0 x -intercept 8 3π ,1 16
1 undefined
.
.
( x, y )
0
4
.
y = tan ( 4 x )
−1
π
vertical asymptote, x =
π 4
Graph the points from the table and connect with a smooth curve. Repeat until reaching the interval endpoints.
335
Chapter 4
1 21. We have the following information about the graph of y = tan x on 2 [ −2π , 2π ] :
π
Phase Shift: Reflection about xNone axis? 1 2 No 1 1 x-intercepts occur when sin x = 0. This happens when x = 0, ± π , so 2 2 that x = 0, ± 2π . So, the points ( 0, 0 ) , ( 2π , 0 ) , ( −2π , 0 ) are on the graph. Period:
= 2π
1 Vertical asymptotes occur when cos x = 0. This happens when 2 1 π x = ± , so that x = ±π . 2 2 1 Using this information, we find that the graph of y = tan x on [ −2π , 2π ] is 2 given by:
336
Section 4.3
1 22. We have the following information about the graph of y = cot x on 2 [ −2π , 2π ]: Period:
π 1
2
= 2π
Reflection about xaxis? No
Phase Shift: None
1 π 1 x-intercepts occur when cos x = 0. This happens when x = ± , so 2 2 2 that x = ±π . So, the points ( −π , 0 ) , (π , 0 ) are on the graph. 1 Vertical asymptotes occur when sin x = 0. This happens when 2 1 x = 0, ± π , so that x = 0, ± 2π . 2 1 Using this information, we find that the graph of y = cot x on [ −2π , 2π ] is 2 given by:
337
Chapter 4
23. We have the following information about the graph of y = − cot ( 2π x ) on
[ −1,1]:
Period:
π 1 = 2π 2
Reflection about xaxis? Yes
Phase Shift: None
x-intercepts occur when cos ( 2π x ) = 0. This happens when
3π 1 3 1 3 , so that x = ± , ± . So, the points ± , 0 , ± , 0 4 2 2 4 4 4 are on the graph. Vertical asymptotes occur when sin ( 2π x ) = 0. This happens when 2π x = ±
π
,±
1 2π x = 0, ± π , ± 2π , so that x = 0, ± , ±1. 2 Using this information, we construct the graph of y = − cot ( 2π x ) on [ −1,1] as follows: Step 1: Graph y = cot ( 2π x ) on [ −1,1] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of y = − cot ( 2π x ) on [ −1,1] :
338
Section 4.3
24. We have the following information about the graph of y = − tan ( 2π x ) on
[ −1,1]:
Period:
π 1 = 2π 2
Reflection about xaxis? Yes
Phase Shift: None
x-intercepts occur when sin ( 2π x ) = 0. This happens when
1 2π x = 0, ± π , ±2π , so that x = 0, ± , ± 1. So, the points 2 1 ( 0,0 ) , ± , 0 , ( ±1, 0 ) are on the graph. 2 Vertical asymptotes occur when cos ( 2π x ) = 0. This happens when
3π 1 3 , so that x = ± , ± . 2 2 4 4 Using this information, we construct the graph of y = − tan ( 2π x ) on [ −1,1] as follows: 2π x = ±
π
,±
Step 1: Graph y = tan ( 2π x ) on [ −1,1] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of y = − tan ( 2π x ) on [ −1,1] :
339
Chapter 4
25. We have the following information about the graph of y = 2 tan ( 3x ) on
[ −π ,π ] :
Period:
π
Reflection about xaxis? No
Phase Shift: None
3 x-intercepts occur when sin ( 3x ) = 0 . This happens when
3x = 0, ± π , ± 2π , ± 3π , so that x = 0, ±
π
3
,±
2π , ± π . So, the points 3
2π , 0 , ± , 0 , ( ±π , 0 ) are on the graph. 3 3 Vertical asymptotes occur when cos ( 3x ) = 0 . This happens when
( 0,0 ) , ±
3π 5π π π 5π ,± , so that x = ± , ± , ± . 2 2 2 6 2 6 Stretching: The coefficient 2 has the effect of making the graph of y = 2 tan ( 3x ) approach the vertical asymptotes twice as quickly as the 3x = ±
π
π
,±
graph of y = tan ( 3x ) .
Using this information, we find that the graph of y = 2 tan ( 3x ) on [ −π , π ] is given by:
340
Section 4.3
1 26. We have the following information about the graph of y = 2 tan x on 3 [ −3π , 3π ] : Period:
π 1 3
= 3π
Reflection about xaxis? No
Phase Shift: None
1 x-intercepts occur when sin x = 0. This happens when y = 4 sin ( x + π ) , 3 so that x = 0, ±3π . So, the points ( 0,0 ) , ( ±3π , 0 ) are on the graph. 1 Vertical asymptotes occur when cos x = 0 . This happens when 3 1 π 3π . x = ± , so that x = ± 3 2 2 Stretching: The coefficient 2 has the effect of making the graph of 1 y = 2 tan x approach the vertical asymptotes twice as quickly as the 3 1 graph of y = tan x . 3 1 Using this information, we find that the graph of y = 2 tan x on [ −3π , 3π ] is 3 given by:
341
Chapter 4
π 27. We have the following information about the graph of y = − tan x − on 2 [ −π ,π ] : Period:
π 1
Reflection about xaxis? Yes
=π
Phase Shift: Right
π 2
units
π π x-intercepts occur when sin x − = 0 . This happens when x − = 0, ± π , 2 2 π 3π π so that x = ± , . So, the points ± , 0 are on the graph. 2 2 2 outside
π Vertical asymptotes occur when cos x − = 0. This happens when 2 π π 3π x− =± , ± , so that x = 0, ± π , 2π . 2 2 2 outside
π Using this information, we construct the graph of y = − tan x − on [ −π , π ] as 2 follows: π Step 1: Graph y = tan x − on 2 [ −π ,π ] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of π y = − tan x − on [ −π , π ] : 2
342
Section 4.3
π 28. We have the following information about the graph of y = tan x + on 4 [ −π ,π ] : Period:
π 1
Reflection about xaxis? No
=π
Phase Shift: Left
π
4
units
π π x-intercepts occur when sin x + = 0. This happens when x + = 0, ± π , 4 4 π 3π 5π π 3π so that x = − , , − . So, the points − , 0 , , 0 are on the 4 4 4 4 4 outside
graph.
π Vertical asymptotes occur when cos x + = 0 . This happens when 4 π π 3π π x + = ± , so that x = − , . 4 2 4 4 π Using this information, we construct the graph of y = tan x + on [ −π , π ] as 4 follows:
343
Chapter 4
π 29. We have the following information about the graph of y = cot x − on 4 [ −π ,π ] : Period:
π 1
=π
Reflection about xaxis? No
Phase Shift: Right
π 4
units
π π π x-intercepts occur when cos x − = 0. This happens when x − = ± , 4 2 4 3π π π 3π so that x = , − . So, the points − , 0 , , 0 are on the graph. 4 4 4 4 π Vertical asymptotes occur when sin x − = 0. This happens when 4 π π 3π 5π x − = 0, ± π , so that x = , − , . 4 4 4 4 outside
π Using this information, we construct the graph of y = cot x − on [ −π , π ] as 4 follows:
344
Section 4.3
π 30. We have the following information about the graph of y = − cot x + on 2 [ −π ,π ] : Period:
π 1
=π
Reflection about xaxis? Yes
Phase Shift: Left
π
2
units
π x-intercepts occur when cos x + = 0. This happens when 2 π π 3π x+ =± , ± , so that x = 0, ± π , −2π . So, the points 2 2 2 outside
( 0, 0 ) , ( ±π , 0 ) are on the graph. π Vertical asymptotes occur when sin x + = 0. This happens when 2 π π 3π . x + = 0, ± π , so that x = ± , 2 2 2 outside
π Using this information, we construct the graph of y = − cot x + on [ −π , π ] as 2 follows: π Step 1: Graph y = cot x + on 2 [ −π ,π ] :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of π y = − cot x + on [ −π , π ] : 2
345
Chapter 4
31. We have the following information about the graph of y = tan ( 2 x − π ) on
[ −2π , 2π ] :
Period:
π
2
Reflection about xaxis? No
Phase Shift: Right
π 2
units
π π Step 1: Graph y = tan(2 x ) on − , . 2 2 We have the following information about this particular graph x-intercepts occur when . sin 2 x = 0 .. This happens when 2x = 0, ± π , so that x = 0, ±
π . So, the points ( 0, 0 ) , ± , 0 are on the graph. 2 2
π
Vertical asymptotes occur when cos 2 x = 0 . This happens when 2x = ± x=±
π 4
π 2
so that
.
Step 2: Extend the graph in Step 1 to Step 3: Translate the graph in Step 2 to [ −2π , 2π ] : the right π units. (No visible change 2 since the period is π .) 2
346
Section 4.3
32. We have the following information about the graph of y = cot ( 2 x − π ) on
[ −2π , 2π ] :
Period:
π 2
Reflection about xaxis? No
Phase Shift: Right
π 2
units
π π Step 1: Graph y = cot(2x ) on − , . 2 2 We have the following information about this particular graph x-intercepts occur when cos 2 x = 0. This happens when 2x = ±
π 2
so that
π . So, the points ± , 0 are on the graph. 4 4 Vertical asymptotes occur when sin 2x = 0 . This happens when 2x = 0, ± π , so x=±
π
that x = 0, ±
π
2
.
Step 2: Extend the graph in Step 1 to Step 3: Translate the graph in Step 2 to [ −2π , 2π ] : the right π units. (No visible change 2 since the period is π .) 2
347
Chapter 4
3π 3π 33. In order to graph y = sec ( 2 x ) on − , , follow these steps: 4 4 Step 1: Graph the y = cos 2x with a dashed curve.
Step 2: Draw the asymptotes the correspond the x -intercepts of y = cos 2 x.
Step 3: Draw the U shape between the asymptotes.
348
Section 4.3
34. In order to graph y = csc ( 2 x ) on [ −π , π ] , follow these steps: Step 1: Graph the y = sin 2 x with a dashed curve.
Step 2: Draw the asymptotes the correspond the x -intercepts of y = sin 2x.
Step 3: Draw the U shape between the asymptotes.
349
Chapter 4
1 35. In order to graph y = sec x on [ −2π , 2π ] , proceed as described below: 2 1 Graph the guide function y = cos x on [ −2π , 2π ] using the approach 2 discussed earlier in Chapter 4. 1 π x-intercepts: Occur when x = ± , so that x = ±π 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. 1 The graph of the guide function (dotted) and the given function y = sec x on 2 [ −2π , 2π ] is as follows:
350
Section 4.3
1 36. In order to graph y = csc x on [ −2π , 2π ] , proceed as described below: 2 1 Graph the guide function y = sin x on [ −2π , 2π ] using the approach 2 discussed earlier in Chapter 4. 1 x-intercepts: Occur when x = 0, ± π , so that x = 0, ± 2π 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
1 The graph of the guide function (dotted) and the given function y = csc x on 2 [ −2π , 2π ] is as follows:
351
Chapter 4
37. In order to graph y = − csc ( 2π x ) on [ −1,1] , proceed as described below:
Step 1: Graph y = csc ( 2π x ) on [ −1,1]
Graph the guide function y = sin ( 2π x ) on [ −1,1] using the approach discussed earlier in Chapter 4. 1 x-intercepts: Occur when 2π x = 0, ± π , ± 2π , so that x = 0, ± , ± 1 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function y = csc ( 2π x ) on [ −1,1] is as follows:
Step 2: Now, reflect the graph in Step 1 over the x-axis to obtain the graph of y = − csc ( 2π x ) on [ −1,1] , as shown below:
352
Section 4.3
38. In order to graph y = − sec ( 2π x ) on [ −1,1] , proceed as described below:
Step 1: Graph y = sec ( 2π x ) on [ −1,1]
Graph the guide function y = cos ( 2π x ) on [ −1,1] using the approach discussed earlier in Chapter 4. π 3π 1 3 x-intercepts: Occur when 2π x = ± , ± , so that x = ± , ± 2 2 4 4 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function y = sec ( 2π x ) on [ −1,1] is as follows:
Step 2: Now, reflect the graph in Step 1 over the x-axis to obtain the graph of y = − sec ( 2π x ) on [ −1,1] , as shown below:
353
Chapter 4
39. In order to graph y = 2 sec ( 3x ) on [ 0,2π ] , proceed as described below:
Graph the guide function y = 2 cos ( 3x ) on [ 0,2π ] using the approach discussed earlier in Chapter 4. π 3π 5π 7π 9π 11π x-intercepts: Occur when 3x = , , so that , , , , 2 2 2 2 2 2 π π 5π 7π 3π 11π x= , , , , , 6 2 6 6 2 6 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) and the given function y = 2 sec ( 3x ) on
[0,2π ] is as follows:
354
Section 4.3
1 40. In order to graph y = 2 csc x on [ −3π , 3π ] , proceed as described below: 3 1 Graph the guide function y = 2 sin x on [ −3π , 3π ] using the approach 3 discussed earlier in Chapter 4. 1 x-intercepts: Occur when x = 0, ± π , so that x = 0, ± 3π . 3 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. 1 The graph of the guide function (dotted) and the given function y = 2 csc x on 3 [ −3π , 3π ] is as follows:
355
Chapter 4
41. First, note that the period is 2π and the phase shift is
π 2
units to the right.
π 5π Hence, one period of this function would occur on the interval , , for instance. 2 2 π π 5π In order to graph y = −3csc x − on , , proceed as described below: 2 2 2 π π 5π Step 1: Graph y = 3csc x − on , . 2 2 2 π π 5π Graph the guide function y = 3sin x − on , using the approach 2 2 2 discussed earlier in Chapter 4. π π 3π 5π x-intercepts: Occur when x − = 0, π , 2π , so that x = , , 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function π y = 3csc x − on 2 π 5π 2 , 2 is as follows:
Step 2: Now, reflect the graph in Step 1 over the x-axis to obtain the graph of π π 5π y = −3csc x − on , , as 2 2 2 shown below:
356
Section 4.3
42. First, note that the period is 2π and the phase shift is
π 4
units to the left. Hence,
π 7π one period of this function would occur on the interval − , , for instance. 4 4 π π 7π In order to graph y = 5sec x + on − , , proceed as described below: 4 4 4 π π 7π Graph the guide function y = 5 cos x + on − , using the 4 4 4 approach discussed earlier in Chapter 4. π π 3π π 5π x-intercepts: Occur when x + = , , so that x = , 4 2 2 4 4 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. π The graph of the guide function (dotted) and the function y = 5sec x + on 4 π 7π − 4 , 4 is as follows:
357
Chapter 4
43. First, note that the period is 2π and the phase shift is π units to the right. Hence, one period of this function would occur on the interval [π ,3π ] , for instance. 1 In order to graph y = sec ( x − π ) on [π ,3π ] , proceed as described below: 2 1 Graph the guide function y = cos ( x − π ) on [π ,3π ] using the approach 2 discussed earlier in Chapter 4. π 3π 3π 5π , so that x = , x-intercepts: Occur when x − π = , 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. 1 The graph of the guide function (dotted) and the function y = sec ( x − π ) on is 2 [π ,3π ] as follows:
358
Section 4.3
44. First, note that the period is 2π and the phase shift is π units to the left. Hence, one period of this function would occur on the interval [ −π , π ] , for instance.
In order to graph y = −4 csc ( x + π ) on [ −π , π ] , proceed as described below:
Step 1: Graph y = 4 csc ( x + π ) on [ −π , π ] .
Graph the guide function y = 4 sin ( x + π ) on [ −π , π ] using the approach discussed earlier in Chapter 4. x-intercepts: Occur when x + π = 0, π , 2π , so that x = −π , 0, π At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) and the function y = 4 csc ( x + π ) on [ −π , π ] is as follows:
Step 2: Now, reflect the graph in Step 1 over the x-axis to obtain the graph of y = −4 csc ( x + π ) on [ −π , π ] , as shown below:
359
Chapter 4
45. First, note that the period is π and the phase shift is
π 2
units to the right. Hence,
π 3π one period of this function would occur on the interval , , for instance. 2 2 In order to graph y = 2 sec ( 2 x − π ) on [ −2π , 2π ] , proceed as described below: π 3π Step 1: Graph y = 2 sec ( 2 x − π ) on , . 2 2 π 3π Graph the guide function y = 2 cos ( 2 x − π ) on , using the approach 2 2 discussed earlier in Chapter 4. π 3π 3π 5π x-intercepts: Occur when 2 x − π = , , so that x = , 2 2 4 4 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function π 3π y = 2 sec ( 2 x − π ) on , is as 2 2 follows:
Step 2: Extend the graph using periodicity to the interval [ −2π , 2π ] , as shown below:
360
Section 4.3
46. First, note that the period is π and the phase shift is
π 2
units to the left. Hence,
π π one period of this function would occur on the interval − , , for instance. 2 2 In order to graph y = 2 csc ( 2 x + π ) on [ −2π , 2π ] , proceed as described below: π π Step 1: Graph y = 2 csc ( 2 x + π ) on − , . 2 2 π π Graph the guide function y = 2 sin ( 2 x + π ) on − , using the approach 2 2 discussed earlier in Chapter 4. x-intercepts: Occur when 2x + π = 0, π , 2π , so that x = −
π
, 0,
π
2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) and the function π π y = 2 csc ( 2 x + π ) on − , is as 2 2 follows:
Step 2: Extend the graph using periodicity to the interval [ −2π , 2π ] , as shown below:
361
Chapter 4
47. y = −2 + 3 tan ( 2 x )
π π First, graph y = 3 tan ( 2 x ) on − , using A = 3, B = 2. 4 4
π
The period
•
Find two vertical asymptotes we use Bx = − Bx =
π
B
x=
=
π
•
2
.
π
π 2
x=−
π 4
and
•
. 2 4 Find the x -intercept between the two asymptotes Bx = 0 x = 0.
•
Divide the period of x
− −
π 4
π 8
0
π 8
π 4
π
4
into four equal parts in steps of
π
16
.
y = tan ( 4 x )
( x, y )
undefined
vertical asymptote, x=0
−3 0 3
undefined
362
π − , −3 8 π ,0 x -intercept 4 π ,3 8 vertical asymptote, x=
π
4
Section 4.3
• Graph the points from the table and connect with a smooth curve.
• Translate the graph vertically down 2 units to obtain the graph.
363
Chapter 4
π 48. First, graph y = −3cot ( 3x ) on 0, using A = −3, B = 3. 3
π
The period
•
Find two vertical asymptotes we use Bx = 0 x = 0 and Bx = π x =
•
Find the x -intercept between the two asymptotes Bx =
•
Divide the period of
B
=
π
•
3
.
π 3
into four equal parts in steps of
x
y = tan ( 4 x )
0
undefined
π
−3
12
π
0
6
π
3
4
π 3
undefined
364
π 2
x=
π
12
π 6
π 3
.
.
( x, y ) vertical asymptote, x=0 π , −3 12 π ,0 x -intercept 6 π ,3 4 vertical asymptote, x=
π
3
.
Section 4.3
• Graph the points from the table and connect with a smooth curve. Notice the graph is reflected over the x-axis because A is negative.
365
Chapter 4
49. First, note that the period is 2π and the phase shift is
π 2
units to the right.
π 5π Hence, one period of this function would occur on the interval , , for instance. 2 2 π π 5π In order to graph y = 3 − 2 sec x − on , , proceed as described below: 2 2 2 π π 5π Step 1: Graph y = −2 sec x − on , . 2 2 2 π π 5π Graph the guide function y = −2 cos x − on , using the 2 2 2 approach discussed earlier in Chapter 4. (Note: You will need to reflect π the graph of y = 2 cos x − over the x-axis.) 2 π π 3π x-intercepts: Occur when x − = , , so that x = π , 2π 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function π π 5π y = −2 sec x − on , is as 2 2 2 follows:
Step 2: Translate the graph is Step 1 vertically up 3 units to obtain the graph π π 5π of y = 3 − 2 sec x − on , , as 2 2 2 shown below:
366
Section 4.3
50. First, note that the period is 2π and the phase shift is
π 2
units to the left. Hence,
π 3π one period of this function would occur on the interval − , , for instance. 2 2 π π 3π In order to graph y = −3 + 2 csc x + on − , , proceed as described below: 2 2 2 π π 3π Step 1: Graph y = 2 csc x + on − , . 2 2 2 π π 3π Graph the guide function y = 2 sin x + on − , using the 2 2 2 approach discussed earlier in Chapter 4. π π π 3π x-intercepts: Occur when x + = 0, π , 2π , so that x = − , , 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function π π 3π y = 2 csc x + on − , is as 2 2 2 follows:
Step 2: Translate the graph is Step 1 vertically down 3 units to obtain the π graph of y = −3 + 2 csc x + on 2 π 3π − 2 , 2 , as shown below:
367
Chapter 4
51. We have the following information about the graph of y =
1 1 π + tan x − : 2 2 2 Phase Shift:
Vertical Translation: Reflection about xaxis? 1 π =π Right units Up unit No 2 1 2 1 π Step 1: Graph y = tan x − on [ 0, π ] . 2 2 We have the following information about this particular graph π π x-intercepts occur when sin x − = 0 . This happens when x − = 0, π , so 2 2 π 3π π 3π that x = , . So, the points , 0 , , 0 are on this graph. (Note: 2 2 2 2 Period:
π
Outisde
These will change in Step 2 when we translate this graph vertically.) π π π Vertical asymptotes occur when cos x − = 0 . This happens when x − = so 2 2 2 that x = π . (Note: These remain unchanged by vertical translation.)
Step 2: Translate the graph in Step 1 vertically up 1 unit. 2
368
Section 4.3
52. We have the following information about the graph of y =
3 1 π − cot x + : 4 4 2 Phase Shift:
Vertical Translation: Reflection about xaxis? 3 π =π Left units Up unit Yes 4 1 2 1 π π π Step 1: Graph y = cot x + on − , . 4 2 2 2 We have the following information about this particular graph π π π x-intercepts occur when cos x + = 0. This happens when x + = so that 2 2 2 x = 0. So, the point ( 0, 0 ) is on this graph. (Note: This will change in Step 2 when we translate this graph vertically.) π π Vertical asymptotes occur when sin x + = 0. This happens when x + = 0, π , 2 2 Period:
π
so that x = ±
π
2
. (Note: These remain unchanged by vertical translation.)
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of 1 π π π y = − cot x + on − , . 4 2 2 2
Step 3: Translate the graph in Step 2 3 vertically up unit to obtain the graph 4 3 1 π π π of y = − cot x + on − , . 4 4 2 2 2
369
Chapter 4
53. First, note that the period is π and the phase shift is
π 2
units to the right. Hence,
π 3π one period of this function would occur on the interval , , for instance. 2 2 π 3π In order to graph y = −2 + 3 csc ( 2 x − π ) on , , proceed as described below: 2 2 π 3π Step 1: Graph y = 3csc ( 2 x − π ) on , . 2 2 π 3π Graph the guide function y = 3sin ( 2 x − π ) on , using the approach 2 2 discussed earlier in Chapter 4. π 3π x-intercepts: Occur when 2x − π = 0, π , 2π , so that x = , π , 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) and the function π 3π y = 3csc ( 2 x − π ) on , is as 2 2 follows:
Step 2: Translate the graph is Step 1 vertically down 2 units to obtain the graph of y = −2 + 3 csc ( 2 x − π ) π 3π on , as shown below: 2 2
370
Section 4.3
54. First, note that the period is π and the phase shift is
π 2
units to the left. Hence,
π π one period of this function would occur on the interval − , , for instance. 2 2 π π In order to graph y = −1 + 4 sec ( 2 x + π ) on − , , proceed as described below: 2 2 π π Step 1: Graph y = 4 sec ( 2 x + π ) on − , . 2 2 π π Graph the guide function y = 4 cos ( 2 x + π ) on − , using the 2 2 approach discussed earlier in Chapter 4. π 3π π π x-intercepts: Occur when 2x + π = , , so that x = − , 2 2 4 4 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) Step 2: Translate the graph is Step 1 vertically down 1 unit to obtain the and the function y = 4 sec ( 2 x + π ) on graph of y = −1 + 4 sec ( 2 x + π ) on π π − , is as follows: 2 2 π π − 2 , 2 as shown below:
371
Chapter 4
55. First, note that the period is 2π
1
= 4π and the phase shift is 2
π 2
units to the right.
π 9π Hence, one period of this function would occur on the interval , , for instance. 2 2 π 1 π 9π In order to graph y = −1 − sec x − on , , proceed as described below: 4 2 2 2 π 1 π 9π Step 1: Graph y = − sec x − on , : 4 2 2 2 π 1 π 9π Graph the guide function y = − cos x − on , using the 4 2 2 2 approach discussed earlier in Chapter 4. 1 π π 3π 3π 7π x-intercepts: Occur when x − = ± , ± , so that x = , 2 4 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) and the function π 1 π 9π y = − sec x − on , is as 4 2 2 2 follows:
Step 2: Translate the graph is Step 1 vertically down 1 unit to obtain the π 1 graph of y = −1 − sec x − on 4 2 π 9π 2 , 2 as shown below:
372
Section 4.3
56. First, note that the period is 2π
1
= 4π and the phase shift is 2
π 2
units to the left.
π 7π Hence, one period of this function would occur on the interval − , , for instance. 2 2 π 1 π 7π In order to graph y = −2 + csc x + on − , , proceed as described below: 4 2 2 2 π 1 π 7π Step 1: Graph y = csc x + on − , . 4 2 2 2 π 1 π 7π Graph the guide function y = sin x + on − , using the 4 2 2 2 approach discussed earlier in Chapter 4. 1 π π 3π 7π x-intercepts: Occur when x + = 0, π , 2π , so that x = − , , 2 4 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of the guide function (dotted) and the function π 1 π 7π y = csc x + on − , is as 4 2 2 2 follows:
Step 2: Translate the graph is Step 1 vertically down 2 units to obtain the π 1 graph of y = −2 + csc x + on 4 2 π 7π − 2 , 2 as shown below:
373
Chapter 4
π 57. Consider the function y = tan π x − . 2 π Domain: All real numbers x for which cos π x − ≠ 0. Observe that 2 π cos π x − = 0 precisely when 2 πx −
π
2
= (2 n +1)
πx =
π
π
2
+ (2 n +1)
π
2 2 π 1 1 π x = + (2 n +1) = ( 2n + 2 ) = n +1, π 2 2 2 where n is an integer. Hence, the domain is the set of all real numbers x such that x ≠ n , where n is an integer. Range: All real numbers.
π 58. Consider the function y = cot x − . 2 π Domain: All real numbers x for which sin x − ≠ 0 . Observe that 2 π π π π sin x − = 0 precisely when x − = nπ , so that x = + nπ = (1 + 2n ) , where n 2 2 2 2 is an integer. Hence, the domain is the set of all real numbers x such that 2n + 1 π , where n is an integer. x≠ 2 Range: All real numbers. 59. Consider the function y = 2 sec(5x ) . Domain: All real numbers x for which cos ( 5x ) ≠ 0. Observe that cos(5x) = 0
π
2n + 1 π , where n is an integer. Hence, 2 10 2n + 1 the domain is the set of all real numbers x such that x ≠ π , where n is an 10 integer. Range: Note that the maximum and minimum y-values for the guide function y = 2 cos(5x ) are 2 and –2, respectively. Hence, the range is ( −∞, −2 ] ∪ [ 2,∞ ) .
precisely when 5x = ( 2 n + 1)
, so that x =
374
Section 4.3
60. Consider the function y = −4 sec(3x ) . Domain: All real numbers x for which cos ( 3x ) ≠ 0 . Observe that cos(3x) = 0
π
2n + 1 π , where n is an integer. Hence, 2 6 2n + 1 the domain is the set of all real numbers x such that x ≠ π , where n is an 6 integer. Range: Note that the maximum and minimum y-values for the guide function y = 4 cos(3x ) are 4 and –4, respectively. Hence, the range is ( −∞, −4 ] ∪ [ 4,∞ ) .
precisely when 3x = ( 2n + 1)
, so that x =
1 61. Consider the function y = 2 − csc x − π . 2 1 Domain: All real numbers x for which sin x − π ≠ 0. Observe that 2 1 1 where n is sin x − π = 0 precisely when x − π = nπ , so that x = 2( n + 1)π 2 2 Any even multiple of π
an integer. Hence, the domain is the set of all real numbers x such that x ≠ 2 nπ , where n is an integer. Range: Note that the maximum and minimum y-values for the guide function 1 y = 2 − sin x − π are 3 and 1, respectively. Hence, the range is ( −∞,1] ∪ [3,∞ ) . 2 1 62. Consider the function y = 1 − 2 sec x + π . 2 1 Domain: All real numbers x for which cos x + π ≠ 0 . Observe that 2 1 2n + 1 1 π , so that x = (2n − 1)π cos x + π = 0 precisely when x + π = 2 2 2 Any odd multiple of π
where n is an integer. Hence, the domain is the set of all real numbers x such that x ≠ (2n − 1)π , where n is an integer. Range: Note that the maximum and minimum y-values for the guide function 1 y = 1 − 2 cos x + π are 3 and – 1, respectively. Hence, the range is ( −∞, −1] ∪ [3,∞ ) . 2
375
Chapter 4
63.
64.
65.
66.
67. We need to graph π 1 y = 3 + csc x − on the interval 3 2 5π 0, 12 .
68. We need to graph π y = 2 − sec 4x + on the interval 3 π π 12 , 6 .
376
Section 4.3
69. Forgot that the amplitude is 3, not 1. So, the guide function should have been y = 3sin(2 x ) . 70. The vertical asymptotes for y = tan x occur at x = ±
π
, while the x-intercept 2 occurs at x = 0. Hence, the quantities used in Steps 2 and 3 are incorrect. (In fact, they coincide with those for cotangent.) 71. The period of cotangent is π . So, the
period of y = cot ( 2x ) is p =
π
=
72. The guide function for secant is cosine. Thus, the graph will be as shown.
π
. B 2 Also, to find the x -intercept, we need to set 2x =
π
2
. Thus, the x-intercept is
π 0, . Then the graph will be as shown. 8
π 73. True. This is a consequence of the fact that cos x − = sin x. Indeed, 2 π 1 1 = = csc x. observe that sec x − = π sin x 2 cos x − 2
377
Chapter 4
π 74. False. This is a consequence of the fact that sin x − ≠ cos x. For instance, 2 π π for x = π , sin π − = sin = 1, while cos π = −1. So, in general, 2 2 π 1 1 ≠ , so that csc x − ≠ sec x, in general. π cos x 2 sin x − 2 75. True. Since the maximum and minimum of the guide function are respectively 3 and −3, and there is no vertical shift, this must be the range of the given function. 76. False. The range of y = A csc ( 2x − π3 ) is ( −∞, −A] ∪ [ A, ∞ ) . Shifting the graph
up k units results in adding k to the endpoints of the range, which gives us ( −∞, −A + k ] ∪ [ A + k, ∞ ) . 77. All integer multiples of n since tangent has period π . 78. All even integer multiples of n since cosecant has period 2π . 79. Consider the following graph.
The x-values of the red points are the solutions of the equation, and they are −π , −
π
2
, 0,
π
2
, π.
80. Consider the following graph.
There are no solutions to the given equation since the graph does not intersect the x-axis.
378
Section 4.3
81. We must discard any x-value for which sin(2 x + π2 ) = 0. This occurs when π 2 x + π2 = nπ , so that x = (2 n−1) , n is an integer . Hence, the domain is all real 4 π numbers x such that x ≠ (2n−1) , n is an integer. 4
82. We must discard any x-value for which sin(4x − π ) = 0. This occurs when 4x − π = nπ , so that x = n4π , n is an integer . Hence, the domain is all real numbers
x such that x ≠ n4π , n is an integer. NOTE: Regarding problems 83–84, the convention for the three graphs displayed in each case is as follows. The dotted graph corresponds to the first term of the sum (including the negative if it is present); the dashed graph corresponds to the second term (no negative, even if it is present), and the solid graph is the sum (or difference if the second term has a negative in front of it) of the two functions. 83.
84.
85. Consider the graphs of the following functions on the interval [ −2π , 2π ] : y1 = cos x (dotted bold), y2 = sin x (dashed bold), y3 = cos x + sin x (solid bold)
379
Chapter 4
In order to determine the amplitude of y3 , we need to determine its maximum and minimum y-values. From the graph, observe that both the maximum and minimum values of y3 occur at the x-values where y1 and y2 intersect. But, π 5π . As such, sin x = cos x when x = , 4 4 2 2 π π π + = 2 y3 = cos + sin = 2 4 4 4 2 2 2 5π 5π 5π + y3 =− 2 = cos + sin = − 2 2 4 4 4 maximum of y3 − minimum of y3 So, the amplitude of y3 is = 2 . 2 86. Consider the graphs of the following functions on the interval [ −2π , 2π ] : y1 = cos x + sin x (dashed bold), y2 = sec x + csc x (solid bold)
Notice that y1 is not the guide function of y2 .
380
Chapter 4 Review Solutions ----------------------------------------------------------------------1. The distance from maximum to next consecutive maximum is 2π . Hence, the period is 2π .
maximum y-value − minimum y-value 4 − ( −4) = =4 2 2 the amplitude is 4. 2. Since A =
3. Assume that the equation of the form y = K + Acos(Bx + C ) . From Exercises 1 and 2, we know that since the general shape of the curve is that of a nonreflected cosine. So, 2π Since the period is 2π , we know that = 2π , so that B = 1. B Since the maximum y-value is 4 and the minimum y-value is −4 , we have: 4 + ( −4) • Vertical shift K = = 0 – so there is no vertical shift 2 • Amplitude A = 4 , and in fact, A = 4 (since there is no reflection). So, the equation thus far is y = 4 cos( x + C ). To find C, use the fact that the point ( 0, 4 ) is on the graph: 4 = 4 cos(π (0) + C )
1 = cos(C ) Though this equation has infinitely many solutions, the simplest one is given by C = 0 . Doing so results in the equation y = 4 cos x . 4. The distance from maximum to next consecutive maximum is π . Hence, the period is π . 5. Since
maximum y-value − minimum y-value 5 − ( −5) = = 5, 2 2 the amplitude is 5. A=
381
Chapter 4
6. Assume that the equation of the form y = K + Asin(Bx + C ) . From Exercises 4 and 5, we know that since the general shape of the curve is that of a reflected sine. So, 2π Since the period is π , we know that = π , so that B = 2 . B Since the maximum y-value is 5 and the minimum y-value is −5 , we have: 5 + ( −5) • Vertical shift K = = 0 – so there is no vertical shift 2 • Amplitude A = 5 , and in fact, A = −5 (to account for the reflection).
So, the equation thus far is y = −5sin(2 x + C ). To find C, use the fact that the point ( 0,0 ) is on the graph: 0 = −5 sin(2(0) +C ) 0 = sin(C ) Though this equation has infinitely many solutions, the simplest one is given by C = 0 . Doing so results in the equation y = −5sin 2x . 7. The distance from maximum to next consecutive maximum is 2π . Hence, the period is 2π . 8. Since 1 1 − − maximum y-value − minimum y-value 4 4 1 A= = = , 2 2 4 1 the amplitude is . 4
382
Chapter 4 Review
9. Assume that the equation of the form y = K + Asin(Bx + C ) . From Exercises 7 and 8, we know that since the general shape of the curve is that of a non-reflected sine. So, 2π Since the period is 2π , we know that = 2π , so that B = 1. B 1 1 Since the maximum y-value is and the minimum y-value is − , we 4 4 have: 1 1 +− 4 4 • Vertical shift K = = 0 – so there is no vertical shift 2 1 1 • Amplitude A = , and in fact, A = (since there is no reflection). 4 4 1 So, the equation thus far is y = sin( x + C ) . To find C, use the fact that the point 4 1 ( 0,0 ) is on the graph: 0 = sin(0 + C ) , so that 0 = sin(C ) 4 Though this equation has infinitely many solutions, the simplest one is given by 1 C = 0 . Doing so results in the equation y = sin x . 4 10. The distance from minimum to next consecutive minimum is 4π . Hence, the period is 4π . 11. Since
maximum y-value − minimum y-value 2 − ( −2) = = 2, 2 2 the amplitude is 2. A=
383
Chapter 4
12. Assume that the equation of the form y = K + Acos(Bx + C ) . From Exercises 10 and 11, we know that since the general shape of the curve is that of a nonreflected cosine. So, 2π Since the period is 4π , we know that = 4π , so that B = 1 . 2 B Since the maximum y-value is 2 and the minimum y-value is −2 , we have: 2 + ( −2) • Vertical shift K = = 0 – so there is no vertical shift 2 • Amplitude A = 2 , and in fact, A = 2 (since there is no reflection).
1 So, the equation thus far is y = 2 cos x + C . To find C, use the fact that the 2 point ( 0,2 ) is on the graph:
1 2 = 2 cos (0) + C 2 1 = cos(C ) Though this equation has infinitely many solutions, the simplest one is given by x C = 0 . Doing so results in the equation y = 4 cos . 2 13. Since A = 2 and B = 2π , we conclude that the amplitude is 2 and 2π the period is p = = 1. 2π
15. Since A =
1 and B = 3 , we 5
conclude that the amplitude is the period is p =
2π . 3
1 π and B = , we conclude 3 2 1 that the amplitude is and the period is 3 2π p= =4. π 2
14. Since A =
7 and B = 6 , we conclude 6 7 that the amplitude is and the period is 6 2π π p= = . 6 3
16. Since A =
1 and 5
384
Chapter 4 Review
1 17. We have the following information about the graph of y = −2 sin x on 2 [ −2π , 2π ] : Amplitude: −2 = 2
Period:
2π = 4π 1 2
1 1 x-intercepts occur when sin x = 0 . This happens when x = 0, ± π , 2 2 and so x = 0, ± 2π . So, the points ( 0, 0 ) , ( −2π , 0 ) , ( 2π , 0 ) are on the graph. 1 The maximum value of the graph is 2 since the maximum of sin x is 1, 2 1 π and the amplitude is 2. This occurs when x = , and so x = π . So, the 2 2 point (π , 2) is on the graph. 1 The minimum value of the graph is – 2 since the minimum of sin x is 2 1 π –1, and the amplitude is 2. This occurs when x = − , and so x = −π . 2 2 So, the point (−π , −2) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 2 sin x over the x-axis. 2 1 Using this information, we find that the graph of y = −2 sin x on [ −2π , 2π ] is 2 given by:
385
Chapter 4
18. In order to construct the graph of y = 3sin ( 3x ) on [ −2π , 2π ] , we first gather
the information essential to forming the graph for one period, namely on the 2π interval 0, . Then, extend this picture to cover the entire interval. Indeed, we 3 have the following: Amplitude: 3 2π Period: 3 x-intercepts occur when sin3x = 0 . This happens when 3x = nπ , and so nπ π 2π , where n is an integer. So, the points ( 0, 0 ) , , 0 , , 0 are x= 3 3 3 on the graph. The maximum value of the graph is 3 since the maximum of sin3x is 1, and the amplitude is 3. This occurs when 3x =
π
2
, so that x =
π
6
. So, the
π point ,3 is on the graph. 6 The minimum value of the graph is – 3 since the minimum of sin3x is –1, 3π π and the amplitude is 3. This occurs when 3x = , so that x = . So, the 2 2 π point , −3 is on the graph. 2 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. 2π Now, using this information, we can form the graph of y = 3sin ( 3x ) on 0, , 3 and then extend it to the interval [ −2π , 2π ] , as shown below:
386
Chapter 4 Review
1 cos ( 2x ) on [ −2π , 2π ] , we first gather 2 the information essential to forming the graph for one period, namely on the interval [ 0, π ] . Then, extend this picture to cover the entire interval. Indeed, we
19. In order to construct the graph of y =
have the following: Amplitude: 1
Period:
2
2π =π 2
(2n +1)π and 2 (2n +1)π π 3π so, x = , where n is an integer. So, the points , 0 , , 0 are 4 4 4 on the graph. The maximum value of the graph is 1 since the maximum of cos 2 x is 1, 2 and the amplitude is 1 . This occurs when 2x = 0, 2π , so that x = 0, π . 2 So, the points 0, 1 , π , 1 are on the graph. 2 2 The minimum value of the graph is – 1 since the minimum of cos 2 x is 2 x-intercepts occur when cos 2 x = 0 . This happens when 2x =
(
)(
)
π
– 1, and the amplitude is 1 . This occurs when 2x = π , so that x = . 2 2 π So, the point , − 1 is on the graph. 2 2 Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. 1 Now, using this information, we can form the graph of y = cos ( 2x ) on [ 0, π ] , 2 and then extend it to the interval [ −2π , 2π ] , as shown below:
387
Chapter 4
1 1 20. We have the following information about the graph of y = − cos x on 4 2 [ −2π , 2π ] : Amplitude: −
1 1 = 4 4
Period:
2π = 4π 1 2
1 π 1 x-intercepts occur when cos x = 0 . This happens when x = ± , and 2 2 2 so x = ±π . So, the points ( −π , 0 ) , (π , 0 ) are on the graph. 1 The maximum value of the graph is 1 since the maximum of cos x is 4 2 1 1, and the amplitude is 1 . This occurs when x = 0 , and so x = 0 . So, 4 2
(
the point 0, 1
4
) is on the graph.
1 The minimum value of the graph is – 1 since the minimum of cos x is 4 2 1 –1, and the amplitude is 1 . This occurs when x = ±π , and so x = ±2π . 4 2
(
So, the points 2π , − 1
)
(
)
and −2π , − 1 on the graph. 4 4 Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 4 cos x over the x-axis. 2 1 1 Now, using this information, we can form the graph of y = − cos x on 4 2 [ −2π , 2π ] :
388
Chapter 4 Review
21.
22.
Amplitude 3, 2π Period = 2π , 1 Phase Shift
Amplitude 1 Period
π
23.
2π = 2π , 1
Phase Shift −
2 Vertical Shift 2 (Up)
2
π
4 Vertical Shift 3 (Up) 24.
Amplitude 4, 2π Period , 3
Amplitude 2, 2π Period = π, 2
Phase Shift −
Phase Shift
π
4 Vertical Shift −2 (Down) 25.
π
3 Vertical Shift −1 (Down) 26.
Amplitude 1, 2π Period = 2π , 1 Phase Shift π 1 (Up) Vertical Shift 2 27.
Amplitude 1, 2π Period = 2π , 1 Phase Shift −π 3 (Up) Vertical Shift 2 28.
Amplitude 1, 2π Period = 4π , 1 2 Phase Shift −
π
1
Amplitude 1, 2π Period = 8π , 1 4
= −2π
Phase Shift
= 4π
4 Vertical Shift 0
2 Vertical Shift 0
29.
π
1
30.
Amplitude 5, 2π π Period = , 4 2
Amplitude 6, 2π Period , 3
Phase Shift
Phase Shift −
π
π
3 Vertical Shift 0
4 Vertical Shift 4 (Up) 389
Chapter 4
π 31. First, we have the following information for the equation y = 3 + sin x + : 2 2π −π = 2π , Phase Shift , Vertical Shift 3 units up Amplitude 1, Period 2 1 π In order to graph y = 3 + sin x + on [ −2π , 2π ] , follow these steps: 2 π Step 1: Graph y = sin x + on 2 π π π 3π π − , − + 2 2 2 = − 2 , 2 :
Step 2: Extend the graph in Step 1 to the interval [ −2π , 2π ] using periodicity:
Step 3: Translate the graph in Step 3 vertically up 3 units to obtain the π graph of y = 3 + sin x + on 2 [ −2π , 2π ] .
390
Chapter 4 Review
π 32. First, we have the following information for the equation y = −2 + sin x − : 2 2π π = 2π , Phase Shift , Vertical Shift 2 units down Amplitude 1, Period 1 2 π In order to graph y = −2 + sin x − on [ −2π , 2π ] , follow these steps: 2 π Step 1: Graph y = sin x − on 2 π π π 5π π , + 2 2 2 = 2 , 2 :
Step 2: Extend the graph in Step 1 to the interval [ −2π , 2π ] using periodicity:
Step 3: Translate the graph in Step 2 vertically down 2 units to obtain the π graph of y = −2 + sin x − on 2 [ −2π , 2π ] :
391
Chapter 4
33. First, we have the following information for the equation y = −2 + cos ( x + π ) :
2π = 2π , Phase Shift −π , Vertical Shift 2 units down 1 In order to graph y = −2 + cos ( x + π ) on [ −2π , 2π ] , follow these steps: Amplitude 1,
Period
Step 1: Graph y = cos ( x + π ) on
[ −π , −π + 2π ] = [ −π , π ] :
Step 2: Extend the graph in Step 1 to the interval [ −2π , 2π ] using periodicity:
Step 3: Translate the graph in Step 2 vertically down 2 units to obtain the graph of y = −2 + cos ( x + π ) on
[ −2π , 2π ] :
392
Chapter 4 Review
34. First, we have the following information for the equation y = 3 + cos ( x − π ) :
2π = 2π , Phase Shift π , Vertical Shift 3 units up 1 In order to graph y = 3 + cos ( x − π ) on [ −2π , 2π ] , follow these steps: Amplitude 1,
Period
Step 1: Graph y = cos ( x − π ) on
[π , π + 2π ] = [π ,3π ] :
Step 2: Extend the graph in Step 1 to the interval [ −2π , 2π ] using periodicity:
Step 3: Translate the graph in Step 2 vertically up 3 units to obtain the graph of y = 3 + cos ( x + π ) on
[ −2π , 2π ] :
393
Chapter 4
π 35. In order to graph y = 3 − sin x on [ −2π , 2π ] , follow these steps: 2 Step 1: Note that the period of π y = sin x is 4. So, following the 2 approach of Section 4.1, we graph π y = sin x on [ 0, 4 ] : 2
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of π y = − sin x on [ 0, 4 ] : 2
Step 3: Extend the graph in Step 2 to [ −2π , 2π ] to obtain the graph of
Step 4: Translate the graph in Step 2 vertically up 3 units to obtain the graph π of y = 3 − sin x on [ −2π , 2π ] : 2
π y = − sin x on [ −2π , 2π ] : 2
394
Chapter 4 Review
π 36. In order to construct the graph of y = 2 cos x on [ −2π , 2π ] , we first 2 gather the information essential to forming the graph for one period, namely on the interval [ 0, 4 ] . Then, extend this picture to cover the entire interval. Indeed, we have the following: Amplitude: 2
2π
Period:
π
=4
2
π (2n +1)π π x-intercepts occur when cos x = 0 . This happens when x = 2 2 2 and so, x = (2n + 1) , where n is an integer. So, the points (1, 0 ) , ( 3, 0 ) are on the graph. π The maximum value of the graph is 2 since the maximum of cos x is 1, 2 and the amplitude is 2. This occurs when
π
2 So, the points ( 0,2 ) , ( 4,2 ) are on the graph.
x = 0, 2π , so that x = 0, 4 .
π The minimum value of the graph is – 2 since the minimum of cos x is 2 –1, and the amplitude is 2. This occurs when the point ( 2, −2 ) is on the graph.
π
2
x = π , so that x = 2 . So,
Since the coefficient used to compute the amplitude is nonnegative, there is no reflection over the x-axis. π Now, using this information, we can form the graph of y = 2 cos x on [ 0, 4 ] , 2 and then extend it to the interval [ −2π , 2π ] , as shown below:
395
Chapter 4
37. d We have the following information about the graph:
Amplitude 2, Period 2π , Phase Shift
π
, Vertical Shift 0 4 So, the maximum and minimum values are 2 and –2, respectively. Also, there is no reflection about the x-axis. 38. b We have the following information about the graph: Amplitude 3, Period 2π , Phase Shift 0, Vertical Shift 2 (down) So, the maximum and minimum values are 1 and –5, respectively. Also, there is no reflection about the x-axis. 39. a We have the following information about the graph: Amplitude 3, Period 2π , Phase Shift 0, Vertical Shift 3 (up) So, the maximum and minimum values are 6 and 0, respectively. Also, there is reflection about the x-axis. 40. c We have the following information about the graph: Amplitude 3, Period 2π , Phase Shift − π , Vertical Shift 0 4 So, the maximum and minimum values are 3 and –3, respectively. Also, there is no reflection about the x-axis. NOTE: Regarding problems 41–44, the convention for the three graphs displayed in each case is as follows. The dotted graph corresponds to the first term of the sum (including the negative if it is present); the dashed graph corresponds to the second term (no negative, even if it is present), and the solid graph is the sum (or difference if the second term has a negative in front of it) of the two functions. 41.
42.
396
Chapter 4 Review
43.
44.
45. b We have the following information about the graph: Same shape as y = tan x and has period π
Vertical asymptotes x =
( 4 ) = 12 tan (π 4 ) = 12
π
2
+ nπ , where n is an integer
y π
46. c We have the following information about the graph: Same shape as y = tan x, but reflected over x-axis
Period
π 1
= 2π
y ( 0 ) = − tan ( 12 ⋅ 0 ) = 0
2 Vertical asymptotes x = π + 2 nπ = (2n +1)π , where n is an integer
47. j We have the following information about the graph:
Same shape as y = cot x and has period Vertical asymptotes x =
π
2
nπ , where n is an integer 2
48. f We have the following information about the graph: Same shape as y = cot x and has period π Vertical asymptotes x = nπ , where n is an integer
( 4 ) = 2 cot (π 4 ) = 2
y π
397
Chapter 4
49. g We have the following information about the graph: Same shape as y = tan x, has period π , shifted vertically up 2 units,
and phase shift − π
4
Vertical asymptotes x =
( 4) = 3
π 4
+ nπ , where n is an integer
y ( 0 ) = 2 + tan π
50. h We have the following information about the graph:
Same shape as y = cot x, has period π , shifted vertically down unit, and phase shift π
4
Vertical asymptotes x =
(
)
π 4
1 2
+ nπ , where n is an integer
1 1 3 y ( 0 ) = − + cot − π = − −1 = − 4 2 2 2 51. a We have the following information about the graph: This graph has the same shape and vertical asymptotes as y = sec x, but with vertices at 0 and 4 rather than ±1 (due to the vertical shift and expansion factors) 52. d We have the following information about the graph: This graph has the same shape as y = sec x, but shifted right π
2 and reflected over the x-axis Vertical asymptotes x = ( n + 1)π , where n is an integer The vertices (which constitute the local minimum and maximum
values) occur at the odd multiples of π
2
53. i We have the following information about the graph: Period 2π , Phase Shift − π , Vertical Shift 2 4
Vertical asymptotes x = nπ −
π
, where n is an integer 4 Vertices occur at 1 and 3
54. e The function has the same shape as y = csc x, but with vertices at ±2 rather than ±1. The vertical asymptotes are x = nπ , where n is an integer.
398
Chapter 4 Review
π 55. Consider the function y = 4 tan x + . 2 π π Domain: All real numbers x for which cos x + ≠ 0 . Observe that cos x + = 0 2 2 precisely when x+
π
2
= (2 n + 1)
x=−
π 2
π
2
+ (2 n + 1)
π 2
x = nπ where n is an integer. Hence, the domain is the set of all real numbers x such that x ≠ nπ , where n is an integer. Range: All real numbers.
π 56. Consider the function y = cot 2 x − . 2 π π Domain: All real numbers x for which sin 2 x − ≠ 0 . Observe that sin 2 x − = 0 2 2 π π nπ π precisely when 2 x − = nπ , so that x = + = (1+ n ) , where n is an integer. 2 2 2 2 nπ , where n is an integer. Hence, the domain is the set of all real numbers x such that x ≠ 2 Range: All real numbers. 57. Consider the function y = 3sec(2x ) . Domain: All real numbers x for which cos ( 2 x ) ≠ 0 . Observe that cos(2x) = 0
π
2n + 1 π , where n is an integer. Hence, the 2 4 2n + 1 π , where n is an integer. domain is the set of all real numbers x such that x ≠ 4 Range: Note that the maximum and minimum y-values for the guide function y = 3cos(2x) are 3 and –3, respectively. Hence, the range is ( −∞, −3] ∪ [3,∞ ) . precisely when 2x = ( 2n + 1)
, so that x =
58. Consider the function y = 1 + 2 csc x . Domain: All real numbers x for which sin x ≠ 0 . Observe that sin x = 0 precisely when x = nπ , where n is an integer. Hence, the domain is the set of all real numbers x such that x ≠ nπ , where n is an integer. Range: Note that the maximum and minimum y-values for the guide function y = 1 + 2 sin x are 3 and – 1, respectively. Hence, the range is ( −∞, −1] ∪ [3,∞ ) .
399
Chapter 4
π 59. We have this information about the graph of y = − tan x − on [ −2π , 2π ] : 4 Period: Reflection about x-axis? Phase Shift: Yes π π =π Right units 1 4 π π x-intercepts occur when sin x − = 0 . This happens when x − = 0, ± π , 4 4 3π π 5π 3π π 5π so that x = − , , . So, the points − , 0 , , 0 , , 0 are on 4 4 4 4 4 4 the graph. π Vertical asymptotes occur when cos x − = 0 . This happens when 4 π π 3π π 3π x− =± , ± , so that x = − , . 4 2 2 4 4 π Using this information, we construct the graph of y = − tan x − on [ −2π , 2π ] : 4 as follows: π Step 1: Graph y = tan x − on 4 π π π 5π 4 , 4 + π = 4 , 4 :
Step 2: Extend the graph in Step 1 to [ −2π , 2π ] :
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of π y = − tan x − on [ −2π , 2π ] : 4
400
Chapter 4 Review
60. We have the following information about the graph of y = 1 + cot ( 2 x ) :
Period:
π
Vertical Translation: Up 1 unit
2
Reflection about xaxis? No
Phase Shift: None
π Step 1: Graph y = cot ( 2 x ) on 0, . 2 We have the following information about this particular graph x-intercepts occur when cos ( 2 x ) = 0 . This happens when 2x =
π 2
so that x =
π 4
.
π So, the point , 0 is on this graph. (Note: This will change in Step 2 when 4 we translate this graph vertically.) Vertical asymptotes occur when sin ( 2 x ) = 0 . This happens when 2x = 0, π , so that x = 0,
π
2
. (Note: These remain unchanged by vertical translation.)
Step 2: Extend the graph in Step 1 to [ −2π , 2π ] :
Step 3: Translate the graph in Step 2 vertically up 1 unit to obtain the graph of y = 1 + cot ( 2 x ) on [ −2π , 2π ] :
401
Chapter 4
61. First, note that the period is 2π and the phase shift is π units to the right. Hence, one period of this function would occur on the interval [π ,3π ] , for instance.
In order to graph y = 2 + sec ( x − π ) on [ −2π , 2π ] , proceed as described below:
Step 1: Graph y = sec ( x − π ) on [π ,3π ] .
Graph the guide function y = cos ( x − π ) on [π ,3π ] using the approach discussed earlier in Chapter 4. π 3π 3π 5π x-intercepts: Occur when x − π = , , so that x = , 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes.
The graph of guide function (dotted) and the function y = sec ( x − π ) on
[π ,3π ] :
Step 2: Extend the graph of Step 1 to [ −2π , 2π ] :
Step 3: Translate the graph is Step 1 vertically up 2 units to obtain the graph of y = 2 + sec ( x − π ) on
[ −2π , 2π ] as shown below
402
Chapter 4 Review
62. First, note that the period is 2π and the phase shift is
π 4
units to the left. Hence,
π 7π one period of this function would occur on the interval − , , for instance. 4 4 π In order to graph y = − csc x + on [ −2π , 2π ] , proceed as described below: 4 π π 7π Step 1: Graph y = csc x + on − , . 4 4 4 π π 7π Graph the guide function y = sin x + on − , using the approach 4 4 4 discussed earlier in Chapter 4. π π 3π 7π x-intercepts: Occur when x + = 0, π , 2π , so that x = − , , 4 4 4 4 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of guide function (dotted) and π π 7π the function y = csc x + on − , : 4 4 4
Step 3: Reflect the graph in Step 2 over the x-axis to obtain the graph of π y = − csc x + on [ −2π , 2π ] : 4
403
Step 2: Extend the graph using periodicity to the interval [ −2π , 2π ] :
Chapter 4
63. First, note that the period is 2π and the phase shift is
π 3
units to the right.
π 7π Hence, one period of this function would occur on the interval , , for instance. 3 3 π In order to graph y = 2 + 3sec x − on [ −2π , 2π ] , proceed as described below: 3 π π 7π Step 1: Graph y = 3sec x − on , . 3 3 3 π π 7π Graph the guide function y = 3cos x − on , using the usual 3 3 3 π π 3π 5π 11π approach. x-intercepts: Occur when x − = , , so that x = , 3 2 2 6 6 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of guide function (dotted) Step 2: Extend the graph is Step 1 to π [ −2π , 2π ] : and the function y = 3sec x − 3 π 7π on , : 3 3
Step 2: Translate the graph is Step 2 vertically up 2 units to obtain the graph π of y = 2 + 3sec x − on [ −2π , 2π ] , as 3 shown below:
404
Chapter 4 Review
64. We have the following information about the graph of y = 1 + tan ( 2 x + π ) :
Period:
Vertical Translation: Up 1 unit
π
2
Reflection about xaxis? No
Phase Shift: Left
π
2
units
π Step 1: Graph y = tan ( 2 x + π ) on − , 0 . 2 We have the following information about this particular graph x-intercepts occur when sin ( 2 x + π ) = 0 . This happens when 2 x + π = 0, π , so π , 0 . So, the points − , 0 , ( 0, 0 ) are on this graph. (Note: 2 2 These will change in Step 2 when we translate this graph vertically.) Vertical asymptotes occur when cos ( 2 x + π ) = 0 . This happens when that x = −
2x + π =
π
π
so that x = −
2 translation.)
π 4
. (Note: These remain unchanged by vertical
Step 2: Extend the graph in Step 1 to [ −2π , 2π ] :
Step 3: Translate the graph in Step 2 vertically up 1 unit to obtain the graph of y = 1 + tan ( 2 x + π ) on [ −2π , 2π ] :
405
Chapter 4
65. a. The graphs of y1 = cos x (solid)
b. The graphs of y1 = cos x (solid)
y2 = cos ( x + π6 ) (dashed) are as follows:
y2 = cos ( x − π6 ) (dashed) are as follows:
The graph of y2 is that of y1 shifted left π6 units.
The graph of y2 is that of y1 shifted right π6 units.
66. a. The graphs of y1 = sin x (solid)
b. The graphs of y1 = sin x (solid)
y2 = sin x + 12 (dashed) are as follows:
y2 = sin x − 12 (dashed) are as follows:
The graph of y2 is that of y1 shifted up 12 unit.
The graph of y2 is that of y1 shifted down 12 unit.
406
Chapter 4 Review
67. Consider the graphs of the following functions on the interval [ −2π ,2π ] : y1 = 4 cos x (dashed bold), y2 = 3sin x (dotted bold), y3 = 4 cos x − 3sin x (solid bold)
The amplitude of y = 4 cos x − 3sin x appears to be approximately 5. 68. Consider the graphs of the following functions on the interval [ −2π ,2π ] : y1 = 3 sin x (dotted bold), y2 = cos x (dashed bold), y3 = 3 sin x + cos x (solid bold)
The amplitude of y = 3 sin x + cos x appears to be approximately 2. 407
Chapter 4 Practice Test Solutions ---------------------------------------------------------------1. Amplitude is 5 and period is 2π . 3 2. We have the following information about the graph:
Period = 23π Amplitude = 1 Phase shift = none Vertical shift = none No reflection over the x-axis
3. We have the following information about the graph:
Period = 2π Amplitude = 1 Phase shift = π4 right Vertical shift = none No reflection over the x-axis
4. We have the following information about the graph:
Period = 2π Amplitude = 4 Phase shift = none Vertical shift = none Reflect over the x-axis
408
Chapter 4 Practice Test
5. We have the following information about the graph:
Period = 4π Amplitude = 1 Phase shift = none Vertical shift = none No reflection over the x-axis
6. We have the following information about the graph:
Period = 2π Amplitude = 1 Phase shift = π6 left Vertical shift = none No reflection over the x-axis
7. We have the following information about the graph:
Period = 2π Amplitude = 1 Phase shift = none Vertical shift = down 2 units No reflection over the x-axis
409
Chapter 4
1 8. In order to construct the graph of y = −2 cos x on [ −4π , 4π ] , we first 2 gather the information essential to forming the graph for one period, namely on the interval [ 0, 4π ] . Then, extend this picture to cover the entire given interval. Indeed, we have: Amplitude: −2 = 2
Period:
2π = 4π 1 2
1 π 3π 1 x-intercepts occur when cos x = 0 . This happens when x = , , 2 2 2 2 and so x = π , 3π . So, the points (π , 0 ) , ( 3π , 0 ) are on the graph. 1 The maximum value of the graph is 2 since the maximum of cos x is 1, 2 1 and the amplitude is 2. This occurs when x = 0, 2π , and so x = 0, 4π . 2 So, the points (0,2), (4π ,2) are on the graph. 1 The minimum value of the graph is – 2 since the minimum of cos x is 2 1 –1, and the amplitude is 2. This occurs when x = π , and so x = 2π . So, 2 the point (2π , −2) is on the graph. Since the coefficient used to compute the amplitude is negative, we reflect 1 the graph of y = 2 cos x over the x-axis. 2 1 Now, using this information, we can form the graph of y = −2 cos x on 2 [0, 4π ] , and then extend it to the interval [ −4π , 4π ] , as shown below:
410
Chapter 4 Practice Test
π 9. We have the following information about the graph of y = tan π x − : 2 Phase Shift: Period: Reflection about xaxis? π π 1 =1 No Right 2 = units π π 2 π 1 3 Step 1: Graph y = tan π x − on , . 2 2 2 We have the following information about this particular graph π π x-intercepts occur when sin π x − = 0. This happens when π x − = 0, ± π , so 2 2 3 5 3 5 that x = , . So, the points , 0 , , 0 are on the graph. 2 2 2 2 π Vertical asymptotes occur when cos π x − = 0. This happens when 2 πx −
π
2
=±
π
2
so that x = 1,2.
π The graph of y = tan π x − on 2 1 3 2 , 2 is as follows:
Step 2: Extend the graph in Step 1 to 1 5 2 , 2 to show two periods:
10. The vertical asymptotes of y = csc ( 2π x ) occur when sin ( 2π x ) = 0 . This happens
when 2π x = nπ , where n is an integer. This simplifies to x =
411
n , where n is an integer. 2
Chapter 4
11. In order to graph y = −2 sec (π x ) on two periods, proceed as described below:
Step 1: Observe that the period is 2, with no phase shift or vertical shift. So, graphing this function on the interval [0,4] would constitute two periods. Graph y = sec (π x ) on [ 0,2 ]
Graph the guide function y = 2 cos (π x ) on [ 0,2 ] using the usual approach.
π 3π
1 3 , so that x = , 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function (dotted) and the function y = sec (π x ) on [ 0,2 ] is:
x-intercepts: Occur when π x =
Step 2: Extend the graph in Step 1 to [0,4]:
,
Step 3: Now, reflect the graph in Step 2 over the x-axis to obtain the graph of y = −2 sec (π x ) on [ 0, 4 ] :
412
Chapter 4 Practice Test
12. There is no amplitude since the range is all real numbers. The period is
and the phase shift is
2π π = . 4 2
π 4
13. First, we have the following information for the equation y = −2 + 3 sin ( 2 x − π ) :
2π π = π , Phase Shift , Vertical Shift 2 units down 2 2 In order to graph y = −2 + 3 sin ( 2 x − π ) on [ 0,2π ] , follow these steps: Amplitude 3 , Period
Step 1: Graph y = 3sin ( 2 x − π ) on
Step 2: Extend the graph in Step 1 to the interval [ 0,2π ] using periodicity:
π π π 3π 2 , 2 + π = 2 , 2 :
Step 3: Translate the graph in Step 2 vertically down 2 units to obtain the graph of y = −2 + 3 sin ( 2 x − π ) on [ 0,2π ] :
413
Chapter 4
π 14. First, we have the following information for the equation y = 1 − cos π x − : 2 Amplitude −1 = 1 , Period
2π
π
= 2 , Phase Shift
π
2 = 1 , Vertical Shift 1 unit π 2
up
π 1 5 In order to graph y = 1 − cos π x − on one period (say , , follow these 2 2 2 steps: π Step 1: Graph y = cos π x − on 2 1 5 2 , 2 :
Step 2: Reflect the graph in Step 1 over the x-axis to obtain the graph of π 1 5 y = cos π x − on , : 2 2 2
Step 3: Translate the graph in Step 2 vertically up 1 unit to obtain the graph π 1 5 of y = 1 − cos π x − on , : 2 2 2
414
Chapter 4 Practice Test
2π
π
2 = 1 units to the π 2 π 1 5 right. Hence, one period of this function would occur on the interval , , for 2 2 instance. π 1 5 In order to graph y = 2 − csc π x − on , , proceed as described below: 2 2 2 π 1 5 Step 1: Graph y = − csc π x − on , . 2 2 2 π 1 5 Graph the guide function y = − sin π x − on , using the approach 2 2 2 discussed earlier in Chapter 4. π 1 3 5 x-intercepts: Occur when π x − = 0, π , 2π , so that x = , , 2 2 2 2 At each x-intercept of this guide function, draw a vertical asymptote. On the intervals where this guide function is above the x-axis, draw a ∪ − shaped graph with vertex at the maximum of the guide function, opening upward toward the asymptotes. On the intervals where this guide function is below the x-axis, draw a ∩ − shaped graph with vertex at the minimum of the guide function, opening downward toward the asymptotes. The graph of the guide function Step 2: Translate the graph is Step 1 (dotted) and the function vertically up 2 units to obtain the graph π π 1 5 1 5 y = − csc π x − on , is as of y = 2 − csc π x − on , , as 2 2 2 2 2 2 follows: shown below:
15. First, note that the period is
= 2 and the phase shift is
415
Chapter 4
1 1 + cot ( 2x ) : 2 2 Reflection about x-axis? Phase Shift: None No
16. We have the following information about the graph of y =
Period:
Vertical Translation: 1 Up unit 2 2 1 π Step 1: Graph y = cot ( 2x ) on one period, say 0, . 2 2 We have the following information about this particular graph
π
x-intercepts occur when cos ( 2 x ) = 0 . This happens when 2x =
π 2
so that x =
π 4
.
π So, the point , 0 is on this graph. (Note: This will change in Step 2 when we 4 translate this graph vertically.) Vertical asymptotes occur when sin ( 2 x ) = 0 . This happens when 2x = 0, π , so
π
that x = 0, . (Note: These remain unchanged by vertical translation.) 2
Step 2: Extend the graph in Step 1 to [0,2π ] :
Step 3: Translate the graph in Step 2 1 unit: vertically up 2
416
Chapter 4 Practice Test
17. The period is
2π
ω
and the phase shift is −
φ φ (which means units left) ω ω
18. We plot the given points on a single set of axes and fit a trigonometric curve π π to them. One possibility is the curve y = 61 − 27 cos x − 6 6
19. We plot the given points on a single set of axes and fit a trigonometric curve π π to them. One possibility is the curve y = 31 + 14 cos t − 6 6
20. (a) Fluctuation on one period = maximum – minimum = 25 – (−25) = 50. So, it fluctuates by 50,000 rabbits over time. 8π 2π (b) We must find the smallest positive value of t such that sin t− = −1. 5 5 2π 8π π t− = − , so that t = 2.75 . So, the first year after 2016 in This occurs when 5 5 2 which the rabbit population attains a maximum is 2018.
417
Chapter 4
21. (a) Fluctuation on one period = maximum – minimum = 35 – (−35) = 70. So, it fluctuates by 70,000 dollars over time. π (b) We must find the smallest positive value of t such that sin t = −1 . This 3 π 3π occurs when t = , so that t = 4.5 . So, the first year after 2015 in which home 3 2 prices attain a maximum is 2020. 22. The vertical asymptotes of y = 2 csc ( 3x − π ) correspond to the x-intercepts of y = 2 sin ( 3x − π ) .
23. The x-intercepts of y = tan 2 x occur when sin 2 x = 0 . This happens when nπ 2 x = nπ , or x = , where n is an integer. 2 24. Observe that lim− ( − tan 2 x ) = −∞ . To see this, note that x = x→ π
4
π 4
is a vertical
π asymptote of y = − tan 2x since cos 2 ⋅ = 0 . So, as you approach this 4 asymptote from the right-side, the y-values become increasingly large without bound, and from the left-side, the y-values become increasingly negative without bound. This is illustrated in the graph below:
418
Chapter 4 Practice Test
25. a. True. Observe that using the even identity for cosine, followed by the π π π reduction formula, we obtain cos x − = cos − − x = cos − x = sin x. 2 2 2 π π π b. False. Take, for instance, x = . Observe that cos + = −1, while 2 2 2 π sin = 1. 2 c. True. Observe that π 1 1 1 csc x + = = = = sec x. π 2 π π cos x sin x + sin x cos + cos x sin 2 2 2 π d. False. This would require cos x + = sin x , which is not true in general by 2 part b. e. True. Since tangent has period π , tan(x − π ) = tan x. 26. Consider the graphs of the following functions on the interval [ −2π ,2π ] : y1 = 2 cos x (dashed bold), y2 = sin 2 x (dotted bold), y3 = 2 cos x − sin 2 x (solid bold)
419
Chapter 4 Cumulative Test Solutions ------------------------------------------------------------1. Let x = length of the shortest leg of a 30° − 60° − 90° triangle, which is 8 in. Then, the length of the longer leg is 8 3 ≈ 13.86 in. and the length of the hypotenuse is 16 in. 2. Angles F and G are both 72° (since they are vertical angles). So, since C and G are corresponding angles, they are equal to 72°. 3. sec θ = cos1 θ = 9115 = 53 1° 4. 33° 44′ 10′′ = 33° + 44′ ( 601°′ ) +10′′ ( 3600 ′′ ) ≈ 33.736°
5. Consider the triangle:
132 ft. tan 64.3° = 132b ft. b = tan64.3 ≈ 63.5 ft.
θ
6. Observe that 1579° = 4 ( 360° ) +139°. So, 139° is the smallest positive angle
coterminal with 1579°.
420
Chapter 4 Cumulative Test
7. Consider the following triangle:
sin θ =
opp 5 5 26 = = hyp 26 26
adj 1 26 = = hyp 26 26 opp tan θ = =5 adj 1 1 cot θ = = tan θ 5 1 = 26 sec θ = cos θ cos θ =
θ
csc θ =
1 26 = sin θ 5
8. Note that sec θ < 0 cos θ < 0 . As such, since also tan θ < 0 , we conclude that sin θ > 0 . Hence, the terminal side of θ is QII. 9. Impossible since the range of y = sec x is ( −∞, −1] ∪ [1,∞ ) 10. θ = 150°, 210°
−2
2 5 sinθ 3 11. tan θ = cos θ = 5 =− 5 3
12. csc θ = − 4 sin θ = csc1 θ = − 41 . Since the terminal side of θ is in QIII, we have the triangle: 41
4 41
13. Consider the following triangle:
θ θ
So, the slope of this line is tan θ = 34 . Hence, cos θ = − 541 = − 5 4141 . 421
Chapter 4
( )
π 14. −105° 180 = − 712π
15. The reference angle of 76π is π − 76π = π6 or 30 . 17. Using the formula s = rθ yields s = π3 ( 6 cm ) = 2π cm .
16. − 22
(
)
(
)
π 18. Using the formula A = 12 r 2 θ ⋅ 180π yields A = 12 (3.4 m)2 35° ⋅ 180 ≈ 3.53 m 2 . 3m 19. Observe that v = st = 4sec
π 20. Observe that ω = θt = 1624sec = 32π rad sec
21. Since May corresponds to x = 5, we have w(5) = 145 + 10 cos ( 56π ) = 145 − 10
( ) ≈ 136 lbs 3 2
22. We have the following information about the graph:
Period = π Amplitude = 3 Phase shift = none Vertical shift = none No reflection over the x-axis
23. Amplitude = 0.007 24. Amplitude = 2
π Frequency = 850 2π = 425 hertz
Period = 2π
Phase shift = π4 right
25. The domain is the set of all real numbers x such that x − π2 ≠ ( 2 n 2+1)π , or π x ≠ π2 + (2n+1) = ( n + 1)π , where n is an integer. This is equivalent to saying the 2 domain is the set of all real numbers for which x ≠ nπ , where n is an integer. The range is the set of all real numbers.
422
CHAPTER 5 Section 5.1 Solutions -------------------------------------------------------------------------------1.
2.
1 sin x csc x = sin x = 1 sin x
3. sec( − x ) cot x =
sin x cos x tan x cot x = = 1 cos x sin x
1 1 cos x 1 cos x = csc x = = cos( − x ) sin x cos( x ) sin x sin( x )
4. tan( − x ) cos( − x ) =
sin( − x ) cos( − x ) = sin( −x ) = − sin x cos( −x )
5.
1 sin2 x ( cot 2 x + 1) = sin2 x csc 2 x = sin2 x = 1 2 sin x 6.
1 cos 2 x ( tan2 x + 1) = cos 2 x sec 2 x = cos 2 x = 1 2 cos x 7.
( sin x − cos x )( sin x + cos x ) = sin2 x − cos 2 x 8. 2 ( sin x + cos x ) = sin2 x + 2 sin x cos x + cos 2 x
= ( sin2 x + cos 2 x ) + 2 sin x cos x = 1 + 2 sin x cos x
423
Chapter 5
9.
10.
1 1 csc x sin x = = sec x = cos x cos x cot x sin x
1 1 sec x cos x = = csc x = sin x sin x tan x cos x
11.
2 2 1 − cos 4 x (1 − cos x ) (1 + cos x ) = = 1 − cos 2 x = sin2 x 2 2 1 + cos x 1 + cos x
12.
1 − sin 4 x (1 − sin x ) (1 + sin x ) = 1 − sin2 x = cos 2 x = 2 1 + sin2 x 1 + sin x 2
2
13.
(1 − cos x ) (1 + cos x ) sin2 x 1 − cos 2 x = 1 − (1 + cos x ) = 1− = 1− 1− 1 − cos x 1 − cos x 1 − cos x = 1 − 1 − cos x = − cos x
14. 1−
(1 + sin x ) (1 − sin x ) cos 2 x 1 − sin2 x = 1 − (1 − sin x ) = 1− = 1− 1 + sin x 1 + sin x 1 + sin x = 1 − 1 + sin x = sin x
15.
sin x − cos x sin x cos x − tan x − cot x sin x cos x + 2 cos 2 x + 2 cos 2 x = cos x sin x + 2 cos 2 x = 2 2 sin x cos x sin x + cos x tan x + cot x + cos x sin x sin x cos x 2
=
2
sin2 x − cos 2 x + 2 cos 2 x = sin2 x − cos 2 x + 2 cos 2 x 2 2 sin + cos x x =1
= sin2 x + cos 2 x = 1
424
Section 5.1
16. sin x − cos x sin x cos x − tan x − cot x sin x cos x + cos 2 x = cos x sin x + cos 2 x = + cos 2 x 2 2 sin x cos x sin x + cos x tan x + cot x + cos x sin x sin x cos x 2
=
2
sin2 x − cos 2 x + cos 2 x = sin2 x − cos 2 x + cos 2 x = sin 2 x 2 2 x x sin + cos =1
17.
( cot x + 1) − csc x = csc2 x − csc2 x = 0 cot 2 x + 1 − csc x = csc x csc x csc x 2
2
18. cos 2 x (1 − cos x ) + cos 2 x (1 + cos x ) 2 cos 2 x cos 2 x cos 2 x + = = 1 + cos x 1 − cos x 1 − cos 2 x (1 + cos x )(1 − cos x )
=
2 cos 2 x = 2 cot 2 x 2 sin x
19. 1 + sin ( − x ) 1 − sin ( − x ) 1 − sin ( x ) 1 + sin ( x ) − = − 1 + cos ( − x ) 1 − cos ( − x ) 1 + cos ( x ) 1 − cos ( x ) = =
(1 − sin ( x ) ) (1 − cos ( x ) ) − (1 + sin ( x ) ) (1 + cos ( x ) ) (1 + cos ( x ) ) (1 − cos ( x ) )
(1 − sin x − cos x + sin x cos x ) − (1 + sin x + cos x + sin x cos x )
1 − cos 2 x −2 sin x − 2 cos x −2 ( sin x + cos x ) −2 sin x + cos x = = = ⋅ 2 2 1 − cos x sin x sin x sin x = −2 csc x [1 + cot x ] = −2 [ csc x + (csc x )(cot x )]
20. cot 2 x cot 2 x − csc 2 x 1 − csc x = =− = − sin x csc x csc x csc x
425
Chapter 5
21.
sec θ ⋅ sin θ =
1 sin θ ⋅ sin θ = = tan θ , as claimed. cos θ cos θ
22.
csc θ ⋅ cos θ ⋅ tan θ =
1 cos θ sin θ ⋅ ⋅ = 1, as claimed. sin θ 1 cos θ
23. tan2 x + sec 2 x = ( sec 2 x − 1) + sec 2 x = 2 sec 2 x − 1, as claimed. 24. csc θ sin θ 1 1 = csc θ sin θ ⋅ = ⋅ sin θ ⋅ tan θ = tan θ cot θ cot θ sin θ 25.
( sin x + cos x ) + ( sin x − cos x ) 2
2
(
) (
)
= sin2 x + 2 sin x cos x + cos 2 x + sin2 x − 2 sin x cos x + cos 2 x = 2, What remains after cancellation = 1
What remains after cancellation = 1
as claimed. 26. (1 − sin x )(1 + sin x ) = 1 − sin2 x = cos 2 x, as claimed. 27. ( csc x + 1)( csc x − 1) = csc 2 x − 1 = (1 + cot 2 x ) − 1 = cot 2 x, as claimed. 28. ( sec x + 1)( sec x − 1) = sec 2 x − 1 = (1 + tan2 x ) − 1 = tan2 x, as claimed. 29. sin x cos x sin2 x + cos 2 x 1 1 1 tan x + cot x = + = = = = csc x sec x, cos x sin x sin x cos x sin x cos x sin x cos x as claimed.
30. csc x − sin x =
1 1 − sin2 x cos 2 x cos x − sin x = = = ( cos x ) = cot x cos x, sin x sin x sin x sin x
as claimed.
426
Section 5.1
31.
2 − sin2 x 1 + (1 − sin x ) 1 + cos 2 x 1 cos 2 x = = = + = sec x + cos x cos x cos x cos x cos x cos x 2
32.
2 − cos 2 x 1 + (1 − cos x ) 1 + sin2 x 1 sin2 x = = = + = csc x + sin x sin x sin x sin x sin x sin x 2
33. 1 1 + = 2 csc x sec 2 x
1 1 + = sin2 x + cos 2 x = 1, as claimed. 1 1 2 sin x cos 2 x
34. 1 1 tan2 x − cot 2 x sin2 x cos 2 x 2 2 x x tan cot − = = − = − 2 x cot 2 x tan 2 x tan cot 2 cos 2 x sin2 x x =1
sin 4 x − cos 4 x = = sin2 x cos 2 x
=
sin2 x 2
2
sin x cos x
−
=1 2 2 2 2 ( sin x − cos x )( sin x + cos x )
sin2 x cos 2 x cos 2 x 2
2
sin x cos x
=
1 1 − 2 = sec 2 x − csc 2 x, 2 cos x sin x
as claimed. 35.
(1 + sin x ) + (1 − sin x ) = 2 = 2 = 2 sec 2 x, as claimed. 1 1 + = 1 − sin x 1 + sin x (1 + sin x )(1 − sin x ) 1 − sin2 x cos2 x
36.
(1 + cos x ) + (1 − cos x ) = 2 = 2 = 2 csc 2 x, as claimed. 1 1 + = 1 − cos x 1 + cos x (1 + cos x )(1 − cos x ) 1 − cos2 x sin2 x
37.
sin2 x 1 − cos 2 x (1 − cos x ) (1 + cos x ) = 1 + cos x, as claimed. = = 1 − cos x 1 − cos x 1 − cos x
427
Chapter 5
38.
cos 2 x 1 − sin2 x (1 − sin x ) (1 + sin x ) = 1 + sin x, as claimed. = = 1 − sin x 1 − sin x 1 − sin x
39.
( sec x + tan x )( sec x − tan x ) = sec 2 x − tan2 x = (1 + tan x ) − tan x sec x + tan x = 2
sec x − tan x
=
sec x − tan x
2
sec x − tan x
1 , as claimed. sec x − tan x
40.
csc x + cot x )( csc x − cot x ) csc 2 x − cot 2 x (1 + cot x ) − cot x ( csc x + cot x = = = 2
csc x − cot x
=
csc x − cot x
csc x − cot x
1 , as claimed. csc x − cot x
41. cos x − sin2 x 1 sin x − csc x − tan x sin x cos x cos x − sin2 x sin x cos x = = = , as claimed. 1 cos x sin x + cos 2 x sin x + cos 2 x sec x + cot x + cos x sin x sin x cos x
42.
1 sin x 1 + sin x + sec x + tan x cos x cos x 1 + sin x sin x sin x ⋅ = = tan x, = = cos x = 1 1 + sin x csc x + 1 cos x cos x 1 sin x + +1 sin x sin x as claimed.
43.
cos 2 x + 1 + sin x (1 − sin x ) + 1 + sin x − sin x − sin x − 2 = = cos 2 x + 3 − sin2 x − 4 (1 − sin2 x ) + 3 2
=
− ( sin x + 1) ( sin x − 2 ) − (sin x − 2) (sin x + 2)
2
=
sin x + 1 1 + sin x = , as claimed. sin x + 2 2 + sin x
428
2
Section 5.1
44.
sin x (sin x + 1) sin x + 1 − cos 2 x sin x + 1 − (1 − sin x ) sin2 x + sin x = = = cos 2 x 1 − sin2 x 1 − sin2 x (1 − sin x ) (1 + sin x ) 2
=
sin x , as claimed. 1 − sin x
45. =1 2 1 sin x cos x 1 sin x + cos 2 x 1 1 + = = sec x ( tan x + cot x ) = cos x cos x sin x cos x cos x sin x cos x cos x sin x csc x 1 1 1 = , as claimed. = ( csc x ) = 2 2 cos 2 x cos x sin x cos x
46. tan x ( csc x − sin x ) =
2 2 sin x 1 sin x 1 − sin x cos x − sin x = = = cos x, cos x sin x cos x sin x cos x
as claimed. 47. 1 + cos 2 ( − x )
1 − csc 2 ( − x )
=
1 + cos 2 ( x ) 1 + cos 2 ( x ) = = − tan2 x − cos 2 x tan2 x 2 2 1 − csc ( x ) − cot x
= − tan2 x − cos 2 x
sin2 x 2
cos x
= − sin 2 x − tan2 x
48.
cos x (1 + cos x ) +
sin x sin x = cos x + cos 2 x + = cos x + ( cos 2 x + sin2 x ) = 1 + cos x 1 csc x sin x
429
Chapter 5
49. 1 sin2 x sin x cos x sin3 x + sin x cos 2 x sin x tan x + + = = tan x cos x sin x sin x cos x =
sin x ( sin2 x + cos 2 x ) sin x cos x
=
1 = sec x cos x
50. 1 + sin x (1 + sin x )(1 − sin x ) 1 − sin2 x cos 2 x cos x = = = = cos x cos x (1 − sin x ) cos x (1 − sin x ) cos x (1 − sin x ) 1 − sin x 51. Observe that the left-side of the equation simplifies as follows: cos 2 x ( tan x − sec x )( tan x + sec x ) = cos 2 x ( tan2 x − sec 2 x )
(
= cos 2 x tan2 x − (1 + tan2 x )
)
= − cos 2 x Since − cos x is, in fact, never equal to 1 (being a non-positive quantity), the given equation is a contradiction. 2
52. Observe that cos 2 x ( tan x − sec x )( tan x + sec x ) = cos 2 x ( tan 2 x − sec 2 x )
(
= cos 2 x tan2 x − (1 + tan2 x )
)
= − cos 2 x = −(1 − sin2 x ) = sin2 x − 1
So, the given equation is an identity. 53. Observe that
1 cot x 1 csc x cot x sin x 1 1 1 = = = cot x cot x cos x 1 sec x tan x tan x 1 tan x sin x tan x sin x cos x cos x cos x 1 = = cot 3 x cot x sin x tan x = cot x
= cot x
So, the given equation is an identity.
430
Section 5.1
54. The equation sin x cos x = 0 is a conditional since, for instance, it does not hold if π 1 x = (since the left-side reduces to in such case). 4 2 55. The equation sin x + cos x = 2 is a conditional since, for instance, it does not hold if x = 0 (since the left-side reduces to 1). 56. The equation sin2 x + cos 2 x = 1 is a known identity. 57. Observe that
tan2 x − sec 2 x = tan2 x − (1 + tan2 x ) = −1 ≠ 1.
Since the equation is never true, the given equation is a contradiction. 58. Observe that
sec 2 x − tan2 x = (1 + tan 2 x ) − tan 2 x = 1.
So, the given equation is an identity. 59. The equation sin x = 1 − cos 2 x is a conditional. To see this, observe that the
right-side reduces as follows:
1 − cos 2 x = sin2 x = sin x . As such, the given
3π , for instance. 2 (Note: The equation IS an identity if one restricts attention to only those x-values for which sin x > 0.)
equation is not true for x =
60. The equation csc x = 1 + cot 2 x is a conditional. To see this, observe that the
right-side reduces as follows:
1 + cot 2 x = csc 2 x = csc x . As such, the given
3π , for instance. 2 (Note: The equation IS an identity if one restricts attention to only those x-values for which sin x > 0 (and hence, csc x > 0).)
equation is not true for x =
61. Observe that
sin2 x + cos 2 x = 1 = 1. So, the given equation is an identity.
431
Chapter 5
62. Observe that
sin2 x + cos 2 x = 1 = 1. So, the equation
sin2 x + cos 2 x = sin x + cos x is an conditional. Indeed, take for instance x =
π 4
.
Then, the left-side = 1, while the right-side is 2. a 2 + b 2 = a + b, which is false.)
(Note: A common algebraic error is to say 63. Let x = a sin θ , where −
π 2
≤θ ≤
π 2
. Then,
a 2 − x 2 = a 2 − ( a sin θ )2 = a 2 (1 − sin2 θ ) = a 2 cos 2 θ = a cos θ .
Note that cos θ = cos θ , for all −
π 2
≤θ ≤
π 2
. Hence, we conclude that
a 2 − x 2 = a cos θ .
64.
Let x = a tan θ , where −
π 2
<θ <
π 2
. Then,
a 2 + x 2 = a 2 + ( a tan θ )2 = a 2 (1 + tan 2 θ ) = a 2 sec 2 θ = a sec θ .
Note that sec θ = sec θ , for all −
π 2
<θ <
π 2
. Hence, we conclude that
a 2 + x 2 = a sec θ .
65. Observe that
sin ( 2t ) cos ( 2t ) + 1 sin ( 2t ) cos ( 2t ) + 1 = −3cos(2t ). y = −3 = −3 t t sin 2 cos 2 + 1 t t sin 2 + sec 2 ( ) ( ) ( ) ( ) cos(2t )
The period is π , so we inspect the graph on the interval [ 0, π ] . The maximum of 3 occurs when 2t = 0, so t = 0, while the minimum of −3 occurs when 2t = π , so t =
As such, it takes
π 2
≈ 1.57 sec. to go from max to min displacement.
432
π
2
.
Section 5.1
66. Observe that
π π sin 2t − + 1 sin 2t − 2 + 1 π 2 = 4 y = 4 = 4 sin 2t − 2 1 + csc 2t − π sin 2t − π + 1 2 2 π sin 2t − 2
The period is π and phase shift
π
units to the right. The maximum displacement 4 from equilibrium is 4, and the first positive occurrence of a max is when 2t −
π
=
π
, so t =
2 2 maximum.
π
2
. So, if starting at t = 0, it takes
π
2
≈ 1.57 sec. to reach the first
67. Simplified the two fractions in Step 2 incorrectly. The correct computations are: cos x cos x cos 2 x = = sin x cos x − sin x cos x − sin x 1− cos x cos x sin x sin x sin2 x = = cos x sin x − cos x sin x − cos x 1− sin x sin x 68. The error is as follows: sec x =
1 1 , not . cos x sin x
69. Need to observe that tan2 x = 1 is a conditional, which does not hold for all values
of x. Verifying the truth of the equation for a particular value of x (here, x =
π
4
) is
not enough to prove that it is an identity. 70. Verifying the truth of the equation for particular values of x (here, odd multiples
of
π
) is not enough to prove that it is an identity. Indeed, the given equation is 2 actually a conditional. To see this, take x = 0. Then, sin 0 − cos 0 = −1 ≠ 1.
433
Chapter 5
71. False. The equation sin2 x = 1 is not an identity since even though it is true for
π
+ nπ (n is an integer), it is false for all other values of x for which sin x is 2 defined. x=
72. False. The equation sin x = 1 has infinitely many solutions, but it is not always true. 73. False. The given conditional does not hold at x = 0, for instance. 74. True. This is due to the identity 1 + cot 2 x = csc 2 x. 75. The equation cos θ = 1 − sin2 θ is true whenever cos θ > 0 (since 1 − sin2 θ = cos 2 θ = cos θ .) This occurs for those angles θ whose terminal side is in QI or
QIV. 76. The equation − cos θ = 1 − sin2 θ is true whenever cos θ < 0 (since 1 − sin2 θ = cos 2 θ = cos θ .) This occurs for those angles θ whose terminal side is
in QII or QIII. 77. The equation sin( A + B ) = sin A + sin B is not true in general. For instance, let
A = 30 and B = 60. Then, sin ( 30 + 60 ) = sin ( 90 ) = 1, whereas
sin ( 30 ) + sin ( 60 ) =
1 3 1+ 3 + = ≠ 1. 2 2 2
1 1 78. The equation cos A = cos A is not true in general. For instance, let A = π2 . 2 2 1 2 1 π π π , whereas cos = 0. Then, cos ⋅ = cos = 2 2 2 4 2 2
434
Section 5.1
79. 2 2 ( a sin x + b cos x ) + ( b sin x − a cos x ) = a 2 sin2 x + 2ab sin x cos x + b 2 cos 2 x + b 2 sin2 x − 2ab sin x cos x + a 2 cos 2 x = ( a 2 + b2 ) sin2 x + ( a 2 + b2 ) cos 2 x = ( a 2 + b2 )( sin2 x + cos 2 x ) = a 2 + b2 =1
80. 1 + cot 3 x 1 + cot 3 x + cot x(1 + cot x ) + cot x = 1 + cot x 1 + cot x 3 1 + cot x + cot x + cot 2 x = 1 + cot x (1 + cot x ) + ( cot3 x + cot 2 x ) = 1 + cot x 1 + cot x ) + cot 2 x ( cot x + 1) ( = 1 + cot x
=
(1 + cot x ) (1 + cot 2 x ) 1 + cot x
= csc 2 x
81.
(1 − sin x ) (1 + sin x ) 1 + sin2 x 1 + sin2 x 1 − sin 4 x = = =1 = 2 2 cos 2 x − cos 4 x 2 − (1 − sin2 x ) 1 + sin x cos 2 x ( 2 − cos 2 x ) 2
2
82. 2 1 − 2 sin x − sin2 x + 2 sin3 x (1 − 2 sin x ) − sin x (1 − 2 sin x ) = cos 2 x cos 2 x
=
(1 − 2 sin x ) (1 − sin2 x ) cos 2 x
435
= 1 − 2 sin x
Chapter 5
83. a + 2c − d sin x + 2 cot x − csc x = = 2−b 2 − cos x
=
2 cos x 1 sin2 x + 2 cos x − 1 − sin x sin x = sin x 2 − cos x 2 − cos x
sin x +
sin2 x + 2 cos x − 1 1 − cos 2 x + 2 cos x − 1 cos x ( − cos x + 2 ) = cot x = = sin x ( 2 − cos x ) sin x ( 2 − cos x ) sin x (2 − cos x )
84. a + b2 1 b 1 cos x = + = + = sec x + cot x ab b a cos x sin x 85. The correct identity is cos( A + B ) = cos A cos B − sin A sin B.
Consider the graphs of the following functions. Note that the graphs of y1 (BOLD) and y3 are the same. (The graph of y2 is dotted.)
π y1 = cos x + 4 π π y2 = cos ( x ) cos + sin ( x ) sin 4 4 π π y3 = cos ( x ) cos − sin ( x ) sin 4 4
436
Section 5.1
86. The correct identity is cos( A − B ) = cos A cos B + sin A sin B.
Consider the graphs of the following functions. Note that the graphs of y1 (BOLD) and y3 are the same. (The graph of y2 is dotted.)
π y1 = cos x − 4 π π y2 = cos ( x ) cos − sin ( x ) sin 4 4 π π y3 = cos ( x ) cos + sin ( x ) sin 4 4
87. The correct identity is sin( A + B ) = sin A cos B + cos A sin B.
Consider the graphs of the following functions. Note that the graphs of y1 (BOLD) and y2 are the same. (The graph of y3 is dotted.)
π y1 = sin x + 4 π π y2 = sin ( x ) cos + cos ( x ) sin 4 4 π π y3 = sin ( x ) cos − cos ( x ) sin 4 4
437
Chapter 5
88. The correct identity is sin( A − B ) = sin A cos B − cos A sin B.
Consider the graphs of the following functions. Note that the graphs of y1 (BOLD) and y3 are the same. (The graph of y2 is dotted.)
π y1 = sin x − 4 π π y2 = sin ( x ) cos + cos ( x ) sin 4 4 π π y3 = sin ( x ) cos − cos ( x ) sin 4 4
438
Section 5.2 Solutions --------------------------------------------------------------------------------1. Using the addition formula sin( A − B ) = sin A cos B − cos A sin B yields
π π π π π π π sin = sin − = sin cos − cos sin 12 3 4 3 4 3 4 3 2 1 2 2 3 1 = − = − = 2 2 2 2 2 2 2
6− 2 4
2. Using the addition formula cos( A − B ) = cos A cos B + sin A sin B yields
π π π π π π π cos = cos − = cos cos + sin sin 12 3 4 3 4 3 4
3 1 2 3 2 2 1 = = + + 2 2 2 = 2 2 2 2
2+ 6 4
3. Using the even identity for cosine, along with the addition formula cos( A + B ) = cos A cos B − sin A sin B yields 5π 5π π π π π π π cos − = cos = cos + = cos cos − sin sin 12 12 6 4 6 4 6 4
3 2 1 2 2 3 1 = − = − = 2 2 2 2 2 2 2
6− 2 4
4. Using the odd identity for sine, along with the addition formula sin( A + B ) = sin A cos B + cos A sin B, yields π 5π 5π π π π π π sin − = − sin = − sin + = − sin cos + cos sin 12 12 6 4 4 6 4 6 1 2 3 2 2 1 3 − 2− 6 = − + + = − = 4 2 2 2 2 2 2 2
439
Chapter 5
5. Using the odd identity for sine, along with the addition formula sin( A + B ) = sin A cos B + cos A sin B, yields 7π π π π π π π sin = sin + = sin cos + cos sin 12 3 4 3 4 3 4
3 2 1 2 2 3 1 = + = + = 2 2 2 2 2 2 2
6+ 2 4
6. Using the even identity for cosine, along with the addition formula cos( A + B ) = cos A cos B − sin A sin B yields 7π π π π π π π cos = cos + = cos cos − sin sin 12 3 4 3 4 3 4
3 1 2 3 2 2 1 = − − = = 2 2 2 2 2 2 2
2− 6 4
7.
π π First, we need to compute both sin and cos . Then, we use the definition 12 12 π of tangent to compute tan . To this end, we have the following: 12 Step 1: Using the odd identity for sine and Exercise 1 yields 6− 2 π π sin − = − sin = − 4 12 12 Step 2: Using the even identity for cosine and the addition formula cos( A − B ) = cos A cos B + sin A sin B yields π π π π π π π π cos − = cos = cos − = cos cos + sin sin 12 12 3 4 3 4 3 4 3 2+ 6 1 2 3 2 2 1 = + = + = 4 2 2 2 2 2 2 2
440
Section 5.2
Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields 6− 2 π π sin − − sin − 4 12 12 π =− 6+ 2 tan − = = = 6+ 2 6+ 2 12 cos − π cos π 12 12 4 =
− 6+ 2 ⋅ 6+ 2
6 − 2 −6 + 2 2 6 − 2 −8 + 4 3 = = = −2 + 3 6−2 4 6− 2
Alternatively, π π tan − tan π π π 4 3 = 1 − 3 = −2 + 3 tan − = tan − = 12 4 3 1 + tan π tan π 1 + 3 4 3 8. Using the periodicity and odd identities of tangent, followed by Exercise 7, yields π 13π π π π tan = tan + π = tan = − − tan = − tan − 12 12 12 12 12
= − −2 + 3 = 2 − 3 9. Using the addition formula sin( A + B ) = sin A cos B + cos A sin B yields
sin (105 ) = sin ( 60 + 45 ) = sin ( 60 ) cos ( 45 ) + cos ( 60 ) sin ( 45 ) 3 2 1 2 = + = 2 2 2 2
2+ 6 4
10. First, applying the addition formula cos( A − B ) = cos A cos B + sin A sin B yields
cos (195 ) = cos ( 225 − 30 ) = cos ( 225 ) cos ( 30 ) + sin ( 225 ) sin ( 30 ) (1)
Next, observe that the reference angle for 225 is 45. Using this observation (and affixing appropriate signs), we further simplify (1) as follows: = cos ( −45 ) cos ( 30 ) + sin ( −45 ) sin ( 30 ) 2 3 − 2 1 2 3 1 − 2 − 6 = − + = + = − 2 2 2 2 2 2 2 4
441
Chapter 5
11. First, we need to compute both sin ( −105 ) and cos ( −105 ) . Then, we use the
definition of tangent to compute tan ( −105 ) . To this end, we have the following:
Step 1: Using the odd identity for sine and Exercise 9 yields 2+ 6 sin ( −105 ) = − sin (105 ) = − 4 Step 2: Using the even identity for cosine and the addition formula cos( A + B ) = cos A cos B − sin A sin B yields
cos ( −105 ) = cos (105 ) = cos ( 60 + 45 ) = cos ( 60 ) cos ( 45 ) − sin ( 60 ) sin ( 45 )
2− 6 1 2 3 2 = − = 4 2 2 2 2 Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields 2+ 6 − sin ( −105 ) 4 =− 6− 2 = tan ( −105 ) = cos ( −105 ) 2− 6 2− 6 4 − 6− 2 2 + 6 −6 − 2 2 6 − 2 −8 − 4 3 = ⋅ = = = 2+ 3 −4 2−6 2− 6 2+ 6 Alternatively, tan ( −60 ) − tan ( 45 ) − 3 − 1 = tan ( −105 ) = tan ( −60 ) − 45 = 1 + tan ( −60 ) tan ( 45 ) 1 − 3
(
)
1+ 3 = − = 2 + 3 1− 3
442
Section 5.2
12.
tan A + tan B yields 1 − tan A tan B tan120 + tan 45 (1) tan(165 ) = tan(120 + 45 ) = 1 − tan120 tan 45 Now, observe that tan 45 = 1, and since the reference angle of 120 is 60 , we have (affixing the appropriate signs) 3 sin120 sin 60 2 tan120 = = = 1 =− 3 − cos 60 cos120 − 2 So, continuing the computation in (1) yields − 3 + 1 − 3 + 1 − 3 + 1 1 − 3 −2 3 + 3 + 1 = = ⋅ = tan(165 ) = 1− 3 1+ 3 1+ 3 1− 3 1− − 3 Using the addition formula tan( A + B ) =
(
=
)
4−2 3 = −2 + 3 −2
13. Using the odd identity of tangent, followed by Exercise 7, yields 1 1 1 1 π = = = cot = π − −2 + 3 π 12 tan π − − tan − tan − 12 12 12
=
1 1 2+ 3 2+ 3 = ⋅ = = 2+ 3 4 −3 2− 3 2− 3 2+ 3
443
Chapter 5
14.
5π 5π First, we need to compute both sin − and cos − . Then, we use the 12 12 5π definition of cotangent to compute cot − . To this end, we have the following: 12 Step 1: Using the odd identity for sine and Exercise 3 yields 5π 6 − 2 cos − = 4 12 Step 2: Using the odd identity for sine, along with the addition formula sin( A + B ) = sin A cos B + cos A sin B, yields π 5π 5π π π π π π sin − = − sin = − sin + = − sin cos + cos sin 12 12 6 4 4 6 4 6 1 2 3 2 2 1 3 − 2− 6 = − + + = − = 4 2 2 2 2 2 2 2 Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields 5π 6− 2 cos − 6− 2 5π 12 = 4 cot − = = 12 sin − 5π − 2 − 6 − 2 − 6 12 4 =
6− 2
−
( 2 + 6)
⋅
6 − 2 6−2 2 6 +2 8−4 3 = = = −2 + 3 − (6 − 2) −4 6− 2
444
Section 5.2
15.
11π First, we need to compute cos − . Then, we use the definition of secant to 12 11π compute sec − . To this end, we have the following: 12 Step 1: Using the even identity for cosine and the addition formula cos( A − B ) = cos A cos B + sin A sin B yields 11π 11π 11π cos − = cos = cos π − 12 12 12 11π 11π π π cos π sin = cos + sin = − cos (1) 12 12 12 = −1 =0 Using the addition formula cos( A − B ) = cos A cos B + sin A sin B (again) now yields π π π π π π π cos = cos − = cos cos + sin sin 12 3 4 3 4 3 4 3 2+ 6 1 2 3 2 2 1 = + = + = 4 2 2 2 2 2 2 2 2+ 6 11π Hence, using this in (1) yields cos − . = − 4 12 Step 2: Using the definition of secant and the computation in Step 1 now yields 1 1 −4 11π sec − = = = 2+ 6 2+ 6 12 cos − 11π − 12 4 =
−4 ⋅ 2+ 6
(
)
2 − 6 −4 2 − 6 = = 2−6 2− 6
445
2− 6
Chapter 5
16.
13π First, we need to compute cos − . Then, we use the definition of secant to 12 13π compute sec − . To this end, we have the following: 12 Step 1: Using the even identity for cosine and the addition formula cos( A + B ) = cos A cos B − sin A sin B yields 11π 13π 13π cos − = cos = cos π + 12 12 12 11π 11π π π cos π sin = cos − sin = − cos (1) 12 12 12 = −1 =0 Using the addition formula cos( A − B ) = cos A cos B + sin A sin B (again) now yields π π π π π π π cos = cos − = cos cos + sin sin 12 3 4 3 4 3 4 3 2+ 6 1 2 3 2 2 1 = + = + = 4 2 2 2 2 2 2 2
2+ 6 13π Hence, using this in (1) yields cos − . = − 4 12 Step 2: Using the definition of secant and the computation in Step 1 now yields 1 1 −4 13π sec − = = = 2+ 6 2+ 6 12 cos − 13π − 12 4 =
−4 ⋅ 2+ 6
(
)
2 − 6 −4 2 − 6 = = 2−6 2− 6
446
2− 6
Section 5.2
17.
17π First, we need to compute sin . Then, we use the definition of cosecant to 12 17π compute csc . To this end, we have the following: 12 Step 1: Using the addition formula sin( A + B ) = sin A cos B + cos A sin B, yields 17π 2π 3π 2π 3π 2π 3π sin + = sin = sin cos + cos sin 4 12 3 3 4 3 4 3 2 1 2 2 1 3 − 2− 6 = − + − = − + = 4 2 2 2 2 2 2 2 Step 2: Using the definition of cosecant and the computations in Step 1 yields 1 1 4 4 6− 2 17π csc = = = ⋅ = 2− 6 = 6− 2 12 sin 17π − 2 − 6 − 2 − 6 − 2 + 6 12 4
(
)
18.
23π First, we need to compute sin . Then, we use the definition of cosecant to 12 23π compute csc . To this end, we have the following: 12 Step 1: Using the addition formula sin( A + B ) = sin A cos B + cos A sin B, yields 23π 2π 5π 2π 5π 2π 5π sin + = sin = sin cos + cos sin 4 12 3 3 4 3 4 3 2 1 2 2 1 3 2− 6 = − + − − = − = 4 2 2 2 2 2 2 2 Step 2: Using the definition of cosecant and the computations in Step 1 yields 1 1 4 4 6+ 2 23π csc = = = ⋅ = − 2− 6 = 2− 6 2− 6 6+ 2 12 sin 23π 2− 6 12 4
(
447
)
Chapter 5
19. First, we need to compute cos ( −195° ) . Then, we use the definition of secant to
compute sec ( −195° ) . To this end, we have the following: Step 1: Using the addition formula cos( A − B ) = cos A cos B + sin A sin B yields cos ( −195° ) = cos ( 30° − 225° ) = cos ( 30° ) cos ( 225° ) + sin ( 30° ) sin ( 225° )
3 2 1 2 2 1 3 − 6 − 2 (1) = + − + − = − = 4 2 2 2 2 2 2 2 Step 2: Using the definition of secant and the computation in Step 1 now yields 1 1 −4 −4 2− 6 sec ( −195° ) = = = = ⋅ cos ( −195° ) 2+ 6 2+ 6 2+ 6 2− 6 − 4 =
−4
( 2 − 6) = 2 − 6 2−6
20. First, we need to compute sin ( 285° ) . Then, we use the definition of cosecant to
compute csc ( 285° ) . To this end, we have the following:
Step 1: Using the addition formula sin( A − B ) = sin A cos B − sin B cos A yields sin ( 285° ) = sin ( 315° − 30° ) = sin ( 315° ) cos ( 30° ) − sin ( 30° ) cos ( 315° ) 2 3 1 2 2 1 3 − 6 − 2 (1) = − − = − + = 2 2 2 2 2 2 2 4 Step 2: Using the definition of cosecant and the computation in Step 1 now yields −4 1 1 = = csc ( 285° ) = sin ( 285° ) 2+ 6 − 2 + 6 4 =
−4 ⋅ 2+ 6
(
)
2 − 6 −4 2 − 6 = = 2−6 2− 6
2− 6
21. Using the addition formula cos( A − B ) = cos A cos B + sin A sin B with A = 2 x and B = 3x, followed by the even identity for cosine at the very end, yields
sin 2 x sin3x + cos 2 x cos3x = cos(2x − 3x ) = cos( − x ) = cos x .
448
Section 5.2
22. Using the addition formula cos( A + B ) = cos A cos B − sin A sin B with A = x and B = 2 x yields sin x sin 2 x − cos x cos 2 x = − [ cos x cos 2 x − sin x sin 2 x ] = − cos( x + 2 x ) = − cos 3x 23. Using the addition formula sin( A − B ) = sin A cos B − cos A sin B with A = x and B = 2x, followed by the odd identity for sine at the very end, yields
sin x cos 2x − cos x sin 2 x = sin( x − 2 x ) = sin( − x ) = − sin x 24. Using the addition formula sin( A + B ) = sin A cos B + cos A sin B with A = 2 x and B = 3x yields sin 2x cos3x + cos 2x sin3x = sin(2x + 3x ) = sin5x
25. Observe that
( sin A − sin B ) + ( cos A − cos B ) − 2 = 2
2
sin2 A − 2 sin A sin B + sin2 B + cos 2 A − 2 cos A cos B + cos 2 B − 2 = sin A + cos A) + ( sin B + cos B ) − 2 [sin A sin B + cos A cos B ] − 2 = ( 2
2
2
2
=1
=1
−2 [sin A sin B + cos A cos B ] = −2 [ cos A cos B + sin A sin B ] = −2 cos( A − B ) 26. Observe that
( sin A + sin B ) + ( cos A + cos B ) − 2 = 2
2
sin2 A + 2 sin A sin B + sin2 B + cos 2 A + 2 cos A cos B + cos 2 B − 2 = sin A + cos A) + ( sin B + cos B ) + 2 [sin A sin B + cos A cos B ] − 2 = ( 2
2
=1
2
2
=1
2 [sin A sin B + cos A cos B ] = 2 [ cos A cos B + sin A sin B ] = 2 cos( A − B ) 27. cos ( 12 x ) sin ( 52 x ) + cos ( 52 x ) sin ( 12 x ) = sin ( 12 x + 52 x ) = sin3x 28. cos ( 50 ) cos x + sin ( 50 ) sin x = cos ( 50 − x )
449
Chapter 5
29. Observe that 2 − ( sin A + cos B ) − ( cos A + sin B ) = 2
2
2 − ( sin2 A + 2 sin A cos B + cos 2 B ) − ( cos 2 A + 2 cos A sin B + sin2 B ) 2 − ( sin2 A + cos 2 A) − ( sin2 B + cos 2 B ) − 2 [sin A cos B + cos A sin B ] = =1
=1
−2 [sin A cos B + cos A sin B ] = −2 sin( A + B )
30. Observe that 2 − ( sin A − cos B ) − ( cos A + sin B ) = 2
2
2 − ( sin2 A − 2 sin A cos B + cos 2 B ) − ( cos 2 A + 2 cos A sin B + sin2 B ) 2 − ( sin2 A + cos 2 A) − ( sin2 B + cos 2 B ) + 2 [sin A cos B − cos A sin B ] = =1
=1
2 [sin A cos B − cos A sin B ] = 2 sin( A − B )
31.
Using the addition formula tan( A − B ) = B = 23 yields
tan 49 − tan 23 = tan(49 − 23 ) = tan(26 ) 1 + tan 49 tan 23
32.
Using the addition formula tan( A + B ) =
B = 23 yields
tan A − tan B with A = 49 and 1 + tan A tan B
tan A + tan B with A = 49 and 1 − tan A tan B
tan 49 + tan 23 = tan(49 + 23 ) = tan(72 ) 1 − tan 49 tan 23
33.
π 3π tan − tan 8 8 = tan π − 3π = tan − π = −1 π 3π 8 8 4 1 + tan tan 8 8
450
Section 5.2
34.
7π π tan + tan 12 6 = tan 7π + π = tan 3π = −1 12 6 7π π 4 1 − tan tan 12 6
35.
1 1 Observe that cos(α + β ) = cos α cos β − sin α sin β = − − − sin α sin β . We need 3 4 to compute both sin α and sin β . We proceed as follows:
Since the terminal side of α is in QIII, we know that sin α < 0. As such, since 1 we are given that cos α = − , the 3 diagram is:
Since the terminal side of β is in QII, we know that sin β > 0. As such, since we 1 are given that cos β = − , the diagram is: 4
β α
From the Pythagorean Theorem, we have z 2 + ( −1)2 = 32 , so that z = −2 2. Thus, sin α = −
From the Pythagorean Theorem, we have z 2 + ( −1)2 = 4 2 , so that z = 15. Thus, sin β =
2 2 . 3
451
15 . 4
Chapter 5
Hence, 1 1 1 1 2 2 15 cos(α + β ) = − − − sin α sin β = − − − − 3 4 3 4 3 4
=
1 2 30 1 + 2 30 + = 12 12 12
36.
1 1 Observe that cos(α − β ) = cos α cos β + sin α sin β = − + sin α sin β . We need 3 4 to compute both sin α and sin β . We proceed as follows:
Since the terminal side of α is in QIV, we know that sin α < 0. As such, since 1 we are given that cos α = , the 3 diagram is as follows:
Since the terminal side of β is in QII, we know that sin β > 0. As such, since we are 1 given that cos β = − , the diagram is as 4 follows:
β
α
From the Pythagorean Theorem, we have z 2 + 12 = 32 , so that z = −2 2. Thus, sin α = −
2 2 . 3
From the Pythagorean Theorem, we have z 2 + ( −1)2 = 4 2 , so that z = 15. Thus, sin β =
452
15 . 4
Section 5.2
Hence, 1 1 1 1 2 2 15 cos(α − β ) = − + sin α sin β = − + − 3 4 3 4 3 4
=−
(
1 2 30 − 1 + 2 30 − = 12 12 12
)
37.
4 1 Observe that cos (α + β ) = cos α cos β − sin α sin β = cos α cos β − . We need to 5 2 compute both cos α and cos β . We proceed as follows:
Since the terminal side of α is in QII, we know that cos α < 0. As such, since we 4 are given that sin α = , the diagram is 5
Since the terminal side of β is in QI, we know that cos β > 0. As such, since we 1 are given that sin β = , the diagram is 2
From the Pythagorean Theorem, we have z 2 + 4 2 = 52 , so that z = −3. Thus, 3 cos α = − . 5
From the Pythagorean Theorem, we have z 2 + 12 = 2 2 , so that z = 3. Thus, cos β = −
Hence,
3 . 2
(
3 4 1 −3 3 − 4 − 3 3 + 4 4 1 3 cos (α + β ) = cos α cos β − = − = − = 10 10 5 2 5 2 5 2
453
)
Chapter 5
38.
−24 3 Observe that sin (α + β ) = sin α cos β + cos α sin β = sin α + sin β . We need 25 5 to compute both sin α and sin β . We proceed as follows:
Since the terminal side of α is in QI, we know that sin α > 0. As such, since we 3 are given that cos α = , the diagram is 5
From the Pythagorean Theorem, we have z 2 + 32 = 52 , so that z = 4. Thus, 4 sin α = . 5
Since the terminal side of β is in QII, we know that sin β > 0. As such, since we are −24 given that cos β = , the diagram is 25
From the Pythagorean Theorem, we have z 2 + 24 2 = 252 , so that z = 7. Thus, 7 sin β = . 25
−24 3 4 −24 3 7 −3 Hence, sin (α + β ) = sin α + sin β = + = 25 5 5 25 5 25 5
454
Section 5.2
3 1 39. Observe that sin(α − β ) = sin α cos β − cos α sin β = cos β − cos α . We need 5 5 to compute both cos α and cos β . We proceed as follows:
Since the terminal side of α is in QIII, we know that cos α < 0. As such, since 3 we are given that sin α = − , the 5 diagram is as follows:
Since the terminal side of β is in QI, we know that cos β > 0. As such, since we are 1 given that sin β = , the diagram is as 5 follows:
β
α
From the Pythagorean Theorem, we have z 2 + 32 = 52 , so that z = −4. 4 Thus, cos α = − . 5
From the Pythagorean Theorem, we have z 2 + 12 = 52 , so that z = 24 = 2 6. Thus, cos β =
2 6 . 5
Hence, −6 6 + 4 3 1 3 2 6 4 1 −6 6 4 + = sin(α − β ) = − cos β − cos α = − − − = 25 25 25 5 5 5 5 5 5
455
Chapter 5
3 1 40. Observe that sin(α + β ) = sin α cos β + cos α sin β = − cos β + cos α . We 5 5 need to compute both cos α and cos β . We proceed as follows:
Since the terminal side of α is in QIII, we know that cos α < 0. As such, since 3 we are given that sin α = − , the 5 diagram is as follows:
Since the terminal side of β is in QII, we know that cos β < 0. As such, since we are 1 given that sin β = , the diagram is as 5 follows:
β
α
From the Pythagorean Theorem, we have z 2 + ( −3)2 = 52 , so that z = −4. 4 Thus, cos α = − . 5
From the Pythagorean Theorem, we have z 2 + 12 = 52 , so that z = − 24 = −2 6. Thus, cos β = −
2 6 . 5
Hence, 6 6 −4 3 1 3 2 6 4 1 6 6 4 . − = sin(α + β ) = − cos β + cos α = − − + − = 5 5 5 25 25 25 5 5 5
456
Section 5.2
41.
tan α + tan β , we need to determine the values of 1 − tan α tan β cos α and sin β so that we can then use this information in combination with the given values of sin α and cos β in order to compute tan α and tan β . We proceed as follows:
Observe that since tan(α + β ) =
Since the terminal side of α is in QIII, we know that cos α < 0. As such, since 3 we are given that sin α = − , the 5 diagram is:
Since the terminal side of β is in QII, we know that sin β > 0. As such, since we are 1 given that cos β = − , the diagram is: 4
β α
From the Pythagorean Theorem, we have z 2 + ( −3)2 = 52 , so that z = −4. 4 As such, cos α = − , and so, 5 3 sin α − 5 3 = = tan α = cos α − 4 4 5
From the Pythagorean Theorem, we have z 2 + ( −1)2 = 4 2 , so that z = 15. 15 , and so 4 15 sin β 4 = − 15 tan β = = 1 cos β − 4
As such, sin β =
457
Chapter 5
Hence, tan α + tan β tan(α + β ) = = 1 − tan α tan β
=
3 3 − 4 15 − 15 4 4 = 3 4 + 3 15 1 − − 15 4 4
(
)
3 − 4 15 4 − 3 15 12 − 16 15 − 9 15 + 12(15) 192 − 25 15 ⋅ = = −119 16 − 9(15) 4 + 3 15 4 − 3 15
42. Using the values of tan α and tan β obtained in Exercise 41, we obtain
tan α − tan β tan(α − β ) = = 1 + tan α tan β
=
3 3 + 4 15 + 15 4 4 = 3 − 4 3 15 1 + − 15 4 4
(
)
3 + 4 15 4 + 3 15 12 + 16 15 + 9 15 + 12(15) 192 + 25 15 ⋅ = = −119 16 − 9(15) 4 − 3 15 4 + 3 15
43. cos( A + B ) cos A cos B − sin A sin B cos A cos B sin A sin B = = − sin A cos B sin A cos B sin A cos B sin A cos B cos A sin B = − = cot A − tan B sin A cos B 44. sin ( A + B ) sin A cos B + cos A sin B sin A cos B cos A sin B = = + cos A cos B cos A cos B cos A cos B cos A cos B sin A sin B = + = tan A + tan B cos A cos B 45.
π π π sin + θ = sin cos θ + cos sin θ = 1 ⋅ cos θ + 0 ⋅ sin θ = cos θ 2 2 2
46.
3π 3π 3π + θ = cos cos cos θ − sin sin θ = 0 ⋅ cos θ − ( −1) sin θ = sin θ 2 2 2
458
Section 5.2
47.
tan (π − θ ) =
tan π − tan θ 0 − tan θ − tan θ = = = − tan θ 1+ tan π tan θ 1+ 0 ⋅ tan θ 1
48.
π sin x tan x + tan +1 tan 1 cos x cos x + sinx x + π 4 = tan x + = = cos x ⋅ = 4 π 1 − tan x ⋅1 1 − sin x cos x cos x − sin x 1 − tanx tan cos x 4
49. cos ( 2 x ) = cos ( x + x ) = cos x cos x − sin x sin x = cos2 x − sin2 x
= (1 − sin2 x ) − sin2 x = 1 − 2 sin2 x
50. cos ( 2 x ) = cos ( x + x ) = cos x cos x − sin x sin x = cos2 x − sin2 x
= cos 2 x − (1 − cos 2 x ) = 2 cos 2 x − 1
51. Observe that sin( A + B ) + sin( A − B ) = (sin A cos B + sin B cos A ) + (sin A cos B − sin B cos A )
= 2 sin A cos B So, the given equation is an identity. 52. Observe that cos( A + B ) + cos( A − B ) = (cos A cos B − sin A sin B ) + (cos A cos B + sin A sin B )
= 2 cos A cos B So, the given equation is an identity.
π π π 53. The equation sin x − = cos x + is a conditional since it is false for x = , 2 2 2 for instance. π π π 54. The equation sin x + = cos x + is a conditional since it is false for x = , 2 2 2 for instance.
459
Chapter 5
55. Observe that sin 2 x = sin( x + x ) = sin x cos x + sin x cos x = 2 sin x cos x. So, the given equation is an identity. 56. Observe that cos 2x = cos( x + x ) = cos x cos x − sin x sin x = cos 2 x − sin2 x. So, the given equation is an identity. 57. The equation sin( A + B ) = sin A + sin B is a conditional since it is false for
A=B =
π 2
, for instance.
58. The equation cos( A + B ) = cos A + cos B is a conditional since it is false for
A=B =
π 2
, for instance.
59. Observe that
tan π + tan β 0 + tan β = = tan β . 1 − tan π tan β 1 − ( 0 ) tan β So, the given equation is an identity. tan(π + β ) =
60. Observe that
tan A − tan π tan A − 0 = = tan A. 1 + tan A tan π 1 + tan A ( 0 ) So, the given equation is an identity. tan( A − π ) =
460
Section 5.2
61. Using the addition formula sin( A + B ) = sin A cos B + cos A sin B with A = x and
B=
π 3
yields
π π π y = cos sin x + sin cos x = sin x + . 3 3 3 So, the graph is the graph of y = sin x shifted to the left
π
3
units, as seen below.
62. Using the addition formula sin( A − B ) = sin A cos B − cos A sin B with A = x and
B=
π 3
yields
π π π y = cos sin x − sin cos x = sin x − . 3 3 3 So, the graph is the graph of y = sin x shifted to the right
461
π
3
units, as seen below.
Chapter 5
63. Using the addition formula cos( A − B ) = cos A cos B + sin A sin B with A = x and
B=
π 4
yields
π π π π π y = sin x sin + cos x cos = cos x cos + sin x sin = cos x − . 4 4 4 4 4 So, the graph is the graph of y = cos x shifted to the right
π
4
units, as seen below.
64. Using the addition formula cos( A + B ) = cos A cos B − sin A sin B with A = x and
B=
π 4
yields
π π π π π y = sin x sin − cos x cos = − cos x cos − sin x sin = − cos x + . 4 4 4 4 4 So, the graph is the graph of y = cos x shifted to the left over the x-axis, as seen below.
462
π 4
units and then reflected
Section 5.2
65. Using the addition formula sin( A + B ) = sin A cos B + cos A sin B with A = x and B = 3x yields y = − sin x cos 3x − cos x sin3x = − [sin x cos 3x + cos x sin3x ] = − sin ( x + 3x ) = − sin 4 x
So, the general shape of the graph is that of y = sin x , but with period is
2π π = and 4 2
then reflected over the x-axis, as seen below.
66. Using the addition formula cos( A − B ) = cos A cos B + sin A sin B with A = x and B = 3x yields y = sin x sin3x + cos x cos 3x = cos x cos 3x + sin x sin3x = cos ( x − 3x )
= cos( −2x ) = cos 2 x So, the general shape of the graph is that of y = cos x, but the period is seen below.
463
2π = π , as 2
Chapter 5
67. Observe that sin( x + h ) − sin x sin x cos( h ) + sin( h ) cos x − sin x ( sin x cos( h ) − sin x ) + sin( h ) cos x = = h h h sin x ( cos( h ) − 1) + sin( h ) cos x sin x ( cos( h ) − 1) sin( h ) cos x = = + h h h cos( h ) − 1 sin( h ) sin( h ) 1 − cos( h ) = sin x + cos x = cos x − sin x h h h h 68. Observe that cos( x + h ) − cos x cos x cos( h ) − sin x sin( h ) − cos x = h h ( cos x cos( h ) − cos x ) − sin x sin( h ) = h cos x ( cos( h ) − 1) − sin x sin( h ) = h cos( h ) − 1 sin( h ) = cos x − sin x h h 1 − cos( h ) sin( h ) = − cos x − sin x h h sin( h ) 1 − cos( h ) = − sin x − cos x h h 69.
Observe that tan (θ 2 − θ1 ) =
tan θ 2 − tan θ1 m − m1 = 2 . 1 + tan θ 2 tan θ1 1 + m2 m1
70.
Observe that tan (θ1 − θ 2 ) =
tan θ1 − tan θ 2 m − m2 = 1 . 1 + tan θ1 tan θ 2 1 + m1m2
464
Section 5.2
71. E = A cos ( kz − ct ) = A cos ( kz ) cos ( ct ) + sin ( kz ) sin ( ct )
2π = n, an integer, then z = nλ , so that kz = knλ = kn = 2nπ . λ k Hence, sin( kz ) = 0 . So, the formula for E simplifies to E = A cos ( kz − ct ) = A cos ( kz ) cos ( ct ) .
If
z
72. E = A cos ( kz − ct ) = A cos ( kz ) cos ( ct ) + sin ( kz ) sin ( ct )
So, in particular, E (0) = A cos ( kz ) . 73. Consider this triangle:
a) h (θ ) = 15sin θ b) We have h (θ + 10° ) = 15sin (θ + 10° )
= 15 ( sin θ cos10° + cos θ sin10° ) = 15sin θ cos10° + 15 cos θ sin10°
θ 74. Consider this triangle:
a) x (θ ) = 15 cos θ b) We have x (θ + 10° ) = 15 cos (θ + 10° )
= 15 ( cos θ cos10° − sin θ sin10° ) = 15 cos θ cos10° − 15sin θ sin10°
θ 75.
π π π π π π T (t ) = 38 − 2.5 cos t − = 38 − 2.5 cos t cos + sin t sin 2 6 6 2 6 3 π = 38 − 2.5sin t 6
465
Chapter 5
76.
π π π π π C (t ) = 15.4 − 4.7 sin t cos + sin cos t = 15.4 − 4.7 cos t 2 2 4 4 4 77. tan( A + B ) ≠ tan A + tan B. Should have used tan( A + B ) =
tan A + tan B . 1 − tan A tan B
78. Tangent is an odd function, not an even function. Hence, the very first step 7π 7π should have been: tan − = − tan . The remaining calculations are correct. 6 6 79. False. Observe that cos (15 ) = cos ( 45 − 30 ) = cos ( 45 ) cos ( 30 ) + sin ( 45 ) sin ( 30 ) 2 3 2 1 6+ 2 = + = 4 2 2 2 2
while
2 3 2− 3 cos ( 45 ) − cos ( 30 ) = . − = 2 2 2 80. 3 π π π 1 , and sin = and it is clear that False. Since sin = 1, sin = 2 3 2 6 2 3 +1 1≠ , the given equality is not true. 2
81. True. Observe that sin ( 90 + x ) = sin 90 cos x + sin x cos 90 = cos x + 0 = cos x. 82. True. Observe that cos ( 90 + x ) = cos 90 cos x − sin 90 sin x = − sin x. 83. cos (α x ) cos ( β x ) + sin (α x ) sin ( β x ) = cos(α x − β x ) = cos (α − β ) x 84. cos (α x ) sin ( β x ) + sin (α x ) cos ( β x ) = sin (α + β ) x
466
Section 5.2
85. Approaches may vary slightly here, but they will result in equivalent answers as long as the identities are applied correctly. The final form of the simplification depends on how you initially choose to group the input quantities. For instance, the following is a legitimate simplification: sin ( A + B + C ) = sin ( ( A + B ) + C ) = sin( A + B ) cos C + sin C cos( A + B )
= [sin A cos B + sin B cos A] cos C + sin C [ cos A cos B − sin A sin B ]
= sin A cos B cos C + sin B cos A cos C + cos A cos B sin C − sin A sin B sin C 86. Approaches may vary slightly here, but they will result in equivalent answers as long as the identities are applied correctly. The final form of the simplification depends on how you initially choose to group the input quantities. For instance, the following is a legitimate simplification: cos ( A + B + C ) = cos ( ( A + B ) + C ) = cos( A + B ) cos C − sin C sin( A + B )
= [ cos A cos B − sin B sin A] cos C − sin C [ sin A cos B + sin B cos A]
= cos A cos B cos C − sin B sin A cos C − sin A cos B sin C − sin B cos A sin C 87. Observe that sin( A − B ) = sin A cos B − sin B cos A. We seek values of A and B such that the right-side equals sin A − sin B. Certainly, if we choose values of A and B such that cos B = cos A = 1, then this occurs. And this occurs precisely when A and B are integer multiples of 2π (and they need not be the same multiple of 2π !). So, the solutions are B = 2 mπ , A = 2nπ , where n and m are integers. 88. Observe that sin( A + B ) = sin A cos B + sin B cos A. We seek values of A and B such that the right-side equals sin A + sin B. Certainly, if we choose values of A and B such that cos B = cos A = 1, then this occurs. And this occurs precisely when A and B are integer multiples of 2π (and they need not be the same multiple of 2π !). So, the solutions are B = 2 mπ , A = 2nπ , where n and m are integers.
467
Chapter 5
89. 2 2 ( sin x + cos y ) − 1 + cot x tan y = ( sin x + cos y ) − sin x cos y + sin x cos y cot x tan y sin x cos y sin x cos y cos x sin y 2 ( sin x + cos y ) − sin x cos y + sin x cos y sin x cos y = sin x cos y
( sin x + cos y ) − sin x cos y + cos x sin y = 2
sin x cos y
sin x + 2 sin x cos y + cos 2 y − sin x cos y + cos x sin y = sin x cos y 2
sin( x + y ) sin2 x + cos 2 y + sin x cos y + cos x sin y = sin x cos y
=
sin2 x cos 2 y sin( x + y ) + + sin x cos y sin x cos y sin x cos y
=
sin x cos y sin ( x + y ) + + cos y sin x sin x cos y
90. cot β − cos (α + β ) sec α csc β = cot β − ( cos α cos β − sin α sin β ) 1 1 cos α sin β 1 1 + sin α sin β ⋅ cos α sin β
= cot β − cos α cos β ⋅
= cot β − cot β + tan α = tan α
468
1 1 cos α sin β
Section 5.2
91. a. The following is the graph of sin1 1 − cos1 y1 = cos x − sin x , 1 1 along with the graph of y = cos x (dotted graph):
b. The following is the graph of sin 0.1 1 − cos 0.1 y2 = cos x − sin x , along 0.1 0.1 with the graph of y = cos x (dotted graph):
sin 0.01 1 − cos 0.01 c. The following is the graph of y3 = cos x − sin x , along with 0.01 0.01 the graph of y = cos x (dotted graph):
Refer back to Exercise 67 for a development of this difference quotient formula. Looking at the sequence of graphs in parts a – c, it is clear that this sequence of graphs better approximates the graph of y = cos x as h → 0.
469
Chapter 5
92. a. The following is the graph of sin1 1 − cos1 y1 = − sin x − cos x , 1 1 along with the graph of y = − sin x (dotted graph):
b. The following is the graph of sin 0.1 1 − cos 0.1 y2 = − sin x − cos x , 0.1 0.1 along with the graph of y = − sin x (dotted graph):
sin 0.01 1 − cos 0.01 c. The following is the graph of y3 = − sin x − cos x , along 0.01 0.01 with the graph of y = − sin x (dotted graph):
Refer back to Exercise 68 for a development of this difference quotient formula. Looking at the sequence of graphs in parts a – c, it is clear that this sequence of graphs better approximates the graph of y = − sin x as h → 0.
470
Section 5.3 Solutions -----------------------------------------------------------------------------------1.
2.
1 and cos x < 0, the 5 terminal side of x must be in QII. The diagram is as follows:
1 and cos x < 0, the 5 terminal side of x must be in QII. The diagram is as follows:
Since sin x =
Since sin x =
5
5
From the Pythagorean Theorem, we have z 2 + 12 =
( 5 ) , so that z = −2. 2
From the Pythagorean Theorem, we have z 2 + 12 =
2 . Therefore, 5 sin 2 x = 2 sin x cos x
As such, cos x = −
( 5 ) , so that z = −2. 2
2 . Therefore, 5 cos 2 x = cos 2 x − sin2 x
As such, cos x = −
4 1 2 = 2 − = −5 5 5
2
2
4 1 3 2 1 = − − = 5 −5 = 5 5 5
471
Chapter 5
3.
4.
5 and sin x < 0, the terminal 13 side of x must be in QIV. The diagram is as follows:
5 and sin x < 0, the 13 terminal side of x must be in QIII. The diagram is as follows:
Since cos x =
From the Pythagorean Theorem, we have z 2 + 52 = 132 , so that z = −12. 12 As such, sin x = − , so that 13 12 − sin x 12 tan x = = 13 = − . 5 cos x 5 13 Therefore, 12 24 2− − 2 tan x 5 5 tan 2x = = = 2 2 119 1 − tan x 12 − 1− − 25 5 =
120 119
Since cos x = −
From the Pythagorean Theorem, we have z 2 + ( −5)2 = 132 , so that z = −12. 12 As such, sin x = − , so that 13 12 − sin x 12 tan x = = 13 = . 5 cos x − 5 13 Therefore, 12 24 2 2 tan x 5 tan 2x = = 2 = 5 2 119 1 − tan x 12 − 1− 25 5 = −
472
120 119
Section 5.3
6.
5.
−12 12 3π −12 12 = and π < x < , the Since tan x = = and −5 5 2 −5 5 diagram is as follows: 3π π < x < , the diagram is as follows: 2 Since tan x =
From the Pythagorean Theorem, we have ( −5)2 + ( −12)2 = z 2 , so that z = 13. 12 5 As such, sin x = − and cos x = − . 13 13 Therefore, 12 5 sin 2 x = 2 sin x cos x = 2 − − 13 13 =
120 169
From the Pythagorean Theorem, we have ( −5)2 + ( −12)2 = z 2 , so that z = 13. 12 5 As such, sin x = − and cos x = − . 13 13 Therefore, cos 2 x = cos 2 x − sin2 x 2
2
119 5 12 = − −− = − 169 13 13
473
Chapter 5
7. Since sec x = 5, it follows that 1 1 . Since also = 5 and so, cos x = cos x 5 sin x > 0, the terminal side of x must be in QI. The diagram is as follows:
8. Since sec x = 3 , it follows that 1 1 . Since = 3 and so, cos x = cos x 3 also sin x < 0, the terminal side of x must be in QIV. The diagram is as follows:
5
3
From the Pythagorean Theorem, we have z 2 + 12 =
( 5 ) , so that z = 2. 2
2 , so that 5 2 sin x tan x = = 5 = 2. 1 cos x 5 Therefore, 2 tan x 2(2) 4 4 tan 2 x = = = = − . 2 2 1 − tan x 1 − 2 −3 3
From the Pythagorean Theorem, we have z 2 + 12 =
As such, sin x =
( 3 ) , so that z = − 2. 2
2 , so that 3 2 − sin x 3 = − 2. tan x = = 1 cos x 3 Therefore,
As such, sin x = −
tan 2 x =
= 2 2.
474
( (
) )
2 − 2 2 tan x −2 2 = = 2 2 1 − tan x 1 − − 2 1− 2
Section 5.3
9. Since csc x = −2 5, it follows that
1 1 5 . =− = −2 5 and so, sin x = − 10 sin x 2 5 Since also cos x < 0 , the terminal side of x must be in QIII. The diagram is as follows:
10. Since csc x = − 13 , it follows that 1 1 = − 13 and so, sin x = − . sin x 13 Since also cos x > 0 , the terminal side of x must be in QIV. The diagram is as follows:
− 5
13
From the Pythagorean Theorem, we have
(
z + − 5 2
) = (10 ) , so that z = − 95 . 2
2
z 2 + ( −1) = 2
95 . Therefore, 10 5 95 sin 2 x = 2 sin x cos x = 2 − − 10 10
As such, cos x = −
=
From the Pythagorean Theorem, we have
10 19 19 = 100 10
2
12 . Therefore, 13 1 12 sin 2 x = 2 sin x cos x = 2 − 13 13
As such, cos x =
=−
475
( 13 ) , so that z = 12 .
2 12 4 3 = − 13 13
Chapter 5
11. Since csc x < 0 , it follows that 12 sin x < 0 . Since also cos x = − , the 13 terminal side of x must be in QIII. The diagram is as follows:
12 , it 13 follows that cos x < 0 , and so the terminal side of x must be in QII. The diagram is as follows:
From the Pythagorean Theorem, we have 2 z 2 + ( −12 ) = 132 , so that z = −5 .
From the Pythagorean Theorem, we have z 2 + 12 2 = 132 , so that z = −5 . 5 As such, cos x = − . Therefore, 13 1 1 csc 2 x = = sin 2 x 2 sin x cos x 1 1 169 = = = − 120 120 12 5 2 − − 169 13 13
12. Since cot x < 0 and sin x =
5 , so that 13 5 − sin x 5 tan x = = 13 = . cos x − 12 12 13 Therefore, 1 1 1 − tan2 x cot 2 x = = = 2 tan x tan 2 x 2 tan x 2 1 − tan x As such, sin x = −
2
5 1− 119 6 119 12 = = ⋅ = 144 5 120 5 2 12
476
Section 5.3
13. Using tan 2A =
2 tan A with 1 − tan2 A
14. Using tan 2A =
A = 15 yields 2 tan15 = tan ( 2 ⋅15 ) = tan30 1 − tan2 15 1 sin30 2 = 3 = = cos 30 3 3 2
A=
yields
π 2 sin 4 = 2 =1 = π 2 cos 4 2 16. Using sin 2 A = 2 sin A cos A with A = 15 yields 1 sin (15 ) cos (15 ) = 2 sin (15 ) cos (15 ) 2 1 = sin ( 2 ⋅15 ) 2 1 = sin ( 30 ) 2 11 1 = = 22 4
π
yields 8 π π 1 π π sin cos = 2 sin cos 8 8 2 8 8 1 π 1 2 = sin = 2 4 2 2
=
8
π 2 tan 8 = tan 2 ⋅ π = tan π 8 π 4 1 − tan2 8
15. Using sin 2 A = 2 sin A cos A with
A=
π
2 tan A with 1 − tan2 A
2 4
17. Using cos 2A = cos 2 A − sin2 A with A = 2 x yields cos 2 2x − sin2 2x = cos(2 ⋅ 2x ) = cos 4 x
18. Using cos 2 A = cos 2 A − sin2 A with A = x + 2 yields cos 2 ( x + 2) − sin2 ( x + 2) = cos(2 ⋅ ( x + 2))
= cos(2 x + 4) 19. 1 − 2 sin2 (105° ) = cos ( 2 ⋅105° ) = −
3 2
477
Chapter 5
20. −4 sin
21.
2 3π 3π 3π 3π 3π cos = −2 2 sin cos = −2 sin 2 ⋅ = −2 = − 2 8 8 8 8 8 2
tan ( 4x ) tan ( 8x ) 1 2 tan ( 4x ) 1 = = tan ( 2 ⋅ 4 x ) = 2 2 1 − tan ( 4 x ) 2 1 − tan ( 4x ) 2 2
22. 1 − 2 cos 2 (195° ) = − cos ( 2 ⋅195° ) = − cos ( 390° ) = − cos30° = −
3 2
23. ( sinθ − cos θ ) = sin θ − 2 sin θ cos θ + cos θ 2
2
2
2 = sin θ + cos 2 θ − 2 sin θ cos θ = 1 − sin 2θ sin2θ
1
x sin 2 − − 1 cos x sin x ( ) = 1− cos x + sin2 x = 2 sin2 x = sin x = tan x 1 − cos 2x = 24. sin 2x 2 sin x cos x 2 sin x cos x 2 sin x cos x cos x 2
2
25. Observe that
csc 2 A =
2
1 1 1 1 1 1 = = = csc A sec A . sin 2 A 2 sin A cos A 2 sin A cos A 2
26. Observe that 1 1 1 − tan2 A 1 1 tan2 A 1 cot 2 A = = = = − = [ cot A − tan A] . 2 tan A tan 2A 2 tan A 2 tan A tan A 2 1 − tan2 A 27. Observe that ( sin x − cos x )( cos x + sin x ) = sin2 x − cos 2 x = − cos 2 x − sin2 x = − cos 2x . 28. Observe that 2 sin x cos x ( sin x + cos x ) = sin2 x + 2 sin x cos x + cos2 x = ( sin2 x + cos2 x ) + 2 = sin2 x =1
= 1 + sin 2 x
478
Section 5.3
29. Note that starting with the right-side is easier this time. Indeed, observe that 2 2 2 2 1 + cos 2 x 1 + ( cos x − sin x ) (1 − sin x ) + cos x cos 2 x + cos 2 x 2 cos 2 x = = = = = cos 2 x 2 2 2 2 2 30. Note that starting with the right-side is easier this time. Indeed, observe that 2 2 2 2 1 − cos 2 x 1 − ( cos x − sin x ) (1 − cos x ) + sin x sin2 x + sin2 x 2 sin2 x = = = = = sin2 x 2 2 2 2 2 31. Observe that cos 4 x − sin 4 x = ( cos 2 x − sin2 x )( cos 2 x + sin2 x ) = cos 2x =1
32. Note that starting with the right-side is easier this time. Indeed, observe that 1 1 2 1 − sin2 (2 x ) = 1 − [ 2 sin x cos x ] 2 2 = 1 − 2 sin2 x cos 2 x
= ( sin2 x + cos 2 x ) − 2 sin2 x cos 2 x 2
= sin 4 x + 2 sin2 x cos 2 x + cos 4 x − 2 sin2 x cos 2 x = sin 4 x + cos 4 x 33. Observe that 8sin2 x cos 2 x = 2 ⋅ 4 sin2 x cos 2 x
= 2 [ 2 sin x cos x ] = 2 [sin 2x ] 2
2
= 2 1 − cos 2 2 x = 2 − 2 cos 2 2 x = 1 + 1 − 2 cos 2 2 x
= ( cos 2 2 x + sin2 2 x ) + 1 − 2 cos 2 2 x = 1 − cos 2 2x + sin2 2 x = 1 − cos 2 2 x − sin2 2x = 1 − cos(2 ⋅ 2x ) = 1 − cos 4x
479
Chapter 5
34. Observe that
2 cos 2 ( 2x ) − 2 cos 4 ( 2x ) + 2 cos 2 ( 2 x ) sin2 ( 2x ) = 2 cos 2 ( 2 x ) 1 − cos 2 ( 2x ) + 2 cos 2 ( 2x ) sin2 ( 2x ) = 2 cos 2 ( 2 x ) sin2 ( 2 x ) + 2 cos 2 ( 2 x ) sin2 ( 2 x ) = 4 cos 2 ( 2 x ) sin2 ( 2 x ) = 2 cos ( 2 x ) sin ( 2x )
2
= [sin(4x )]
2
= sin2 (4 x )
35. Note that starting with the right-side is easier this time. Indeed, observe that −2 sin2 x −2 sin2 x −2 sin2 x 1 1 −2 sin x csc 2 x = = = =− = − sec 2 x 2 2 2 2 2 sin 2 x [ 2 sin x cos x ] 2 cos x 2 4 sin x cos x 2
2
36. Note that starting with the right-side is easier this time. Indeed, observe that sec x csc x 1 1 1 = = = 1 1 cos 2 x cos x sin x cos 2 x [2 cos x sin x ] cos 2x sin 2x cos 2x 2 2 1 1 4 = = = = 4 csc 4 x 1 1 1 x sin 4 ⋅ ⋅ [ 2 sin 2x cos 2 x ] sin 4 x 2 2 4 37. Note that starting with the right-side is easier this time. Indeed, observe that sin x ( 4 cos 2 x − 1) = sin x ( 3cos 2 x + cos 2 x − 1)
(
= sin x 3cos 2 x − (1 − cos 2 x ) = sin x ( 3cos 2 x − sin2 x )
)
= 3sin x cos 2 x − sin3 x
= sin x ( cos 2 x − sin 2 x ) + 2 sin x cos 2 x = sin x cos 2 x + (2 sin x cos x ) cos x = sin x cos 2 x + sin 2 x cos x = sin( x + 2 x ) = sin3x
480
Section 5.3
38. Observe that
tan x + tan 2 x = tan3x = tan( x + 2 x ) = 1 − tan x tan 2 x tan x (1 − tan2 x ) + 2 tan x =
1 − tan2 x (1 − tan2 x ) − 2 tan2 x
2 tan x 1 − tan2 x 2 tan x 1 − tan x 2 1 − tan x tan x +
3tan x − tan3 x tan x ( 3 − tan x ) = = 1 − 3tan2 x 1 − 3tan2 x 2
1 − tan2 x 39. Note that starting with the right-side is easier this time. Indeed, observe that 2 2 2 sin x cos x − 4 sin3 x cos x = 2 sin x cos x (1 − 2 sin2 x ) = 2 sin x cos x 1 − sin x − sin x =cos2 x 1 = sin 2 x ( cos 2 x − sin2 x ) = sin 2 x cos 2 x = ( 2 sin 2 x cos 2 x ) 2 1 = sin 4 x 2 40. Note that starting with the right-side is easier this time. Indeed, observe that ( cos 2x − sin 2x )( cos 2x + sin 2x ) = ( cos2 2x − sin2 2x ) = cos ( 2 ⋅ 2x ) = cos 4x 41. sin 2 x ( 2 − 4 sin2 x ) = 2 sin 2 x (1 − 2 sin2 x ) = 2 sin 2 x cos 2 x = sin 4x 42. 1 − sin2 2 x − 4 ( sin x cos x ) = 1 − sin2 2 x − ( 2 sin x cos x ) = 1 − sin2 2 x − ( sin 2 x ) 2
2
= 1 − 2 sin2 2 x = cos 4 x
43. tan 4 x =
sin 4 x sin ( 2 ⋅ 2x ) 2 sin 2 x cos 2 x 4 sin x cos x cos 2 x = = = cos 4 x cos ( 2 ⋅ 2 x ) 1 − 2 sin2 2 x 1 − 2 sin2 2 x
481
2
Chapter 5
44.
cos 6x = cos ( 2 ⋅ 3x ) = 1 − 2 sin2 3x = 1 − 2 [sin3x ] = 1 − 2 [sin( x + 2x )] 2
2
= 1 − 2 [ cos x sin 2 x + cos 2 x sin x ] = 1 − 2 [ cos x(2 sin x cos x ) + cos 2 x sin x ] 2
= 1 − 2 2 sin x cos 2 x + cos 2 x sin x 45. Before graphing, observe that sin 2 x 2 sin x cos x y= = 1 − cos 2 x 1 − ( cos 2 x − sin2 x )
=
2
2
46. Before graphing, observe that 2 tan x 2 tan x y= = 2 2 − sec x 2 − (1 + tan2 x )
2 sin x cos x ( cos x + sin2 x ) − ( cos2 x − sin2 x )
=
2
2 sin x cos x cos x = = = cot x 2 sin2 x sin x So, a snapshot of the graph is as follows:
2 tan x = tan 2x 1 − tan2 x
π
and 2 the vertical asymptotes occur when (2 n + 1)π cos 2 x = 0 , namely when 2 x = , 2 (2 n + 1)π and so x = . The graph is as 4 follows: So, the period of this function is
482
Section 5.3
47. Before graphing, observe that =1 cos x sin x cos x + sin x + 2 2 cot x + tan x sin x cos x cos x + sin x 1 sin x cos x = = y= = = = sec 2 x 2 2 2 2 cot x − tan x cos x − sin x cos x − sin x cos x − sin x cos 2x = cos2 x sin x cos x sin x cos x 2
So, the period of this function is
2
2π = π . The graph is as follows: 2
48. Before graphing, observe that 1 1 1 1 1 ⋅ = = csc 2 x y = tan x cot x sec x csc x = ⋅ 2 2 cos x sin x sin 2 x =1 2π So, the period of this function is = π . The graph is as follows: 2
483
Chapter 5
49.
1 − cos 2t 1 cos 2t C (t ) = 12 − 20 = 12 − 20 − = 12 − 10 + 10 cos 2t = 2 + 10 cos 2t 2 2 2
50. 1 + cos 2t 1 cos 2t S (t ) = 0.098 + 0.387 = 0.098 + + 0.387 2 2 2 = 0.049 + 0.049 cos 2t + 0.387 = 0.436 + 0.049 cos 2t
51.
d = 13 − 12 cos 2θ = 13 − 12 ( 2 cos 2 θ − 1) = 25 − 24 cos 2 θ
52.
d = 41 − 40 cos 2θ = 41 − 40 (1 − 2 sin2 θ ) = 1 + 80 sin2 θ
53. Observe that T = 98.6 + 4 sin2 t
1 − cos 2t = 98.6 + 4 2 = 100.6 − 2 cos 2t
484
Section 5.3
54. Observe that T = 98.6 + 6 sin t cos t
= 98.6 + 3 ( 2 sin t cos t ) = 98.6 + 3sin 2t
55. Observe that 1 F = 500 sin 60 + (200)2 750 (1 − cos120 ) + 3.75sin120 2 3 40,000 3 1 = 500 750 1 − − + 3.75 + 2 2 2 2 = 250 3 + 22,500,000 + 37,500 3 ≈ 22,565,385 lbs.
56. Observe that
1 F = 500 sin 75 + (200)2 750 (1 − cos150 ) + 3.75sin150 . 2 You can immediately enter this into the calculator to get the estimate indicated below. However, we simplify the various quantities using addition formulas first in order to provide the exact (not just approximate) answer. To this end, observe that sin 75 = sin ( 30 + 45 ) = sin30 cos 45 + cos 30 sin 45 2+ 6 1 2 3 2 = + = 4 2 2 2 2
485
Chapter 5
cos 75 = cos ( 30 + 45 ) = cos30 cos 45 − sin30 sin 45
3 2 1 2 6− 2 = − = 4 2 2 2 2 As such, we use this information to now compute the following: cos150 = cos ( 2 ⋅ 75 ) = cos 2 75 − sin2 75 2
6− 2 6+ 2 = − 4 4 =
2
(6 − 2 12 + 2 ) − (6 + 2 12 + 2 ) = −4 12 = − 3
16 16 sin150 = sin ( 2 ⋅ 75 ) = 2 sin 75 cos 75
2
6 + 2 6 − 2 2(6 − 4) 1 = 2 = = 4 4 16 4
So, we have
2+ 6 1 3 1 2 F = 500 + (200) 750 1 + + 3.75 4 2 4 2 = 125 2 + 125 6 + 7,500,000 3 + 15,018,750 ≈ 28,009,614 lbs. 57. Should use sin x = −
assuming that sin x < 0 .
59. False. Let A =
π 4
2 2 since we are 3
58. Note that tan 2 x =
sin 2 x , not cos 2 x
sin 2x . cos x 60. False. Let A = π . Observe that cos ( 2 ⋅ π ) = 1
. Observe that
π
π sin 2 ⋅ = sin = 1 4 2 π sin 4 ⋅ = sin (π ) = 0 4 Since 1 + 1 ≠ 0 , the statement is false.
cos ( 4 ⋅ π ) = 1 Since 1 − 1 ≠ 1 , the statement is false.
486
Section 5.3
61. False. Note that tan 2x =
62. False. Note that sin 2 x = 2 sin x cos x . For instance, there exist x values for which sin x > 0 and cos x < 0 (namely
2 tan x . 1 − tan2 x
π π Take any x ∈ , . For such x, 4 2 tan x > 1 , so that 1 − tan2 x < 0 . So, for these values of x, tan 2 x < 0 .
π
< x < π ). Thus, for these x values, 2 sin 2 x < 0 .
63. For any integer n, we have 64. Observe that nπ nπ nπ 1 nπ nπ 1 − cos 2 ⋅ sin cos = 2 sin cos 2 2 2 2 2 2 sin2 nπ = 2 2 1 nπ = sin 2 ⋅ 1 − cos ( nπ ) 0, n is even 2 2 = = 2 1 1, n is odd = sin ( nπ ) = 0 2 65. Observe that
2 tan 2 x 1 − tan2 2 x 2 tan x 4 tan x 2 2 1 − tan x 1 − tan2 x = = 2 2 2 tan x 1 − tan2 x ) − 4 tan2 x ( 1− 2 2 1 − tan x 1 − tan2 x
tan 4 x = tan(2 ⋅ 2 x ) =
=
4 tan x (1 − tan2 x )
(
)
(1 − tan x ) − 4 tan x 2
2
2
66. Using the odd identity for tangent, together with Exercise 65, we find that 4 tan x (1 − tan2 x ) 4 tan x ( tan2 x − 1) = tan ( −4 x ) = − tan(4 x ) = − 2 2 (1 − tan2 x ) − 4 tan2 x (1 − tan2 x ) − 4 tan2 x 67. One cannot verify the identity for this value of x since it does not belong to the domains of the functions involved.
487
Chapter 5
68. One cannot verify the identity for this value of x since it does not belong to the domains of the functions involved. 69. Observe that
sin 2 x = sin x 2 sin x cos x = sin x 2 sin x cos x − sin x = 0 sin x ( 2 cos x − 1) = 0 sin x = 0 or 2 cos x − 1 = 0
π 5π The x-values in [ 0,2π ] that satisfy these equations are x = 0, , π , . 3 3 70. Observe that
cos 2 x = cos x 2 cos 2 x − 1 = cos x 2 cos 2 x − cos x − 1 = 0
( 2 cos x + 1)( cos x − 1) = 0 1 or cos x = 1 2 2π 4π The x-values in [ 0,2π ] that satisfy these equations are x = 0, , . 3 3 cos x = −
71. Consider the following graphs:
72. Consider the following graphs:
Observe that y1 (dotted graph) is a good approximation of y2 on [ −1,1].
Observe that y1 (dotted graph) is a good approximation of y2 on [ −1,1].
488
Section 5.3
73. Consider the following graphs:
Since the graphs do not coincide, the given equation is not an identity. 74. Consider the following graphs:
From the picture, it seems as though the given equation is an identity. But, one cannot prove an identity by inspection alone. The formal proof is as follows: sin2 x + cos 2 x ) − 2 sin2 x − 2 sin x cos x ( 1 − 2 sin2 x − 2 sin x cos x = 2 sin x cos x ( cos 2 x − sin2 x ) 2 sin x cos x ( cos 2 x − sin2 x )
( cos x − sin x ) − 2 sin x cos x = ( 2 sin x cos x ) ( cos x − sin x ) 2
2
2
=
cos 2 x − sin 2 x ( sin 2x )( cos 2x )
2
= csc 2 x sec 2 x ( cos 2 x − sin 2x )
489
Section 5.4 Solutions ----------------------------------------------------------------------------------1. 1 − cos A A Use sin = with A = 30 to obtain 2 2 3 1− − 30 1 cos 30 2 = 2− 3 = sin (15 ) = sin = = 2 2 4 2
2− 3 . 2
2. 1 + cos A A Use cos = with A = 45 to obtain 2 2 2 1+ + 45 1 cos 45 2 = 2+ 2 = cos ( 22.5 ) = cos = = 2 2 4 2
2+ 2 . 2
3. 1 + cos A 11π A Use cos = − with A = to obtain 2 6 2 11π 3 11π 1 + cos 1+ 11π 6 =− 6 =− 2 = − 2+ 3 = − 2+ 3 . cos = cos 2 2 4 2 12 2
4.
π 1 − cos A A Use sin = with A = to obtain 2 4 2 π 2 π 1 − cos 1− π 4 = 2 = 2− 2 = sin = sin 4 = 2 2 4 8 2
490
2− 2 . 2
Section 5.4
5. 1 + cos A A Use cos = with A = 150 to obtain 2 2 3 1− + 150 1 cos150 2 = 2− 3 = cos ( 75 ) = cos = = 2 2 4 2
2− 3 . 2
6. 1 − cos A A Use sin = with A = 150 to obtain 2 2 3 1+ − 150 1 cos150 2 = 2+ 3 = sin ( 75 ) = sin = = 2 2 4 2
2+ 3 . 2
Alternatively, you could use Exercise 5 together with the identity sin A = 1 − cos 2 A with A = 75 to obtain 2
2− 3 2− 3 4−2+ 3 2+ 3 = 1− sin 75 = 1 − cos 75 = 1 − = = 2 4 4 2
2
7. 1 + cos A 23π A Use cos = with A = to obtain 2 6 2 3 1+ 23π 23π + 1 cos 6 2 = 2+ 3 = cos ( 2312π ) = cos 6 = = 2 2 4 2
2+ 3 . 2
8. 1 − cos A 9π A Use sin = − with A = to obtain 2 4 2
1 − cos sin ( 98π ) = sin = − 2 2 9π 4
9π 4
=−
491
1−
2 2 = − 2− 2 = − 2− 2 . 2 4 2
Chapter 5
9. 1 + cos A 3π A Use cos = with A = to obtain 2 4 2
2 1− 3π 3π + 1 cos 4 2 = 2− 2 = = cos ( 38π ) = cos 4 = 2 2 4 2
2− 2 . 2
10. 1 − cos A 7π A Use sin = with A = to obtain 2 6 2 1 − cos 7π sin = sin = 2 12 2 7π 6
7π 6
=
3 2 = 2+ 3 = 2 4
1+
2+ 3 . 2
11. 1 − cos A A Use tan = with A = 135 to obtain 1 + cos A 2
135 1 − cos135 tan ( 67.5 ) = tan . = 1 + cos135 2 Since cos135 = − cos 45 = −
2 , the above further simplifies to 2
2 1− − 2 tan ( 67.5 ) = = 2 1+ − 2
2+ 2 2+ 2 2+ 2 4+4 2 +2 2 = ⋅ = = 4−2 2− 2 2+ 2 2− 2 2
492
3+2 2
Section 5.4
12. 1 − cos A A Use tan = − with A = 202.5 to obtain 1 + cos A 2
405 1 − cos 405 tan ( 202.5 ) = tan . = − 1 + cos 405 2 Since cos 405 = cos ( 360 + 45 ) = cos ( 45 ) =
tan ( 202.5 ) =
2 2 = 2 1+ 2
=
3−2 2
13. Observe that
1−
2− 2 2 = 2+ 2 2
2 , the above further simplifies to 2
2− 2 2− 2 = ⋅ 2+ 2 2− 2
4−4 2 +2 4−2
1 1 9π 9π sec − = (1) = sec = 9π 8 8 cos 9π 4 8 cos 2
π 2 9π π . Hence, using the formula Also, note that cos = cos 2π + = cos = 4 4 4 2 9π 1 + cos A A cos = − with A = gives 2 4 2 9π 2 1 + cos 1+ 9π 4 =− 2 = − 2+ 2 . cos 4 = − 2 2 2 2 1 2 9π Using this in (1) then yields: sec − = − = 8 2+ 2 2+ 2 − 2
493
Chapter 5
14. 1 + cos A 9π A First, note that using the formula cos = − with A = yields 2 4 2 9π 2 1 + cos 9π 1+ 9π 4 =− 4=− 2 = − 2+ 2 . cos = cos 2 2 2 2 8 9π Now, using sin A = − 1 − cos 2 A with A = subsequently yields 8 1 1 −1 9π csc = = = 2 8 sin 9π 9π 2+ 2 − 1 − cos 2 1 − − 8 8 2
−1
= 1−
2+ 2 4
=
−1 2− 2 4
=
−2 2− 2
15. 1 − cos A 13π A , we obtain First, note that using tan = with A = 1 + cos A 4 2 1 1 1 13π cot = = (1) = 13π 8 tan 13π 13 π 1 − cos 4 4 8 tan 2 1 + cos 13π 4
( (
494
) )
Section 5.4
(
Next, observe that cos 13π subsequently yields: 13π cot = 8
=
4
) = cos ( 2π + 5π 4 ) = cos (5π 4 ) = − 22 . Using this in (1)
−1
2 1− − 2 2 1+ − 2 −1 4+4 2 +2 4−2
=
=
−1 2+ 2 2 2− 2 2
=
−1 2+ 2 2− 2
−1
=
2+ 2 2+ 2 ⋅ 2− 2 2+ 2
−1 3+2 2
1 − cos A A Note: Recall that there are two other formulae equivalent to tan = , 1 + cos A 2 and either one could be used in place of this one to obtain an equivalent (but perhaps sin A 13π A with A = yields different looking) result. For instance, using tan = 4 2 1 + cos A 1 1 1 13π cot = = (2) = 13π 8 tan 13π sin 13π 4 4 8 tan 2 13π 1 + cos 4 Since 2 cos 13π = cos 2π + 5π = cos 5π = − 4 4 4 2 2 sin 13π = sin 2π + 5π = sin 5π = − 4 4 4 2 we can use these in (2) to obtain 1 2− 2 2− 2 2 2 2 −2 13π = = ⋅ = = 1− 2 . cot = −2 − 2 − 2 2 2 8 − 2 2 1− 2 Even though the forms of the two answers are different, they are equivalent.
(
( (
) )
( (
) )
495
( ) ( )
(
)
)
Chapter 5
16. 1 − cos A 7π A First, note that using tan = − with A = , we obtain 1 + cos A 4 2 1 1 1 7π (1) cot = = = 7π 8 tan 7π 7π 1 cos − 4 4 8 tan 2 − 7 π 1 + cos 4
( (
(
Next, observe that cos 7π 7π cot = 8
)4 = cos (π 4 ) = 22 . Using this in (1) subsequently yields:
1 2 1− 2 − 2 1+ 2
=
1
= −
) )
4−4 2 +2 4−2
1 2− 2 2 − 2+ 2 2 =
1
= −
2− 2 2+ 2
1
= −
2− 2 2− 2 ⋅ 2+ 2 2− 2
−1 3−2 2
1 − cos A A Note: Recall that there are two other formulae equivalent to tan = − , 1 + cos A 2 and either one could be used in place of this one to obtain an equivalent (but perhaps sin A 7π A with A = yields different looking) result. For instance, using tan = 4 2 1 + cos A 1 1 1 7π (2) cot = = = 7π 8 tan 7π sin 7π 4 4 8 tan 2 7 π 1 + cos 4
(
496
(
)
)
Section 5.4
Since
(
)
( )
(
)
( )
2 2 = cos π = sin 7π = − sin π = − 4 4 4 4 2 2 we can use these in (2) to obtain 1 2+ 2 2+ 2 2 2 2 +2 7π = = ⋅ = = − 1+ 2 . cot = −2 − 2 − 2 2 2 8 − 2 2 1+ 2 Even though the forms of the two answers are different, they are equivalent. cos 7π
(
)
17.
5 3π and < x < 2π , the terminal side of x lies in QIV. Hence, the 13 2 x x lies in QII, and so sin > 0 . Therefore, we have terminal side of 2 2
Since cos x =
1 − cos x x sin = = 2 2
5 13 = 13 − 5 = 8 = 2 = 2 13 . 2 26 26 13 13
1−
18.
5 3π and π < x < , the terminal side of x lies in QIII. Hence, the 13 2 x x lies in QII, and so cos < 0 . Therefore, we have terminal side of 2 2
Since cos x = −
5 1+ − 1 + cos x x 13 = − 13 − 5 = − 8 = − 2 = − 2 13 . =− cos = − 2 2 26 26 13 13 2
497
Chapter 5
19.
12 3π and π < x < , we 5 2 know that sin x < 0 and cos x < 0 . Also, π x 3π x , sin > 0 . note that since < < 2 2 4 2 Consider the following diagram: Since tan x =
We see from the diagram that 5 cos x = − . So, we have 13
20.
12 3π and π < x < , we 5 2 know that sin x < 0 and cos x < 0 . Also, π x 3π x , cos < 0 . note that since < < 2 2 4 2 Consider the following diagram: Since tan x =
We see from the diagram that 5 cos x = − . So, we have 13 5 1+ − 1 + cos x x 13 cos = − =− 2 2 2
5 1− − 1 − cos x x 13 sin = = 2 2 2
=−
3 3 13 = = 13 13
498
2 2 13 = − 13 13
Section 5.4
21. Since sec x = 5 , we know that 1 π . This, together with 0 < x < , cos x = 2 5 implies that the terminal side of x is in QI. x is also in QI. Hence, the terminal side of 2 x Thus, tan > 0 . 2 Consider the following diagram:
5
499
We see from the diagram that 2 . So, we have sin x = 5 2 sin x x 5 = 2 = tan = 5 +1 2 1 + cos x 1 + 1 5 =
2
( 5 − 1) = 5 − 1
5 −1 2 Alternatively, we could have used one of the other two equivalent formulae x for tan . Specifically, the 2 following computation is also valid: 1 1− 1 − cos x x 5 = tan = 1 1 + cos x 2 1+ 5 =
5 −1 = 5 +1
5 −1 5 −1 ⋅ 5 +1 5 −1
=
6−2 5 = 4
3− 5 2
Chapter 5
22. Since sec x = 3, we know that 1 cos x = . This, together with 3 3π < x < 2π , implies that the terminal 2 side of x is in QIV. Hence, the terminal x x is in QII. Thus, tan < 0. side of 2 2 Consider the following diagram:
We see from the diagram that 2 sin x = − . So, we have 3 2 − sin x x 3 = − 2 = tan = 3 +1 2 1 + cos x 1 + 1 3 =
(
2− 6 2 Alternatively, we could have used one of the other two equivalent formulae for x tan . Specifically, the following 2 computation is also valid: 1 1− 1 − cos x x 3 =− tan = − 1 1 + cos x 2 1+ 3 =
3
− 2
)
− 2 3 −1 − 2 3 −1 ⋅ = 3 −1 3 +1 3 −1
500
=−
3 −1 =− 3 +1
3 −1 3 −1 ⋅ 3 +1 3 −1
=−
4−2 3 = − 2− 3 2
Section 5.4
23. Since csc x = 3, we know that 1 π sin x = . This, together with < x < π , 2 3 implies that the terminal side of x is in x is in QII. Hence, the terminal side of 2 x QI. So, sin > 0. 2 Consider the following diagram:
24. Since csc x = −3 , we know that 1 sin x = − . This, together with 3 3π < x < 2π , implies that the terminal 2 side of x is in QIV. Hence, the terminal x x is in QII. Thus, cos < 0 . side of 2 2 Consider the following diagram:
2 2 −2 2
We see from the diagram that 2 2 cos x = − . So, we have 3
2 2 1− − 3 1 − cos x x sin = = 2 2 2 =
We see from the diagram that 2 2 . So, we have cos x = 3
2 2 1+ 3 1 + cos x x cos = − =− 2 2 2
3+2 2 6
=−
501
3+2 2 1 2 = − + 6 2 3
Chapter 5
3π , it follows that 2 sin x < 0. This, together with 1 cos x = − , implies that the terminal side 4 of x is in QIII. Hence, the terminal side x x is in QII. Thus, cot < 0 . of 2 2 Consider the following diagram: 25. Since π < x <
3π 1 < x < 2π and cos x = , it 2 4 follows that sin x < 0 . So, the terminal side of x is in QIV, and the terminal side x x is in QII. So, sin > 0 and of 2 2 x csc > 0 . 2 Consider the following diagram:
26. Since
− 15
− 15
We see from the diagram that 15 . So, we have sin x = − 4 1 1 x = cot = 1 − cos x 2 tan x − 2 1 + cos x =
1 1 1− − 4 − 1 1+ − 4
= −
=
−1 3 =− 5 5 4 3 4
We see from the diagram that 15 . So, we have sin x = − 4 1 1 x csc = = 1 − cos x 2 sin x 2 2 1 1 1 = = = 1 3 3 1− 4 8 2 2 2 =
15 5
502
4 4 6 2 6 = = 6 3 6
Section 5.4
1 + cos A 5π A 27. Using the formula cos = with A = yields 2 6 2
5π 1 + cos 5π 6 = cos 6 = cos 5π . 2 2 12 1 − cos A π A 28. Using the formula sin = with A = yields 2 4 2 π 1 − cos π 4 = sin 4 = sin π . 2 2 8
sin A A 29. Using the formula tan = with A = 150 yields 2 1 + cos A 150 sin150 = tan = tan ( 75 ) . 1 + cos150 2 30. Observe that 1 − cos150 1 − cos150 1 + cos150 1 − cos 2 150 = ⋅ = sin150 sin150 1 + cos150 ( sin150 )(1 + cos150 )
=
sin 2 150 sin150 = ( sin150 )(1 + cos150 ) 1 + cos150
sin A A Now, using the formula tan = with A = 150 yields + A 2 1 cos 150 sin150 = tan = tan ( 75 ) . 1 + cos150 2 31. Use the known Pythagorean identity sin2 θ + cos 2 θ = 1 with θ =
that the given equation is in fact an identity.
503
x to conclude 2
Chapter 5
x x x 32. Observe that cos 2 − sin2 = cos 2 ⋅ = cos x 2 2 2 33. 1 cos x − csc x − cot x sin x sin x sin x 1 − cos x x = ⋅ = = sin2 2 2 csc x sin x 2 2 sin x
34.
1 = csc x + cot x
1 1 sin x x = = = tan 1 cos x 1+ cosx 1 + cos x 2 + sin x sin x sin x
35. 1 − sin ( 90° − x ) 2
=
1 − [sin 90° cos x − sin x cos 90°] 2
2
1 − cos x 1 − cos x 2 x = = = sin 2 2 2
36.
x x x 2 sin cos = sin 2 ⋅ = sin x 2 2 2 37.
x sin 2 = 1 + tan x = 1 + 1 − cos x = sin x + 1 − cosx = 1 − cosx + sinx 1+ sin x sin x sin x sin x x 2 cos 2 38. 2
2 1 + cos x x x 1 + cos x = 2 cos = 2 ± 2 cos = 2 = 1 + cos x 2 2 2 2 2
= 1 ± cos 2 x = 1 ± 1 − sin2 x 2
x x 1 − cos x 1 − cos x 39. Observe that tan = tan = ± . = 2 2 1 + cos x 1 + cos x 2
2
504
Section 5.4
40. Note that starting with the right-side is easier. Indeed, observe that
(1 − cos x )(1 − cos x ) 1 cos x (1 − cos x ) − = ( csc x − cot x ) = = 2 sin x 1 − cos 2 x sin x sin x 2
2
2
(1 − cos x ) (1 − cos x ) 1 − cos x x 2 x = = tan = tan2 = 2 (1 − cos x ) (1 + cos x ) 1 + cos x 2 41. Observe that sin A A A tan + cot = + 2 2 1 + cos A =
1 sin A 1 + cos A = + sin A 1 + cos A sin A 1 + cos A
sin2 A + (1 + cos A)
2
(1 − cos A) + (1 + cos A) = 2
2
( sin A)(1 + cos A) ( sin A)(1 + cos A) 2 1 − cos A)(1 + cos A) + (1 + cos A) ( = ( sin A)(1 + cos A) (1 + cos A) (1 − cos A) + (1 + cos A) 2 = = = 2 csc A sin A ( sin A) (1 + cos A) 42. Observe that A A cot − tan = 2 2
1 sin A 1 + cos A sin A − = − sin A 1 + cos A sin A 1 + cos A 1 + cos A
2 1 + cos A) − sin2 A (1 + cos A) − (1 − cos A) ( = = ( sin A)(1 + cos A) ( sin A)(1 + cos A) 2 (1 + cos A) − (1 − cos A)(1 + cos A) = ( sin A)(1 + cos A) (1 + cos A) (1 + cos A) − (1 − cos A) 2 cos A = = = 2 cot A sin A ( sin A) (1 + cos A) 2
2
505
Chapter 5
43. Observe that 1 1 2 2 1 + cos A 2 (1 + cos A) A csc 2 = = = = ⋅ = 2 1 − cos A 1 − cos A 1 + cos A 1 − cos 2 A 2 sin2 A 1 − cos A 2 2 2 (1 + cos A) = sin2 A 44. Observe that 1 1 2 A = = sec 2 = 2 1 + cos A 2 cos 2 A 1 + cos A 2 2 2 1 − cos A 2 (1 − cos A) 2 (1 − cos A) = ⋅ = = 1 + cos A 1 − cos A 1 − cos 2 A sin2 A
45. Using Exercise 43, we see that 2 (1 + cos A) A csc = ± = ± csc 2 A ( 2 + 2 cos A) = ± csc A 2 + 2 cos A . 2 sin A 2
46. Using Exercise 44, we see that 2 (1 − cos A) A sec = ± = ± csc 2 A ( 2 − 2 cos A) = ± csc A 2 − 2 cos A 2 sin A 2
47. Before graphing, observe that 2
1 + cos x x y = 4 cos = 4 2 2 . 2
1 + cos x = 4 = 2 + 2 cos x 2 Observe that the period is 2π , the amplitude is 2, there is no phase shift, and the vertical shift is up 2 units. So, following the approach in Chapter 4, we see that the graph is as follows:
506
Section 5.4
48. Before graphing, observe that 2
1 − cos x x y = −6 sin = −6 2 2 . 2
1 − cos x = −6 = −3 + 3cos x 2 Observe that the period is 2π , the amplitude is 3, there is no phase shift, and the vertical shift is down 3 units. So, following the approach in Chapter 4, we see that the graph is as follows: 49. Before graphing, observe that
(1 + cos x ) − (1 − cos x ) x 1 − cos x 1 − tan 1 − (1 + cos x ) 2 cos x 2 = 1 + cos x = y= = = cos x. 2 1 − cos x (1 + cos x ) + (1 − cos x ) 2x 1 + tan 1 + 2 1 + cos x (1 + cos x ) 2
So, the graph is as follows:
507
Chapter 5
50. Before graphing, observe that 2
x x x x x x y = 1 − sin + cos = 1 − sin2 + 2 sin cos + cos 2 2 2 2 2 2 2 2x x x x = 1 − sin + cos 2 + 2 sin cos = − sin x 2 2 2 2 =1 x = sin 2⋅ 2 So, the graph is that of y = sin x reflected over the x-axis, as shown below:
51. Observe that
θ θ 1 θ θ a2 A = bh = a sin a cos = a 2 ⋅ ⋅ 2 sin cos = sin θ 2 2 2 2 2 2 52. Consider the following triangle:
θ
Since the sum of the interior angles in a triangle must be 180 , we see that θ = 30. So, using Exercise 43, the area of this triangle is:
( 7 in.) sin30 = 49 in.2 A= 2
2
508
4
Section 5.4
53. Consider the triangle:
54. Consider the triangle:
34 2 2
θ
θ
1 − 334 1 − cos θ tan = = = 2 1 + cos θ 1 + 334
θ
= =
( 34 − 3)( 34 + 3) = 34 + 3
34 − 3 34 + 3
tan
θ 2
5 34 + 3
=
1 − 13 1 − cos θ = = 1 + cos θ 1 + 13
=
2 ≈ 0.71 2
5 34 − 3 34 − 3 ⋅ = ≈ 0.57 5 34 + 3 34 − 3
55. We have the following two triangles depicting Bin A and Bin B: Bin A Bin B
α
α
2
9 . Also, using the second triangle we have 41 1 − 419 1 − cos α 32 α h = 75sin = 75 = 75 = 75 ≈ 46.9 ft. 2 2 82 2
Observe that cos α =
509
2 3 4 3
Chapter 5
56. We have the following two triangles depicting Bin A and Bin B: Bin A Bin B
α
α
2
7 . Also, using the second triangle we have 25 1 − 257 1 − cos α 18 α h = 75sin = 75 = 75 = 75 ≈ 45 ft. 2 2 50 2
Observe that cos α =
3,000 3 = , so 4,000 4 that we have the following right triangle:
57. First, note that tan θ =
58. Using the following triangle, we have s p cos θ = , sin θ = . 2 2 2 s +p s + p2
θ 4 From this, we see that cos θ = . Now, 5 using the half-angle formula for tangent, we see that 1 1 − 54 1 − cos θ θ 5 tan = = = 9 1 + cos θ 1 + 54 2 5 =
s 2 + p2
p
θ
s So, using the half-angle formula for tangent yields p s 2 + p2 sin θ tan θ = = s 1 + cos θ 1 + 2 s + p2
1 1 = . 9 3
=
510
p s 2 + p2 + s
.
Section 5.4
59.
If π < x <
3π π x 3π , then < < , so that 2 2 2 4
x sin is positive, not negative. 2
60.
x sin2 x 2 , not Note that tan2 = 2 cos 2 x 2 x sin2 2. cos 2 x
61. False. Use A = π , for instance. Indeed, observe that π π sin + sin = 2 , while sin π = 0 . 2 2
62. False. Use A = π , for instance. Indeed, observe that π π cos + cos = 0 , while cos π = −1 . 2 2
5π , for instance. 4 5π Observe that tan = 1 > 0 , while 4 5π tan < 0 since the terminal side of 8 5π is in QII. 8
64. False. Use x = 450 , for instance. Observe that sin ( 450 ) = sin ( 360 + 90 )
63. False. Use x =
= sin ( 90 ) = 1 > 0,
450 while sin = sin ( 225 ) < 0 since the 2 terminal side of 225 is in QIII.
65. Observe that 1 − cos A 1 − cos A 1 − cos A A tan = ± =± ⋅ =± 1 + cos A 1 + cos A 1 − cos A 2
(1 − cos A)
2
1 − cos 2 A
=±
(1 − cos A) = ± 1 − cos A = ± 1 − cos A ⋅ 1 + cos A
=±
1 − cos 2 A sin2 A sin A =± =± 1 + cos A ( sin A)(1 + cos A) ( sin A)(1 + cos A)
2
sin2 A
sin A
sin A
1 + cos A
A is in QI or QIII. 2 sin A A Here, we use tan = . Further, for such angles A, both sin A > 0 and 2 1 + cos A Case 1: The terminal side of
511
Chapter 5
sin A sin A sin A A > 0 , so that we have tan = = . 1 + cos A 2 1 + cos A 1 + cos A A is in QII or QIV. Case 2: The terminal side of 2 sin A A Here, we use tan = − . Further, for such angles A, both sin A < 0 and 1 + cos A 2 sin A 1 + cos A ≥ 0 . Hence, < 0 , so that we have 1 + cos A sin A sin A sin A A tan = − = −− . = 1 + cos A 2 1 + cos A 1 + cos A sin A A So, in either case, we conclude that tan = . 2 1 + cos A One cannot evaluate the identity at A = π since the denominator on the right-side would 1 + cos A ≥ 0 . Hence,
be 0 (hence, the fraction is not well-defined) and
π
2
is not in the domain of tangent.
66. Observe that
1 − cos A 1 − cos A 1 − cos A A tan = ± =± ⋅ =± 1 + cos A 1 + cos A 1 − cos A 2
(1 − cos A) = ± 1 − cos A
(1 − cos A)
2
1 − cos 2 A
2
=±
sin2 A sin A A is in QI or QIII. Case 1: The terminal side of 2 A 1 − cos A Here, we use tan = . Further, for such angles A, both sin A > 0 and sin A 2 1 − cos A A 1 − cos A 1 − cos A 1 − cos A ≥ 0 . Hence, = . > 0 , so that we have tan = sin A sin A sin A 2 A is in QII or QIV. Case 2: The terminal side of 2 1 − cos A A Here, we use tan = − . Further, for such angles A, both sin A < 0 and sin A 2
512
Section 5.4
1 − cos A < 0 , so that we have sin A 1 − cos A A 1 − cos A 1 − cos A tan = − = − − . = sin A sin A sin A 2 A 1 − cos A . So, in either case, we conclude that tan = sin A 2 One cannot evaluate the identity at A = π since the denominator on the right-side would 1 − cos A ≥ 0 . Hence,
be 0 (hence, the fraction is not well-defined) and
π
2
is not in the domain of tangent.
67. Way 1:
3 2 1 sin15° = sin ( 45° − 30° ) = sin 45° cos 30° − cos 45° sin30° = 2 − 2 2 2 2 =
6− 2 4
Way 2: 1 − cos30° 30° sin = = 2 2
=
3 2 2 3 6 2 6 2 − − − 2 = = = 2 2 4 4
1−
8−4 3 2− 3 = 16 2
68. Way 1: tan 45° − tan30° tan15° = tan ( 45° − 30° ) = = 1 + tan 45° tan30°
3 3 = 3− 3 = 2− 3 3 3+ 3 1+ 3 1−
Way 2: sin30° 30° tan = = 2 1 + cos 30°
1 2 1+
513
3 2
=
1 =2− 3 2+ 3
Chapter 5
69.
1 − cos ( x2 ) 1 + cos ( x2 ) 1 − cos 2 ( x2 ) 1 − cos ( x2 ) x2 x tan = tan = ± =± =± 1 + cos ( x2 ) 1 + cos ( x2 ) 1 + cos ( x2 ) 1 + cos ( x2 ) 4 2 ± 1 − cos x ± 1 − cos x 2 = 2 ± 1 + cos x 2 ± 1 + cos x 2
1 − cos x 2 = = = x 1 + cos ( 2 ) 1 + cos x 1± 2
± sin2 ( x2 )
±
70.
x sin = sin = ± 4 2 x 2
1 − cos ( x2 ) 2
=±
1 + cos x 2 =± 2
1±
71. Consider the graphs of the following functions: 3 5 x x 2 2 x x y2 = sin y1 = − + 3! 5! 2 2
2 ± 1 + cos x 2 2
72. Consider the graphs of the following functions: 2 4 x x 2 2 x y2 = cos y1 = 1 − + 2! 4! 2
Observe that y1 (dotted graph) is a good approximation of y2 on [ −1,1] .
Observe that y1 (dotted graph) is a good approximation of y2 on [ −1,1] .
514
Section 5.4
73. Consider the graphs of the following functions: x x y1 = csc 2 + sec 2 y2 = 4 csc 2 x 2 2
Since the graphs coincide, this suggests that the given equation is an identity. The proof of this fact is as follows: x x cos 2 + sin2 1 1 1 x x 2 2 = csc 2 + sec 2 = + = x x x x 2 2 sin 2 x cos 2 x cos 2 sin2 cos 2 sin2 2 2 2 2 2 2 1 4 4 = = = 2 1 x x x sin2 2 ⋅ ⋅ 4 cos 2 sin2 2 cos x sin x 4 2 2 2 2 2 =
4 = 4 csc 2 x 2 sin x
74. Consider the graphs of the following functions: x x y1 = tan2 + cot 2 2 2 2 y2 = −2 cot x sec x Since the graphs do not coincide, the given equation is not an identity.
515
Section 5.5 Solutions ----------------------------------------------------------------------------------1. sin 2 x cos x =
1 1 [sin(2x + x ) + sin(2x − x )] = [sin3x + sin x ] 2 2
2. 1 [sin(5x + 10x ) + sin(5x − 10x )] 2 1 1 = [sin15x + sin( −5x )] = [sin15x − sin5x ] 2 2
cos10 x sin5x =
3. 1 [cos(4x − 6x ) − cos(4x + 6x )] 2 5 5 = [ cos( −2x ) − cos10 x ] = [ cos 2 x − cos10 x ] 2 2
5sin 4 x sin 6 x = 5 ⋅
4. 1 [ cos(2x − 4x ) − cos(2x + 4x )] 2 3 3 = − [ cos( −2 x ) − cos 6x ] = − [ cos 2x − cos 6x ] 2 2
−3sin 2x sin 4 x = −3 ⋅
5. 4 cos( − x ) cos 2 x = 4 ⋅
1 [ cos( −x + 2x ) + cos( −x − 2x )] 2
= 2 [ cos x + cos( −3x )] = 2 [ cos x + cos 3x ]
6. −8cos 3x cos 5x = −8 ⋅
1 [cos(3x + 5x ) + cos(3x − 5x )] 2
= −4 [ cos 8x + cos( −2 x )] = −4 [ cos 8x + cos 2 x ]
516
Section 5.5
7. 3x 5x 1 3x 5 x 3x 5 x sin sin = cos − − cos + 2 2 2 2 2 2 2 =
1 1 [cos( −x ) − cos 4x ] = [cos x − cos 4x ] 2 2
8. π x 5π x 1 π x 5π x π x 5π x sin − + sin = cos − cos 2 2 2 2 2 2 2 =
9.
1 1 [cos( −2π x ) − cos 3π x ] = [cos 2π x − cos 3π x ] 2 2
4 4 2 4 1 2 2 cos x cos x = cos x + x + cos x − x 3 3 3 3 2 3 3
=
1 2 1 2x cos 2 x + cos − x = cos 2 x + cos 2 3 2 3
10.
π π π π 1 π π sin − x cos − x = sin − x − x + sin − x − − x 2 4 2 2 4 2 4
=
11.
sin 97.5 sin 22.5 =
1 3π π 1 π 3π sin − x + sin x = sin x − sin x 2 4 4 2 4 4
1 1 1 cos 75 − cos120 ) = cos 75 + ( 2 2 2
12.
cos85.5 cos 4.5 =
1 1 cos 90 + cos81 ) = ( cos81 ) ( 2 2
517
Chapter 5
13.
5 x + 3x 5 x − 3x cos5x + cos3x = 2 cos cos = 2 cos 4x cos x 2 2
14.
2x + 4x 2x − 4x cos 2x − cos 4x = −2 sin sin = −2 sin(3x )sin( −x ) = 2 sin3x sin x 2 2
15.
3x − x 3x + x sin3x − sin x = 2 sin cos = 2 sin x cos 2x 2 2
16.
10 x + 5x 10 x − 5x 15 5 sin10x + sin5x = 2 sin cos = 2 sin x cos x 2 2 2 2 17. sin8x − sin 6 x = 2 sin x cos 7 x 18. cos 2 x − cos 6 x = −2 sin 4 x sin ( −2 x ) = 2 sin 4 x sin 2 x 19. x 5x x 5x − + x 5x 3x 3x sin − sin = 2 sin 2 2 cos 2 2 = 2 sin( −x ) cos = −2 sin x cos 2 2 2 2 2 2
20.
x 5x x 5x + − x 5x 3x cos − cos = −2 sin 2 2 sin 2 2 = −2 sin sin ( − x ) 2 2 2 2 2 3x = 2 sin sin x 2
518
Section 5.5
21. 7 7 2 2 3x+ 3x 3x−3x 2 7 cos x + cos x = 2 cos cos 2 2 3 3
3 5 3 5 = 2 cos x cos − x = 2 cos x cos x 2 6 2 6 22.
7 7 2 2 x+ x x− x 2 7 3 cos 3 3 sin x + sin x = 2 sin 3 2 2 3 3
3 5 3 5 = 2 sin x cos − x = 2 sin x cos x 2 6 2 6 23.
0.4 x + 0.6 x 0.4 x − 0.6x sin ( 0.4 x ) + sin ( 0.6x ) = 2 sin cos 2 2 = 2 sin ( 0.5x ) cos ( −0.1x ) = 2 sin ( 0.5x ) cos ( 0.1x ) 24.
0.3x + 0.5x 0.3x − 0.5x cos ( 0.3x ) − cos ( 0.5x ) = −2 sin sin 2 2 = −2 sin(0.4 x )sin( −0.1x ) = 2 sin(0.4x )sin(0.1x ) 25. cos3x − cos x = sin3x + sin x
3x + x 3x − x − 2 sin sin 2 2
3x − x = − tan = − tan x 2 3x − x 3x + x 2 sin cos 2 2
519
Chapter 5
26. sin 4 x + sin 2 x = cos 4 x − cos 2 x
4x + 2x 4x − 2x 2 sin cos 2 2
4x − 2x = − cot = − cot x 2 4x − 2x 4x + 2x − 2 sin sin 2 2
27.
x + 3x x − 3x − 2 sin sin cos x − cos 3x 2 2 − sin 2 x sin( −x ) sin 2 x sin x = tan 2 x = = = sin x cos 2 x sin3x − sin x 3x − x 3x + x sin x cos 2 x 2 sin cos 2 2
28. sin 4 x + sin 2 x = cos 4 x + cos 2 x
4x + 2x 4x − 2x 2 sin cos 2 2
4x + 2x = tan = tan3x 2 4x + 2x 4x − 2x 2 cos cos 2 2
29. sin x − sin ( 3x ) = cos x + cos ( 3x )
x − 3x x + 3x 2 sin cos 2 2
x − 3x = tan = tan ( −x ) = − tan x 2 x − 3x x + 3x 2 cos cos 2 2
30. cos5x + cos 3x = sin5x − sin3x
5 x + 3x 5 x − 3x 2 cos cos 2 2
5x − 3x = cot = cot x 2 5x − 3x 5x + 3x 2 sin cos 2 2
520
Section 5.5
31. sin A + sin B = cos A + cos B
A+ B A− B 2 sin cos 2 2
A+ B = tan 2 A+ B A− B 2 cos cos 2 2
32. sin A − sin B = cos A + cos B
A− B A+ B 2 sin cos 2 2
A− B = tan 2 A− B A+ B cos 2 cos 2 2
33. cos A − cos B = sin A + sin B
A+ B A− B − 2 sin sin 2 2
A− B = − tan 2 A− B A+ B cos 2 sin 2 2
34. cos A − cos B = sin A − sin B
35.
A+ B A− B − 2 sin sin 2 2
A+ B = − tan 2 A+ B A− B cos 2 sin 2 2
A+ B A− B A+ B A− B 2 sin cos sin cos sin A + sin B 2 2 = 2 ⋅ 2 = sin A − sin B A− B A+ B A+ B A− B 2 sin cos cos sin 2 2 2 2 A+ B A− B = tan cot 2 2
521
Chapter 5
36.
A+ B A− B A+ B A− B − 2 sin sin sin sin cos A − cos B 2 2 2 2 =− ⋅ = cos A + cos B A+ B A− B A+ B A− B 2 cos cos cos cos 2 2 2 2 A+ B A− B = − tan tan 2 2
37.
1 − cos( A + B ) 1 + cos( A − B ) A+ B A− B sin A + sin B = 2 sin ± cos = 2 ± 2 2 2 2 =
± 2 1 − cos( A + B ) 1 + cos( A − B ) 2⋅ 2
= ± (1 − cos ( A + B ) ) (1 + cos ( A − B ) )
38. cos A sin B + sin ( A − B ) = cos A sin B + sin A cos B − sin B cos A = sin A cos B 39.
A B A B A B A B −2 sin cos + cos sin sin cos − cos sin 2 2 2 2 2 2 2 2 2 2 A B A B = −2 sin cos − cos sin 2 2 2 2 2 2 1 A+ B A − B 1 B + A B − A = −2 sin + sin − sin + sin 2 2 2 2 2 2 2 2 2 A+ B A− B A+ B A− B = − sin + sin − sin − sin 4 2 2 2 2
1 A + B A − B A+ B A−B = − 4 sin sin = −2 sin sin 2 2 2 2 2 ( cos A − cos B )( cos A + cos B ) = cos 2 A − cos 2 B = cos A − cos B = cos A + cos B cos A + cos B
522
Section 5.5
40.
A+ B A− B A− B A+ B A+ B A− B sin + cos + sin cos = sin = sin A 2 2 2 2 2 2
41.
5π π π R(t ) = sin t + − sin t + π 3 6 6 π 5π 5π π π π = sin t cos + sin cos t − sin t cos (π ) + sin(π ) cos t 6 3 3 6 6 6 3 π 1 π π = sin t + − cos t + sin t 6 2 2 6 6 3 3 π π cos t = sin t − 2 6 2 6 3 π 1 π = 3 sin t − cos t 6 2 6 2 4π π = 3 cos t + 3 6
π π π 42. Profit = Revenue – Cost = cos t − cos t + . 3 3 3 π x+y x−y Using the formula cos x − cos y = −2 sin sin with x = t and 3 2 2 y=
π 3
t+
π 3
, we obtain
π π π π π 1 π π Profit = −2 sin t + sin − = −2 − sin t + = sin t + . 6 6 6 6 3 2 3 3
523
Chapter 5
43. Description of G note: cos ( 2π (392)t )
Description of B note: cos ( 2π (494)t )
Combining the two notes: cos ( 784π t ) + cos ( 988π t ) . Using the sum-to-product identity then yields: 784π t + 988π t 784π t − 988π t cos ( 784π t ) + cos ( 988π t ) = 2 cos cos 2 2 = 2 cos ( 886π t ) cos ( −102π t )
= 2 cos ( 886π t ) cos (102π t ) (since cosine is even) The beat frequency is 494 – 392 = 102 Hz. 494 + 392 = 443 Hz. The average frequency is 2 44. Description of F note: cos ( 2π (349)t )
Description of A note: cos ( 2π (440)t )
Combining the two notes: cos ( 698π t ) + cos ( 880π t ) . Using the sum-to-product identity then yields: 698π t + 880π t 698π t − 880π t cos ( 698π t ) + cos ( 880π t ) = 2 cos cos 2 2 = 2 cos ( 789π t ) cos ( −91π t )
= 2 cos ( 789π t ) cos ( 91π t ) (since cosine is even) The beat frequency is 440 – 349 = 91 Hz. 440 + 349 = 394.5 Hz. The average frequency is 2
524
Section 5.5
45. The resulting signal is 2π tc 2π tc sin + sin −6 −6 1.55 × 10 0.63 ×10 1 1 1 1 + − −6 −6 −6 −6 1.55 × 10 0.63 × 10 1.55 × 10 0.63 × 10 cos 2π tc = 2 sin 2π tc 2 2 6 6 6 6 10 10 10 10 + − 1.55 0.63 1.55 0.63 cos 2π tc = 2 sin 2π tc 2 2 1 6 1 6 2π tc 1 2π tc 1 = 2 sin + − 10 cos 10 2 1.55 0.63 2 1.55 0.63 46. Using the same model as in Exercise 45, we find that: c c 1 6 1 − =c − 10 Hz The beat frequency is −6 −6 1.55 × 10 0.63 × 10 1.55 0.63 c c + −6 0.63 × 10 −6 = c 1 + 1 106 Hz The average frequency is 1.55 × 10 2 2 1.55 0.63 47.
1540π t + 2418π t 1540π t − 2418π t sin ( 2π (770)t ) + sin ( 2π (1209)t ) = 2 sin cos 2 2 = 2 sin (1979π t ) cos ( −439π t ) = 2 sin (1979π t ) cos ( 439π t )
48.
1394π t + 2954π t 1394π t − 2954π t sin ( 2π (697)t ) + sin ( 2π (1477)t ) = 2 sin cos 2 2 = 2 sin ( 2174π t ) cos ( −780π t ) = 2 sin ( 2174π t ) cos ( 780π t )
525
Chapter 5
49. Note that A + 52.5 + 7.5 = 180 , so that A = 120 . So, the area is 1 2 (10 ft.) sin (52.5 ) sin ( 7.5 ) 100 ⋅ 2 cos (52.5 − 7.5 ) − cos (52.5 + 7.5 ) 2 ft. = 2 sin (120 ) 2 sin (120 )
2 1 25 − 50 cos ( 45 ) − cos ( 60 ) 2 2 2 2 ft. = ft. = 2 sin (120 ) 3 2
= =
25
( 2 − 1) ft. = 25 ( 2 − 1) 3 ft. 2
2
3
3
25
( 6 − 3 ) ft. ≈ 5.98 ft. 2
2
3
50. Note that A + 75 + 45 = 180 , so that A = 60 . So, the area is 1 2 (12 in.) sin ( 75 ) sin ( 45 ) 144 ⋅ 2 cos ( 75 − 45 ) − cos ( 75 + 45 ) 2 in. = 2 sin ( 60 ) 2 sin ( 60 )
3 1 36 + 36 cos ( 30 ) − cos (120 ) 2 2 2 2 in. = in. = sin ( 60 ) 3 2
=
36
(
( 3 + 1) in. = 36 ( 3 + 1) 3 in. 2
3
2
3
)
= 36 + 12 3 in.2 ≈ 56.78 in.2 51.
sin A sin B + sin A cos B + cos A cos B = ( cos A cos B + sin A sin B ) + sin A cos B = cos ( A − B ) +
526
1 ( sin ( A + B ) + sin ( A − B ) ) 2
Section 5.5
52. A− B A+ B A + B 2 cos sin + cos 2 2 2 A− B A+ B A− B A+ B = 2 cos sin + 2 cos cos 2 2 2 2 1 1 = 2 ( sin A + sin B ) + 2 ( cos A + cos B ) 2 2 = sin A + sin B + cos A + cos B
53. In the final step of the computation, note that cos A cos B ≠ cos AB and sin A sin B ≠ sin AB . Should have used the product-to-sum identities. 54. In general, sin A cos B − sin B cos A ≠ 0 . Should have used the product-to-sum identity to simplify this. 55. False. From the product-to-sum identities, we have 1 cos A cos B = [ cos( A + B ) + cos( A − B )] , 2 and the right-side is not, in general, expressible as the cosine of a product. 56. False. From the sum-to-product identities, we have 1 sin A sin B = [ cos( A − B ) − cos( A + B )] , 2 and the right-side is not, in general, expressible as the sine of a product. 57. True. From the product-to-sum identities, we have 1 cos A cos B = [ cos( A + B ) + cos( A − B )] . 2 58. True. From the product-to-sum identities, we have 1 sin A sin B = [ cos( A − B ) − cos( A + B )] . 2
527
Chapter 5
59. Observe that sin A sin B sin C = [sin A sin B ] sin C
1 = ( cos ( A − B ) − cos( A + B ) ) sin C 2 1 = cos ( A − B ) sin C − cos( A + B )sin C 2 1 1 = sin (C + A − B ) + sin (C − ( A − B ) ) 2 2 1 − sin (C + A + B ) + sin (C − ( A + B ) ) 2 1 [sin( A − B + C ) + sin(C − A + B ) − sin( A + B + C ) − sin( A + B − C )] 4 At this point, depending on which terms you decide to apply the odd identity for sine, the answer can take on a different form. =
60. Observe that cos A cos B cos C = [ cos A cos B ] cos C 1 = ( cos ( A + B ) + cos( A − B ) ) cos C 2 1 = cos ( A + B ) cos C + cos( A − B ) cos C 2 1 1 = cos ( A + B + C ) + cos ( A + B − C ) 2 2 1 + cos ( A − B + C ) + cos ( A − B − C ) 2 =
1 cos ( A + B + C ) + cos ( A + B − C ) 4 + cos ( A − B + C ) + cos ( A − B − C )
At this point, depending on which terms you decide to apply the odd identity for sine, the answer can take on a different form.
528
Section 5.5
61. Applying an addition formula on the right-side yields the equivalent equation cos A cos B = cos A cos B − sin A sin B 0 = sin A sin B Hence, at least one of A or B is equal to nπ , for some integer n. 62. Applying an addition formula on the right-side yields the equivalent equation sin A cos B = sin A cos B + sin B cos A 0 = sin B cos A (2 n + 1)π or B = nπ (or both), for some integer n. Hence, either A = 2 63.
1 1 1 sin ( A + A) + sin ( A − A) = sin ( 2A) + sin 0 = sin ( 2A) 2 2 2 1 = ( 2 sin A cos A) = sin A cos A 2 64.
1 1 1 cos ( A − A) − cos ( A + A) = cos ( 0 ) − cos ( 2 A) = 1 − cos ( 2 A) 2 2 2 1 1 = 1 − (1 − 2 sin2 A) = 2 sin2 A = sin2 A 2 2 A+ B A− B 65. First, we know that sin A + sin B = 2 sin cos . Since A and B are 2 2 assumed to be complementary angles, we continue this equality to get 2 1 + cos( A − B ) π A− B A− B = 2 sin cos = 1 + cos( A − B ) cos = 2 = 2 2 2 2 4 2 A+ B A− B 66. First, we know that cos A − cos B = −2 sin sin . Since A and B are 2 2 assumed to be complementary angles, we continue this equality to get 2 A− B 1 − cos( A − B ) π A− B = −2 sin sin sin = −2 = − 2 ± 2 4 2 2 2 = ± 1 − cos( A − B )
529
Chapter 5
67. Consider the graph of y = 4 sin x cos x cos 2x , as seen below:
From the graph, it seems as though this function is equivalent to sin 4 x . We prove this identity below: 4 sin x cos x cos 2 x = 2 ( 2 sin x cos x ) cos 2x = 2 sin 2x cos 2x = sin(2 ⋅ 2 x ) = sin 4x = sin2 x
68. Consider the graph of y = 1 + tan x tan 2x , as seen below:
From the graph, it seems as though this function is equivalent to sec 2 x . We prove this identity below: sin x 2 sin x cos x 2 sin 2 x cos 2 x + 2 sin2 x sin x sin 2 x ⋅ =1 + =1 + = 1 + tan x tan 2 x = 1 + cos 2 x cos 2 x cos x cos 2 x cos x cos 2 x
(
( cos x − sin x ) + 2 sin x = cos x + sin x = 2
=
)
2
2
2
cos 2 x
cos 2 x
530
2
1 = sec 2 x cos 2 x
Chapter 5 Review Solutions --------------------------------------------------------------------------1. tan x ( cot x + tan x ) = tan x cot x + tan2 x = 1 + tan2 x = sec 2 x 2. ( sec x + 1)( sec x − 1) = sec 2 x − 1 = (1 + tan2 x ) − 1 = tan2 x 3.
2 2 tan 4 x − 1 ( tan x − 1) ( tan x + 1) = tan2 x + 1 = sec 2 x = 2 2 tan x − 1 ( tan x − 1)
4. sec 2 x ( cot 2 x − cos 2 x ) =
5.
1 cos 2 x 1 1 − sin2 x cos 2 x 2 − cos x = − 1 = = = cot 2 x 2 2 2 2 2 cos x sin x sin x sin x sin x
cos x ( cos( − x ) − tan( − x ) ) − sin x = cos x(cos x + tan x ) − sin x sin x = cos 2 x + cos x − sin x cos x = cos 2 x + sin x − sin x = cos 2 x
6. tan2 x + 1 sec 2 x 1 = = 2 2 2 sec x 2 sec x 2
7.
cot x + cot 2 x − 2 = tan 2 x + 2 + cot 2 x − 2 = tan 2 x + cot 2 x ( tan x + cot x ) − 2 = tan2 x + 2 tan x 2
=1
8. 1 cos 2 x 1 − cos 2 x sin2 x − = = =1 csc x − cot x = sin2 x sin2 x sin2 x sin2 x 2
2
531
Chapter 5
9. 1 1 1 cos 2 x 1 − cos 2 x sin2 x − = − = = =1 sin2 x tan2 x sin2 x sin 2 x sin2 x sin2 x
10.
( csc x − 1) + ( csc x + 1) = 2 csc x 1 1 + = csc x + 1 csc x − 1 ( csc x − 1)( csc x + 1) csc 2 x − 1 1 2⋅ 2 2 csc x 2 csc x sin x = 2 ⋅ sin x = = = (1 + cot 2 x ) − 1 cot 2 x cos22 x sin x cos2 x sin x 2 sin x sin x 1 2 tan x = = 2⋅ ⋅ = cos x cos x cos x cos x cos x
11. tan2 x − 1 tan2 x − 1 tan2 x − 1 = = sec 2 x + 3tan x + 1 (1 + tan2 x ) + 3tan x + 1 tan2 x + 3tan x + 2
=
( tan x − 1) ( tan x + 1) ( tan x − 1) = ( tan x + 2 ) ( tan x + 1) ( tan x + 2 )
12. cot x ( sec x − cos x ) =
2 2 cos x 1 cos x 1 − cos x cos x sin x − = = x cos = sin x sin x cos x sin x cos x sin x cos x
13. 1 − sin2 x cos 2 x cos x cos x = = ⋅ = cos x cot x sin x sin x 1 sin x
14. cos 2 x 1 − sin2 x 1 sin2 x = = − = csc x − sin x sin x sin x sin x sin x
15. 2 tan2 2x + cos 2x + sin2 2 x = tan2 2x + 1 = sec 2 2x 1
532
Chapter 5 Review
16.
cos 2 x − sin2 x cos 2 x − sin2 x cos 2 x − sin2 x cos 2 x − sin2 x = = = sin2 x cos 2 x sin2 x cos 2 x − sin2 x 1 − tan2 x 1− − cos 2 x cos 2 x cos 2 x cos 2 x cos 2 x − sin2 x cos 2 x = cos 2 x ⋅ = 2 2 1 cos x − sin x
17. Observe that
sin2 x 2 sin2 x + cos 2 x sin x + ( sin x + cos x ) sin2 x + 1 1 + sin2 x 2 tan x + 1 = 2 ⋅ +1 = = = = cos 2 x cos 2 x cos 2 x cos 2 x cos 2 x So, the given equation is an identity. 2
2
2
2
18. The equation sin x − cos x = 0 is a conditional since, for instance, it does not hold if x = π (since the left-side reduces to 1 in such case). 19. The equation cot 2 x − 1 = tan2 x is a conditional since, for instance, it does not hold if x = π4 (since the left-side reduces to 0, while the right-side equals 1 in such case). 20. Observe that
cos 2 x (1 + cot 2 x ) = cos 2 x csc 2 x = cos 2 x ⋅
1 = cot 2 x . 2 sin x
So, the given equation is an identity. 21. Using the addition formula cos( A + B ) = cos A cos B − sin A sin B yields 7π π π π π π π cos = cos + = cos cos − sin sin 12 3 4 3 4 3 4
3 1 2 3 2 2 1 = − − = = 2 2 2 2 2 2 2
2− 6 4
22. Using the addition formula sin( A − B ) = sin A cos B − cos A sin B yields
π π π π π π π sin = sin − = sin cos − cos sin 12 3 4 3 4 3 4 3 2 1 2 2 3 1 = − = − = 2 2 2 2 2 2 2
533
6− 2 4
Chapter 5
23.
tan A − tan B yields 1 + tan A tan B tan30 − tan 45 (1) tan( −15 ) = tan(30 − 45 ) = 1 + tan30 tan 45 1 1 3 Now, observe that tan30 = 2 = = and tan 45 = 1 . So, continuing the 3 3 3 2 computation in (1) yields 3 −3 3 −1 3 − 3 3 − 3 3 3 − 9 − 3 + 3 3 6 3 − 12 3 = = ⋅ = = = 3 −2 tan( −15 ) = 3 9−3 6 3 3+ 3 3+ 3 3− 3 1+ 3 3 Using the addition formula tan( A − B ) =
24.
tan A + tan B to compute tan105 : 1 − tan A tan B tan 60 + tan 45 (1) tan(105 ) = tan(60 + 45 ) = 1 − tan 60 tan 45 3 sin 60 Now, observe that tan 45 = 1 , and tan 60 = = 2 = 3 . So, continuing the 1 cos 60 2 computation in (1) yields 3 +1 3 +1 1+ 3 1+ 2 3 + 3 4 + 2 3 tan(105 ) = = ⋅ = = = −2 − 3 . −2 1− 3 1− 3 1− 3 1+ 3 First, we use the addition formula tan( A + B ) =
Thus, cot105 =
1 1 1 −2 + 3 −2 + 3 = ⋅ = = −2 + 3 = tan105 4 −3 −2 − 3 −2 − 3 −2 + 3
25. Using the addition formula sin( A − B ) = sin A cos B − cos A sin B with A = 4 x and B = 3x yields sin 4 x cos3x − cos 4 x sin3x = sin(4 x − 3x ) = sin x .
534
Chapter 5 Review
26. Using the even/odd identities for cosine and sine, followed by the addition formula cos( A − B ) = cos A cos B + sin A sin B with A = x and B = 2 x , and finally one more application of the even identity of cosine, yields sin( −x )sin( −2 x ) + cos( −x ) cos( −2x ) = sin x sin 2 x + cos x cos 2 x = cos x cos 2 x + sin x sin 2 x = cos( x − 2 x ) = cos( −x ) = cos x .
27. Using the addition formula tan( A − B ) =
yields
tan5x − tan 4x = tan(5x − 4 x ) = tan x . 1 + tan5x tan 4x
28. Using the addition formula tan( A + B ) = tan
yields
tan A − tan B with A = 5x and B = 4 x 1 + tan A tan B
π
+ tan
π
tan A + tan B π π with A = and B = 1 − tan A tan B 4 3
3 = tan π + π = tan 7π . π π 4 3 12 1 − tan tan 4 3 4
tan α − tan β , we need to determine the values of 1 + tan α tan β cos α and sin β so that we can then use this information in combination with the given values of sin α and cos β in order to compute tan α and tan β . We proceed as follows:
29. Observe that since tan(α − β ) =
Since the terminal side of α is in QIV, we know that cos α > 0 . As such, since 3 we are given that sin α = − , the 5 diagram is as follows:
Since the terminal side of β is in QIII, we know that sin β < 0 . As such, since 24 we are given that sin β = − , the 25 diagram is as follows:
α
β
535
Chapter 5
From the Pythagorean Theorem, we have z 2 + ( −3)2 = 52 , so that z = 4. 4 As such, cos α = , and so, 5 3 sin α − 5 3 = =− tan α = 4 cos α 4 5
From the Pythagorean Theorem, we have z 2 + ( −24)2 = 252 , so that z = −7. 7 As such, cos β = − , and so 25 24 sin β − 25 24 = = tan β = cos β 7 −7 25
Hence, −21 − 96 3 24 − − 117 tan α − tan β 28 4 7 = . tan(α − β ) = = = 44 1 + tan α tan β 3 24 28 − 72 1+ − 28 4 7
5 7 30. Observe that cos(α + β ) = cos α cos β − sin α sin β = − cos β − sin α . 13 25 We need to compute both sin α and sin β . We proceed as follows: Since the terminal side of α is in QII, we Since the terminal side of β is in QII, know that sin α > 0 . As such, since we we know that cos β < 0 . As such, since 5 are given that cos α = − , the diagram is we are given that sin β = 7 , the 13 25 as follows: diagram is as follows:
α
β
From the Pythagorean Theorem, we have z 2 + ( −5)2 = 132 , so that z = 12. 12 Thus, sin α = . 13
From the Pythagorean Theorem, we have z 2 + 7 2 = 252 , so that z = −24. 24 Thus, cos β = − . 25
536
Chapter 5 Review
Hence,
5 24 12 7 36 . cos(α + β ) = cos α cos β − sin α sin β = − − − = 13 25 13 25 325 9 7 31. Observe that cos(α − β ) = cos α cos β + sin α sin β = + sin α sin β . 41 25 We need to compute both sin α and sin β . We proceed as follows: Since the terminal side of α is in QIV, we know that sin α < 0 . As such, since we 9 are given that cos α = , the diagram is 41 as follows:
Since the terminal side of β is in QI, we know that sin β > 0 . As such, since 7 , the we are given that cos β = 25 diagram is as follows:
β
α
From the Pythagorean Theorem, we have z 2 + 92 = 412 , so that z = −40. 40 Thus, sin α = − . 41
From the Pythagorean Theorem, we have z 2 + 7 2 = 252 , so that z = 24. 24 Thus, sin β = . 25
Hence, 897 9 7 40 24 . cos(α − β ) = cos α cos β + sin α sin β = + − = − 1,025 41 25 41 25
537
Chapter 5
5 4 32. Observe that sin(α − β ) = sin α cos β − cos α sin β = − − − cos α sin β . We 13 5 need to compute both cos α and sin β . We proceed as follows: Since the terminal side of α is in QIII, we know that cos α < 0 . As such, since we 5 are given that sin α = − , the diagram is 13 as follows:
Since the terminal side of β is in QII, we know that sin β > 0 . As such, since we 4 are given that cos β = − , the diagram is 5 as follows:
β
α
From the Pythagorean Theorem, we have z 2 + ( −5)2 = 132 , so that z = −12. 12 Thus, cos α = − . 13
From the Pythagorean Theorem, we have z 2 + ( −4)2 = 52 , so that z = 3. 3 Thus, sin β = . 5
Hence,
5 4 12 3 56 sin(α − β ) = sin α cos β − cos α sin β = − − − − = . 13 5 13 5 65 33. Observe that cos( A + B ) + cos( A − B ) = (cos A cos B − sin A sin B ) + (cos A cos B + sin A sin B )
= 2 cos A cos B So, the given equation is an identity.
538
Chapter 5 Review
34. Observe that cos( A − B ) − cos( A + B ) = ( cos A cos B + sin A sin B ) − ( cos A cos B − sin A sin B )
= 2 sin A sin B So, the given equation is an identity. 35. Before graphing, observe that π π y = cos cos x − sin sin x = − sin x . 2 2 =0
=1
So, the graph is simply the graph of y = sin x reflected over the x-axis, as seen below:
36. Using the addition formula sin( A + B ) = sin A cos B + cos A sin B with A = x and 2π 2π 2π 2π yields y = cos x sin B= + sin x cos = sin x + . 3 3 3 3 2π So, the graph is the graph of y = sin x shifted to the left units, as seen below. 3
539
Chapter 5
3 3π and π < x < , the 5 2 terminal side of x must be in QII. The diagram is as follows:
7 3π < x < 2π , and 25 2 the terminal side of x must be in QIV. The diagram is as follows:
37. Since sin x =
38. Since cos x =
From the Pythagorean Theorem, we have z 2 + 32 = 52 , so that z = −4. 4 As such, cos x = − . Therefore, 5 cos 2 x = cos 2 x − sin2 x
From the Pythagorean Theorem, we have z 2 + 7 2 = 252 , so that z = −24. 24 As such, sin x = − . Therefore, 25 sin 2 x = 2 sin x cos x
2
336 24 7 = 2 − = − 625 25 25
2
7 4 3 16 9 = − − = − = 5 5 25 25 25
39. Observe that since cot x = −
11 61 , it follows that tan x = − . 61 11
So, 61 122 2− − 2 tan x 11 11 = 671 = = tan 2 x = 2 2 3,600 1 − tan x 1,800 61 − 1− − 121 11
540
Chapter 5 Review
12 π and < x < π , the 2 −5 diagram is as follows:
40. Since tan x =
From the Pythagorean Theorem, we have ( −5)2 + 12 2 = z 2 , so that z = 13. 12 5 As such, sin x = and cos x = − . 13 13 Therefore, 2 2 5 12 2 2 cos 2 x = cos x − sin x = − − 13 13 = −
119 169
25 , it follows that 24 1 25 24 = and so, cos x = . Since cos x 24 25 also 0 < x < π2 , the terminal side of x must be in QI. The diagram is as follows: 41. Since sec x =
From the Pythagorean Theorem, we have z 2 + 24 2 = 252 , so that z = 7 . 7 As such, sin x = . Therefore, 25 7 24 336 sin 2 x = 2 sin x cos x = 2 = 25 25 625
541
Chapter 5
From the Pythagorean Theorem, we have z 2 + 4 2 = 52 , so that z = −3. 4 As such, tan x = − . Therefore, 3 4 2 − 2 tan x 24 3 tan 2 x = . = = 2 2 1 − tan x 7 4 1− − 3
5 42. Since csc x = , it follows that 4 1 5 4 = and so, sin x = . Since also sin x 4 5 π 2 < x < π , the terminal side of x must be in QII. The diagram is as follows:
43. Using cos 2 A = cos 2 A − sin2 A with A = 15 yields cos 2 15 − sin2 15 = cos(2 ⋅15 ) = cos ( 30 ) =
3 . 2
2 tan A π yields with A = − 2 1 − tan A 12 π π π 1 2 tan − sin − − sin π π 12 6 6 = tan 2 ⋅ − = =− 2 = − 3 = tan − = 3 π 3 6 cos − π cos π 12 1 − tan2 − 12 6 6 2
44. Using tan 2A =
45. Using sin 2 A = 2 sin A cos A with A =
π
yields 12 π π π π π π 1 3 6 sin cos = 3 2 sin cos = 3sin 2 ⋅ = 3sin = 3 = . 12 12 12 12 12 6 2 2
542
Chapter 5 Review
46. Using cos 2A = 1 − 2 sin2 A with A =
π 8
yields
2 π π π . 1 − 2 sin2 = cos 2 ⋅ = cos = 2 8 8 4
47. Note that starting with the right-side is easier this time. Indeed, observe that 1 1 ( sin A − cos A) 1 + sin 2A = ( sin A − cos A) 1 + ⋅ 2 sin A cos A 2 2 2 = sin A + sin A cos A − cos A − sin A cos 2 A
= sin A + (1 − cos 2 A) cos A − cos A − sin A (1 − sin2 A) = sin A + cos A − cos3 A − cos A − sin A + sin3 A = sin3 A − cos3 A
48. Observe that 2 sin A cos3 A − 2 sin3 A cos A = ( 2 sin A cos A) ( cos 2 A − sin 2 A) = ( sin 2 A)( cos 2 A) . 49. Observe that sin 2A A = tan 2 ⋅ = tan A . 1 + cos 2A 2
50. Observe that 1 − cos 2 A A = tan 2 ⋅ = tan A . sin 2A 2
51. Observe that
2 cos 2 θ − sin2 θ 2 cos 2 θ − sin2 θ 2 cos 2 θ − sin2 θ 2 cos 2θ = = = V= 1 + cos 2θ 1 + ( cos 2 θ − sin2 θ ) 1 + cos 2 θ − (1 − cos 2 θ ) 2 cos 2 θ =
cos 2 θ sin2 θ − = 1 − tan2 θ 2 2 cos θ cos θ
52. Using Exercise 51, observe that 2
2 π sin 1 1 2 6 π π V = 1 − tan2 = 1 − = 1 − 2 = 1 − = . 3 3 3 6 6 cos π 2 6
543
Chapter 5
1 − cos A A 53. Use sin = − with A = −45 to obtain 2 2
−45 sin ( −22.5 ) = sin =− 2
1 − cos ( −45 ) 2
=−
1−
2 2 = − 2− 2 = − 2− 2 . 2 4 2
1 + cos A A 54. Use cos = with A = 135 to obtain 2 2
2 1+ − 2 135 1 + cos135 2− 2 cos ( 67.5 ) = cos = = = = 2 2 4 2
2− 2 . 2
1 − cos A 3π A 55. First, note that using tan = with A = , we obtain 4 1 + cos A 2 1 1 1 3π (1) = = cot = 3π 8 tan 3π 3π − 1 cos tan 4 4 8 2 3 π 1 + cos 4
( ) ( )
( 4 ) = − 22 in (1) subsequently yields:
Next, using cos 3π
3π cot = 8
=
1
2 1− − 2 2 1+ − 2 1 4+4 2 +2 4−2
=
=
1 2+ 2 2 2− 2 2
=
1 3+2 2
544
1 2+ 2 2− 2
=
1 2+ 2 2+ 2 ⋅ 2− 2 2+ 2
Chapter 5 Review
1 − cos A A , Note: Recall that there are two other formulae equivalent to tan = 1 + cos A 2 and either one could be used in place of this one to obtain an equivalent (but perhaps sin A 3π A with A = yields different looking) result. For instance, using tan = 4 2 1 + cos A 1 1 1 3π = = (2) cot = 3π 8 tan 3π sin 3π 4 tan 4 8 2 1 + cos 3π 4
( ) ( )
( 4 ) = − 22 and sin (3π 4 ) = 22 , we can use these in (2) to obtain
Since cos 3π
13π cot = 8
1 2 2
=
2− 2 2− 2 2 2 2 −2 = ⋅ = = 2 2 2 2
2 −1 .
2 2 Even though the forms of the two answers are different, they are equivalent. 1−
7π 1 + cos A A 56. First, note that using the formula cos = − with A = yields 4 2 2 7π 2 1 + cos 1+ 7π π 7 4 =− 4=− 2 = − 2+ 2 . cos = cos 2 2 2 2 8 7π subsequently yields Now, using sin A = − 1 − cos 2 A with A = 8 1 1 −1 −1 7π = = = csc = 2 8 sin 7π 7π 2+ 2 2+ 2 − 1 − cos 2 − 1 8 1 − − 8 4 2 =
−1 2− 2 4
=
−2 2− 2
545
Chapter 5
7 3π and π < x < , 25 2 the terminal side of x lies in QIII. The diagram is as follows:
57. Since sin x = −
Hence, the terminal side of
x lies in QII, 2
x and so sin > 0. Therefore, we have 2 24 1− − 1 − cos x x 25 = sin = 2 2 2 =
49 7 7 2 = = . 50 5 2 10
4 π 58. Since cos x = − and < x < π , the terminal side of x lies in QII. Hence, the 5 2 x x terminal side of lies in QI, and so cos > 0. Therefore, we have 2 2 4 1+ − 1 + cos x x 5 = 1 = 10 . cos = = 2 2 10 10 2 59. The diagram is as follows:
x
Since the terminal side of x is in QIII, the x terminal side of lies in QII, and so 2 x tan < 0 . Therefore, we have 2 9 1− − 1 − cos x x 41 =− tan = − 1 + cos x 9 2 1+ − 41 =−
Note that cos x = −
9 . 41
546
50 25 5 =− = − 32 16 4
Chapter 5 Review
59. The diagram is as follows:
x
9 . 41 Since the terminal side of x is in QIII, the x terminal side of lies in QII, and so 2 x tan < 0 . Therefore, we have 2 Note that cos x = −
9 1− − 1 − cos x x 41 =− tan = − 1 + cos x 9 2 1+ − 41
=−
50 25 5 =− = − 32 16 4
17 15 , we know that cos x = . This, together with sin x < 0 , implies 15 17 x that the terminal side of x is in QIV. Hence, the terminal side of is in QII. 2 x Thus, sin > 0 . 2 So, we have 60. Since sec x =
15 1− 1 − cos x 2 17 x 17 = sin = = = 2 2 2(17) 17 2 1 − cos A π A 61. Using the formula sin = with A = yields 6 2 2
π 1 − cos π 6 = sin 6 = sin π . 2 2 12
547
Chapter 5
1 − cos A 11π A 62. Using the formula tan = with A = yields 6 1 + cos A 2 11π 11π 1 − cos 6 = tan 6 = tan 11π . 11π 12 2 1 + cos 6
63. 2
A A A A A A A A A 2 A + 2 sin cos + cos 2 = sin2 + cos 2 + 2 sin cos sin + cos = sin 2 2 2 2 2 2 2 2 2 2 A = 1 + sin 2 ⋅ = 1 + sin A 2
64. sec 2
65. Since cot 2
A A 1 A 1 1 − cos A + tan 2 = + tan2 = + 2 2 cos 2 A 2 1 + cos A 1 + cos A 2 2 2 1 − cos A 3 − cos A = + = 1 + cos A 1 + cos A 1 + cos A
1 1 A = = , we see that A 1 − cos A 2 tan2 2 1 + cos A A A 1 A 1 1 + cos A csc 2 + cot 2 = + cot 2 = + 2 2 sin 2 A 2 1 − cos A 1 − cos A 2 2 2 1 + cos A 3 + cos A = + = 1 − cos A 1 − cos A 1 − cos A
66. Use the known identity 1 + tan2 θ = sec 2 θ with θ =
that 1 + tan2
A A = sec 2 . 2 2
548
A to immediately conclude 2
Chapter 5 Review
67. Before graphing, observe that π π 1 − cos x 12 = sin x = sin π x . y= 12 2 24 2
This graph has the same shape as the graph of y = sin x with period
π
seen below:
68. Before graphing, observe that x x x y = cos 2 − sin2 = cos 2 ⋅ = cos x 2 2 2 So, this graph is simply the graph of y = cos x, as seen below:
69. 6 sin5x cos 2 x = 6 ⋅
1 [sin(5x + 2x ) + sin(5x − 2x )] = 3 [sin 7x + sin3x ] 2
549
π 24
= 24 , as
Chapter 5
70. 3sin 4 x sin 2x = 3 ⋅
1 3 [cos(4x − 2x ) − cos(4x + 2x )] = [ cos 2x − cos 6x ] 2 2
5x + 3x 5x − 3x 71. cos5x − cos3x = −2 sin sin = −2 sin(4 x )sin( x ) 2 2 3 3 5 5 2x+ 2x 2x−2x 5 3 x 72. sin x + sin x = 2 sin cos = 2 sin ( 2 x ) cos 2 2 2 2 2
4x 2x 4x 2x − 3 + 3 4x 2x x 3 3 73. sin cos = 2 sin cos x − sin = 2 sin 2 2 3 3 3
7x + x 7x − x 74. cos 7 x + cos x = 2 cos cos = 2 cos ( 4 x ) cos ( 3x ) 2 2
75.
76.
77.
cos8x + cos 2 x = sin8x − sin 2 x
sin5x + sin3x = cos 5x + cos 3x
sin A + sin B = cos A − cos B
8x + 2 x 8x − 2 x 2 cos cos 2 2
8x − 2 x = cot = cot 3x 2 8x − 2 x 8x + 2 x 2 sin cos 2 2 5 x + 3x 5 x − 3x 2 sin cos 2 2
5 x + 3x = tan = tan 4 x 2 5 x + 3x 5 x − 3x 2 cos cos 2 2
A+ B A− B 2 sin cos 2 2
A− B = − cot 2 A− B A+ B − 2 sin sin 2 2
550
Chapter 5 Review
78.
sin A − sin B = cos A − cos B
A− B A+ B 2 sin cos 2 2
A+ B = − cot 2 A+ B A− B − 2 sin sin 2 2
π π π 79. The correct identity is sin 2 x + = sin 2 x cos + cos 2 x sin . 4 4 4 Consider the graphs of the following functions. Note that the graphs of y1 (BOLD) and y2 are the same. (The graph of y3 is dotted.)
π y1 = sin 2 x + 4 π π y2 = sin ( 2x ) cos + cos ( 2x ) sin 4 4 π π y3 = sin ( 2 x ) cos − cos ( 2 x ) sin 4 4
π π π 80. The correct identity is cos 2 x − = cos 2 x cos + sin 2x sin . 3 3 3 Consider the graphs of the following functions. Note that the graphs of y1 (BOLD) and y3 are the same. (The graph of y2 is dotted.) π y1 = cos 2x − 3 π π y2 = cos ( 2 x ) cos − sin ( 2 x ) sin 3 3 π π y3 = cos ( 2 x ) cos + sin ( 2x ) sin 3 3
551
Chapter 5
81. Observe that sin ( 3( x + h ) ) − sin3x h
= =
sin3x cos(3h ) + sin(3h ) cos 3x − sin3x h sin3x ( cos(3h ) − 1) + sin(3h ) cos 3x
h sin3x ( cos(3h ) − 1) sin(3h ) cos 3x = + h h sin(3h ) 1 − cos(3h ) = cos 3x − sin3x h h
sin(3) 1 − cos(3) a. The following is the graph of y1 = cos3x − sin3x , along with 1 1 the graph of y = 3cos3x (dotted graph):
sin(0.3) 1 − cos(0.3) b. The following is the graph of y2 = cos3x − sin3x , along 0.1 0.1 with the graph of y = 3cos3x (dotted graph):
552
Chapter 5 Review
sin(0.03) 1 − cos(0.03) c. The following is the graph of y3 = cos3x − sin3x , along 0.01 0.01 with the graph of y = 3cos3x (dotted graph):
Looking at the sequence of graphs in parts a – c, it is clear that this sequence of graphs better approximates the graph of y = 3cos3x as h → 0 . 82. Observe that cos ( 3( x + h ) ) − cos 3x cos 3x cos(3h ) − sin3x sin(3h ) − cos 3x = h h cos 3x ( cos(3h ) − 1) − sin3x sin(3h ) = h cos(3h ) − 1 sin(3h ) = cos 3x − sin3x h h 1 − cos(3h ) sin(3h ) = − cos 3x − sin3x h h
553
Chapter 5
1 − cos(3) sin(3) a. The following is the graph of y1 = − cos3x − sin3x , along with 1 1 the graph of y = −3sin3x (dotted graph):
1 − cos(0.3) sin(0.3) b. The following is the graph of y2 = − cos3x − sin3x , along 0.1 0.1 with the graph of y = −3sin3x (dotted graph):
554
Chapter 5 Review
1 − cos(0.03) sin(0.03) c. The following is the graph of y2 = − cos3x − sin3x , 0.01 0.01 along with the graph of y = −3sin3x (dotted graph):
Looking at the sequence of graphs in parts a – c, it is clear that this sequence of graphs better approximates the graph of y = −3sin3x as h → 0 . 83. To the right are the graphs of the following functions: y1 = tan 2 x (solid) y2 = 2 tan x (dashed) 2 tan x 1 − tan2 x Note that the graphs of y1 and y3 are the same.
y3 =
555
Chapter 5
84. To the right are the graphs of the following functions: y1 = cos 2 x (solid)
y2 = 2 cos x (dashed) y3 = 1 − 2 sin2 x Note that the graphs of y1 and y3 are the same.
85. To the right are the graphs of the following functions: x y1 = cos (solid) 2 1 y2 = cos x (dashed) 2 1 + cos x y3 = − 2 Note that the graphs of y1 and y3 are the same. 86. To the right are the graphs of the following functions: x y1 = sin (solid) 2 1 y2 = sin x (dashed) 2 1 − cos x y3 = − 2 Note that the graphs of y1 and y3 are the same.
556
Chapter 5 Review
87. To the right are the graphs of the following functions: y1 = sin5x cos 3x (solid) y2 = sin 4 x (dashed) 1 [sin8x + sin 2x ] 2 Note that the graphs of y1 and y3 are the same.
y3 =
88. To the right are the graphs of the following functions: y1 = sin3x cos 5x (solid) y2 = cos 4x (dashed) 1 [sin8x − sin 2x ] 2 Note that the graphs of y1 and y3 are the same.
y3 =
557
Chapter 5 Practice Test Solutions ------------------------------------------------------------------1.
sec x − tan x = ( sec x − tan x ) =
( sec x + tan x ) = sec 2 x − tan2 x = 1 + tan2 x − tan2 x sec x + tan x ( sec x + tan x ) sec x + tan x
1 sec x + tan x
2. The identity tan x = 3. Observe that
sin x (2 n + 1)π does not hold for x = , where n is an integer. cos x 2
sin2 x + cos 2 x = 1 = 1 . So, the equation
sin2 x + cos 2 x = sin x + cos x is a conditional. Indeed, take for instance x =
π
. 4 Then, the left-side = 1, while the right-side is 2 . (Note: A common algebraic error is to say
a 2 + b 2 = a + b , which is false.)
1 − cos A π A 4. Use the odd identity for sine, followed by sin = with A = to 4 2 2 obtain
π 2 π 1 − cos 1− 4 π π 4 =− 2 = − 2− 2 = − 2− 2 . sin − = − sin = − sin = − 2 2 4 2 8 8 2 1 + cos A 11π A 5. Using the formula cos = − with A = yields 6 2 2
11π 3 1 + cos 11π 1+ 11π 6 =− 6=− 2 = − 2+ 3 . cos = cos 2 2 2 2 12
558
Chapter 5 Practice Test
1 − cos A 13π A 6. Using the formula sin = − with A = yields 4 2 2
13π 2 13π 1 − cos 1+ 4 4 13π 2 = − 2+ 2 . =− sin =− = sin 2 2 2 8 2 7. Using the formula tan( A + B ) =
tan A + tan B π π with A = , B = yields 4 3 1 − tan A tan B
π
π
tan + tan 7π π π 4 3 . (1) tan = tan + = π 12 4 3 1 − tan tan π 4 3
Since tan
π 4
= 1 and tan
π 3
=
sin
π
cos
π
3 =
3 1
2 = 3 , we can substitute these into (1) to 2
3 7π 1 + 3 1 + 3 1 + 3 1 + 2 3 + 3 = ⋅ = = −2 − 3 . further obtain tan = 1− 3 12 1 − 3 1 − 3 1 + 3
2 3π and < x < 2π , the terminal side of x lies in QIV. Hence, the 5 2 x x terminal side of lies in QII, and so sin > 0 . Therefore, we have 2 2 8. Since cos x =
1 − cos x x sin = = 2 2
559
1− 2
2 5 =
3 30 . = 10 10
Chapter 5
1 3π and π < x < , the 5 2 terminal side of x must be in QIII. The diagram is as follows: 9. Since sin x = −
From the Pythagorean Theorem, we have z 2 + ( −1)2 = 52 , so that z = −2 6 . As such, cos x = −
2 6 . Therefore, 5 2
2 6 1 2 cos 2 x = cos x − sin x = − −− 5 5 . 2
10. Consider the following diagram:
2
11. Consider the following diagram:
41
− 7
Observe that sin 2 x = 2 sin x cos x 40 4 5 = 2 − = − 41 41 41
Observe that 2 tan x tan 2 x = 1 − tan2 x 7 2 7 2− − 3 = 3 = −3 7 = 2 7 7 1− 1− − 9 3
560
Chapter 5 Practice Test
12. Using cos( A + B ) = cos A cos B − sin A sin B with A = 3x, B = 7x yields
cos3x cos 7x − sin3x sin 7 x = cos(3x + 7x ) = cos(10 x ) . 13. Consider the following two triangles:
− 5
−2 2
Hence, we have 5 2 1 2 2 − sin( A + B ) = sin A cos B + sin B cos A = − + − 3 3 3 3 2 2 10 2 10 − 2 =− + = 9 9 9
14. Consider the following two triangles:
− 7
2 5 − 5
561
Chapter 5
Hence, we have sec( A + B ) =
=
1 1 1 = = cos( A + B ) cos A cos B − sin A sin B 7 5 3 2 5 − − − − 4 5 4 5 1 35 6 5 + 20 20
15. Observe that −
=
20
5
( 7 + 6)
=
4 5 = 7 +6
2 tan x 2 tan x = − = − tan 2x . 2 2 1 − tan x 1 − tan x
16. If the terminal side of a + b is in QII, then
terminal side of
( 7 − 6 ) = 24 5 − 4 35 29 ( 7 + 6 )( 7 − 6 ) 4 5
1 + cos( a + b ) a+b = cos , since the 2 2
a+b is in QI. 2
17. Observe that 1 x +3 x −3 x +3 x −3 x + 3 x − 3 + − 2 sin cos = 2 ⋅ sin + sin = sin x + sin3 2 2 2 2 2 2 2 18. Observe that 10 cos(3 − x ) + 10 cos( x + 3) = 10 [ cos(3 − x ) + 10 cos( x + 3)] (3 − x ) + ( x + 3) (3 − x ) − ( x + 3) = 10 2 cos cos 2 2 = 20 cos 3cos( − x ) = 20 cos x cos 3
19. Let u = 3sin x , where −
π 2
≤x≤
π 2
. Then,
9 − u 2 = 9 − (3sin x )2 = 9 − 9sin2 x = 9(1 − sin2 x ) = 9 cos 2 x = 3 cos x .
20.
(1 + cos x )(1 − cos x ) = 1 − cos2 x = sin2 x = sin2 x sec 2x ( cos x + sin x )( cos x − sin x ) cos 2 x − sin2 x cos 2x
562
Chapter 5 Practice Test
21. Consider the following triangle:
Observe that cos ( x2 ) = −
=−
− 33
1 + cos x 2 1 − 633 6 − 33 = − 2 12
− 3
22.
1 1 − 2 sin2 A 1 − sin2 A = = cos 2 A 2 2 2
2 5π 5π 5π 2 5π 23. cos 2 − sin = cos 2 ⋅ = cos = − 2 8 8 8 4
24. cos ( 45 + x ) = cos 45 cos x − sin 45 sin x =
2 ( cos x − sin x ) 2
25. 1 2π π 2π π 1 π + + cos − = cos π + cos cos 2 3 3 3 3 2 3 =
1 1 4π 2π 1 1 2 cos cos = − = − 2 4 6 6 2 2
563
Chapter 5 Cumulative Review Solutions ------------------------------------------------------1.
360 ( 30 min.) 240 360 = x= = 45 min. x 30 min. 240
2. A = D = 110 (since they are vertical angles) and D = H = 110 (since they are corresponding angles) 1 3. First, note that 57 39′ = 57 + 39′ = 57.65 . So, sin ( 57.65 ) ≈ 0.8448 . ′ 60
4. Consider the following triangle:
α
Observe that 54 54 α = cos −1 ≈ 42 cos α = 73 73
5. 45 + 360 k , where k is any integer.
564
Chapter 5 Cumulative Review
6. Consider the following diagram:
θ
2 − 3
5
sin θ = −
3 15 =− 5 5
cos θ =
2 10 = 5 5
tan θ = −
3 6 =− 2 2
cot θ = −
2 6 =− 3 3
sec θ =
5 10 = 2 2
csc θ = −
5 15 =− 3 3
7. First, note that cot ( −450 ) = cot ( −90 ) = 0 and sin ( −450 ) = sin ( −90 ) = −1.
Hence,
cot ( −450 ) − sin ( −450 ) = 1 .
8. Observe that sin θ = −0.3424 θ = sin −1 ( −0.3424 ) ≈ −20 . Hence, the angle with
smallest positive measure coterminal with the given angle is 340 . 9. Consider the following triangle:
sec θ =
17
565
1 1 = 1 = − 17 cos θ − 17
Chapter 5
10.
17π 180 ⋅ = 153 20 π
3 11π 11. cos = 6 2
π 12. Using the formula s = rθ d yields 180° 5π 5π 180° π = r ( 25° ) ⋅ = 2 ft. r= 8 18 25° π 180° 1 2 π r θd yields 2 180° 1 2 π 2 A = ( 21 ft.) ( 50 ) ≈ 192 ft. 2 180°
13. Using the formula A =
5π rad 5π in. 14. v = rω = ( 6 in.) sec. = 2 12 sec. π rad 3π cm 15. First, note that v = rω = ( 6 cm ) sec. Then, = 4 sec. 2 s 3π cm v = s = vt = sec. ( 8 sec. ) = 12π cm ≈ 37.7 cm t 2 16. Amplitude
7 3
Period
2π 5
17. The following is some information regarding the given function:
2π 3 Amplitude 2 Reflection over x-axis No phase or vertical shift Period
2π We sketch the graph on 0, . 3
566
Chapter 5 Cumulative Review
18. The following is some information regarding the given function:
Period 2 Amplitude 2 No reflection over x-axis Phase shift 1 unit to the right No vertical shift We sketch the graph on [ 0,2 ] . 19. We must identify A, B, C, and D in the equation y = A cos( Bx + C ) + D . First, note that since the maximum is 3 and the minimum is 1, the amplitude A = 1. Also, inherent in this is a vertical shift of the standard cosine curve UP 1 unit. 2π = π , so that B = 2. Next, the period is B
Next, since the maximum occurs at x =
π
2
, we shift the standard cosine graph
π
2
C π = , so that C = π . B 2 Consequently, the equation we seek is y = cos(2x + π ) + 2 , which is equivalent to units to the right. As such, the phase shift is
y = 2 − cos 2 x if you apply the addition formula on the right-side of the previous equation. 20. The solid curve is the desired graph.
21.
567
Chapter 5
sec 4 x − 1 ( sec x + 1) ( sec x − 1) 22. = sec 2 x − 1 = tan2 x = 2 sec 2 x + 1 sec x + 1 2
23.
2
11π π 2π π 2π π 2π cos − = cos − − = cos − cos + sin − sin 12 4 3 4 3 4 3 2 1 2 3 − 2 − 6 2+ 6 = = − − + − = 4 4 2 2 2 2
24. Consider the following triangle:
Observe that cos 2 x = cos 2 x − sin2 x 2
2
527 24 7 = − −− = 625 25 25
1 − cos A 3π A 25. Using the formula tan = − with A = − yields 4 1 + cos A 2
3π 3π 1 − cos − − 2 3π 4 = − 1+ 2 = − 2 + 2 tan − = tan 4 = − 3π 1 − 22 2− 2 8 2 1 + cos − 4 =−
( 2 + 2 )( 2 + 2 ) = − 2 + 2 = −2 2 − 2 = − 2 − 1. 2 2 ( 2 − 2 )( 2 + 2 )
568
CHAPTER 6 Section 6.1 Solutions ------------------------------------------------------------------------------------
2 1. The equation arccos = θ is 2
2 2. The equation arccos − = θ is 2
2 . Since the 2 range of arccosine is [ 0, π ] , we
2 . Since the range 2 of arccosine is [ 0, π ] , we conclude that
equivalent to cos θ =
conclude that θ =
π 4
.
3 3. The equation arcsin − = θ is 2 3 . Since the 2 π π range of arcsine is − , , we 2 2 equivalent to sin θ = −
conclude θ = −
π 3
θ=
5. The equation cot −1 ( −1) = θ is equivalent to 2 2 = cos θ . Since the cot θ = −1 = − sin θ 2 2 range of inverse cotangent is ( 0, π ) , we 3π . 4
3π . 4
1 4. The equation arcsin = θ is 2 1 equivalent to sin θ = . Since the range of 2 π π arcsine is − , , we conclude that 2 2
θ=
.
conclude that θ =
equivalent to cos θ = −
π 6
.
3 6. The equation tan −1 = θ is 3 1 3 sin θ . = 2 = equivalent to tan θ = 3 cos θ 3 2 Since the range of inverse tangent is
π π π − , , we conclude that θ = . 6 2 2
569
Chapter 6
2 3 7. The equation arc sec = θ is 3 2 3 , which is 3 3 3 . = further the same as cos θ = 2 2 3 Since the range of inverse secant is π π 0, 2 ∪ 2 , π , we conclude
equivalent to sec θ =
that θ =
π 6
8. The equation arc csc ( −1) = θ is equivalent to csc θ = −1, which is further the same as sin θ = −1. Since the range of π π inverse cosecant is − ,0 ∪ 0, , we 2 2
conclude that θ = −
π 2
.
.
9. The equation csc −1 ( 2 ) = θ is
equivalent to csc θ = 2, which is further 1 the same as sin θ = . Since the range 2 π π of inverse cosecant is − ,0 ∪ 0, , 2 2
we conclude that θ =
π 6
.
(
)
11. The equation arc tan − 3 = θ is
equivalent to
10. The equation sec −1 ( −2 ) = θ is
equivalent to sec θ = −2, which is further 1 the same as cos θ = − . Since the range of 2 π π inverse secant is 0, ∪ , π , we 2 2 2π conclude that θ = . 3 12. The equation arc cot
( 3 ) = θ is 3
3
2 = sin θ . Since tan θ = − 3 = − 1 cos θ 2 the range of inverse tangent is
π π π − , , we conclude that θ = − . 3 2 2
2 = cos θ . sin θ 2 Since the range of inverse cotangent is
equivalent to cot θ = 3 =
1
( 0, π ) , we conclude that θ =
570
π 6
.
Section 6.1
1 13. The equation cos −1 = θ is 2 1 equivalent to cos θ = . Since the 2 range of arccosine is [ 0, π ] , we
conclude that θ =
π
3
, which
3 14. The equation cos −1 − = θ is 2
3 . Since the range 2 of arccosine is [ 0, π ] , we conclude that equivalent to cos θ = −
θ=
corresponds to θ = 60 . 2 15. The equation sin = θ is 2 −1
2 . Since the 2 π π range of arcsine is − , , we 2 2 equivalent to sin θ =
conclude that θ =
π
4
5π , which corresponds to θ = 150 . 6
16. The equation sin −1 ( 0 ) = θ is equivalent to sin θ = 0. Since the range of arcsine is π π − 2 , 2 , we conclude that θ = 0, which
corresponds to θ = 0 .
, which
corresponds to θ = 45 . 3 17. The equation cot −1 − = θ is 3 equivalent to 1 3 cos θ cot θ = − . Since =− 2 = 3 sin θ 3 2 the range of inverse cotangent is ( 0, π ) ,
we conclude that θ =
2π , which 3
18. The equation tan −1
( 3 ) = θ is 3
2 = sin θ . cos θ 2 Since the range of inverse tangent is π π π − , , we conclude that θ = , which 3 2 2
equivalent to tan θ = 3 =
corresponds to θ = 60 .
corresponds to θ = 120 .
571
1
Chapter 6
3 19. The equation arc tan = θ is 3 equivalent to 1 3 sin θ tan θ = . Since the = 2 = 3 cos θ 3 2 π π range of inverse tangent is − , , 2 2
we conclude that θ =
π
6
20. The equation arc cot (1) = θ is
2
2 = cos θ . sin θ 2 2 Since the range of inverse cotangent is
equivalent to cot θ = 1 =
( 0, π ) , we conclude that θ =
π
4
, which
corresponds to θ = 45 .
, which
corresponds to θ = 30 . 21. The equation arc csc ( −2 ) = θ is equivalent to csc θ = −2, which is 1 further the same as sin θ = − . Since 2 the range of inverse cosecant is π π − 2 ,0 ∪ 0, 2 , we conclude that
π θ = − , which corresponds to 6
θ = −30 .
2 3 22. The equation csc −1 − = θ is 3 2 3 equivalent to csc θ = − , which is 3 3 3 =− . further the same as sin θ = − 2 2 3 Since the range of inverse cosecant is π π − 2 ,0 ∪ 0, 2 , we conclude that
π θ = − , which corresponds to θ = −60 . 3
572
Section 6.1
(
)
(
)
23. The equation arc sec − 2 = θ is
24. The equation arc csc − 2 = θ is
equivalent to sec θ = − 2, which is 1 further the same as cos θ = − . 2 Since the range of inverse secant is π π 0, 2 ∪ 2 , π , we conclude that 3π θ = , which corresponds to 4
equivalent to csc θ = − 2, which is further 1 the same as sin θ = − . Since the range 2 π π of inverse cosecant is − ,0 ∪ 0, , we 2 2
θ = 135 .
conclude that θ = −
π
4
which corresponds
to θ = −45 .
25. cos −1 ( 0.5432 ) ≈ 57.10
26. arccos ( −0.3245 ) = 108.94o
27. arcsin ( 0.1223 ) ≈ 7.02o
28. sin −1 ( 0.7821) ≈ 51.45 .
29. tan −1 (1.895 ) ≈ 62.18
30. tan −1 ( 3.2678 ) ≈ 72.99
31.
32.
33.
34.
1 sec −1 (1.4973 ) = cos −1 ≈ 48.10 1.4973
1 csc −1 ( −3.7893 ) = sin −1 −3.7893
1 sec −1 ( 2.7864 ) = cos −1 ≈ 68.97 2.7864
1 csc −1 ( −6.1324 ) = sin −1 ≈ −9.39 −6.1324
≈ −15.30
35.
36.
1 cot −1 ( −4.2319 ) = π + tan −1 −4.2319 ≈ 180 − 13.30
1 cot −1 ( −0.8977 ) = π + tan −1 −0.8977 ≈ 180 − 48.09 = 131.91
= 166.70
573
Chapter 6
37. sin −1 ( −0.5878 ) ≈ −0.63
38. sin −1 ( 0.8660 ) ≈ 1.05
39. cos −1 ( 0.1423 ) ≈ 1.43
40. arccos ( 0.7469 ) ≈ 0.73
41. tan −1 (1.3242 ) ≈ 0.92
42. tan −1 ( −0.9279 ) ≈ −0.75
43.
44
1 cot −1 ( −0.5774 ) = π + tan −1 −0.5774
1 cot −1 ( 2.4142 ) = tan −1 ≈ 0.39 2.4142
≈ 2.09 45. 1 arc sec(3.462) = arccos ( 3.462 ) ≈ 1.28
46.
47.
48.
1 csc −1 ( 3.2361) = sin −1 ≈ 0.31 3.2361
1 csc −1 ( −2.9238) = sin −1 ≈ −0.35 −2.9238
1 sec −1 ( −1.0422 ) = cos −1 ≈ 2.86 −1.0422
5π 5π π 5π π 49. sin −1 sin since − ≤ ≤ = 2 12 2 12 12 5π 5π π 5π π 50. sin −1 sin − since − ≤ − ≤ = − 12 2 12 2 12 π π π 51. cos −1 cos = since 0 ≤ ≤ π 6 6 6 5π 5π 5π 52. cos −1 cos since 0 ≤ ≤π = 6 6 6 53. sin ( sin −1 (1.03 ) ) is undefined since 1.03 is not in the domain of inverse sine. 54. sin ( sin −1 (1.1) ) is undefined since 1.1 is not in the domain of inverse sine.
574
Section 6.1
π π 7π 55. Note that we need to use the angle θ in − , such that sin θ = sin − . 6 2 2 7π π π −1 To this end, observe that sin −1 sin − = sin sin = . 6 6 6
π π 7π 56. Note that we need to use the angle θ in − , such that sin θ = sin . To 2 2 6 7π π π −1 this end, observe that sin −1 sin = sin sin − = − . 6 6 6 4π 57. Note that we need to use the angle θ in [ 0, π ] such that cos θ = cos . To 3 4π 2π 2π . this end, observe that arccos cos = arccos cos = 3 3 3 5π 58. Note that we need to use the angle θ in [ 0, π ] such that cos θ = cos − . To 3 5π π π this end, observe that arccos cos − = arccos cos = . 3 3 3
11π 59. arccos cos = arccos 6
( ) = π6 3 2
4π 2π 1 60. arccos cos = arccos ( − 2 ) = 3 3
61. cos[cos −1 ( 2 )] is undefined since 2 is not in the domain of inverse cosine. 62. cos[cos −1 ( −1.5 )] is undefined since –1.5 is not in the domain of inverse cosine.
(
)
63. Since cot ( cot −1 x ) = x for all −∞ < x < ∞, we see that cot cot −1 3 =
3.
5π 64. Note that we need to use the angle θ in [ 0, π ] such that cot θ = cot . To this 4 5π π π end, observe that cot −1 cot = cot −1 cot = . 4 4 4
575
Chapter 6
π π π 65. Note that we need to use the angle θ in 0, ∪ , π such that sec θ = sec − . 2 2 3 π π π To this end, observe that sec −1 sec − = sec −1 sec = . 3 3 3 1 1 66. sec sec −1 is undefined since is not in the domain of inverse secant. 2 2 1 1 67. csc csc −1 is undefined since is not in the domain of inverse cosecant. 2 2 π π 7π 68. Note that we need to use the angle θ in − ,0 ∪ 0, such that csc θ = csc . 2 2 6 π 7π π −1 To this end, observe that csc −1 csc = csc csc − = − . 6 6 6 69. Since cot ( cot −1 x ) = x for all −∞ < x < ∞, we see that cot ( cot −1 0 ) = 0 .
π 70. Note that we need to use the angle θ in [ 0, π ] such that cot θ = cot − . To 4 π 3π 3π . this end, observe that cot −1 cot − = cot −1 cot = 4 4 4 71. Since tan −1 ( tan x ) = x for all −
π
72. Since tan −1 ( tan x ) = x for all −
π
2
2
<x<
<x<
π π , we see that tan −1 tan − = − . 2 4 4
π
π π , we see that tan −1 tan = . 2 4 4
π
576
Section 6.1
3 3 73. Let θ = sin −1 . Then, sin θ = , 4 4 as shown in the diagram:
2 2 74. Let θ = cos −1 . Then, cos θ = , as 3 3 shown in the diagram:
θ
θ
Using the Pythagorean Theorem, we see that z 2 + 32 = 4 2 , so that z = 7.
Using the Pythagorean Theorem, we see that z 2 + 2 2 = 32 , so that z = 5.
7 3 . Hence, cos sin −1 = cos θ = 4 4
5 2 . Hence, sin cos −1 = sin θ = 3 3
12 75. Let θ = tan −1 . Then, 5 12 tan θ = , as shown in the diagram: 5
7 7 76. Let θ = tan −1 . Then, tan θ = , 24 24 as shown in the diagram:
θ
θ
Using the Pythagorean Theorem, we see that 12 2 + 52 = z 2 , so that z = 13. 12 12 Hence, sin tan −1 = sin θ = . 13 5
Using the Pythagorean Theorem, we see that 7 2 + 24 2 = z 2 , so that z = 25. 24 7 Hence, cos tan −1 = cos θ = . 25 24
577
Chapter 6
3 3 77. Let θ = sin −1 . Then, sin θ = , 5 5 as shown in the diagram:
2 2 78. Let θ = cos −1 . Then, cos θ = , as 5 5 shown in the diagram:
θ
θ
Using the Pythagorean Theorem, we see that z 2 + 32 = 52 , so that z = 4. 3 3 Hence, tan sin −1 = tan θ = . 4 5 2 79. Let θ = sin −1 . Then, 5 sin θ =
2 , as shown in the diagram: 5
θ
Using the Pythagorean Theorem, we see that z 2 + 2 2 = 52 , so that z = 21. 2 Hence, tan cos −1 = tan θ = 5
21 . 2
7 7 80. Let θ = cos −1 , . Then, cos θ = 4 4 as shown in the diagram:
θ
2
7
578
Section 6.1
Using the Pythagorean Theorem, we see that z 2 +
( 2 ) = 5 , so that 2
that z 2 +
2
z = 23. So, 2 1 sec sin −1 = sec θ = cos θ 5
=
Using the Pythagorean Theorem, we see
( 7 ) = 4 , so that z = 3. So, 2
2
7 1 sec cos −1 sec = = θ cos θ 4 . .
5 5 23 = 23 23
4 4 7 = 7 7 Alternatively, you can use the formula 1 sec −1 x = cos −1 here. Indeed, x 7 1 −1 −1 sec cos = sec cos 4 4 7 =
4 7 4 4 = sec sec −1 = = 7 7 7
1 1 81. Let θ = cos −1 . Then, cos θ = , 4 4 as shown in the diagram:
1 1 82. Let θ = sin −1 . Then, sin θ = , as 4 4 shown in the diagram:
θ
θ
579
Chapter 6
Using the Pythagorean Theorem, we see that z 2 + 12 = 4 2 , so that z = 15. Hence, 1 1 csc cos −1 = csc θ = sin θ 4 . 4 4 15 = = 15 15
Using the Pythagorean Theorem, we see that z 2 + 12 = 4 2 , so that z = 15. Hence, 1 1 = 4. csc sin −1 = csc θ = sin θ 4 Alternatively, you can use the formula 1 csc −1 x = sin −1 here. Indeed, observe x that 1 −1 1 −1 csc sin = csc sin 1 4 4 = csc ( csc −1 ( 4 ) ) = 4
60 83. Let θ = sin −1 . Then, 61 60 sin θ = , as shown in the diagram: 61
41 41 84. Let θ = sec −1 . Then, sec θ = , 9 9 9 so that cos θ = , as shown in the 41 diagram:
θ
Using the Pythagorean Theorem, we see that z 2 + 60 2 = 612 , so that z = 11. Hence, 11 60 cot sin −1 = cot θ = . 60 61
θ
Using the Pythagorean Theorem, we see that z 2 + 92 = 412 , so that z = 40. Hence, 9 41 cot sec −1 = cot θ = . 40 9
580
Section 6.1
85. Observe that A = tan −1 ( 34 ) tan( A) = 34 and B = sin −1 ( 54 ) sin( B ) = 54 .
These relationships give us the following right triangles, where the third side in both cases is found using the Pythagorean theorem.
So,
cos( A − B ) = cos A cos B + sin A sin B = ( 54 )( 53 ) + ( 53 )( 54 ) = 24 25 .
86. Observe that A = tan −1 ( 125 ) tan( A) = 125 and B = sin −1 ( 35 ) sin( B ) = 53 .
These relationships give us the following right triangles, where the third side in both cases is found using the Pythagorean theorem.
So,
3 16 cos( A + B ) = cos A cos B − sin A sin B = ( 135 )( 54 ) − ( 12 13 )( 5 ) = − 65 .
87. Observe that A = cos −1 ( 135 ) cos( A) = 135 and B = tan −1 ( 34 ) tan( B ) = 34 .
These relationships give us the following right triangles, where the third side in both cases is found using the Pythagorean theorem.
So,
3 5 56 4 sin( A + B ) = sin A cos B + sin B cos A = ( 12 13 )( 5 ) + ( 5 )( 13 ) = 65 .
581
Chapter 6
88. Observe that A = cos −1 ( 35 ) cos( A) = 35 and B = tan −1 ( 125 ) tan( B ) = 125 .
These relationships give us the following right triangles, where the third side in both cases is found using the Pythagorean theorem.
So,
5 3 33 sin( A − B ) = sin A cos B − sin B cos A = ( 54 )( 12 13 ) − ( 13 )( 5 ) = 65 .
89. Observe that A = cos −1 ( 35 ) cos( A) = 35 . This relationship give us the following right triangle, where the third side is found using the Pythagorean theorem.
So, sin(2 A) = 2 sin( A) cos( A) = 2 ( 54 )( 53 ) = 24 25 .
90. Observe that A = sin −1 ( 53 ) sin( A) = 53 . This relationship give us the following right triangle, where the third side is found using the Pythagorean theorem.
So, cos(2 A) = cos 2 ( A) − sin2 ( A) = ( 54 ) − ( 53 ) 2
= 257 .
582
2
Section 6.1
91. Observe that A = sin −1 ( 135 ) sin( A) = 135 .
92. Observe that A = cos −1 ( 135 ) cos( A) = 135 .
This relationship give us the following right triangle, where the third side is found using the Pythagorean theorem.
This relationship give us the following right triangle, where the third side is found using the Pythagorean theorem.
So,
2 ( 125 ) 2 tan( A) 120 tan(2 A) = . = = 2 2 1 − tan ( A) 1 − ( 125 ) 119
So,
93. Let θ = sin −1 ( u ). Then, sin θ = u. Using the Pythagorean theorem to get the third side, we have the following right triangle:
94. Let θ = cos −1 ( u ). Then, cos θ = u. Using the Pythagorean theorem to get the third side, we have the following right triangle:
So,
So,
1 − u2 cos θ = = 1 − u2 . 1
sin θ =
tan(2 A) =
583
2 ( 125 ) 2 tan( A) 120 . = =− 2 2 1 − tan ( A) 1 − ( 125 ) 119
1 − u2 = 1 − u2 . 1
Chapter 6
95. Let θ = cos −1 ( u ). Then, cos θ = u. Using the Pythagorean theorem to get the third side, we have the following right triangle:
So, 1 − u2 tan θ = . u
97. Solve the following equation in [0, 11]: 4.3cos ( π6 t ) + 56.2 = 56.2 4.3cos ( π6 t ) = 0
96. Let θ = sin −1 ( u ). Then, sin θ = u. Using the Pythagorean theorem to get the third side, we have the following right triangle:
So, tan θ =
6
1 − u2
=
u 1 − u2 . 1 − u2
98. Solve the following equation in [0, 11]: 4.3cos ( π6 t ) + 56.2 = 51.9 4.3cos ( π6 t ) = −4.3
cos ( π6 t ) = 0 π
u
cos ( π6 t ) = −1
t = π2 or 32π
π 6
t =π
t = 3 or 9 So, this occurs in April or October.
So, this occurs in July.
99. Solve the following equation in [0, 6]: 5.5 + 1.5sin ( π6 t ) = 7.0
100. Solve the following equation in [0, 6]: 5.5 + 1.5sin ( π6 t ) = 6.25
1.5sin ( π6 t ) = 1.5
t=6
1.5sin ( π6 t ) = 0.75
sin ( π6 t ) = 1 π 6
sin ( π6 t ) = 0.5
t = π2
t =3 In month 3, the monthly average pollen level is 7.0.
π 6
t = π6 or 56π
t = 1 or 5 In months 1 and 5, the monthly average pollen level is 6.25.
584
Section 6.1
101. Use i = I sin ( 2π f t ) with f = 5
102. Use i = I sin ( 2π f t ) with f = 100 and
and I = 115. Find the smallest positive value of t for which i = 85. To this end, observe 115sin ( 2π ⋅ 5t ) = 85
I = 240. Find the smallest positive value of t for which i = 100. To this end, observe 240 sin ( 2π ⋅100t ) = 100
85 sin (10π t ) = 115 85 sin −1 115 t= ≈ 0.026476 10π So, t ≈ 0.026476 sec. = 26 ms .
sin ( 200π t ) =
100 240
100 sin −1 240 ≈ 0.000684 t= 200π So, t ≈ 0.000684 sec. = 0.68 ms .
103. Given that H (t ) = 12 + 2.4 sin(0.017t − 1.377), we must find the value of t for which H (t ) = 14.4. To this end, we have 12 + 2.4 sin(0.017t − 1.377) = 14.4 sin(0.017t − 1.377) = 1 0.017t − 1.377 = sin −1 (1) = π t=
2
π + 1.377 2
≈ 173.4 0.017 Now, note that 173.4 − 151 = 22.4. As such, this corresponds to June 22.
104. Given that H (t ) = 12 + 2.4 sin(0.017t − 1.377), we must find the value of t for which H (t ) = 9.6. To this end, we have 12 + 2.4 sin(0.017t − 1.377) = 9.6 sin(0.017t − 1.377) = −1 0.017t − 1.377 = sin −1 ( −1) = − π
2
− π + 1.377 2 ≈ −11.4 0.017 So, counting backward into December 12 days implies this corresponds to Dec.19.
t=
585
Chapter 6
105. We need to find the smallest value 106. We need to find the smallest value of t of t for which larger than 11.3 for which 12.5 cos ( 0.157t ) + 2.5 = 0, 12.5 cos ( 0.157t ) + 2.5 = 15. and the graph of the left-side is decreasing prior to this value. We solve this graphically. The solid graph corresponds to the left-side of the equation. We have:
We approach this graphically, where the solid graph corresponds to the left-side of the equation. We have:
This occurs at approximately 40 years. Notice that the solution is approximately t = 11.3, or about 11 years. 107. Consider the following diagram:
Let θ = α + β . Then, tan α + tan β 1 − tan α tan β 1 7 8 + 8x x x = = 2x = 2 1 7 x − 7 x − 7 1− x2 x x
tan θ = tan (α + β ) =
β
α
108. Using Exercise 103, we see that tan θ =
8x 8x , so that θ = tan −1 2 . Thus, x −7 x −7 2
we have the following specific calculations: 80 x = 10: θ = tan −1 ≈ 0.71 radians ≈ 41 93 160 x = 20: θ = tan −1 ≈ 0.39 radians ≈ 22 393
586
Section 6.1
k 2 tan −1 f d with 109. Use the formula M = 1 − π 2 f = 2 m = 0.002 km and d = 4 km. We have the following specific calculation: k = 2 km: 2 2 tan −1 0.002 4 = 0.001 π − 2 tan −1 1 ≈ 0.0007048 km ≈ 0.70 m 1 − M= 2 π π 2 k = 10 km: 10 2 tan −1 0.002 4 = 0.001 π − 2 tan −1 5 ≈ 0.00024 km ≈ 0.24 m 1 − M= 2 π π 2
k 2 tan −1 f d 110. Use the formula M = 1 − 2 π with f = 3 m = 0.003 km and d = 2.5 km . We have the following specific calculation: k = 5 km: 5 2 tan −1 0.003 2.5 0.0015 π − 2 tan −1 ( 2 ) ≈ 0.00044 km ≈ 0.44 m 1 − = M= 2 π π
k = 10 km: 10 2 tan −1 0.003 2.5 = 0.0015 π − 2 tan −1 4 ≈ 0.00023 km ≈ 0.23 m 1 − M= ( ) 2 π π
587
Chapter 6
111. Consider the following diagram:
300 , we have 200 − x 300 α = tan −1 . 200 − x 150 Also, since tan β = , we have x 150 β = tan −1 . x Therefore, since α + β + θ = π , we see that
Since tan α =
300 −1 150 − tan . 200 − x x
θ = π − tan−1
α θβ
112. Consider the following diagram:
α θβ
280 , we have 200 − x 280 α = tan −1 . 200 − x 130 Also, since tan β = , we have x 130 β = tan −1 . x Therefore, since α + β + θ = π , we see that
Since tan α =
280 −1 130 − tan . 200 − x x
θ = π − tan−1 750 o 113. θ = tan −1 ≈ 53.7 550
1000 o 114. θ = tan −1 ≈ 59.0 600
6 115. θ = cos −1 ≈ 68.7 13 21
−11 116. θ = cos −1 ≈ 164.7 10 13
588
Section 6.1
π π 117. The identity sin −1 ( sin x ) = x is valid only for x in the interval − , , not 2 2 [0, π ]. 118. The identity cos −1 ( cos x ) = x is valid only for x in the interval [ 0, π ] , not
π π − 2 , 2 . 119. In general, cot −1 x ≠
1 . tan −1 x
120. In general, csc −1 x ≠
1 . sin −1 x
121. False. The portion to the right of the y-axis, when reflected over the y-axis does not match up identically with the left portion, as seen below: 1 More precisely, note that for instance sec −1 (1) = cos −1 = 0, while 1 1 sec −1 ( −1) = cos −1 − = π . As such, sec −1 ( x ) ≠ sec −1 ( − x ) , for all x in the domain of 1 inverse secant.
1 122. True. First, judging from the graph of y = csc −1 x = sin −1 , it seems as x though the inverse cosecant function is odd. To prove this, observe that since the inverse sine function is odd, we obtain 1 −1 1 −1 csc −1 ( − x ) = sin −1 = − sin = − csc ( x ) . −x x From the definition of odd, we are done. (Note: In general, if f and g are both odd functions, then the composition f g is odd on its domain.) 123. False. It should be 120.
124. False. It should be −30.
1 1 125. sec −1 does not exist since is not in the domain of the inverse secant 2 2 function (which coincides with the range of the secant function).
589
Chapter 6
1 1 126. csc −1 does not exist since is not in the domain of the inverse cosecant 2 2 function (which coincides with the range of the cosecant function).
π 127. Consider the function f ( x ) = 2 − 4 sin x − . 2 a. Note that this function has a phase shift of π
2
units to the right. So, we take the
π π π interval used to define sin −1 x, namely − , and add to both endpoints to 2 2 2 get the interval [ 0, π ] . Note that f is, in fact, one-to-one on this interval. b. Now, we determine a formula for f −1, along with its domain:
π y = 2 − 4 sin x − 2 y−2 π = sin x − −4 2 π y−2 sin −1 = x− 2 −4 π 2−y + sin −1 =x 2 4
π
2−x + sin −1 2 4 The domain of f −1 is equal to the range of f. Since the amplitude of f is 4 and there is a vertical shift up of 2 units, we see that the range of f is [ −2,6 ] . Hence, So, the equation of the inverse of f is given by: f −1 ( x ) =
the domain of f −1 is [ −2,6 ].
590
Section 6.1
π 128. Consider the function f ( x ) = 3 + cos x − . 4 a. Note that this function has a phase shift of π
4
units to the right. So, we
take the interval used to define cos −1 x, namely [ 0, π ] and add
π 4
to both
π 5π endpoints to get the interval , . Note that f is, in fact, one-to-one on 4 4 this interval. b. Now, we determine a formula for f −1, along with its domain:
π y = 3 + cos x − 4 π y − 3 = cos x − 4 cos −1 ( y − 3 ) = x −
π 4
π 4
+ cos −1 ( y − 3 ) = x
π
+ cos −1 ( x − 3 ) 4 The domain of f −1 is equal to the range of f. Since the amplitude of f is 1 and there is a vertical shift up of 3 units, we see that the range of f is [ 2, 4 ]. Hence, the domain So, the equation of the inverse of f is given by: f −1 ( x ) =
of f −1 is [ 2, 4 ].
129. Consider the following triangle:
Let θ = cos −1 ( x1 ) . Then, x1 = cos θ . So, using the triangle, we see that x2 − 1 sin θ = x
x −1 2
θ
591
Chapter 6
Let θ = sin −1 ( x1 ) . Then, x1 = sin θ . So,
130. Consider the following triangle:
using the triangle, we see that cos θ =
x2 − 1 x
θ x2 − 1 131. The graphs of the following two functions on the interval [ −3,3] is below:
Y1 = sin ( sin −1 x ) ,
132. The graphs of the following two functions on the interval [ −3,1] is below:
Y1 = cos ( cos −1 x ) ,
Y2 = x
The graphs are different outside the interval [ −1,1] because the identity
sin ( sin −1 x ) = x only holds for −1 ≤ x ≤ 1.
Y2 = x
The results are different outside the interval [ −1,1] because the identity cos ( cos −1 x ) = x only holds for −1 ≤ x ≤ 1.
592
Section 6.1
133. The graphs of the following two π π functions on the interval − , is 2 2 −1 below: Y1 = csc ( csc x ) , Y2 = x
Observe that the graphs do indeed coincide on this interval. This occurs since csc −1 ( csc x ) = x holds when π π x ∈ − ,0 ∪ 0, . 2 2
134. The graphs of the following two functions on the interval [ 0, π ] is below:
Y1 = sec −1 ( sec x ) ,
Y2 = x
Observe that the graphs do indeed coincide on this interval. This occurs since sec −1 ( sec x ) = x holds when 0≤ x ≤π .
593
Section 6.2 Solutions -----------------------------------------------------------------------------------1. The values of θ in [ 0,2π ) that satisfy cos θ = −
2 3π 5π , . are θ = 2 4 4
2. The values of θ in [ 0,2π ) that satisfy sin θ = −
2 5π 7π , . are θ = 2 4 4
3. The values of θ in [ 0,2π ) that satisfy sin θ =
3 π 2π are θ = , 2 3 3
4. The values of θ in [ 0,2π ) that satisfy cos θ = − 5. The values of θ in [ 0,2π ) that satisfy cos θ =
3 5π 7π , are θ = 2 6 6
1 π 5π are θ = , 2 3 3
6. The values of θ in [ 0,2π ) that satisfy sin θ = −
1 7π 11π , are θ = 2 6 6
1 7. First, observe that csc θ = −2 is equivalent to sin θ = − . The values of θ in 2 1 [0, 4π ) that satisfy sin θ = − are 2 7π 7π 11π 11π 7π 11π 19π 23π θ= , , , , , . + 2π , + 2π = 6 6 6 6 6 6 6 6 1 8. First, observe that sec θ = −2 is equivalent to cos θ = − . The values of θ in 2 1 [0, 4π ) that satisfy cos θ = − are 2 2π 2π 4π 4π 2π 4π 8π 10π θ= , , , , , . + 2π , + 2π = 3 3 3 3 3 3 3 3
594
Section 6.2
9. The only way tan θ = 0 is for sin θ = 0 . The values of θ in ( −∞, ∞ ) that
satisfy sin θ = 0 (and hence, the original equation) are θ = n π , where n is an integer . 10. The only way cot θ = 0 is for cos θ = 0 . The values of θ in ( −∞, ∞ ) that satisfy cos θ = 0 (and hence, the original equation) are θ =
(2 n + 1) π , where n is an integer . 2
1 must satisfy 2 7π 7π 11π 11π 7π 11π 19π 23π 2θ = , , , , , + 2π , + 2π = . 6 6 6 6 6 6 6 6 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 7π 11π 19π 23π , , , θ= 12 12 12 12 11. The values of θ in [ 0,2π ) that satisfy sin 2θ = −
3 must satisfy 2 π π 11π 11π π 11π 13π 23π 2θ = , , , , + 2π , + 2π = , . 6 6 6 6 6 6 6 6 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: π 11π 13π 23π , , , θ= 12 12 12 12 12. The values of θ in [ 0,2π ) that satisfy cos 2θ =
1 θ 13. The values of θ in ( −∞, ∞ ) that satisfy sin = − must satisfy 2 2 θ 7π 11π = + 2nπ , + 2nπ , where n is an integer. 2 6 6 So, multiplying all values by 2 yields the following values of θ which satisfy the original equation: 11π 7π 11π 7π θ =2 + 2 nπ , 2 + 2nπ = + 4 nπ , + 4nπ 3 6 6 3 =
( 7 + 12n ) π , (11 + 12n ) π , where n is an integer . 3
3
595
Chapter 6
θ 14. The values of θ in ( −∞, ∞ ) that satisfy cos = −1 must satisfy 2
θ
= ( 2n + 1) π , where n is an integer. 2 So, multiplying by 2 yields the following values of θ which satisfy the original equation: θ = 2 ( 2 n + 1) π , where n is an integer . 2 must satisfy 2 π π 3π 3π π 3π 9π 11π 3θ = , + 2π , , . + 2π = , , , 4 4 4 4 4 4 4 4 So, dividing all values by 3 yields the following values of θ , which satisfy the original equation: π π 3π 11π θ= , , , 12 4 4 12
15. The values of θ in [ 0, π ) that satisfy sin ( 3θ ) =
2 must satisfy 2 3π 3π 5π 5π 3π 5π 11π 13π 3θ = , , , , . + 2π , , + 2π , = 4 4 4 4 4 4 4 4 So, dividing all values by 3 yields the following values of θ , which satisfy the original equation: π 5π 11π θ= , , 4 12 12 16. The values of θ in [ 0, π ) that satisfy cos ( 3θ ) = −
596
Section 6.2
17. The values of θ in [ −2π ,2π ) that satisfy tan 2θ = 3 must satisfy
4π 4π 2π 2π 5π 5π + 2π , − − 2π , − − 2π , , − , − 3 3 3 3 3 3 3 3 . π 4π 7π 10π 2π 5π 8π 11π = , , , ,− ,− ,− ,− 3 3 3 3 3 3 3 3 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: π 4π 7π 10π 2π 5π 8π 11π θ= , , , ,− ,− ,− ,− 6 6 6 6 6 6 6 6 5π 4π 11π π 2π 7π 5π π , , ,− ,− ,− ,− = , 6 3 6 3 3 6 3 6 2θ =
π
,
π
+ 2π ,
18. The values of θ in ( −∞, ∞ ) that satisfy tan 2θ = − 3 must satisfy
2θ =
2π 5π + 2 nπ , + 2nπ , where n is an integer. 3 3
This reduces to
2π + nπ , where n is an integer. 3 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 2θ =
θ=
π 3
+
nπ π ( 2 + 3n ) , where n is an integer . = 2 6
1 19. First, observe that sec θ = −2 is equivalent to cos θ = − . The values of θ in 2 1 2π 4π [ −2π ,0 ) that satisfy cos θ = − are θ = − , − . 2 3 3 20. First, observe that csc θ =
2 3 3 3 is equivalent to sin θ = = . The values of 2 3 2 3
θ in [ −π , π ) that satisfy sin θ =
3 π 2π are θ = , . 2 3 3
597
Chapter 6
θ 21. First, observe that cot = 1 is 2 equivalent to θ cos 2 θ 2 2 = cot = 1 = θ 2 2 2 sin 2 The value of θ in [ 0,2π ) that satisfies this equation must satisfy
θ=
π 2
θ
.
2
=
π
4
3 θ 22. First, observe that cot = − is 3 2 equivalent to θ cos 1 − 3 1 θ 2 2 = cot = − =− = 3 θ 3 3 2 2 sin 2 The value of θ in [ 0,2π ) that satisfies
, so that
this equation must satisfy that θ =
4π . 3
23. First, observe that 2 sin 2θ = 3 is equivalent to sin 2θ =
θ
2
=
2π , so 3
3 . The values of θ in 2
3 must satisfy 2 π π 2π 2π π 2π 7π 8π + 2π , + 2π = , . 2θ = , , , , 3 3 3 3 3 3 3 3 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: π 2π 7π 8π π π 7π 4π = , , . θ= , , , , 6 6 6 6 6 3 6 3
[0,2π ) that satisfy sin 2θ =
θ θ − 2 24. First, observe that 2 cos = − 2 is equivalent to cos = . The values 2 2 2 θ 3π 3π θ − 2 of θ in [ 0,2π ) that satisfy cos = must satisfy = , so that θ = . 2 2 2 4 2
598
Section 6.2
25. First, observe that 3tan 2θ − 3 = 0 is equivalent to tan 2θ =
3 . The values of 3
3 must satisfy 3 π π 7π 7π π 7π 13π 19π + 2π , + 2π = , . 2θ = , , , , 6 6 6 6 6 6 6 6 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: π 7π 13π 19π , , , . θ= 12 12 12 12
θ in [ 0,2π ) that satisfy tan 2θ =
θ θ 26. First, observe that 4 tan − 4 = 0 is equivalent to tan = 1 . The values of θ 2 2
θ π π θ in [ 0,2π ) that satisfy tan = 1 must satisfy = , so that θ = . 2 4 2 2 1 27. First, observe that 2 cos ( 2θ ) + 1 = 0 is equivalent to cos ( 2θ ) = − . The values of 2 1 θ in [ 0,2π ) that satisfy cos ( 2θ ) = − must satisfy 2 2π 2π 4π 4π 2π 4π 8π 10π + 2π , + 2π = . 2θ = , , , , , 3 3 3 3 3 3 3 3 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 2π 4π 8π 10π π 2π 4π 5π = , θ= , , , , , 6 6 6 6 3 3 3 3
599
Chapter 6
28. First, observe that 4 csc ( 2θ ) + 8 = 0 is equivalent to csc ( 2θ ) = −2 , which can be
1 further simplified to sin ( 2θ ) = − . The values of θ in [ 0,2π ) that satisfy 2 1 sin ( 2θ ) = − must satisfy 2 7π 7π 11π 11π 7π 11π 19π 23π + 2π , + 2π = . 2θ = , , , , , 6 6 6 6 6 6 6 6 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 7π 11π 19π 23π , , , θ= 12 12 12 12
θ 29. First, observe that 3 cot − 3 = 0 is equivalent to 2 θ cos 3 θ 3 2. 2 = cot = = 1 θ 3 2 sin 2 2 The value of θ in [ 0,2π ) that satisfy this equation must satisfy
30. First, observe that
θ 2
=
π 6
, so that θ =
3 sec ( 2θ ) + 2 = 0 is equivalent to sec ( 2θ ) = −
be further simplified to cos ( 2θ ) = −
π 3
.
2 , which can 3
3 . The values of θ in [ 0,2π ) that satisfy 2
3 must satisfy 2 5π 5π 7π 7π 5π 7π 17π 19π + 2π , + 2π = . 2θ = , , , , , 6 6 6 6 6 6 6 6 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 5π 7π 17π 19π , , , θ= 12 12 12 12 cos ( 2θ ) = −
600
Section 6.2
31. Factoring the left-side of tan2 θ − 1 = 0 yields the equivalent equation ( tan θ − 1)( tan θ + 1) = 0 which is satisfied when either tan θ − 1 = 0 or tan θ + 1 = 0 .
The values of θ in [ 0,2π ) that satisfy tan θ = 1 are θ = satisfy tan θ = −1 are θ =
π 5π 4
,
4
, and those which
3π 7π . Thus, the solutions to the original equation are , 4 4 π 3π 5π 7π , , . θ= , 4 4 4 4
32. Factoring the left-side of sin2 θ + 2 sin θ + 1 = 0 yields the equivalent equation 2 ( sin θ + 1) = 0 which is satisfied when sin θ + 1 = 0 . The value of θ in [0,2π ) that
satisfies sin θ = −1 is θ =
3π . 2
33. Factoring the left-side of 2 cos 2 θ − cos θ = 0 yields the equivalent equation cos θ ( 2 cos θ − 1) = 0 which is satisfied when either cos θ = 0 or 2 cos θ − 1 = 0 . The
values of θ in [ 0,2π ) that satisfy cos θ = 0 are θ = cos θ =
π 3π 2
,
2
, and those which satisfy
1 π 5π are θ = , . Thus, the solutions to the original equation are 2 3 3 π 3π π 5π θ= , . , , 2 2 3 3
34. Factoring the left-side of tan2 θ − 3 tan θ = 0 yields the equivalent equation
(
)
tan θ tan θ − 3 = 0 which is satisfied when either tan θ = 0 or tan θ − 3 = 0 . The
values of θ in [ 0,2π ) that satisfy tan θ = 0 are θ = 0, π ,2π , and those which satisfy 3 tan θ = 3 =
1
2 are θ = π , 4π . Thus, the solutions to the original equation are 3 3 2 π 4π θ = 0, π , , . 3 3
601
Chapter 6
35. Observe that −2 sin2 θ = sin θ is equivalent to 0 = 2 sin2 θ + sin θ . Factoring the right-side of this equation yields the equivalent equation sin θ ( 2 sin θ + 1) = 0 which is
satisfied when either sin θ = 0 or 2 sin θ + 1 = 0. The values of θ in [ 0,2π ) that satisfy sin θ = 0 are θ = 0, π , and those which satisfy sin θ = −
solutions to the original equation are θ = 0,π ,
1 7π 11π , are . Thus, the 2 6 6
7π 11π , . 6 6
36. Observe that sec 2 θ = 3sec θ − 2 is equivalent to sec 2 θ − 3sec θ + 2 = 0 . Factoring the left side of this equation yields the equivalent equation ( sec θ − 1)( sec θ − 2 ) = 0 ,
which is satisfied when either sec θ = 1 or sec θ = 2 . The value of θ in [ 0,2π ) that satisfies sec θ = 1 (or, equivalently, cos θ = 1 ) is θ = 0 , and those which satisfy 1 π 5π sec θ = 2 (or equivalently cos θ = ) are θ = , . Thus, the solutions to the original 3 3 2 π 5π equation are θ = 0, , . 3 3 37. Factoring the left-side of csc 2 θ + 3csc θ + 2 = 0 yields the equivalent equation ( csc θ + 2 )( csc θ + 1) = 0 which is satisfied when either csc θ + 2 = 0 or csc θ + 1 = 0 .
1 The values of θ in [ 0,2π ) that satisfy csc θ = −2 (or equivalently sin θ = − ) are 2 7π 11π , and those which satisfy csc θ = −1 (or equivalently sin θ = −1 ) are θ= , 6 6 3π 7π 11π 3π θ= , , . Thus, the solutions to the original equation are θ = . 2 6 6 2 38. Observe that cot 2 θ = 1 is equivalent to cot 2 θ − 1 = 0 . Factoring the left-side of this equation yields the equivalent equation ( cot θ − 1)( cot θ + 1) = 0 which is satisfied
when either cot θ − 1 = 0 or cot θ + 1 = 0 . The values of θ in [ 0,2π ) that satisfy cot θ = 1 are θ =
π 5π 4
,
4
and those which satisfy cot θ = −1 are θ =
the solutions to the original equation are θ =
602
π 3π 5π 7π 4
,
4
,
4
,
4
.
3π 7π , . Thus, 4 4
Section 6.2
39. Factoring the left-side of sin2 θ + 2 sin θ − 3 = 0 yields the equivalent equation ( sin θ + 3)( sinθ − 1) = 0 which is satisfied when either sin θ + 3 = 0 or sinθ − 1 = 0 . Note that the equation sin θ + 3 = 0 has no solution since –3 is not in the range of sine.
The value of θ in [ 0,2π ) that satisfies sin θ = 1 is θ =
π 2
.
40. Factoring the left-side of 2 sec 2 θ + sec θ − 1 = 0 yields the equivalent equation ( 2 sec θ − 1)( sec θ + 1) = 0 which is satisfied when either 2 sec θ − 1 = 0 or secθ + 1 = 0 .
1 (or equivalently cos θ = 2 ) has no solution since 2 is 2 not in the range of cosine. The value of θ in [ 0,2π ) that satisfies sec θ = −1 is Note that the equation sec θ =
θ=π . 41. Observe that 4 cos 2 θ − 3 = 0 is equivalent to cos 2 θ = 34 , which implies cos θ = ± 23 . This is satisfied when θ =
π 5π 7π 11π
, , , . 6 6 6 6
42. Factoring the left-side of 4 sin2 θ − 4 sin θ − 3 = 0 yields the equivalent equation
( 2 sin θ + 1)( 2 sinθ − 3) = 0 , which implies that either sin θ = − 12 or sin θ = 32 . The
latter equation has no solution, but the former is satisfied when θ =
7π 11π , . 6 6
1 43. Observe that 9tan2 θ − 3 = 0 is equivalent to tan2 θ = , which implies 3 3 π 5π 7π 11π tan θ = ± . This is satisfied when θ = , , , . 3 6 6 6 6 44. Observe that 2 cos 2 θ − 7 cos θ = −3 is equivalent to 2 cos 2 θ − 7 cos θ + 3 = 0. Factoring the left-side of this equation yields the equivalent equation 1 ( 2 cos θ − 1)( cos θ − 3) = 0, which is satisfied when either cos θ = or cos θ = 3. Note 2 that the equation cos θ = 3 has no solution since 3 is not in the range of cosine. The 1 π 2π values of θ in [ 0,2π ) that satisfy cos θ = are θ = , . 2 3 3
603
Chapter 6
45. In order to find all of the values of θ in 0 ,360 that satisfy sin 2θ = −0.7843 ,
we proceed as follows: Step 1: Find the values of 2θ whose sine is −0.7843. Indeed, observe that one solution is sin −1 ( −0.7843 ) ≈ −51.655 , which is in QIV. Since the angles we seek have positive measure, we use the representative 360 − 51.655 ≈ 308.345. A second solution occurs in QIII, and has value 180 + 51.655 = 231.655. Step 2: Use periodicity to find all values of θ that satisfy the original equation. Using periodicity with the solutions obtained in Step 1, we see that 2θ = 308.345 , 308.345 + 360 , 231.655 , 231.655 + 360 = 308.345 , 668.345 , 231.655 , 591.655 and so, the solutions to the original equation are approximately:
θ ≈ 115.83 , 295.83 , 154.17 , 334.17 46. In order to find all of the values of θ in 0 ,360 that satisfy cos 2θ = 0.5136,
we proceed as follows: Step 1: Find the values of 2θ whose cosine is 0.5136. Indeed, observe that one solution is cos −1 ( 0.5136 ) ≈ 59.096 , which is in QI. A second solution occurs in QIV, and has value 360 − 59.096 = 300.904.
Step 2: Use periodicity to find all values of θ that satisfy the original equation. Using periodicity with the solutions obtained in Step 1, we see that 2θ = 59.096 , 59.096 + 360 , 300.904 ,300.904 + 360 = 59.096 , 419.096 , 300.904 , 660.904 and so, the solutions to the original equation are approximately:
θ ≈ 29.55 , 209.55 , 150.45 , 330.45
604
Section 6.2
θ 47. In order to find all of the values of θ in 0 ,360 that satisfy tan = −0.2343, 2 we proceed as follows:
θ
whose tangent is −0.2343. 2 Indeed, observe that one solution is tan −1 ( −0.2343 ) ≈ −13.187 , which is in QIV. Since the angles we seek have positive measure, we use the representative 360 − 13.187 ≈ 346.813. A second solution occurs in QII, and has value 346.813 − 180 = 166.813.
Step 1: Find the values of
Step 2: Use Step 1 to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside the interval 0 ,360 . The solutions obtained in Step 1 are
θ
= 166.813 , 346.813.
2 When multiplied by 2, the solution corresponding to 346.813 will no longer be in the interval. So, the solution to the original equation in 0 ,360 is approximately θ ≈ 333.63 . θ 48. In order to find all of the values of θ in 0 ,360 that satisfy sec = 1.4275 , 2 we proceed as follows:
Step 1: Find the values of
θ
2
whose secant is 1.4275 .
1 Indeed, observe that one solution is sec −1 (1.4275 ) = cos −1 ≈ 45.531 , 1.4275 which is in QI. A second solution occurs in QIV, and has value 360 − 45.531 ≈ 314.469 .
Step 2: Use Step 1 to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside the interval 0 ,360 . The solutions obtained in Step 1 are
θ
2
= 45.531 , 314.469 .
605
Chapter 6
When multiplied by 2, the solution corresponding to 314.469 will no longer be in the interval. So, the solution to the original equation in 0 ,360 is approximately θ ≈ 91.06 . 49. In order to find all of the values of θ in 0 ,360 that satisfy 5 cot θ − 9 = 0 , we
proceed as follows: Step 1: First, observe that the equation is equivalent to cot θ = Step 2: Find the values of θ whose cotangent is
9 . 5
9 . 5
1 9 = tan −1 5 ≈ 29.0546 , Indeed, observe that one solution is cot −1 = tan −1 9 5 9 5 which is in QI. A second solution occurs in QIII, and has value 29.0546 + 180 = 209.0546 . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are
approximately θ ≈ 29.05 , 209.05 . 50. In order to find all of the values of θ in 0 ,360 that satisfy 5sec θ + 6 = 0 , we
proceed as follows: 6 Step 1: First, observe that the equation is equivalent to sec θ = − . 5 6 Step 2: Find the values of θ whose secant is − . 5 1 6 = cos −1 − 5 ≈ 146.4427 , Indeed, note that one solution is sec −1 − = cos −1 −6 5 6 5 which is in QII. A second solution occurs in QIII, and has value 360 − 146.4427 = 213.5573 . Since the input of secant is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately θ ≈ 146.44 , 213.56 .
606
Section 6.2
51. In order to find all of the values of θ in 0 ,360 that satisfy 4 sin θ + 2 = 0 ,
we proceed as follows: Step 1: First, observe that the equation is equivalent to sin θ = −
2 . 4
2 . 4 2 Indeed, observe that one solution is sin −1 − ≈ −20.7048 , which is in QIV. 4 Since the angles we seek have positive measure, we use the representative 360 − 20.7048 ≈ 339.30 . A second solution occurs in QIII, and has value 20.7048 + 180 = 200.70 . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are
Step 2: Find the values of θ whose sine is −
approximately θ ≈ 200.70 , 339.30 . 52. In order to find all of the values of θ in 0 ,360 that satisfy 3cos θ − 5 = 0 ,
we proceed as follows: Step 1: First, observe that the equation is equivalent to cos θ =
5 . 3
5 . 3 5 Indeed, observe that one solution is cos −1 ≈ 41.8103 , which is in QI. A 3 second solution occurs in QIV, and has value 360 − 41.8103 = 318.19 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately
Step 2: Find the values of θ whose cosine is
θ ≈ 41.81 , 318.19 .
607
Chapter 6
53. In order to find all of the values of θ in 0 ,360 that satisfy 4 cos 2 θ + 5 cos θ − 6 = 0,
we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: ( 4 cos θ − 3)( cos θ + 2 ) = 0 4 cos θ − 3 = 0 or cos θ + 2 = 0 3 =− cos θ = or cos θ 2 4 No solution 3 Step 2: Find the values of θ whose cosine is . 4 3 Indeed, observe that one solution is cos −1 ≈ 41.4096 , which is in QI. A second 4 solution occurs in QIV, and has value 360 − 41.4096 = 318.59. Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately θ ≈ 41.41 , 318.59 . 54. In order to find all of the values of θ in 0 ,360 that satisfy 6 sin2 θ − 13sin θ − 5 = 0,
we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: (3sin θ + 1)( 2 sin θ − 5 ) = 0 3sin θ + 1 = 0 or 2 sin θ − 5 = 0 sin θ = −
1 5 or sin θ = 3 2 No solution
608
Section 6.2
1 Step 2: Find the values of θ whose sine is − . 3 1 Indeed, observe that one solution is sin −1 − ≈ −19.4712 , which is in QIV. 3 Since the angles we seek have positive measure, we use the representative 360 − 19.4712 ≈ 340.53. A second solution occurs in QIII, and has value 180 + 19.47 = 199.47. Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately θ ≈ 199.47 , 340.53 . 55. In order to find all of the values of θ in 0 ,360 that satisfy
6 tan2 θ − tan θ − 12 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: (3tanθ + 4 )( 2 tan θ − 3) = 0 3tan θ + 4 = 0 or 2 tan θ − 3 = 0 4 3 tan θ = − or tan θ = 3 2 4 Step 2: Solve tan θ = − . 3 4 To do so, we must find the values of θ whose tangent is − . 3 4 Indeed, observe that one solution is tan −1 − ≈ −53.13 , which is in QIV. Since 3 the angles we seek have positive measure, we use the representative 360 − 53.13 ≈ 306.87 . A second solution occurs in QII, and has value 180 − 53.13 = 126.87 . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 126.87 , 306.87.
609
Chapter 6
Step 3: Solve tan θ =
3 . 2
To do so, we must find the values of θ whose tangent is
3 . 2
3 Indeed, observe that one solution is tan −1 ≈ 56.31 , which is in QI. A second 2 solution occurs in QIII, and has value 180 + 56.31 = 236.31 . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 56.31 , 236.31 . Step 4: Conclude that the solutions to the original equation are
θ ≈ 56.31 , 126.87 , 236.31 , 306.87 . 56. In order to find all of the values of θ in 0 ,360 that satisfy
6 sec 2 θ − 7 sec θ − 20 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: ( 2 sec θ − 5 )( 3sec θ + 4 ) = 0 2 sec θ − 5 = 0 or 3sec θ + 4 = 0 5 4 sec θ = or sec θ = − 2 3 Step 2: Solve sec θ =
5 2 , or equivalently cos θ = . 2 5
To do so, we must find the values of θ whose cosine is
2 . 5
2 Indeed, observe that one solution is cos −1 ≈ 66.42 , which is in QI. A second 5 solution occurs in QIV, and has value 360 − 66.42 = 293.58 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 66.42 , 293.58 .
610
Section 6.2
4 3 Step 3: Solve sec θ = − , or equivalently, cos θ = − . 3 4 3 To do so, we must find the values of θ whose cosine is − . 4 3 Indeed, observe that one solution is cos −1 − ≈ 138.59 , which is in QII. A 4 second solution occurs in QIII, and has value 360 − 138.59 = 221.41 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 138.59 , 221.41 . Step 4: Conclude that the solutions to the original equation are
θ ≈ 66.42 , 138.59 , 221.41 , 293.58 . 57. In order to find all of the values of θ in 0 ,360 that satisfy 15sin2 2θ + sin 2θ − 2 = 0, we proceed as follows:
Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: (5sin 2θ + 2 )(3sin 2θ − 1) = 0 5sin 2θ + 2 = 0 or 3sin 2θ − 1 = 0 2 1 sin 2θ = − or sin 2θ = 5 3 2 Step 2: Solve sin 2θ = − . 5 2 Step a: Find the values of 2θ whose sine is − . 5 2 Indeed, observe that one solution is sin −1 − ≈ −23.578 , which is in QIV. 5 Since the angles we seek have positive measure, we use the representative 360 − 23.578 ≈ 336.42 . A second solution occurs in QIII, namely 180 + 23.578 = 203.578 .
611
Chapter 6
Step b: Use periodicity to find all values of θ that satisfy sin 2θ = −
2 . 5
Using periodicity with the solutions obtained in Step a, we see that 2θ = 203.578 , 203.578 + 360 , 336.42 , 336.42 + 360 = 203.578 , 563.578 , 336.42 , 696.42
and so, the solutions are approximately: θ ≈ 101.79 , 281.79 , 168.21 , 348.21 1 Step 3: Solve sin 2θ = . 3 1 Step a: Find the values of 2θ whose sine is . 3 1 Indeed, observe that one solution is sin −1 ≈ 19.4712 , which is in QI. 3 A second solution occurs in QII, and has value 180 − 19.4712 = 160.528 . 1 Step b: Use periodicity to find all values of θ that satisfy sin 2θ = . 3 Using periodicity with the solutions obtained in Step a, we see that 2θ = 19.4712 , 19.4712 + 360 , 160.528 , 160.528 + 360 = 19.4712 , 160.528 , 379.4712 , 520.528
and so, the solutions are approximately: θ ≈ 9.74 , 189.74 , 80.26 , 260.26 Step 4: Conclude that the solutions to the original equation are
θ ≈ 101.79 , 281.79 , 168.21 , 348.21 , 9.74 , 189.74 , 80.26 , 260.26 . 58. In order to find all of the values of θ in 0 ,360 that satisfy
θ θ 12 cos 2 − 13cos + 3 = 0 , we proceed as follows: 2 2 Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: θ θ 3cos 2 − 1 4 cos 2 − 3 = 0
θ θ θ 1 θ 3 3cos − 1 = 0 or 4 cos − 3 = 0 or equivalently cos = or cos = 2 2 2 3 2 4
612
Section 6.2
θ 1 Step 2: Solve cos = . 2 3 Step a: Find the values of
θ 2
whose cosine is
1 . 3
1 Indeed, observe that one solution is cos −1 ≈ 70.5288 , which is in QI. 3 A second solution occurs in QIV, and has value 360 − 70.5288 = 289.47 . Step b: Use Step a to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside 0 ,360 . The solutions obtained is Step a are
θ
= 70.5288 , 289.47 . 2 When multiplied by 2, the solution corresponding to 289.47 will no longer be in the interval. So, the solution to the equation in 0 ,360 is approximately
θ ≈ 141.06 .
θ 3 Step 3: Solve cos = . 2 4 Step a: Find the values of
θ 2
whose cosine is
3 . 4
3 Indeed, observe that one solution is cos −1 ≈ 41.4096 , which is in QI. A 4 second solution occurs in QIV, and has value 360 − 41.4096 = 318.59 . Step b: Use Step a to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside 0 ,360 . The solutions obtained is Step a are
θ
= 41.4096 , 318.59 .
2 When multiplied by 2, the solution corresponding to 318.59 will no longer be in the interval. So, the solution to the original equation in 0 ,360 is approximately θ ≈ 82.82 . Step 4: Conclude that the solutions to the original equation are θ ≈ 82.82 , 141.06 .
613
Chapter 6
59. In order to find all of the values of θ in 0 ,360 that satisfy
cos 2 θ − 6 cos θ + 1 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating cos θ as the variable): cos θ =
− ( −6 ) ±
( −6 ) − 4 (1)(1) 6 ± 4 2 = = 3 ± 2 2 So, θ 2 (1) 2 2
is a solution to the original equation if cos θ = 3+2 2
or cos θ = 3 − 2 2 .
No solution since 3 + 2 2 >1
Step 2: Find the values of θ whose cosine is 3 − 2 2 . Indeed, observe that one solution is cos −1 3 − 2 2 ≈ 80.1207 , which is in QI. A
(
)
second solution occurs in QIV, and has value 360 − 80.1207 = 279.88 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 80.12 , 279.88 . Step 3: Conclude that the solutions to the original equation are θ ≈ 80.12 , 279.88 . 60. In order to find all of the values of θ in 0 ,360 that satisfy
sin2 θ + 3sin θ − 3 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating cos θ as the variable): sin θ =
−3 ± 32 − 4 (1)( −3 )
2 (1) So, θ is a solution to the original equation if −3 − 21 −3 + 21 sin θ = or sin θ = . 2 2 No solution since
−3− 21 < −1 2
614
=
−3 ± 21 2
Section 6.2
−3 + 21 . 2 −3 + 21 Indeed, observe that one solution is sin −1 ≈ 52.306 , which is in QI. 2 A second solution occurs in QII, and has value 180 − 52.306 = 127.69 . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 52.306 , 127.69 .
Step 2: Find the values of θ whose sine is
Step 3: Conclude that the solutions to the original equation are θ ≈ 52.306 , 127.69 . 61. In order to find all of the values of θ in 0 ,360 that satisfy
2 tan2 θ − tan θ − 7 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating cos θ as the variable): tan θ =
− ( −1) ±
( −1) − 4 ( 2 )( −7 ) 1 ± 57 = 2 (2) 4 2
So, θ is a solution to the original equation if either 1 − 57 1 + 57 . tan θ = or tan θ = 4 4 Step 2: Solve tan θ =
1 − 57 . 4
To do so, we must find the values of θ whose tangent is
1 − 57 . 4
1 − 57 Indeed, observe that one solution is tan −1 ≈ −58.587 , which is in QIV. 4 Since the angles we seek have positive measure, we use the representative 360 − 58.587 ≈ 301.41 . A second solution occurs in QII, and has value 180 − 58.587 = 121.41 . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 121.41 , 301.41 .
615
Chapter 6
Step 3: Solve tan θ =
1 + 57 . 4
To do so, we must find the values of θ whose tangent is
1 + 57 . 4
1 + 57 Indeed, observe that one solution is tan −1 ≈ 64.93 , which is in QI. A 4 second solution occurs in QIII, and has value 180 + 64.93 = 244.93 . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 64.93 , 244.93 . Step 4: Conclude that the solutions to the original equation are
θ ≈ 64.93 , 121.41 , 244.93 , 301.41 . 62. In order to find all of the values of θ in 0 ,360 that satisfy
3cot 2 θ + 2 cot θ − 4 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating cos θ as the variable): −2 ± 2 2 − 4 ( 3 )( −4 )
−2 ± 52 −1 ± 13 = 2 (3) 6 3 So, θ is a solution to the original equation if either −1 − 13 −1 + 13 cot θ = or cot θ = . 3 3 cot θ =
Step 2: Solve cot θ =
=
−1 − 13 . 3
To do so, we must find the values of θ whose cotangent is
−1 − 13 . 3
Indeed, observe that one solution is −1 − 13 3 −1 ≈ 146.92 , which is in QII. cot −1 = 180 + tan 3 −1 − 13
616
Section 6.2
A second solution occurs in QIV, and has value 180 + 146.92 = 326.92 . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 146.92 , 326.92 . Step 3: Solve cot θ =
−1 + 13 . 3
−1 + 13 . 3 −1 + 13 3 −1 Indeed, observe that one solution is cot −1 ≈ 49.025 , = tan 3 −1 + 13 which is in QI. A second solution occurs in QIII, and has value 180 + 49.025 = 229.025 . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 49.03 , 229.03 . To do so, we must find the values of θ whose cotangent is
Step 4: Conclude that the solutions to the original equation are
θ ≈ 49.03 , 146.92 , 229.03 , 326.92 . 63. Solve the following equation: 0.472 = 0.12 sin(0.79t − 2.38) + 0.387 sin(0.79t − 2.38) = 0.70833 0.79t − 2.38 = sin −1 (0.70833) + 2nπ
The QI value for sin −1 (0.70833) is π4 and the value in QII is 34π . We determine the t-values that work in each case separately. QI: 0.79t − 2.38 = π4 + 2nπ t ≈ 4 + 7.95n So, when n = 0, we get 4 and when n = 1, we get 12. QII: 0.79t − 2.38 = 34π + 2 nπ t ≈ 6 + 7.95n So, when n = 0, we get 6. So, this occurs in the fourth quarter of 2016, second quarter of 2017, and fourth quarter of 2018.
617
Chapter 6
64. Solve the following equation: 0.507 = 0.12 sin(0.79t − 2.38) + 0.387 sin(0.79t − 2.38) = 1 0.79t − 2.38 = sin −1 (1) + 2nπ = π2 + 2nπ So, when n = 0, we get 5 and when n = 1, we get 13, which is discarded. So, this occurs in the first quarter of 2017.
65. Solve the following equation: 41 = 35 − 26 cos ( 12π t − 76π )
cos ( 12π t − 76π ) = −0.23077
t − 76π = cos −1 ( −0.23077) or 12π t − 76π = 2π − cos −1 ( −0.23077) So, t ≈ 20.889 ≈ 21 hours or about 31 hours, which is discarded. Hence, this occurs around 9pm. π
12
66. Solve the following equation: 17 = 35 − 26 cos ( 12π t − 76π )
cos ( 12π t − 76π ) = 139
t − 76π = cos −1 ( 139 ) + 2 nπ or 12π t − 76π = 2π − cos −1 ( 139 ) + 2 nπ So, when n = 0, we get 17 and when n = 1, we get 41, which is discarded. So, this occurs around 5pm. π
12
67. Solving for x yields:
π 2, 400 = 400 sin x + 2,000 6 π 400 = 400 sin x 6 π 1 = sin x 6
This equation is satisfied when
π 6
x=
π 2
, so that x =
2,400 in March.
618
π 6 ⋅ = 3 . So, the sales reach 2 π
Section 6.2
68. Solving for x yields:
π 1,800 = 400 sin x + 2,000 6 π −200 = 400 sin x 6 1 π − = sin x 2 6 π 7π 11π This equation is satisfied when x = , so that x = 7, 11 . So, the sales reach , 6 6 6 1,800 during in July and November. 69. Consider the following diagram:
α
β
θ
θ
Let h = height of the trapezoid, and x = length of one base and two edges of the trapezoid, as labeled above. Note that α = β = θ since they are alternate interior angles. As such, h sin θ = , so that h = x sin θ . x Furthermore, using the Pythagorean Theorem enables us to find z:
z 2 + h2 = x 2 , so that z = x 2 − h 2 = x 2 − ( x sin θ )2 = x 2 (1 − sin2 θ ) = x cos θ . = cos2 θ
Since 0 ≤ θ ≤ π , we conclude that . z = x cos θ . Hence, b1 = x and b2 = x + 2 z = x + 2 x cos θ = x (1 + 2 cos θ ) .
619
Chapter 6
Thus, the area A of the cross-section of the rain gutter is 1 1 A = h ( b1 + b2 ) = ( x sin θ ) x + x (1 + 2 cos θ ) 2 2 = 2 x (1+ cos θ )
= ( x sin θ ) [ x(1 + cos θ )] = x 2 [sin θ + sin θ cos θ ]
1 = x 2 sin θ + ( 2 sin θ cos θ ) 2 sin 2θ = x 2 sin θ + 2
70. Observe that
cos 2 θ − sin2 θ + cos θ = 0
cos 2 θ − (1 − cos 2 θ ) + cos θ = 0 2 cos 2 θ + cos θ − 1 = 0
( 2 cos θ − 1)( cos θ + 1) = 0
So, θ satisfies the original equation if either 2 cos θ − 1 = 0 or cos θ + 1 = 0 . 1 Observe that 2 cos θ − 1 = 0 is equivalent to cos θ = , which is satisfied when 2 π 5π θ= , . 3 3 Also, cos θ + 1 = 0 is equivalent to cos θ = −1, which is satisfied when θ = π . So, we conclude that the solutions to the original equation are θ =
620
π 3
, π,
5π . 3
Section 6.2
π 71. Solving the equation 200 + 100 sin x = 300 , for x > 2,016 yields: 2 π 200 + 100 sin x = 300 2 π 100 sin x = 100 2 π sin x = 1 2 Observe that this equation is satisfied when
π 2
x=
π 2
+ 2 nπ , where n is an integer, so
2 that x = 1 + 2nπ = 1 + 4n, where n is an integer. So, the first value of n for which π 1 + 4 n > 2,016 is n = 504 . The resulting year is x = 1 + 4(504) = 2017 . π 72. Solving the equation 200 + 100 sin x = 150 yields: 6 π 200 + 100 sin x = 150 6 π 100 sin x = −50 6 1 π sin x = − 2 6 π 7π 11π , so that , Observe that this equation is satisfied when x = 6 6 6 6 7π 6 11π 6 19π = 7, 11, 19 . Hence, the first year after 2016 in which x = , , π 6 π 6 π 6 this occurs is 2023.
621
Chapter 6
73. Use ni sin (θi ) = nr sin (θ r ) with the
74. Use ni sin (θi ) = nr sin (θ r ) with the
given information to obtain 1.00 sin ( 75 ) = 2.417 sin (θ r )
given information to obtain 2.417 sin (15 ) = 1.00 sin (θ r )
so that 1.00 sin ( 75 )
so that 2.417 sin (15 )
2.417
= sin (θ r )
1.00
1.00 sin ( 75 ) = θr 24 ≈ sin 2.417
75.
cos 2 x = 0 2 x =
π
2.417 sin (15 ) = θr 39 ≈ sin 1.00
−1
2 n +1 2
= sin (θ r )
76.
−1
sin 2 x = 0 2 x = nπ
x = n2π
x = 2 n4+1 π = π4 + n2π where n is an integer.
where n is an integer.
77. Consider the following triangle:
78. Consider the following triangle:
θ
θ
tan θ = 68 θ = tan −1 ( 68 ) ≈ 36.9
tan θ = 107 θ = tan −1 ( 107 ) ≈ 35
79. The given solution is correct, but the second solution in QIV (namely 360 − 53.85 = 306.15 ), was not given. 80. The given solution is correct, but the second solution in QII (namely 180 − 36.87 = 143.13 ), was not given. 81. The value θ = 3π
2
does not satisfy the original equation. Indeed, observe that
3π 3π 2 + sin = 2 − 1 = 1 , while sin = −1 . So, this value of θ is an extraneous 2 2 solution.
622
Section 6.2
82. The value θ = π
2
does not satisfy the original equation. Indeed, observe that
π π 3sin − 2 = 3 − 2 = 1 , while − sin = −1 . So, this value of θ is an extraneous 2 2 solution.
83. False. For instance, sin θ =
3 π 2π has two solutions on [ 0,2π ] , namely θ = , . 2 3 3
84. False. For instance, sin2 θ =
3 has four solutions on [ 0,2π ] , namely 2
θ=
π 2π 4π 5π 3
,
3
,
3
,
3
.
85. False. Observe that sin2 x = 0.5sin x sin2 x − 0.5sin x = 0 sin x ( sin x − 0.5 ) = 0 sin x = 0 or sin x = 0.5
The values x = 0, π are also solutions. 86. True. Observe that sin 4 x − 1 = 0 ( sin2 x − 1)( sin2 x + 1) = 0 sin x = ±1 x = π2 , 32π Never zero
π 87. Let k be a given nonzero integer. The solutions of sin kx + = 1 satisfy 2 π π 2nπ , where n = 0,1,2,..., k . kx + = + 2nπ x = 2 2 k Hence, there are k + 1 solutions. π 2 88. Let k be a given nonzero integer. The solutions of cos kx − = satisfy 2 2 π 2 π π 7π cos kx − = kx − = or . 2 2 2 4 4 So, in intervals of length 2kπ , there are two solutions. Further, since [ 0,2π ] is comprised of k such intervals, there are 2k solutions.
623
Chapter 6
89. Solving the equation 16 sin 4 θ − 8sin2 θ + 1 = 0 on [ 0,2π ] yields
16 sin 4 θ − 8sin2 θ + 1 = 0
( 4 sin θ − 1) = 0 2
2
( 2 sin θ − 1)( 2 sinθ + 1) = 0
So, θ satisfies the original equation if either 2 sin θ − 1 = 0 or 2 sin θ + 1 = 0 . 1 π 5π . Observe that 2 sin θ − 1 = 0 is equivalent to sin θ = , which is satisfied when θ = , 2 6 6 1 7π 11π Also, 2 sin θ + 1 = 0 is equivalent to sin θ = − , which is satisfied when θ = . , 2 6 6 π 5π 7π 11π . , , So, we conclude that the solutions to the original equation are θ = , 6 6 6 6
π 3 90. Solving the equation cos θ + = on is equivalent to 4 2 π 3 π 3 cos θ + = or cos θ + = − . 4 2 4 2 π 3 π π 11π is satisfied when θ + = + 2nπ , + 2 nπ , so that Observe that cos θ + = 4 2 4 6 6 π π 11π π π 19π − + 2 nπ = − + 2 nπ , + 2 nπ , where n is an integer. θ = − + 2 nπ , 6 4 6 4 12 12 π 3 π 5π 7π Also, cos θ + = − is satisfied when θ + = + 2nπ , + 2nπ , so that 4 2 4 6 6 5π π 7π π 7π 11π − + 2 nπ , − + 2 nπ = + 2 nπ , + 2nπ , where n is an integer. θ= 6 4 6 4 12 6 So, we conclude that the solutions to the original equation are π 19π 7π 11π + 2nπ , + 2 nπ , + 2nπ , where n is an integer, θ = − + 2nπ , 12 12 12 6 π 7π + 2nπ , where n is an integer. which can be further simplified as: θ = − + nπ , 12 12
624
Section 6.2
91. The terminal sides of the angles that 92. The terminal sides of the angles that 1 1 satisfy the inequality sin x < must satisfy the inequality cos x ≥ must land 2 2 land in the scribbled region marked in the scribbled region marked below: below:
As such, the solution set is 0, π3 ∪ 23π , 43π ∪ 53π ,2π )
As such, the solution set is: π 5π 7 π 11π 0, 6 ) ∪ ( 6 , 6 ) ∪ ( 6 ,2π ) 93.
2 cos3 x − cos 2 x − 2 cos x + 1 = 0 cos 2 x ( 2 cos x − 1) − ( 2 cos x − 1) = 0
( cos x − 1) ( 2 cos x − 1) = 0 2
cos x = ±1, 12 5π π x = 0, , π , 3 3
94. tan x = tan x + 2 tan2 x = tan x + 2 tan2 x − tan x − 2 = 0
( tan x + 1)( tan x − 2 ) = 0 tan x = −1 ,2
So, the solutions are x = tan (2), tan (2) + π ≈ 1.11, 4.25 . −1
−1
625
Chapter 6
95. Consider the graphs below of y1 = sin θ , y2 = cos 2θ .
96. Consider the graphs below of y1 = csc θ , y2 = sec θ .
Observe that the solutions of the equation sin θ = cos 2θ on [ 0, π ] are approximately P1( 0.524,0.5 ) , P 2 ( 2.618,0.5 ) .
Observe that the approximate solution to this equation is 0.785. The exact solution is
The exact solutions are
π 5π 6
,
6
.
π
4
.
97. Consider the graphs below of y1 = sin θ , y2 = sec θ .
98. Consider the graphs below of y1 = cos θ , y2 = csc θ .
Since the curves never intersect, there are no solutions of the equation sin θ = sec θ on [ 0, π ] .
Since the curves never intersect, there are no solutions of the equation sin θ = sec θ on [ 0, π ] .
626
Section 6.2
99. Consider the graphs below of y1 = sin θ , y2 = eθ .
100. Consider the graphs below of y1 = cos θ , y2 = eθ .
First, note that while there are no First, note that while there are no positive positive solutions of the equation solutions of the equation cos θ = eθ , the sin θ = eθ , there are infinitely many two curves do intersect at θ = 0 . negative solutions (at least one between each consecutive pair of x-intercepts). They are all irrational, and there is no apparent closed-form formula that can be used to generate them.
627
Section 6.3 Solutions -----------------------------------------------------------------------------------1. By inspection, the values of x in [0,2π ] that satisfy the equation sin x = cos x are x =
π 5π 4
,
4
.
3. Observe that sec x + cos x = −2 1 + cos x = −2 cos x 1 + cos 2 x = −2 cos x
2. By inspection, the values of x in [ 0,2π ] that satisfy the equation sin x = − cos x are 3π 7π x= , . 4 4 4. Observe that sin x + csc x = 2 1 =2 sin x + sin x sin2 x + 1 = 2 sin x
cos 2 x + 2 cos x + 1 = 0 (cos x + 1)2 = 0 cos x + 1 = 0 cos x = −1 The value of x in [ 0,2π ] that satisfies
the equation cos x = −1 is x = π . Substituting this value into the original equation shows that it is, in fact, a solution to the original equation.
sin2 x − 2 sin x + 1 = 0 (sin x − 1)2 = 0 sin x − 1 = 0 sin x = 1 The value of x in [ 0,2π ] that satisfies the
equation sin x = 1 is x = π
. Substituting 2 this value into the original equation shows that it is, in fact, a solution to the original equation.
628
Section 6.3
5. Observe that
6. Observe that sec x + tan x = 1 1 sin x + =1 cos x cos x 1 + sin x =1 cos x
3 3
sec x − tan x =
1 sin x 3 − = cos x cos x 3 1 − sin x 3 = cos x 3
(1 + sin x ) = 1 2
cos 2 x (1 + sin x )(1 + sin x ) = 1 1 − sin2 x
(1 − sin x ) = 1 2
cos 2 x 3 (1 − sin x )(1 − sin x ) = 1 1 − sin2 x 3
(1 + sin x ) (1 + sin x ) =1 (1 + sin x ) (1 − sin x )
(1 − sin x ) (1 − sin x ) 1 = (1 − sin x ) (1 + sin x ) 3
1 − sin x = 1 + sin x
3(1 − sin x ) = 1 + sin x 4 sin x = 2 1 sin x = 2 The values of x in [ 0,2π ] that satisfy
1 π 5π . are x = , 2 6 6 Substituting these values into the the equation sin x =
original equation shows that while
5π is extraneous. is a solution, 6 Indeed, note that 1 5π 5π sec − tan = 6 6 − 3
=−
So, the only solution is
− 2
π
6
2 sin x = 0 sin x = 0 The values of x in [ 0,2π ) that satisfy the
equation sin x = 0 are x = 0, π .
Substituting these values into the original equation shows that while 0 is a solution, π is extraneous. Indeed, note that 1 0 + = −1 ≠ 1 sec (π ) + tan (π ) = −1 −1 So, the only solution is 0 .
1
2 − 3
2
3 3 ≠ 3 3
π 6
.
629
Chapter 6
8. Observe that
7. Observe that csc x + cot x = 3 1 cos x + = 3 sin x sin x 1 + cos x = 3 sin x
3 3 1 cos x 3 − = sin x sin x 3 1 − cos x 3 = sin x 3 csc x − cot x =
(1 + cos x ) = 3 2
(1 − cos x ) = 1 2
sin2 x (1 + cos x )(1 + cos x ) = 3 1 − cos 2 x
sin2 x 3 (1 − cos x )(1 − cos x ) = 1 1 − cos 2 x 3
(1 + cos x ) (1 + cos x ) =3 (1 + cos x ) (1 − cos x ) 3 (1 − cos x ) = 1 + cos x
(1 − cos x ) (1 − cos x ) 1 = (1 − cos x ) (1 + cos x ) 3 3 (1 − cos x ) = 1 + cos x
3 − 3cos x = 1 + cos x 4 cos x = 2 1 2 The values of x in [ 0,2π ] that satisfy cos x =
1 π 5π are x = , . 2 3 3 Substituting this value into the original the equation cos x =
equation shows that while solution,
π
3
is a
5π is extraneous. Indeed, 3
=− So, the only solution is
+ 2
3
1
2 − 3
2
3 =− 3≠ 3 3
π 3
.
1 π 5π are x = , . 2 3 3 Substituting this value into the original 5π equation shows that while is a solution, 3 equation cos x =
π
note that 1 5π 5π csc + cot = 3 3 − 3
3 − 3cos x = 1 + cos x 4 cos x = 2 1 cos x = 2 The values of x in [ 0,2π ] that satisfy the
is extraneous. Indeed, note that 1 π π csc − cot = 3 3 3
1
2 3 2 2 3 3 =− ≠ 3 3 5π So, the only solution is . 3
630
−
Section 6.3
9. Observe that 2 sin x − csc x = 0 1 =0 2 sin x − sin x 2 sin2 x − 1 =0 sin x 2 sin2 x − 1 = 0
10. Observe that 2 sin x + csc x = 3
1 2
2 sin2 x − 3sin x + 1 = 0
sin2 x =
1 1 sin x = − or sin x = 2 2 The solutions to these equations in π 3π 5π 7π [0,2π ] are x = , , , . 4 4 4 4 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
11. Observe that
sin 2 x = 4 cos x
1 =3 sin x 2 sin2 x + 1 =3 sin x 2 sin2 x + 1 = 3sin x
2 sin x +
( 2 sin x − 1)( sin x − 1) = 0
2 sin x − 1 = 0 or sin x − 1 = 0
1 or sin x = 1 2 The solutions to these equations in [ 0,2π ] sin x =
are x =
π 5π π
, , . Substituting these into 6 6 2 the original equation shows that they are all, in fact, solutions to the original equation. 12. Observe that sin 2 x = 3 sin x
2 sin x cos x = 4 cos x 2 cos x ( sin x − 2 ) = 0 cos x = 0 or sin x − 2 = 0 cos x = 0 or sin x =2 No solution
The solutions to these equations in π 3π [0,2π ] are x = , . Substituting 2 2 these into the original equation shows that they are all, in fact, solutions to the original equation.
(
2 sin x cos x = 3 sin x
)
sin x 2 cos x − 3 = 0 sin x = 0 or 2 cos x − 3 = 0 3 2 The solutions to these equations in [ 0,2π ) sin x = 0 or cos x =
are x = 0, π ,
π 11π
, . Substituting these 6 6 into the original equation shows that they are all, in fact, solutions to the original equation.
631
Chapter 6
13. Observe that
sin x
14. Observe that 2 sin x = tan x
cos 2x = sin x
sin x 2 sin x = cos x 2 sin x cos x = sin x
cos 2 x − sin2 x = sin x
(1 − sin x ) − sin x = sin x 2
2 sin2 x + sin x − 1 = 0
( 2 cos x − 1) = 0
sin x = 0 or
( 2 sin x − 1)( sin x + 1) = 0
2 sin x − 1 = 0 or sin x + 1 = 0
2 cos x − 1 = 0
1 2 The solutions to these equations in π 7π [0,2π ) are x = 0, π , , . 4 4 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation. sin x = 0 or cos x =
15. Observe that
2
1 or sin x = −1 2 The solutions to these equations in [ 0,2π ] sin x =
are x =
π 5π 3π
, , . Substituting these 6 6 2 into the original equation shows that they are all, in fact, solutions to the original equation.
tan 2x = cot x sin 2 x cos x = cos 2 x sin x cos x ( cos 2x ) − ( sin 2 x ) sin x = 0 cos ( x + 2x ) = 0 cos3x = 0
Note that the solutions of cos 3x = 0 are 3x =
π
2
,
π
2
+ 2π ,
π
2
+ 4π ,
3π 3π 3π + 2π , + 4π , 2 2 2
so that 5π 3π π 7π 11π , , , , . 6 6 2 2 6 6 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
x=
π
,
632
Section 6.3
16. Observe that
3cot 2 x = cot x 3cos 2 x cos x = sin 2 x sin x 3cos 2 x sin x = cos x sin 2 x 3cos 2x sin x − cos x sin 2 x = 0
2 cos 2x sin x + ( cos 2 x sin x − cos x sin 2 x ) = 0 2 cos 2 x sin x + sin( x − 2 x ) = 0 2 cos 2 x sin x + sin( −x ) = 0 2 cos 2 x sin x − sin x = 0 sin x ( 2 cos 2 x − 1) = 0 sin x = 0 or 2 cos 2 x − 1 = 0 1 2 Note that the solutions of sin x = 0 in [ 0,2π ) are x = 0, π . However, substituting sin x = 0 or cos 2 x =
these values into the original equation show that NONE of them are solutions since the right-side is undefined at each of these values. 1 Next, the solutions of cos 2 x = are 2 π π 5π 5π + 2π 2x = , + 2π , , 3 3 3 3 so that π 5π 7π 11π x= , , , . 6 6 6 6 Substituting all of these into the original equation shows that they are all, in fact, solutions to the original equation.
633
Chapter 6
17. Observe that 3 sec x = 4 sin x
18. Observe that 3 tan x = 2 sin x
3 = 4 sin x cos x 3 = 4 sin x cos x
3 sin x = 2 sin x cos x 3 sin x = 2 sin x cos x
3 = 2 ( 2 sin x cos x )
3 sin x − 2 sin x cos x = 0
3 = 2 sin 2 x
sin x
3 = sin 2 x 2
sin x = 0 or
3 are Next, the solutions of sin 2 x = 2 π π 2π 2π 2x = , + 2π , , + 2π 3 3 3 3 so that π π 7π 4π x= , , , . 6 3 6 3 Substituting all of these into the original equation shows that they are all, in fact, solutions to the original equation. 19. Observe that 2 cos 2 x − 3sin x = 0
2 (1 − sin x ) − 3sin x = 0 2
2 − 2 sin2 x − 3sin x = 0
− ( 2 sin2 x + 3sin x − 2 ) = 0
( 3 − 2 cos x ) = 0
3 − 2 cos x = 0
3 2 The solutions to these equations in [ 0,2π ) sin x = 0 or cos x =
are x = 0, π ,
π 11π
, . Substituting these 6 6 into the original equation shows that they are all, in fact, solutions to the original equation.
The solutions to these equations in [ 0,2π ) are x =
π 5π
, . Substituting these into the 6 6 original eqution shows that they are all, in fact, solutions to the original equation.
− ( 2 sin x − 1)( sin x + 2 ) = 0 2 sin x − 1 = 0 or sin x + 2 = 0 1 sin x = or sin =− x 2 2 no solution
634
Section 6.3
20. Observe that cos 2 x − cos x = 0
( 2 cos2 x − 1) + cos x = 0 2 cos 2 x + cos x − 1 = 0
(2 cos x + 1) ( cos x − 1) = 0
The solutions to these equations in [ 0,2π )
2π 4π . Substituting these into , 3 3 the original equation shows that they are all, in fact, solutions to the original equation. are x = 0,
2 cos x + 1 = 0 or cosx − 1 = 0 2 cos x = −1 or cos x = 1 1 cos x = − or cos x = 1 2 21. Observe that
1 2 1 2 ( sin x cos x ) = 2 2 sin 2 x = 1 Note that the solutions of sin 2 x = 1 are sin x cos x =
2x =
π π
π
, + 2π , + 4π 2 2 2
so that x=
π 5π ,
.
4 4 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
22. Observe that cos x − sin x = −1 (cos x − sin x )2 = ( −1)
2
cos 2 x − 2 sin x cos x + sin2 x = 1 2 + sin2 x x −2 sin x cos x = 1 cos 1
−2 sin x cos x = 0 sin x = 0 or cos x = 0 The solutions to these equations in [ 0,2π )
π 3π are x = 0, ,π , . Substituting these into 2 2 the original equation shows that only π
,π are solutions to the original 2 equation. x=
635
Chapter 6
23. Observe that 1 sin2 x − cos 2x = − 4 1 sin2 x − ( cos 2 x − sin2 x ) = − 4 1 sin2 x − (1 − sin2 x − sin2 x ) = − 4 1 3sin2 x − 1 = − 4 1 sin2 x = 4 1 1 sin x = − or sin x = 2 2 The solutions to these equations in π 5π 7π 11π . [0,2π ] are x = , , , 6 6 6 6 Substituting these into the original equation shows that they are all solutions to original equation.
25. Observe that cos 2 x + 2 sin x + 2 = 0
(1 − sin2 x ) + 2 sin x + 2 = 0 sin2 x − 2 sin x − 3 = 0
( sin x + 1)( sin x − 3) = 0
24. Observe that sin2 x − 2 sin x = 0 sin x ( sin x − 2 ) = 0 sin x = 0 or sin x − 2 = 0
sin x = 0 or sin x =2 No solution
The solutions to these equations in [ 0,2π ) are x = 0, π . Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
The solution to these equations in [ 0,2π ] is 3π . Substituting this into the original 2 equation shows that it is, in fact, a solution to the original equation. x=
sin x + 1 = 0 or sin x − 3 = 0 sin x = −1 or sin x =3 No solution
636
Section 6.3
26. Observe that 2 cos x = sin x + 1 2
2 (1 − sin2 x ) = sin x + 1 2 − 2 sin2 x = sin x + 1 2 sin2 x + sin x − 1 = 0
( sin x + 1)( 2 sin x − 1) = 0
The solutions to these equations in [ 0,2π ] are x =
π 5π 3π
, , . Substituting these 6 6 2 into the original equation shows that they are, in fact, solutions to the original equation.
sin x + 1 = 0 or 2 sin x − 1 = 0 sin x = −1
or sin x =
1 2
27. Observe that 2 sin2 x + 3cos x = 0
28. Observe that 4 cos 2 x − 4 sin x = 5
2 (1 − cos 2 x ) + 3cos x = 0
4 (1 − sin2 x ) − 4 sin x = 5
2 − 2 cos 2 x + 3cos x = 0
4 − 4 sin2 x − 4 sin x = 5
2 cos 2 x − 3cos x − 2 = 0
4 sin2 x + 4 sin x + 1 = 0
( 2 cos x + 1)( cos x − 2 ) = 0
( 2 sin x + 1) = 0
2 cos x + 1 = 0 or cos x − 2 = 0 1 cos x = − or cos x =2 2 No solution The solutions to these equations in 2π 4π [0,2π ] are x = , . Substituting 3 3 these into the original equation shows that they are, in fact, solutions to the original equation.
2
2 sin x + 1 = 0 1 2 The solutions to this equation in [ 0,2π ] sin x = −
7π 11π , . Substituting these into 6 6 the original equation shows that they are, in fact, solutions to the original equation.
are x =
637
Chapter 6
29. Observe that
cos 2x + cos x = 0
30. Observe that
( cos x − sin x ) + cos x = 0 ( cos x − (1 − cos x )) + cos x = 0 2
2
2 cos x 1 = sin x sin x 2 cos x sin x = sin x
2
2
sin x ( 2 cos x − 1) = 0 sin x = 0 or 2 cos x − 1 = 0
2 cos 2 x + cos x − 1 = 0
( 2 cos x − 1)( cos x + 1) = 0
2 cos x − 1 = 0 or cos x + 1 = 0 1 cos x = or cos x = −1 2 The solutions to these equations in π 5π [0,2π ] are x = , , π . Substituting 3 3 these into the original equation shows that they are, in fact, solutions to the original equation.
1 2 Note that the solutions of sin x = 0 in [0,2π ] are x = 0, π , 2π . However, sin x = 0 or cos x =
substituting these values into the original equation show that NONE of them are solutions since both the left- and rightsides are undefined at each of these values. 1 Next, the solutions of cos x = are 2 π 5π x= , 3 3 Substituting all of these into the original equation shows that they are all, in fact, solutions to the original equation. 32.
31.
2 cot x = csc x
2 sin x + sin 2 x + cos 2 x = 0 2
2 sin2 x + sin 2 x + (1 − 2 sin2 x ) = 0
sin 2 x − cos 2 x = 0 sin 2 x = cos 2 x 2 x = π4 , 54π , 94π , 134π
sin 2 x = −1
x = π8 , 58π , 98π , 138π
2 x = 32π , 72π x = 34π , 74π
638
Section 6.3
33.
tan 2 x + 1 = sec 2 x sin 2 x 1 +1 = cos 2 x cos 2 x sin 2 x + cos 2 x 1 = cos 2 x cos 2 x sin 2 x + cos 2 x = 1
( 2 sin x cos x ) + ( 1 − 2 sin2 x ) = 1 2 sin x ( cos x − sin x ) = 0
sin x = 0 or cos x = sin x x = 0, π ,
π 5π 4 , 4 Omit since makes denominator zero
34.
tan 2x − tan x = 0 2 tan x − tan x = 0 1 − tan2 x 2 tan x − tan x (1 − tan2 x ) =0 1 − tan2 x tan x + tan3 x =0 1 − tan2 x tan x (1 + tan2 x ) =0 1 − tan2 x tan x sec 2 x = 0 2 x =0 tan x = 0 or sec No solution
sin x = 0 x = 0, π
639
Chapter 6
35.
cos 2x =0 cos x + sin x cos 2 x − sin 2 x =0 cos x + sin x ( cos x − sin x )( cos x + sin x ) = 0 cos x + sin x cos x = sin x x = π4 , 54π 36. 1 csc 2 x + sec x = −1 − csc x 2 1 1 1 + = −1 − sin 2x cos x 2 sin x 1 1 1 + +1+ =0 2 sin x cos x cos x 2 sin x 1 + 2 sin x + 2 sin x cos x + cos x =0 2 sin x cos x (1 + 2 sin x ) + cos x (1 + 2 sin x ) = 0 2 sin x cos x (1 + 2 sin x )(1 + cos x ) = 0 2 sin x cos x sin x = − 12 , cos x = −1 x = 76π , 116π , π
37. Use the addition formula for cos( A − B ) to simplify the left-side of the equation and then, solve for x as follows: cos 3x cos 2 x + sin3x sin 2x = 1 cos(3x − 2 x ) = 1 cos x = 1 x=0
38. Use the subtraction formula for sin ( A − B ) to simplify the left-side of the equation and then, solve for x as follows: sin3x cos 2 x − cos 3x sin 2 x = 1 sin(3x − 2 x ) = 1 sin x = 1
640
x=
π 2
Section 6.3
39. Use the addition formula for sin( A − B ) to simplify the left-side of the equation and then, solve for x as follows: sin 7 x cos5x − cos 7 x sin5x = 12
sin(7 x − 5x ) = 12 sin 2 x = 12 2 x = π6 , 56π , 136π , 176π x = 12π , 512π , 1312π , 1712π 40. Use the addition formula for cos( A − B ) to simplify the left-side of the equation and then, solve for x as follows: sin 7 x sin5x + cos 7 x cos 5x = −1 cos(7 x − 5x ) = −1 cos 2x = −1 2x = π , 3π x = π2 , 32π
41. In order to find all of the values of x in 0 ,360 that satisfy
1 cos(2x ) + sin x = 0 , we proceed as follows: 2 Step 1: Simplify the equation algebraically. 1 1 cos(2 x ) + sin x = 0 cos 2 x − sin2 x + sin x = 0 2 2 1 (1 − sin2 x ) − sin2 x + sin x = 0 2 2 4 sin x − sin x − 2 = 0 sin x =
Step 2: Solve sin x =
−( −1) ± ( −1)2 − 4(4)( −2) 1 ± 33 = 2(4) 8
1 + 33 . 8
To do so, we must find the values of x whose sine is
641
1 + 33 . 8
Chapter 6
1 + 33 Indeed, observe that one solution is sin −1 ≈ 57.47 , which is in QI. A 8 second solution occurs in QII, and has value 180 − 57.47 = 122.53. Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 57.47 , 122.53. Step 3: Solve sin x =
1 − 33 . 8
To do so, we must find the values of x whose sine is
1 − 33 . 8
1 − 33 Indeed, observe that one solution is sin −1 ≈ −36.38 , which is in QIV. 8 Since the angles we seek have positive measure, we use the representative 360 − 36.38 = 323.62 . A second solution occurs in QIII, and has value 180 + 36.38 = 216.38 . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 216.38 , 323.62 . Step 4: Conclude that the solutions to the original equation are x ≈ 57.47 , 122.53 , 216.38 , 323.62 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation. 42. Observe that
sec 2 x = tan x + 1 1 + tan2 x = tan x + 1 tan x ( tan x − 1) = 0 tan x = 0 or tan x − 1 = 0 so that tan x = 0 or tan x = 1 The solutions to these equations in 0 ,360 ) are x = 0 , 180 , 45 , 225 . Substituting these into the original equation shows that they are all solutions.
642
Section 6.3
43. In order to find all of the values of x in 0 ,360 that satisfy 6 cos 2 x + sin x = 5,
we proceed as follows: Step 1: Simplify the equation algebraically. 6 cos 2 x + sin x = 5 6 (1 − sin2 x ) + sin x = 5 6 − 6 sin2 x + sin x = 5
(3sin x + 1)( 2 sin x − 1) = 0 3sin x + 1 = 0 or 2 sin x − 1 = 0 so that sin x = −
1 1 or sin x = 3 2
1 Step 2: Solve sin x = − . 3 1 To do so, we must find the values of x whose sine is − . 3 1 Indeed, observe that one solution is sin −1 − ≈ −19.47 , which is in QIV. 3 Since the angles we seek have positive measure, we use the representative 360 − 19.47 = 340.53 . A second solution occurs in QIII, and has value 180 + 19.47 = 199.47 . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are x ≈ 199.47 , 340.53 . 1 Step 3: Solve sin x = . 2
1 Indeed, observe that one solution is sin −1 = 30 , which is in QI. A second 2 solution occurs in QII, and has value 180 − 30 = 150. Since the input of sine is simply x, and not some multiple thereof, we see that the solutions are x = 30 , 150. Step 4: Conclude that the solutions to the original equation are x ≈ 30 , 150 , 199.47 , 340.53 .
Substituting these into the original equation shows that they are all solutions.
643
Chapter 6
44. In order to find all of the values of x in 0 ,360 that satisfy sec 2 x = 2 tan x + 4,
we proceed as follows: Step 1: Simplify the equation algebraically. sec 2 x = 2 tan x + 4 1 + tan2 x = 2 tan x + 4 tan2 x − 2 tan x − 3 = 0
( tan x − 3)( tan x + 1) = 0
tan x − 3 = 0 or tan x + 1 = 0 tan x = 3
or tan x = −1
Step 2: Solve tan x = 3. To do so, we must find the values of x whose tangent is 3. Indeed, observe that one solution is tan −1 (3) ≈ 71.57°, which is in QI. A second solution occurs in QIII, and has value 180 + 71.57 = 251.57. Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 71.57 , 251.57. Step 3: Solve tan x = −1. To do so, we must find the values of x whose tangent is −1. Indeed, observe that one solution is tan −1 ( −1) = 135 , which is in QII.
A second solution occurs in QIV, and has value 180 + 135 = 315. Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x = 135 , 315.
Step 4: Conclude that the solutions to the original equation are x ≈ 135 , 315 , 71.57 , 251.57 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation.
644
Section 6.3
45. In order to find all of the values of x in 0 ,360 that satisfy cot 2 x − 3csc x − 3 = 0,
we proceed as follows: Step 1: Simplify the equation algebraically. cot 2 x − 3csc x − 3 = 0
( csc x − 1) − 3csc x − 3 = 0 2
csc 2 x − 3csc x − 4 = 0
( csc x + 1)( csc x − 4 ) = 0 csc x + 1 = 0 or csc x − 4 = 0 csc x = −1 or csc x = 4 sin x = −1
or sin x =
1 4
Step 2: Solve sin x = −1. To do so, we must find the values of x whose sine is −1. Indeed, the only solution is sin −1 ( −1) = 270 , which is in QIII. 1 Step 3: Solve sin x = . 4
1 Indeed, observe that one solution is sin −1 ≈ 14.48 , which is in QI. A second 4 solution occurs in QII, and has value 180 − 14.48 = 165.52. Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are x ≈ 14.48 , 165.52. Step 4: Conclude that the solutions to the original equation are x ≈ 14.48 , 165.52 ,270 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation.
645
Chapter 6
46. In order to find all of the values of x in 0 ,360 that satisfy csc 2 x + cot x = 7,
we proceed as follows: Step 1: Simplify the equation algebraically: csc 2 x + cot x = 7
(1 + cot x ) + cot x = 7 2
cot 2 x + cot x − 6 = 0
( cot x + 3)( cot x − 2 ) = 0
cot x + 3 = 0 or cot x − 2 = 0 cot x = −3
or cot x = 2
Step 2: Solve cot x = −3. To do so, we must find the values of x whose cotangent is −3. 1 Indeed, observe that one solution is cot −1 ( −3 ) = 180 + tan −1 − ≈ 161.57 , 3 which is in QII. A second solution occurs in QIV, and has value 180 + 161.57 = 341.57. Since the input of cotangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 161.57 , 341.57. Step 3: Solve cot x = 2. To do so, we must find the values of x whose cotangent is 2. 1 Indeed, observe that one solution is cot −1 ( 2 ) = tan −1 ≈ 26.57 , 2 which is in QI. A second solution occurs in QIII, and has value 180 + 26.57 = 206.57. Since the input of cotangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 26.57 , 206.57. Step 4: Conclude that the solutions to the original equation are x ≈ 26.57 , 206.57 , 161.57 , 341.57 .
646
Section 6.3
47. In order to find all of the values of x in 0 ,360 that satisfy 2 sin2 x + 2 cos x − 1 = 0,
we proceed as follows: Step 1: Simplify the equation algebraically. 2 sin2 x + 2 cos x − 1 = 0 2 (1 − cos 2 x ) + 2 cos x − 1 = 0 2 − 2 cos 2 x + 2 cos x − 1 = 0 2 cos 2 x − 2 cos x − 1 = 0 cos x = cos x =
−( −2) ± ( −2)2 − 4(2)( −1) 2 ± 2 3 1 ± 3 = = 2(2) 4 2
1− 3 2
1+ 3 cos x = 2
or
No solution since
Step 2: Solve cos x =
.
1+ 3 >1 2
1− 3 . 2
To do so, we must find the values of x whose cosine is
1− 3 . 2
1− 3 Indeed, the only solution is cos −1 ≈ 111.47 , which is in QII. A second 2 solution occurs in QIII and has value 360 − 111.47 = 248.53. Since the input of cosine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 111.47 , 248.53. Step 3: Conclude that the solutions to the original equation are x ≈ 111.47 , 248.53 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation.
647
Chapter 6
48. In order to find all of the values of x in 0 ,360 that satisfy sec 2 x + tan x − 2 = 0,
we proceed as follows: Step 1: Simplify the equation algebraically. sec 2 x + tan x − 2 = 0
(1 + tan x ) + tan x − 2 = 0 2
tan2 x + tan x − 1 = 0 tan x = Step 2: Solve tan x =
−1 ± 12 − 4(1)( −1) −1 ± 5 = 2(1) 2
−1 − 5 . 2
To do so, we must find the values of x whose tangent is
−1 − 5 . 2
−1 − 5 Indeed, observe that one solution is tan −1 ≈ −58.28 , which is in QIV. 2 Since the angles we seek have positive measure, we use the representative 360 − 58.28 = 301.72 A second solution occurs in QII, and has value 301.72 − 180 = 121.72. Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 121.72 , 301.72. Step 3: Solve tan x =
−1 + 5 . 2
To do so, we must find the values of x whose tangent is
−1 + 5 . 2
−1 + 5 Indeed, observe that one solution is tan −1 ≈ 31.72 , which is in QI. A 2 second solution occurs in QIII, and has value 180 + 31.72 = 211.72. Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 31.72 , 211.72. Step 4: Conclude that the solutions to the original equation are x ≈ 31.72 , 211.72 , 121.72 , 301.72 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation.
648
Section 6.3
49.
cos 2x + sin x + 1 = 0
(1 − 2 sin x ) + sin x + 1 = 0 2
2 sin2 x − sin x − 2 = 0 1 ± 1 + 16 1 ± 17 = 4 4 1 + 17 1 − 17 sin x = or sin x = 4 4 sin x =
No solution.
1 − 17 x = sin −1 4 So, the solutions are x ≈ 308.67 ,231.33. 50.
sin2 2 x + sin 2 x − 2 = 0
( sin 2x − 1)( sin 2x + 2 ) = 0 sin 2 x = 1, −2 2 x = 90 , 450 x = 45 , 225
51.
52. tan x + sec x = −1
sin2 x − 2 sin x =0 tan x sin x ( sin x − 2 ) = 0
2
( tan x + 1) + sec x = 0 2
sec 2 x + sec x = 0
x =2 sin x = 0 or sin
sec x ( sec x + 1) = 0
No solution
sec x =0 or sec x = −1
x=
No solution
cos x = −1 x = 180
0 , 180 Omit since both make denominator zero.
Hence, there is no solution.
649
Chapter 6
53. Observe that using the identity
54. Observe that using the identity
sin 2 A = 2 sin A cos A with A =
sin 2 A = 2 sin A cos A with A =
π
3
x yields
π
3
x yields
π π π π 2 sin x cos x + 3 = 4 2 sin x cos x + 3 = 2 3 3 3 3 π π sin 2 ⋅ x = 1 sin 2 ⋅ x = −1 3 3 This equation is satisfied when 2π π This equation is satisfied when x= , 2π 3π 9 3 2 x= , so that x = . So, it takes 3 2 4 3 3 sec. for so that x = . So, it takes 9 4 4 sec. for the volume of air to equal 2 4 the volume of air to equal 4 liters. liters. 55. First, we solve −2 sin x + 2 sin 2 x = 0 on ( 0,2π ) : −2 sin x + 2 sin 2 x = 0
−2 sin x + 2 [ 2 sin x cos x ] = 0
2 sin x ( −1 + 2 cos x ) = 0 2 sin x = 0 or − 1 + 2 cos x = 0 2 sin x = 0 or cos x =
These equations are satisfied when x = π ,
π 5π 3
,
3
1 2
. We now need to determine the
corresponding y-coordinates. x y( x ) = 2 cos x − cos 2x π π π π 1 1 3 y = 2 cos − cos 2 ⋅ = 2 − − = 3 3 3 3 2 2 2 5π 5π 5π 5π 1 1 3 y = 2 cos − cos 2 ⋅ = 2 − − = 3 3 3 3 2 2 2
π
y (π ) = 2 cos (π ) − cos ( 2 ⋅ π ) = 2( −1) − 1 = −3
π 3 5π 3 So, the turning points are , , , , and (π , − 3 ) . 3 2 3 2
650
Point π 3 , 3 2 5π 3 , 3 2
(π , − 3 )
Section 6.3
56. First, we solve −2 sin x + 2 sin 2 x = 0 on ( −2π ,0 ) : −2 sin x + 2 sin 2 x = 0
−2 sin x + 2 [ 2 sin x cos x ] = 0
2 sin x ( −1 + 2 cos x ) = 0 2 sin x = 0 or − 1 + 2 cos x = 0 1 2 π 5π . We now need to determine These equations are satisfied when x = −π , − , − 3 3 the corresponding y-coordinates. 2 sin x = 0 or cos x =
x
y( x ) = 2 cos x − cos 2x
Point
π
π π π y − = 2 cos − − cos −2 ⋅ 3 3 3 1 1 3 = 2 −− = 2 2 2
π 3 − , 3 2
5π 3
5π 5π 5π y − = 2 cos − − cos −2 ⋅ 3 3 3 1 1 3 = 2 −− = 2 2 2
5π 3 − , 3 2
−π
y ( −π ) = 2 cos ( −π ) − cos ( −2 ⋅ π ) = 2( −1) − 1 = −3
( −π , − 3)
−
−
3
π 3 5π 3 So, the turning points are − , , − , , ( −π , − 3 ) . 3 2 3 2 57. We solve the equation
58. We solve the equation
5 + 3sin t = 5 − 2 cos t sin t cos t
5.5 + 4 sin t = 5.5 − 3cos t
= − 32
sin t cos t
tan t = − 23 t = tan −1 ( − 23 ) ≈ −0.588 So, the desired times are t = −0.588 + nπ , where n is an integer. Those occurring in the interval [0,10] are 2.55 minutes, 5.70 minutes, and 8.84 minutes.
= − 34
tan t = − 34 t = tan −1 ( − 34 ) ≈ −0.6435
So, the desired times are t = −0.6435 + nπ , where n is an integer. Those occurring in the interval [0,10] are 2.50 minutes, 5.64 minutes, and 8.78 minutes.
651
Chapter 6
59. We solve the equation tan x + 1 = 2 sin x + 1 on the interval [0, 1.5]. tan x + 1 = 2 sin x + 1 sin x − 2 sin x = 0 cos x sin x − 2 sin x cos x =0 cos x sin x (1 − 2 cos x ) = 0 sin x = 0 or cos x = 12 x = 0, π3 ≈ 1.05 So, the point of intersection of the two trails is (1.05, 2.73).
60. We solve the equation sin x + 1 = 2 cot x + 1 on the interval [0, 1.5]. sin x + 1 = 2 cot x + 1
2 cos x =0 sin x sin2 x − 2 cos x =0 sin x (1 − cos2 x ) − 2 cos x = 0 sin x 2 cos x + 2 cos x − 1 = 0 sin x −
−2 ± 4 + 4 = −1 ± 2 2 −1 − cos 2 or cos x = −1 + 2 x = cos x =
No solution
(
)
x = cos −1 −1 + 2 ≈ 1.14 So, the point of intersection of the two trails is approximately (1.14, 1.91).
652
Section 6.3
61. Solve the equation R(t ) − C (t ) = 0 as follows:
( 2.3 + 0.5 cos ( t ) ) − ( 2.3 + 0.25sin ( t ) ) = 0 π
π
6
6
0.5 cos ( 6 t ) − 0.25sin ( π6 t ) = 0 π
0.5 cos ( π6 t ) = 0.25sin ( π6 t ) tan ( π6 t ) = 2 π 6
t = tan −1 2, π + tan −1 2 t = π6 ( tan −1 2 ) , π6 (π + tan −1 2 ) ≈ 2.11, 8.11
So, breaks even in March and September. 62. Solve the equation R(t ) − C (t ) = 0 as follows:
( 25.7 + 9.6 cos ( t ) ) − ( 25.7 + 0.2 sin ( t ) ) = 0 π
π
6
6
9.6 cos ( π6 t ) − 0.2 in ( π6 t ) = 0
9.6 os ( π6 t ) = 0.2 sin ( π6 t ) tan ( π6 t ) = 48 π 6
t = tan −1 48, π + tan −1 48 t = π6 ( tan −1 48 ) , π6 (π + tan −1 48 ) ≈ 2.96, 8.96
So, breaks even in April and October. 63. Solve the equation as follows, using the double-angle formula for sin 2A : π π t ) cos ( 24 t ) = 37.28 37.10 + 1.40 sin ( 24 π π t ) cos ( 24 t ) = 0.18 1.40 sin ( 24
π π t ) cos ( 24 t ) = 0.18 0.7 2 sin ( 24
sin ( 12π t ) =
0.18 0.7
0.18 −1 0.18 t = sin −1 , π − sin 0.7 0.7 0.18 12 −1 0.18 t = 12π sin −1 , π π − sin ≈ 1, 11 0.7 0.7 So, this happens around 1 am and 11 am. π
12
653
Chapter 6
64. Solve the equation as follows, using the double-angle formula for sin 2A : π t ) = 36.75 37.10 + 1.40 sin ( 24π t ) cos ( 24 π t ) = −0.35 1.40 sin ( 24π t ) cos ( 24
π π t ) cos ( 24 t ) = −0.35 0.7 2 sin ( 24
1 2 π 7π 11π 12 t = 6 , 6
sin ( 12π t ) = −
t = 14,22
So, this happens around 2 pm and 10 pm. 65. Cannot divide by cos x since it could be zero. Rather, should factor as follows: 6 sin x cos x = 2 cos x 6 sin x cos x − 2 cos x = 0 2 cos x ( 3sin x − 1) = 0 Now, proceed…
66. Forgot to check for extraneous solutions. Note that for x = π , we have 1 + sin π = 1 = 1 , while cos π = −1 . Hence, x = π is not a solution to the equation. The remaining values ARE solutions.
67. True. This follows by definition of an identity.
68. False. This is not sufficient. For instance, the equation sin x = 1 has infinitely many solutions, but there are values of x in the domain for which it is not true (for example, x = 0 ).
69. True, because sin x sin x =1 ⇔ −1 = 0 sec x sec x sin x − sec x ⇔ =0 sec x ⇔ sin x − sec x = 0
70. False. The first equation has x = nπ as a solution, while these values render terms in the second equation as undefined.
654
Section 6.3
71. Consider the following graphs:
72. Consider the following graphs:
Since the graphs intersect twice on the given interval, there are 2 solutions of the given equation.
Since the graphs do not intersect on the given interval, there is no solution to the given equation.
73. First, observe that using the addition formulae for sin( A ± B ) yields
π π π π sin x + + sin x − = sin x cos + cos x sin 4 4 4 4 π π + sin x cos − cos x sin 4 4
2 ( 2 sin x ) 2 = 2 sin x We substitute this for the left-side of the given equation to obtain: π π 2 2 1 sin x + + sin x − = 2 sin x = sin x = 4 4 2 2 2 =
The smallest positive value of x for which this is true is x =
655
π 6
or 30 .
Chapter 6
74. First, observe that using the addition formula cos( A − B ) = cos A cos B + sin A sin B yields cos x cos (15 ) + sin x sin (15 ) = 0.7
75. 1 cos x = cos x 2 1 + cos x = cos x 2 1 + cos x = cos 2 x 2 1 + cos x − 2 cos 2 x =0 2 2 cos 2 x − cos x − 1 = 0 ±
cos ( x − 15 ) = 0.7
We now find the solutions to this equation in 0 , 360 : One solution is x = 15 + cos −1 ( 0.7 ) ≈ 15 + 45.57 = 60.57 , which is in QI. The second solution is in QIV and has value 15 + 314.43 = 329.43. Observe that the smallest positive solution is
( 2 cos x + 1)( cos x − 1) = 0 cos x = − 12 , 1 x = 0, 23π , 43π , 2π
approximately x ≈ 60.57 . 76.
1 tan x = sin x 2 1 − cos x = sin x 1 + cos x 1 − cos x = sin2 x 1 + cos x 2 1 − cos x − sin x (1 + cos x ) =0 1 + cos x (1 − cos x ) − (1 − cos 2 x ) (1 + cos x ) =0 1 + cos x
±
(1 − cos x ) − (1 − cos x )(1 + cos x ) = 0 2
1 + cos x 2 (1 − cos x ) 1 − (1 + cos x ) =0 1 + cos x
656
Section 6.3
1 − cos x = 0 or 1 − (1 + cos x ) = 0 2
=− cos x = 1 or cos x = 0 or cos x 2 No solution π 3π x = 0, , 2 2 77. Observe that 1 1 − ≤ sin x cos x ≤ 4 4 1 1 1 − ≤ ⋅ ( 2 sin x cos x ) ≤ 4 2 4 1 1 1 − ≤ ⋅ sin 2 x ≤ 4 2 4 1 1 − ≤ sin 2 x ≤ 2 2 Consider the diagram to the right. We see that 0 ≤ 2 x ≤ π6 or 56π ≤ 2 x ≤ 76π or 116π ≤ 2 x ≤ 2π Hence, π 5π 7π 11π ∪ x ∈ 0, ∪ , , π . 12 12 12 12 78. Observe that cos 2 x − sin2 x ≤ cos 2x ≤
1 2
1 2
1 1 ≤ cos 2 x ≤ 2 2 Consider the diagram to the right. We see that π 2π 4π 5π 3 ≤ 2 x ≤ 3 or 3 ≤ 2 x ≤ 3 Hence, π π 2π 5π x∈x = , ∪ , . 6 3 3 6 −
657
Chapter 6
79. To determine the smallest positive solution (approximately) of the equation sec 3x + csc 2 x = 5 graphically, we search for the intersection points of the graphs of the following two functions: y1 = sec 3x + csc 2 x, y2 = 5
The x-coordinate of the intersection point is in radians. Observe that the smallest positive solution, in degrees, is approximately 7.39° . 80. To determine the smallest positive solution (approximately) of the equation cot 5x + tan 2 x = −3 graphically, we search for the intersection points of the graphs of the following two functions: y1 = cot 5x + tan 2x, y2 = −3
The x-coordinate of the intersection point is in radians. Observe that the smallest positive solution, in degrees, is approximately 33.92 .
658
Section 6.3
81. To determine the smallest positive solution (approximately) of the equation e x − tan x = 0 graphically, we search for the intersection points of the graphs of the following two functions: y1 = e x − tan x, y2 = 0
Observe that the graphs seem to intersect at approximately 1.3 radians, which is about 79.07 . Answers may vary if the solutions in radians is found first and then converted to degrees and vice versa due to rounding. 82. To determine the smallest positive solution (approximately) of the equation e x + 2 sin x = 1 graphically, we search for the intersection points of the graphs of the following two functions: y1 = e x + 2 sin x, y2 = 1
Observe that the two graphs intersect at 0, but never intersect at any positive x-value.
659
Chapter 6 Review Solutions --------------------------------------------------------------------------1. The equation arc tan (1) = θ is
sin θ . Since the cos θ π π range of inverse tangent is − , , we 2 2 equivalent to tan θ = 1 =
conclude that θ =
π 4
.
3. The equation cos −1 ( 0 ) = θ is equivalent to cos θ = 0 . Since the range of arccosine is [ 0, π ] , we conclude that
θ=
π 2
.
2. The equation arc csc ( −2 ) = θ is equivalent to csc θ = −2 , which is further 1 the same as sin θ = − . Since the range 2 π π of inverse cosecant is − ,0 ∪ 0, , 2 2
we conclude that θ = −
6
.
4. The equation sin −1 ( −1) = θ is equivalent to sin θ = −1 . Since the range π π of arcsine is − , , we conclude that 2 2
θ= −
5. The equation csc −1 ( −1) = θ is
π
π 2
.
6. The equation arc tan ( −1) = θ is
equivalent to csc θ = −1 , which is further sin θ equivalent to tan θ = −1 = . Since the same as sin θ = −1 . Since the range of cos θ π π π π inverse cosecant is − ,0 ∪ 0, , we the range of inverse tangent is − , , 2 2 2 2 conclude that θ = −
π
2
, which
corresponds to θ = −90 .
we conclude that θ = −
π 4
, which
corresponds to θ = −45 .
660
Chapter 6 Review
3 7. The equation arc cot = θ is 3 equivalent to 1 3 1 cos θ cot θ = = = 2 = . Since 3 sin θ 3 3 2 the range of inverse cotangent is ( 0, π ) ,
we conclude that θ =
π 3
2 8. The equation cos −1 = θ is 2
2 . Since the 2 range of arccosine is [ 0, π ] , we conclude equivalent to cos θ =
that θ =
π
4
, which corresponds to
θ = 45 .
, which
corresponds to θ = 60 . 9. sin −1 ( −0.6088 ) ≈ −37.50
10. tan −1 (1.1918 ) ≈ 50.00
11.
12.
1 sec −1 (1.0824 ) = cos −1 ≈ 22.50 1.0824
1 cot −1 ( −3.7321) = 180 + tan −1 −3.7321 ≈ 165.00
13. cos −1 ( −0.1736 ) ≈ 1.75
14. tan −1 ( 0.1584 ) ≈ 0.16
15.
16.
1 csc −1 ( −10.0167 ) = sin −1 −10.0167
1 sec −1 ( −1.1223 ) = cos −1 ≈ 2.67 −1.1223
≈ −0.10
π π π 17. sin −1 sin − = − since −1 ≤ − ≤ 1 . 4 4 4 2 2 2 18. cos cos −1 − since −1 ≤ − ≤ 1. = − 2 2 2
(
19. Since tan ( tan −1 x ) = x for all −∞ < x < ∞ , we see that tan tan −1
661
( 3 )) = 3 .
Chapter 6
11π 20. Note that we need to use the angle θ in [ 0, π ] such that cot θ = cot . To 6 11π 5π 5π −1 . this end, observe that cot −1 cot = cot cot = 6 6 6 π π 21. Note that we need to use the angle θ in − ,0 ∪ 0, such that 2 2 2π 2π π π −1 −1 . csc θ = csc . To this end, observe that csc csc = csc csc = 3 3 3 3 22. Since sec ( sec −1 θ ) = θ for all θ ∈ ( −∞, −1] ∪ [1, ∞ ) , we see that
2 3 2 3 = − . sec sec −1 − 3 3 11 23. Let θ = cos −1 . Then, 61 11 cos θ = , as shown in the diagram: 61
40 24. Let θ = tan −1 . Then, 9 40 , as shown in the diagram: tan θ = 9
θ
θ
Using the Pythagorean Theorem, we see that z 2 + 112 = 612 , so that z = 60 . 60 11 Hence, sin cos −1 = sin θ = . 61 61
Using the Pythagorean Theorem, we see that 92 + 40 2 = z 2 , so that z = 41 . 9 40 Hence, cos tan −1 = cos θ = . 41 9
662
Chapter 6 Review
6 25. Let θ = cot −1 . Then, 7 6 cot θ = . Hence, 7 1 7 6 = . tan cot −1 = tan θ = cot θ 6 7
25 25 , so 26. Let θ = sec −1 . Then, sec θ = 7 7 7 , as shown in the diagram: that cos θ = 25
θ
Using the Pythagorean Theorem, we see that z 2 + 72 = 252 , so that z = 24 . 7 25 Hence, cot sec −1 = cot θ = . 24 7
1 27. Let θ = sin −1 . Then, 6 1 sin θ = , as shown in the diagram: 6
5 5 28. Let θ = cot −1 . Then, cot θ = , as 12 12 shown in the diagram:
θ θ
663
Chapter 6
Using the Pythagorean Theorem, we see that z 2 + 12 = 62 , so that z = 35 . So, 1 1 sec sin−1 = sec θ = cos θ 6 . 6 6 35 = = 35 35
Using the Pythagorean Theorem, we see that 52 + 122 = z 2 , so that z = 13 . So, 1 13 5 csc cot −1 = csc θ = = . sin θ 12 12
29. Given that T ( m ) = 47 + 26 sin(0.48m − 1.84) , we must find the value of t for which T ( m ) = 73 . To this end, we have 47 + 26 sin(0.48m − 1.84) = 73 sin(0.48m − 1.84) = 1 0.48m − 1.84 = sin −1 (1) = π
m=
2
π + 1.84 2
≈ 7.105 0.48 As such, the month during which the average is 73 is July.
30. Given that T ( m ) = 47 + 26 sin(0.48m − 1.84) , we must find the value of t for which T ( m ) = 21 . To this end, we have 47 + 26 sin(0.48m − 1.84) = 21 sin(0.48m − 1.84) = −1 0.48m − 1.84 = sin −1 ( −1) = 3π 2 3π + 1.84 2 ≈ 13.6 m= 0.48 As such, using periodicity, the month during which the average is 21 is January.
664
Chapter 6 Review
3 must satisfy 2 4π 4π 5π 5π 4π 10π 5π 11π + 2π , + 2π = . 2θ = , , , , , 3 3 3 3 3 3 3 3 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 2π 5π 5π 11π θ= , , , 3 3 6 6 31. The values of θ in [ 0,2π ] that satisfy sin 2θ = −
θ θ 1 32. First, observe that sec = 2 is equivalent to cos = . The values of θ in 2 2 2 θ 1 θ π 5π [ −2π ,2π ] that satisfy cos = must satisfy = ± , ± . 2 3 3 2 2 So, multiplying all values by 2 yields the following values of θ which satisfy the original equation: 2π 10π θ= ± , ± 3 3 outside interval
2 θ 33. The values of θ in [ −2π ,2π ] that satisfy sin = − must satisfy 2 2 θ π 3π 5π 7π =− , − , . , 2 4 4 4 4 So, multiplying all values by 2 yields the following values of θ which satisfy the original equation: π 3π 5π 7π . θ=− ,− , , 2 2 2 2 Outside interval
665
Chapter 6
1 34. First, observe that csc ( 2θ ) = 2 is equivalent to sin ( 2θ ) = . The values of θ in 2 1 [0,2π ] that satisfy sin ( 2θ ) = must satisfy 2 π π 5π 5π π 13π 5π 17π + 2π , + 2π = , 2θ = , , , , . 6 6 6 6 6 6 6 6 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: π 13π 5π 17π θ= , , , 12 12 12 12 1 35. First, observe that 4 cos ( 2θ ) + 2 = 0 is equivalent to cos ( 2θ ) = − . The values 2 1 of θ in [ 0,2π ] that satisfy cos ( 2θ ) = − must satisfy 2 2π 2π 4π 4π 2π 4π 8π 10π + 2π , + 2π = 2θ = , , , , , . 3 3 3 3 3 3 3 3 So, dividing all values by 2 yields the following values of θ which satisfy the original 2π 4π 8π 10π π 2π 4π 5π = , , , , , , equation: θ = 6 6 6 6 3 3 3 3 θ 3 tan − 1 = 0 is equivalent to 2 θ sin 1 θ 1 2 . The values of θ in 0,2π that satisfy this must tan = = 2 = [ ] θ 3 3 2 2 cos 2 θ π 7π π 7π satisfy = , , so that θ = , . 2 6 6 3 3
36. First, observe that
Outside interval
666
Chapter 6 Review
37. First, observe that 2 tan 2θ + 2 = 0 is equivalent to tan 2θ = −1 . The values of θ in [ 0,2π ] that satisfy tan 2θ = −1 must satisfy
3π 3π 7π 7π 3π 11π 7π 15π + 2π , + 2π = , , , , , . 4 4 4 4 4 4 4 4 So, dividing all values by 2 yields the following values of θ which satisfy the original equation: 3π 11π 7π 15π θ= , , , . 8 8 8 8 2θ =
38. Factoring the left-side of 2 sin2 θ + sin θ − 1 = 0 yields ( sin θ + 1)( 2 sinθ − 1) = 0 sin θ + 1 = 0 or 2 sin θ − 1 = 0 1 sin θ = −1 or sin θ = 2
The values of θ in [ 0,2π ] that satisfy these equations are θ =
π 5π 3π 6
,
6
,
2
.
39. Factoring the left-side of tan2 θ + tan θ = 0 yields the equivalent equation tan θ ( tan θ + 1) = 0 which is satisfied when either tan θ = 0 or tan θ + 1 = 0. The
values of θ in [ 0,2π ) that satisfy tan θ = 0 are θ = 0, π , and those which satisfy tan θ = −1 are θ =
3π 7π , . Thus, the solutions to the original equation are 4 4 3π 7π θ = 0, π , , . 4 4
40. Factoring the left-side of sec 2 θ − 3sec θ + 2 = 0 yields the equivalent equation ( sec θ − 2 )( sec θ − 1) = 0 which is satisfied when either sec θ − 2 = 0 or secθ − 1 = 0.
Note that the values of θ in [ 0,2π ) that satisfy the equation sec θ = 2 (or equivalently 1 π 5π ) are θ = , . Also, the values of θ in [ 0,2π ) that satisfy sec θ = 1 are 2 3 3 θ = 0, 2π . Thus, the solutions to the original equation are cos θ =
θ=
π 5π 3
,
3
667
,0
Chapter 6
41. In order to find all of the values of θ in 0 ,360 that satisfy tan ( 2θ ) = −0.3459 , we proceed as follows:
Step 1: Find the values of 2θ whose tangent is −0.3459. Indeed, observe that one solution is tan −1 ( −0.3459 ) ≈ −19.081 , which is in QIV. Since the angles we seek have positive measure, we use the representative 360 − 19.081 ≈ 340.92 . A second solution occurs in QII, and has value 340.92 − 180 = 160.92 . Step 2: Use Step 1 to find all values of θ that satisfy the original equation. The solutions obtained in Step 1 are 2θ = 160.92 , 160.92 + 360 , 340.92 , 340.92 + 360 . When divided by 2, we see that the solutions to the original equation in 0 ,360 are approximately θ ≈ 80.46 , 260.46 , 170.46 , 350.46 . 42. In order to find all of the values of θ in 0 ,360 that satisfy 6 sin θ − 5 = 0 , we
proceed as follows: Step 1: First, observe that the equation is equivalent to sin θ =
5 . 6
5 . 6 5 Indeed, observe that one solution is sin −1 ≈ 56.44 , which is in QI. A second 6 solution occurs in QII, and has value 180 − 56.443 = 123.56 . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to
Step 2: Find the values of θ whose sine is
the original equation are approximately θ ≈ 56.44 , 123.56 . 43. In order to find all of the values of θ in 0 ,360 that satisfy
4 cos 2 θ + 3cos θ = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: cos θ ( 4 cos θ + 3 ) = 0 3 4 Step 2: Observe that the values of θ that satisfy cos θ = 0 are θ = 90 , 270 . cos θ = 0 or 4 cos θ + 3 = 0 so that cos θ = 0 or cos θ = −
668
Chapter 6 Review
3 Step 3: Find the values of θ whose cosine is − . 4 3 Indeed, observe that one solution is cos −1 − ≈ 138.59 , which is in QII. A 4 second solution occurs in QIII, and has value 360 − 138.59 = 221.41 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately θ ≈ 138.59 , 221.41 . Step 4: Conclude the solutions in 0 ,360 are θ ≈ 90 , 270 , 138.59 , 221.41 . 44. In order to find all of the values of θ in 0 ,360 that satisfy
12 cos 2 θ − 7 cos θ + 1 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: ( 4 cos θ − 1)( 3cos θ − 1) = 0 4 cos θ − 1 = 0 or 3cos θ − 1 = 0 so that cos θ =
1 1 or cos θ = 4 3
1 . 4 1 Indeed, observe that one solution is cos −1 ≈ 75.52 , which is in QI. A second 4 solution occurs in QIV, and has value 360 − 75.52 = 284.48 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately θ ≈ 75.52 , 284.48 .
Step 2: Find the values of θ whose cosine is
1 . 3 1 Indeed, observe that one solution is cos −1 ≈ 70.53 , which is in QI. A second 3 solution occurs in QIV, and has value 360 − 70.53 = 289.47 . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately θ ≈ 70.53 , 289.47 .
Step 3: Find the values of θ whose cosine is
669
Chapter 6
45. In order to find all of the values of θ in 0 ,360 that satisfy
csc 2 θ − 3csc θ − 1 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating csc θ as the variable): −( −3) ± ( −3)2 − 4 (1)( −1) 3 ± 13 csc θ = = 2 (1) 2 So, θ is a solution to the original equation if either 3 + 13 3 − 13 csc θ = or csc θ = , 2 2 which is equivalent to 2 2 sin θ = or sin θ = 3 + 13 3 − 13 No solution since
Step 2: Solve sin θ =
2 3 + 13
2 < −1 3− 13
.
To do so, we must find the values of θ whose sine is
2 3 + 13
.
2 Indeed, observe that one solution is sin −1 ≈ 17.62 , which is in QI. 3 + 13 A second solution occurs in QII, and has value 180 − 17.62 = 162.38 . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 17.62 , 162.38 .
Step 3: Conclude that the solutions to the original equation are
θ ≈ 17.62 , 162.38 . 46. In order to find all of the values of θ in 0 ,360 that satisfy
2 cot 2 θ + 5 cot θ − 4 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating cot θ as the variable): cot θ =
−5 ± 52 − 4 ( 2 )( −4 ) 2 (2)
670
=
−5 ± 57 4
Chapter 6 Review
So, θ is a solution to the original equation if either −5 − 57 −5 + 57 cot θ = or cot θ = . 4 4 −5 − 57 . Step 2: Solve cot θ = 4 To do so, we must find the values of θ whose cotangent is
−5 − 57 . 4
Indeed, observe that one solution is −5 − 57 4 −1 ≈ 162.32 , which is in QII. A second cot −1 = 180 + tan 4 −5 − 57 solution occurs in QIV, and has value 180 + 162.32 = 342.32 . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 162.32 , 342.32 . Step 3: Solve cot θ =
−5 + 57 . 4
−5 + 57 . 4 −5 + 57 4 −1 Indeed, observe that one solution is cot −1 = tan ≈ 57.48 , 4 −5 + 57 which is in QI. A second solution occurs in QIII, and has value 180 + 57.48 = 237.48 . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately θ ≈ 57.48 , 237.48 . To do so, we must find the values of θ whose cotangent is
Step 4: Conclude that the solutions to the original equation are
θ ≈ 57.48 , 237.48 , 162.32 , 342.32 . 47. Observe that
sec x = 2 sin x 1 = 2 sin x cos x 1 = 2 sin x cos x 1 = sin 2 x
671
Chapter 6
5π . 4 4 2 2 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
Note that the solutions of sin 2 x = 1 are 2x =
π
,
π
π
+ 2π , so that x =
,
48. Observe that 3tan x + cot x = 2 3 sin x cos x + =2 3 cos x sin x 3sin2 x + cos 2 x =2 3 sin x cos x 3
3sin2 x + cos 2 x = 2 3 sin x cos x
(3sin x + cos x ) = ( 2 3 sin x cos x ) 2
2
2
2
9sin 4 x + 6 sin2 x cos 2 x + cos 4 x = 12 sin2 x cos 2 x 9sin 4 x − 6 sin2 x cos 2 x + cos 4 x = 0
(3sin x − cos x ) = 0 2
2
2
3sin2 x − cos 2 x = 0
3sin2 x − (1 − sin2 x ) = 0 4 sin2 x = 1
sin x = ±
1 2
The values of x in [ 0,2π ] that satisfy these equations are x =
π 5π 7π 11π 6
,
6
,
Substituting these values into the original equation shows that while solutions,
π 7π 6
,
6
6
,
π 7π 6
,
6
6
.
are
5π 11π is extraneous. So, the only solutions to the original equation are , 6 6
.
672
Chapter 6 Review
49. Observe that 3 tan x − sec x = 1 1 sin x 3 =1 − cos x cos x 3 sin x − 1 = cos x
( 3 sin x − 1) = cos x 2
2
3sin2 x − 2 3 sin x + 1 = 1 − sin2 x 4 sin2 x − 2 3 sin x = 0
(
)
2 sin x 2 sin x − 3 = 0 sin x = 0 or 2 sin x − 3 = 0 sin x = 0 or sin x =
3 2
The values of x in [ 0,2π ) that satisfy these equations are x = 0, π ,
π 2π 3
,
3
Substituting these values into the original equation shows that while π , solutions, 0,
π,
π 3
π 3
. are
2π is extraneous. So, the only solutions to the original equation are 3
.
50. Observe that
2 sin 2 x = cot x cos x sin x 2 4 sin x cos x = cos x
2 ( 2 sin x cos x ) =
4 sin2 x cos x − cos x = 0
cos x ( 4 sin2 x − 1) = 0
673
Chapter 6
cos x = 0 or 4 sin2 x − 1 = 0 1 4 1 cos x = 0 or sin x = ± 2 cos x = 0 or sin2 x =
Note that the solutions of these equations are x =
π 3π π 5π 7π 11π
. , , , , , 2 2 6 6 6 6 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation. 51. Observe that 3 tan x = 2 sin x sin x = 2 sin x 3 cos x 3 sin x = 2 sin x cos x
52. Observe that 3cot x = 2 sin x cos x 3 = 2 sin x sin x 3cos x = 2 sin2 x
3 sin x − 2 sin x cos x = 0
3cos x = 2 (1 − cos 2 x )
sin x
3cos x = 2 − 2 cos 2 x
( 3 − 2 cos x ) = 0
sin x = 0 or
3 − 2 cos x = 0
3 sin x = 0 or cos x = 2 Note that the solutions of these π 11π . equations are x = 0, π , , 6 6 Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
2 cos 2 x + 3cos x − 2 = 0
( 2 cos x − 1)( cos x + 2 ) = 0
2 cos x − 1 = 0 or cos x + 2 = 0 1 cos x = or cos =− x 2 2 No solution Note that the solutions of these equations π 5π are x = , . Substituting these into 3 3 the original equation shows that they are all, in fact, solutions to the original equation.
674
Chapter 6 Review
53. Observe that cos x + sin x + 1 = 0 2
(1 − sin x ) + ( sin x + 1) = 0
54. Observe that 2 cos 2 x − 3 cos x = 0
(
2
(1 − sin x )(1 + sin x ) + ( sin x + 1) = 0 (1 + sin x ) [1 − sin x + 1] = 0 (1 + sin x )( 2 − sin x ) = 0 1 + sin x = 0 or sin x = −1 or
2 − sin x = 0 sin x = 2 No solution
Note that the solution of these 3π . Substituting equations is x = 2 into the original equation shows it is, in fact, a solution. 55. Observe that cos 2 x + 4 cos x + 3 = 0
( 2 cos x − 1) + 4 cos x + 3 = 0 2
cos x = 0 or 2 cos x − 3 = 0 3 2 Note that the solutions of these equations π 3π π 11π are x = , , , . Substituting 2 2 6 6 these into the original equation shows that they are all, in fact, solutions to the original equation. cos x = 0 or cos x =
56. Observe that sin 2 x + sin x = 0 2 sin x cos x + sin x = 0 sin x ( 2 cos x + 1) = 0 sin x = 0 or 2 cos x + 1 = 0
2 cos 2 x + 4 cos x + 2 = 0 2 ( cos x + 1) = 0 2
cos x + 1 = 0 cos x = −1 The solution to this equation is x = π . Substituting it back into the original equation shows that it is, in fact, a solution of the original equation.
)
cos x 2 cos x − 3 = 0
1 2 Note that the solutions of these equations 2π 4π are x = 0, π , , . Substituting these 3 3 into the original equation shows that they are all solutions to the original equation. sin x = 0 or cos x = −
675
Chapter 6
57. In order to find all of the values of x in 0 ,360 that satisfy csc 2 x + cot x = 1,
we proceed as follows: Step 1: Simplify the equation algebraically: csc 2 x + cot x = 1
(1 + cot x ) + cot x = 1 2
cot 2 x + cot x = 0 cot x ( cot x + 1) = 0 cot x = 0 or cot x = −1 Step 2: Observe that the solutions to these equations are given by x = 90 , 270 , 135 , 315 . Substituting them back into the original equation shows
that they are all, in fact, solutions of the original equation. 58. In order to find all of the values of x in 0 ,360 that satisfy
8cos 2 x + 6 sin x = 9 , we proceed as follows: Step 1: Simplify the equation algebraically. 8cos 2 x + 6 sin x = 9
8 (1 − sin2 x ) + 6 sin x = 9 8 − 8sin2 x + 6 sin x = 9 8sin2 x − 6 sin x + 1 = 0
( 4 sin x − 1)( 2 sin x − 1) = 0
4 sin x − 1 = 0 or 2 sin x − 1 = 0 sin x =
1 4
or sin x =
1 2
1 Step 2: Solve sin x = . 4 To do so, we must find the values of x whose sine is
1 . 4
1 Indeed, observe that one solution is sin −1 ≈ 14.48 , which is in QI. 4 A second solution occurs in QII, and has value 180 − 14.48 = 165.52 . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 14.48 , 165.52 .
676
Chapter 6 Review
Step 3: Solve sin x =
1 . 2
1 Indeed, observe that one solution is sin −1 = 30 , which is in QI. A second 2 solution occurs in QII, and has value 180 − 30 = 150 . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are x = 30 , 150 .
Step 4: Conclude that the solutions to the original equation are x ≈ 30 , 150 , 14.48 , 165.52 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation. 59. Observe that sin2 x + 2 = 2 cos x
60. Observe that
1 − 2 sin2 x = 3sin x − 1
1 − cos 2 x + 2 = 2 cos x
2 sin2 x + 3sin x − 2 = 0
cos 2 x + 2 cos x − 3 = 0
( 2 sin x − 1)( sin x + 2 ) = 0
( cos x − 1)( cos x + 3) = 0
2 sin x − 1 = 0 or sin x + 2 = 0 1 sin x = or sin x = −2 2 No solution
cos x − 1 = 0 or cos x + 3 = 0 cos x = 1
or cos = −3 x No solution
Note that the solution of these equations is x = 0 . Substituting it into the original equation shows that it is, in fact, a solution to the original equation. 61. Observe that
cos 2 x = 3sin x − 1
Note that the solutions of these equations are x = 30 , 150 . Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.
cos x − 1 = cos 2x cos x − 1 = 2 cos 2 x − 1 2 cos 2 x − cos x = 0 cos x ( 2 cos x − 1) = 0
677
Chapter 6
cos x = 0 or 2 cos x − 1 = 0 cos x = 0 or cos x =
1 2
Note that the solutions of these equations are x = 90 , 270 , 60 , 300 . Substituting these into the original equation shows that they are all, in fact, solutions to the original equation. 62. In order to find all of the values of x in 0 ,360 that satisfy
12 cos 2 x + 4 sin x = 11 , we proceed as follows: Step 1: Simplify the equation algebraically. 12 cos 2 x + 4 sin x = 11
12 (1 − sin2 x ) + 4 sin x = 11
12 − 12 sin2 x + 4 sin x − 11 = 0 12 sin2 x − 4 sin x − 1 = 0
( 6 sin x + 1)( 2 sin x − 1) = 0
6 sin x + 1 = 0 or 2 sin x − 1 = 0 sin x = −
1 6
or sin x =
1 2
1 Step 2: Solve sin x = − . 6 1 To do so, we must find the values of x whose sine is − . 6 1 Indeed, observe that one solution is sin −1 − ≈ −9.59 , which is in QIV. Since 6 the angles we seek have positive measure, we use the representative 360 − 9.59 ≈ 350.41 . A second solution occurs in QIII, and has value 180 + 9.59 = 189.59 . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately x ≈ 189.59 , 350.41 .
678
Chapter 6 Review
Step 3: Solve sin x =
1 . 2
1 Indeed, observe that one solution is sin −1 = 30 , which is in QI. A second 2 solution occurs in QII, and has value 180 − 30 = 150 . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are x = 30 , 150 .
Step 4: Conclude that the solutions to the original equation are x ≈ 30 , 150 , 189.59 , 350.41 .
Substituting these into the original equation shows that they are, in fact, solutions to the original equation. 63. From the given information, we have the following diagram:
a.
cos 2x = cos 2 x − sin2 x
( ) −( ) = − b. x = cos ( − ) ≈ 2.0344 . So, = − 15
−1
5
2
2 5
2
3 5
1 5
cos 2 x = cos ( 4.0689 ) ≈ −0.6
c. Yes, the results in parts a. and b. are the same.
679
Chapter 6
64. From the given information, we have the following diagram:
a. Since x is in QIV, x2 is in QII. Hence, 1 + 125 1 + cos x 17 =− =− 2 2 24 −1 5 b. x = cos ( 12 ) ≈ 1.1410 . So, cos ( x2 ) = −
cos ( x2 ) = cos ( 0.5705 ) ≈ 0.8416
c. Yes, the results in parts a. and b. are the same. 119
65. To determine the smallest positive solution (approximately) of the equation ln x − sin x = 0 graphically, we search for the intersection points of the graphs of the following two functions: y1 = ln x − sin x, y2 = 0
66. To determine the smallest positive solution (approximately) of the equation ln x − cos x = 0 graphically, we search for the intersection points of the graphs of the following two functions: y1 = ln x − cos x, y2 = 0
Observe that the graphs seems to intersect at approximately 2.219
Observe that the graphs seems to intersect at approximately 1.303 radians, which is
radians, which is about 127.1 .
about 74.66 .
680
Chapter 6 Practice Test Solutions ------------------------------------------------------------------1. The identity sin ( sin −1 x ) = x holds on
the interval −1 ≤ x ≤ 1.
2. The identity tan −1 ( tan x ) = x holds on
the interval −
π 2
<x<
π 2
.
3. The identity cos −1 ( cos x ) = x holds on the interval 0 ≤ x ≤ π . 4. The equation csc −1
( 2 ) = θ is
equivalent to csc θ = 2, which is the 1 . Since the range of same as sin θ = 2 π π inverse cosecant is − ,0 ∪ 0, , we 2 2 conclude that θ =
π 4
.
(
)
5. The equation cot −1 − 3 = θ is
equivalent to − 3
2 = cos θ . Since the 1 sin θ 2 range of inverse cotangent is ( 0, π ) , we cot θ = − 3 =
conclude that θ =
5π . 6
π
3 π 6. sin −1 − = − 3 2
7. cos −1 0 =
4π 1 2π −1 8. cos −1 cos = cos − = 3 2 3
5π 1 π −1 9. tan −1 tan =− = tan − 6 6 3
3π 3π 3π 10. sin −1 sin = since −1 ≤ ≤1. 4 4 4
681
2
Chapter 6
11. Let θ = tan −1 2. Then, tan θ = 2, as shown in the diagram:
12. Since sec ( sec −1 θ ) = θ for all
θ ∈ ( −∞, −1] ∪ [1, ∞ ) , we see that sec ( sec −1 ( 2 ) ) = 2 .
θ 13. Observe that 2 sin θ = − 3 is
3 . The values of 2 θ in that satisfy this equation are 4π 5π θ= + 2 nπ , + 2 nπ , where n is an integer. 3 3
equivalent to sin θ = − Using the Pythagorean Theorem, we see that 12 + 2 2 = z 2 , so that z = 5. Hence, cos ( tan −1 2 ) = cos θ =
14.
1 5 . = 5 5
4 + 4 cos θ = 6
15.
2
4 cos 2 θ − 2 = 0
3 ( 4 + sec θ ) = 0
2 ( 2 cos 2 θ − 1) = 0
sec θ = −4
2 cos 2θ = 0
cos θ = − 14
cos 2θ = 0 2θ = π2 , 32π , 52π , 72π
θ = π4 , 34π , 54π , 74π
2 + 3sec θ = −10 12 + 3sec θ = 0
θ = cos −1 ( − 14 ) ≈ 1.82 There are two solutions in the given interval, namely 1.82 and 2π − 1.82 ≈ 4.46.
682
Chapter 6 Practice Test
16. In order to find all of the values of θ in 0 ,180 that satisfy 3tan θ = 1 , we
proceed as follows: 1 . 3 1 Indeed, observe that one solution is tan −1 ≈ 18.43 , which is in QI. 3 A second solution occurs in QIII, and has value 180 + 18.43 = 198.43 .
Step 1: Find the values of θ whose tangent is
Outside interval
Step 2: We conclude that the solution to the original equation in 0 ,180 is
θ ≈ 18.43 . 17. In order to find all of the values of θ in [ 0,2π ] that satisfy
2 cos 2 θ + cos θ − 1 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side yields ( cos θ + 1)( 2 cos θ − 1) = 0 So, θ is a solution to the original equation if either cos θ = −1 or cos θ = Step 2: Conclude that the solutions to the original equation are θ = π ,
π 5π 3
,
3
1 . 2 .
18. In order to find all of the values of θ in 0 ,360 that satisfy
sin2 θ + 3sin θ − 1 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating sin θ as the variable): −3 ± 32 − 4 (1)( −1)
−3 ± 13 2 (1) 2 So, θ is a solution to the original equation if either −3 − 13 −3 + 13 sin θ = or sin θ = . 2 2 sin θ =
No solution since
−3 − 13 < −1 2
683
=
Chapter 6
−3 + 13 . 2 −3 + 13 Indeed, observe that one solution is sin −1 ≈ 17.63 , which is in QI. A 2 second solution occurs in QII, and has value 180 − 17.62 = 162.38 . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the
Step 2: Find the values of θ whose sine is
solutions to this equation are approximately θ ≈ 17.63 , 162.38 . 19.
20.
sin θ + cos θ = −1
6 tan2 θ + tan θ − 2 = 0
2
(1 − cos θ ) + cos θ = −1
(3tanθ + 2 )( 2 tanθ − 1) = 0
2
tan θ = − 23 , 12 Using the fact that tangent has period π , ( cos θ + 1)( cos θ − 2 ) = 0 we see that the solutions are cos θ = −1 or cos θ =2 θ = tan −1 ( − 23 ) + π , tan −1 ( − 23 ) + 2π ,
cos 2 θ − cos θ − 2 = 0
No solution
tan −1 ( 12 ) , tan −1 ( 12 ) + π
The solutions are θ = π + 2nπ , where n is an integer .
≈ 0.46, 2.55, 3.61, 5.70
1 21. In order to find all of the values of θ in 0 ,360 that satisfy sin 2θ = cos θ , 2 we proceed as follows: Step 1: Simplify the equation algebraically. 1 cos θ 2 1 2 sin θ cos θ = cos θ 2 4 sin θ cos θ = cos θ sin 2θ =
cos θ ( 4 sin θ − 1) = 0 cos θ = 0 or 4 sin θ − 1 = 0 cos θ = 0 or sin θ =
684
1 4
Chapter 6 Practice Test
Step 2: Observe that the solutions of cos θ = 0 are θ = 90 , 270 . 1 Step 3: Solve sin θ = . 4 1 To do so, we must find the values of θ whose sine is . 4 1 Indeed, observe that one solution is sin −1 ≈ 14.48 , which is in QI. A second 4 solution occurs in QIII, and has value 180 − 14.48 = 165.52 . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 14.48 , 165.52 . Step 4: We conclude that the solutions of the original equation in 0 ,360 are approximately θ ≈ 14.48 , 165.52 , 90 , 270 . 22.
cos 2θ + sin2 θ = cos θ
( cos θ − sin θ ) + sin θ = cos θ 2
2
2
cos 2 θ − cos θ = 0 cos θ ( cos θ − 1) = 0 cos θ = 0,1
θ = 0, π2 , 32π 23.
sin 2θ ( sin 2θ + 1) + cos 2 2θ + cos 2θ = 1
( sin 2θ + cos 2θ ) + ( sin 2θ + cos 2θ ) = 1 2
2
1 + ( sin 2θ + cos 2θ ) = 1
sin 2θ + cos 2θ = 0 sin 2θ = − cos 2θ
So,
2θ = 34π , 74π , 114π , 154π
θ = 38π , 78π , 118π , 158π
685
Chapter 6
24. sec 2 ( 2θ ) − 5sec ( 2θ ) − 6 = 0
25.
cot 2 ( 2θ ) + csc 2 ( 2θ ) = 3
cot 2 ( 2θ ) + (1 + cot 2 ( 2θ ) ) = 3
( sec 2θ − 6 )( sec 2θ + 1) = 0 sec 2θ = −1,6
2 cot 2 ( 2θ ) = 2
cos 2θ = −1 or cos 2θ =
cot 2 ( 2θ ) = 1
cot ( 2θ ) = ±1
The first equation yields 2θ = 180 , 540 θ = 90 , 270 . The second equation yields 2θ = cos −1 ( 61 ) , 360 − cos −1 ( 61 ) ,
So,
2θ =
360 + cos −1 ( 61 ) , 720 − cos −1 ( 61 )
θ=
θ ≈ 40.20 , 139.80 ,220.20 , 319.80
π 3π 5π 7π 9π 11π 13π 15π 4
,
4
,
4
,
4
,
4
,
,
4
,
4
4
π 3π 5π 7π 9π 11π 13π 15π 8
,
8
,
8
,
8
,
8
,
8
,
8
,
8
26. In order to find all of the values of θ in 0 ,360 that satisfy
sin2 θ − 3cos θ − 3 = 0 , we proceed as follows: Step 1: Simplify the equation algebraically. Observe that sin2 θ − 3cos θ − 3 = 0
(1 − cos θ ) − 3cos θ − 3 = 0 2
cos 2 θ + 3cos θ + 2 = 0 Factoring the left-side now yields ( cos θ + 2 )( cos θ + 1) = 0 .
So, θ is a solution to the original equation if either cos =− or cos θ = −1 . θ 2 No solution since − 2 < −1
Step 2: Conclude that the solution to this equation is θ = 180 . 27. The equation sin x + cos x = −1 has no solution since the left-side is always nonnegative (for any value of x), while the right-side is negative.
686
Chapter 6 Practice Test
28.
( sin θ + cos θ ) + cos ( 90 + θ ) = 1 2
sin 2 θ + 2 sin θ cos θ + cos 2 θ + cos 90 cos θ − sin 90 sin θ = 1 =1 =0 sin θ + cos θ ) + 2 sin θ cos θ − sin θ = 1 ( 2
2
=1
2 sin θ cos θ − sin θ = 0 sin θ ( 2 cos θ − 1) = 0 sin θ = 0 or 2 cos θ − 1 = 0 sin θ = 0 or cos θ = 12
θ = 0 , 60 , 180 , 300 29.
tan θ + 1 = tan θ tan θ = tan θ − 1 tan θ = ( tan θ − 1)
2
tan θ = tan2 θ − 2 tan θ + 1 tan2 θ − 3tan θ + 1 = 0 tan θ =
3± 9−4 3± 5 = 2 2
So,
3− 5 −1 3 − 5 −1 3 + 5 −1 3 + 5 , tan + 180 , tan , tan + 180 2 2 2 2
θ = tan−1
θ ≈ 69.09 , 249.09
687
Chapter 6
30. First, observe that
sin x + cos x =3 cos 2 x sin x + cos x −3 = 0 cos 2 x sin x + cos x − 3cos 2 x =0 cos 2 x ( sin x + cos x ) − 3 ( cos 2 x − sin2 x ) =0 cos 2x ( sin x + cos x ) − 3 ( cos x − sin x )( cos x + sin x ) = 0 cos 2 x ( cos x + sin x ) 1 − 3 ( cos x − sin x ) =0 cos 2 x cos x + sin x = 0 or cos x − sin x = 13 The solutions to the first equation (namely certain odd multiples of π4 ) make the denominator of the original equation equal to zero, and so are not solutions to the original equation. As for the second equation, observe that cos x − sin x = 13
( cos x − sin x ) = 19 2
cos 2 x + 2 sin x cos x + sin2 x = 19 1 + sin 2 x = 19 sin 2 x = 89
As such, the solutions are 2 x = sin −1 ( 89 ) , sin −1 ( 89 ) + 2π , π − sin −1 ( 89 ) , 3π − sin −1 ( 89 ) x ≈ 0.55, 4.16
688
Chapter 6 Cumulative Review Solutions ------------------------------------------------------------1. cot θ = 129 = 34
( ) 1 csc ( 64.65 ) = ≈ 1.1066. sin ( 64.65 )
2. First, note that 6439′ = 64 + 39′ 601 ′ ≈ 64.65. So,
3. Note that csc θ < 0 sin θ < 0. This, coupled with cot θ > 0, implies that cos θ < 0. Thus, the terminal side of θ lies in QIII. 4. sec ( −225 ) =
1 1 1 = = 2 =− 2 cos ( −225 ) cos ( 225 ) − 2
1 5. sec θ = 0.75 = 34
6. cot θ sec 2 θ =
cos θ 1 1 ⋅ = 2 sin θ cos θ sin θ cos θ
7. Using the formula s = rθ , we see that 58 in. = ( 34 in.) θ θ = ( 58 )( 34 ) = 65 rad. 8. Using the formula A = 12 r 2θ d
( ) yields A = ( 4.5 cm ) ( 24 ) ( ) ≈ 4.24 cm . π
1 2
180
2
π
2
180
9. First, 1 revolution corresponds to 2π rad. So, 55 revolutions = 110π rad. Hence, π rad ω = 110π rad. min. Thus, v = rω = ( 2.5 in.) ( 110min. ) = 225π in. min. 10. csc 240 =
1 1 2 3 = 3 =− sin 240 3 − 2
11. cot θ is undefined when sin θ = 0. This occurs when θ = 0, π , 2π , 3π , 4π . 12. The period = 8 (since it replicates every interval of length 8 units). 13. Amplitude = max2− min = 3−(2−3) = 3
14. y = 3cos ( π4 x )
689
Chapter 6
15. Here is some information regarding the graph: Amplitude = ½ Period = 4π Phase shift π2 units to the right Vertical shift up ½ unit No reflection about the x-axis
The graph of the curve on the interval − 72π , 92π is to the right: 16. The graph of the curve on the interval [ −2π ,2π ] , along with its guide
function, is below:
17. cos x 1 cos 2 x 1 cot x ( cos x − sec x ) = − cos x − = sin x cos x sin x sin x
=
− (1 − cos 2 x ) sin x
690
sin2 x =− = − sin x sin x
Chapter 6 Cumulative Review
18. Consider the following two triangles:
β
α
Observe that cos α = 51 , sin α = − 2 5 6 , cos β = − 53 , sin β = 54 . So,
(
)
cos (α − β ) = cos α cos β + sin α sin β = ( 15 )( − 53 ) + − 2 5 6 ( 54 ) = −3−258 6 = − 3+258 6
19.
2 tan ( − π8 )
1 − tan ( − 8 ) 2
π
= tan ( 2 ⋅ ( − π8 ) ) = tan( − π4 ) = −1
20. Consider the following triangle:
Since π < x < 32π π2 < x2 < 34π . So, the terminal side of x2 is in QII, so that cos ( x2 ) < 0 . Hence, we have
1 + ( − 24 1 + cos x 25 ) cos ( ) = − =− 2 2 x 2
=−
21.
1 5 2 2 =− = − 50 50 10
7 sin( −2 x )sin(5x ) = 7 12 ( cos( −2 x − 5x ) − cos( −2 x + 5x ) )
= 72 [ cos( −7 x ) − cos(3x )] = 72 [ cos(7x ) − cos(3x )]
691
Chapter 6
22.
2 sin ( 7 x 2−3 x ) cos ( 7 x 2+3 x ) sin 7 x − sin3x = tan ( 7 x 2−3 x ) = tan 2x = cos 7 x + cos3x 2 cos ( 7 x +2 3 x ) cos ( 7 x 2−3 x )
(
)
23. Let x = sec −1 − 2 3 3 . Then, sec x = − 2 3 3 . Hence, cos x = − 2 3 3 = − 23 , which
occurs when x = 150 . 24. Let x = sin −1 ( 135 ) . Then, sin x = 135 .
25.
2 cos 2 θ − cos θ − 1 = 0
This yields the following triangle:
( 2 cos θ + 1)( cos θ − 1) = 0 cos θ = − 12 , 1
θ = 0, 23π , 43π , 2π x
So, tan x = 125 .
692
CHAPTER 7 Section 7.1 Solutions ----------------------------------------------------------------------------------1. SSA
2. SSA
3. SSS
4. SAS
5. ASA
6. ASA
7. Observe that: γ = 180 − ( 45 + 60 ) = 75
(10 m ) ( sin 60 ) sin α sin β sin 45 sin 60 = = b= = a b 10 m b sin 45
3 2
(10 m )
(10 m ) ( sin 75 ) sin α sin γ sin 45 sin 75 = = c= ≈ 14 m a c 10 m c sin 45
2 2
8. Observe that: α = 180 − ( 75 + 60 ) = 45
( 25 in.) ( sin 45 ) sin α sin β sin 45 sin 75 = = a= ≈ 18 in. a b a 25 in. sin 75 ( 25 in.) ( sin 60 ) sin β sin γ sin 75 sin 60 = = c= ≈ 22 in. b c 25 in. c sin 75
9. Observe that: β = 180 − ( 46 + 72 ) = 62
( 200 cm ) ( sin 46 ) sin α sin β sin 46 sin 62 = = a= ≈ 163 cm a b a 200 cm sin 62 ( 200 cm ) ( sin 72 ) sin β sin γ sin 62 sin 72 = = c= ≈ 215 cm b c 200 cm c sin 62
693
≈ 12 m
Chapter 7
10. Observe that: α = 180 − (100 + 40 ) = 40
(16 ft.) ( sin 40 ) sin α sin β sin 40 sin 40 = = b= ≈ 16 ft. a b 16 ft. b sin 40 (16 ft.) ( sin100 ) sin α sin γ sin 40 sin100 = = c= ≈ 25 ft. a c 16 ft. c sin 40
11. Observe that: β = 180 − (16.3 + 47.6 ) = 116.1
( 211 yd.) ( sin16.3 ) sin α sin γ sin16.3 sin 47.6 = = a= ≈ 80.2 yd. a c a 211 yd. sin 47.6
( 211 yd.) ( sin116.1 ) sin β sin γ sin116.1 sin 47.6 = = b= ≈ 256.6 yd. b c b 211 yd. sin 47.6
12. Observe that: α = 180 − (104.2 + 33.6 ) = 42.2
( 26 in.) ( sin104.2 ) sin α sin β sin 42.2 sin104.2 = = b= ≈ 37.5 in. a b 26 in. b sin 42.2 ( 26 in.) ( sin33.6 ) sin α sin γ sin 42.2 sin33.6 = = c= ≈ 21.4 in. a c 26 in. c sin 42.2
13. Observe that: γ = 180 − ( 30 + 30 ) = 120
(12 m ) ( sin30 ) sin α sin γ sin30 sin120 = = a= ≈7 m a c a 12 m sin120 (12 m ) ( sin30 ) sin β sin γ sin30 sin120 = = b= ≈7 m b c b 12 m sin120
694
Section 7.1
14. Observe that: β = 180 − ( 45 + 75 ) = 60
( 9 in.) ( sin 45 ) sin α sin γ sin 45 sin 75 = = a= ≈ 7 in. a c a 9 in. sin 75 ( 9 in.) ( sin 60 ) sin β sin γ sin 60 sin 75 = = b= ≈ 8 in. b c b 9 in. sin 75
15. Observe that: α = 180 − ( 26 + 57 ) = 97
(100 yd.) ( sin 97 ) sin α sin γ sin 97 sin57 = a= = ≈ 118 yd. a c a 100 yd. sin57
(100 yd.) ( sin 26 ) sin β sin γ sin 26 sin57 = = b= ≈ 52 yd. b c b 100 yd. sin57
16. Observe that: β = 180 − ( 80 + 30 ) = 70
( 3 ft.) ( sin80 ) sin α sin β sin80 sin 70 = = a= ≈ 3 ft. a b a 3 ft. sin 70 (3ft.) ( sin30 ) sin β sin γ sin 70 sin30 = = c= ≈ 2 ft. b c 3 ft. c sin 70
17. Observe that: γ = 180 − (120 +10 ) = 50
(12 cm ) ( sin120 ) sin α sin γ sin120 sin50 = = a= ≈ 14 cm a c a 12 cm sin50 (12 cm ) ( sin10 ) sin β sin γ sin10 sin50 = = b= ≈ 2.7 cm b c b 12 cm sin50
695
Chapter 7
18. Observe that: α = 180 − ( 46 + 37 ) = 97
(10 yd.) ( sin 97 ) sin α sin γ sin 97 sin 46 a= = = ≈ 14 yd. a c a 10 yd. sin 46
(10 yd.) ( sin37 ) sin β sin γ sin 46 sin37 = = b= ≈ 14 yd. b c 100 yd. b sin 46
19. Observe that: β = 180 − (13.4 + 88.6 ) = 78
(57 m ) ( sin88.6 ) sin α sin β sin88.6 sin 78 = = a= ≈ 14 m a b a 57 m sin 78 (57 m ) ( sin13.4 ) sin β sin γ sin 78 sin13.4 = = c= ≈ 58 m b c 57 m c sin 78
20. Observe that: γ = 180 − (15 + 5 ) = 160
( 2.3 mm ) ( sin160 ) sin α sin γ sin5 sin160 = = c= ≈ 9.0 mm a c 2.3 mm c sin5 ( 2.3 m ) ( sin15 ) sin β sin γ sin15 sin5 = = b= ≈ 6.8 mm b c b 2.3 m sin5
21. Step 1: Determine β . 5sin16 sin α sin β sin16 sin β 5sin16 = = sin β = β = sin −1 ≈ 20 a b 4 5 4 4
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − β1 ≈ 160 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: β1 ≈ 20
γ 1 ≈ 180 − (16 + 20 ) = 144
696
Section 7.1
sin α sin γ 1 sin16 sin144 4 sin144 = c1 = ≈9 = a c1 4 c1 sin16
QII triangle: β 2 ≈ 160
γ 2 ≈ 180 − (16 +160 ) = 4
sin α sin γ 2 sin16 sin 4 4 sin 4 = = c2 = ≈1 a c2 4 c2 sin16
22. Step 1: Determine γ . 20 sin 70 sin γ sin β sin γ sin 70 20 sin 70 = = sin γ = γ = sin −1 ≈ 39 c b 20 30 30 30 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of γ is 180 − 39 = 141 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. α ≈ 180 − ( 70 + 39 ) = 71 sin α sin β sin 71 sin 70 30 sin 71 = a= ≈ 30 = a b a 30 sin 70
23. Step 1: Determine α . sin γ sin α sin 40 sin α = = sin α = sin 40 α = 40 c a 12 12 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of α is 180 − 40 = 140 . In such case, note that α + γ = 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are α and γ , must sum to 180 - this would only occur if the third angle β = 0 , in which case there is no triangle). Step 2: Solve for the triangle. β ≈ 180 − ( 40 + 40 ) = 100 sin γ sin β sin 40 sin100 12 sin100 = b= ≈ 18 = c b 12 b sin 40
697
Chapter 7
24. Step 1: Determine β . sin α sin β sin 25 sin β 111sin 25 −1 111sin 25 = = sin β = β = sin ≈ 36 a b 80 111 80 80
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − β1 ≈ 144 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: β1 ≈ 36
γ 1 ≈ 180 − ( 25 + 36 ) = 119
sin α sin γ 1 sin 25 sin119 80 sin119 = c1 = ≈ 166 = a c1 80 c1 sin 25
QII triangle: β 2 ≈ 144
γ 2 ≈ 180 − ( 25 +144 ) = 11
sin α sin γ 2 sin 25 sin11 80 sin11 = c2 = ≈ 36 = 80 c2 sin 25 a c2
25. Step 1: Determine α . 21sin100 sin α sin β sin α sin100 21sin100 = = sin α = α = sin −1 , a b 21 14 14 14 which does not exist. Hence, there is no triangle in this case. 26. Step 1: Determine β . 26 sin120 sin α sin β sin120 sin β 26 sin120 = = sin β = β = sin −1 , a b 13 26 13 13 which does not exist. Hence, there is no triangle in this case.
27. Step 1: Determine β . 500 sin 40 sin γ sin β sin 40 sin β 500 sin 40 = = sin β = β = sin −1 ≈ 77 c b 330 500 330 330
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − β1 ≈ 103 . Both are tenable solutions, so we need to solve for two triangles.
698
Section 7.1
Step 2: Solve for both triangles. QI triangle: β1 ≈ 77
α1 ≈ 180 − ( 77 + 40 ) = 63
sin α1 sin γ sin 63 sin 40 330 sin 63 = = a1 = ≈ 457 a1 330 sin 40 a1 c
QII triangle: β 2 ≈ 103
α 2 ≈ 180 − (103 + 40 ) = 37
sin α 2 sin γ sin37 sin 40 330 sin37 = a2 = ≈ 309 = a2 330 sin 40 a2 c
28. Step 1: Determine α . 9sin137 sin α sin β sin α sin137 9sin137 = = sin α = α = sin −1 ≈ 23 a b 9 16 16 16 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of α is 180 − 23 = 157 . In such case, note that β + α > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and α , must sum to180 ). Step 2: Solve for the triangle. γ ≈ 180 − ( 23 +137 ) = 20 sin γ sin β sin 20 sin137 16 sin 20 = c= ≈8 = c b c 16 sin137
29. Step 1: Determine α . 2 sin106 sin α sin β sin α sin106 2 sin106 = = sin α = α = sin −1 ≈ 31 a b 2 7 7 7 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of α is 180 − 31 = 149 . In such case, note that β + α > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and α , must sum to 180 ). Step 2: Solve for the triangle. γ ≈ 180 − ( 31 +106 ) = 43 sin γ sin β sin 43 sin106 7 sin 43 ≈2 = = c= c b c sin106 7
699
Chapter 7
30. Step 1: Determine β . sin γ sin β sin11.6 sin β 15.3sin11.6 = = sin β = c b 27.2 15.3 27.2 15.3sin11.6 β = sin −1 ≈ 6.5 27.2 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 − 6.5 = 173.5 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. γ ≈ 180 − (11.6 + 6.5 ) = 161.9 sin γ sin α sin11.6 sin161.9 27.2 sin161.9 = = a= ≈ 42.0 c a 27.2 a sin11.6
31. Step 1: Determine γ . 12 sin 27 sin γ sin α sin γ sin 27 = = γ = sin −1 ≈ 51 c a 12 7 7
This is the solution in QI – label it as γ 1 . The second solution in QII is given by
γ 2 = 180 − 51 ≈ 129 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: γ 1 ≈ 51
β1 ≈ 180 − ( 27 + 51 ) = 102
sin102 sin 27 7 sin102 = b1 = ≈ 15 b1 7 sin 27
QII triangle: γ 2 ≈ 129
β 2 ≈ 180 − ( 27 +129 ) = 24
sin 24 sin 27 7 sin 24 = b2 = ≈ 6.3 b2 7 sin 27
32. Step 1: Determine β . 10 sin14.5 sin γ sin β sin14.5 sin β 10 sin14.5 β = sin −1 = = sin β = ≈ 39 c b 4 10 4 4
700
Section 7.1
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − 39 ≈ 141 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: β1 ≈ 39
α1 ≈ 180 − (14.5 + 39 ) = 126.5
sin α1 sin γ sin126.5 sin14.5 4 sin126.5 = = a1 = ≈ 13 a1 c a1 4 sin14.5
QII triangle: β 2 ≈ 141
α 2 ≈ 180 − (14.5 +141 ) = 24.5
sin α 2 sin γ sin 24.5 sin14.5 4 sin 24.5 = = a2 = ≈ 6.6 sin14.5 a2 c a2 4
33. Step 1: Determine α . 20 sin52 sin α sin β sin α sin52 20 sin52 α = sin −1 = = sin α = ≈ 61 a b 20 18 18 18
This is the solution in QI – label it as α1 . The second solution in QII is given by
α 2 = 180 − 61 ≈ 119 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: α1 ≈ 61
γ 1 ≈ 180 − ( 61 + 52 ) = 67
sin 67 sin52 18sin 67 = c1 = ≈ 21 c1 18 sin52
QII triangle: α 2 ≈ 119
γ 2 ≈ 180 − (119 + 52 ) = 9
sin 9 sin52 18sin 9 = c2 = ≈ 3.6 c2 18 sin52
34. Step 1: Determine γ . 22 sin 47 sin γ sin β sin γ sin 47 = = γ = sin −1 ≈ 63 c b 22 18 18
This is the solution in QI – label it as γ 1 . The second solution in QII is given by
γ 2 = 180 − 63 ≈ 117 . Both are tenable solutions, so we need to solve for two triangles.
701
Chapter 7
Step 2: Solve for both triangles. QI triangle: γ 1 ≈ 63
α1 ≈ 180 − ( 63 + 47 ) = 70
sin 70 sin 47 18sin 70 = a1 = ≈ 23 a1 18 sin 47
QII triangle: γ 2 ≈ 117
α 2 ≈ 180 − (117 + 47 ) = 16
sin16 sin 47 18 sin16 = a2 = ≈ 6.9 18 sin 47 a2
35. Step 1: Determine β . sin γ sin β sin 20 sin β 6.2 sin 20 −1 6.2 sin 20 = = sin β = β = sin ≈ 12 c b 10 6.2 10 10 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 − 12 = 168 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. α ≈ 180 − (12 + 20 ) = 148
sin γ sin α sin 20 sin148 10 sin148 = = a= ≈ 15 c a 10 a sin 20
36. Step 1: Determine β . 3.8sin 28 sin α sin β sin 28 sin β 3.8sin 28 = = sin β = β = sin −1 ≈ 21 a b 5 3.8 5 5 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 − 21 = 159 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. γ ≈ 180 − ( 21 + 28 ) = 131 sin γ sin α sin131 sin 28 5sin131 = = c= ≈8 c a c 5 sin 28
37. No triangle
38. No triangle
702
Section 7.1
39. No triangle
40. No triangle
41. Consider the following diagram:
α β
First, using the properties of corresponding and supplementary angles, we see that α = 80 . As such, the information provided yields an AAS triangle. So, we use Law of Sines to solve the triangle: β = 180 − ( 9 + 80 ) = 91
9
sin 9 sin 91 = 195 w 195 ( sin 91 ) w= ≈ 1,246 ft. sin 9
42. Using the same triangle and information as in Exercise 41, we compute x as follows: 195 ( sin80 ) sin80 sin 9 = x= ≈ 1,227.59 ft. x 195 sin 9 Now, we seek the length of the base of the smaller dotted right-triangle, the height of which coincides with the height of the launch pad (see diagram in the text) (and the angle opposite of which is 1 ). Call the base of this triangle y. Then, from this y information, we see that cos1 = , so that 1,227.59
y = 1,227.59 ( cos1 ) ≈ 1,227.40 ≈ 1,227 ft.
43. Consider the following diagram:
β α 20.5
703
25.5
Chapter 7
We need to determine H. To this end, first observe that α = 154.5 , and so
β = 180 − ( 20.5 +154.5 ) = 5 . So, the information provided yields an AAS
situation for triangle T1. So, use Law of Sines to find x: sin 20.5 sin5 sin 20.5 = x= ≈ 4.01818 mi. x 1 mi. sin5 Now, to find H, we turn our focus to triangle T2. Again, from the information we have, the situation is AAS, so we use Law of Sines to find H: sin 90 sin 25.5 1 sin 25.5 = = x H 4.01818 H so that H ≈ 4.01818 ( sin 25.5 ) ≈ 1.7 mi. 44. Consider the following diagram:
β
10
α
15
We need to determine H. To this end, first observe that α = 165 , and so
β = 180 − (10 +165 ) = 5. So, the information provided yields an AAS situation for
triangle T1. So, we use Law of Sines to find x: 2 mi.) sin10 ( sin10 sin5 = x= ≈ 3.984778 mi. x 2 mi. sin5 Now, to find H, we turn our focus to triangle T2. Again, from the information we have, the situation is AAS, so we use Law of Sines to find H: sin 90 sin15 1 sin15 = = x H 3.984778 H so that H ≈ 3.984778 ( sin15 ) ≈ 1.0 mi.
704
Section 7.1
45. Consider the following diagram:
46. Consider the following diagram:
First, we need to find the missing angle: 180° = 80° + 25° + θ θ = 75° We can use the Law of Sines to find x: Then, we can use the Law of Sines to find x: sin30° sin 70° 120 sin30° = x= ≈ 64 ft. sin 75° sin 25° 64 sin 75° x 120 sin 70° ≈ 146 ft. = y= y 64 sin 25° 47. Consider the following diagram:
β
100
30
We seek x. To this end, first note that β = 50 . So, the information provided yields an AAS situation for triangle T1. So, we use Law of Sines to find x: (1 mi.) sin100 ≈ 1.3 mi. sin100 sin50 = x= x 1 mi. sin50 48. Refer to the diagram in Exercise 45. We seek y this time. Again, using the Law (1 mi.) sin30 ≈ 0.7 mi. sin30 sin50 = y= of Sines yields: y 1 mi. sin50
705
Chapter 7
49. Consider the following diagram:
100o
15
65
We seek x. The information provided yields an AAS situation for triangle T1. So, we use Law of Sines to find x: 100 ft.) sin15 ( sin15 sin100 = x= ≈ 26 ft. x 100 ft. sin100 50. Refer to the diagram in Exercise 47. We seek y this time. Again, using the Law of Sines yields: (100 ft.) sin 65 ≈ 91 ft. sin 65 sin 90 = y= y 100 ft. sin 90 51. Consider the following diagram:
sin105 sin30 = x 15 mi. 15 mi.) sin105 ( x= ≈ 29 mi. sin30
45 75 105
706
Section 7.1
52. Consider the following diagram:
sin 65 sin 70 = x 20 mi. ( 20 mi.) sin 65 ≈ 19 mi. x= sin 70
65
70
45
53. Consider the following diagram:
50
We know the hiker travels 850 yd on a course of 130°, but don’t know if he made a sharp turn back toward the car or just angled back toward the road. All we know is he turned in a Westerly direction and traveled 700 yd before finding the road again. Thus we can create two possible scenarios. sin50 sin B Using the law of Sines yields: = 700 850 ° ° Hence, either B = 68.5 or B =111.5 . sin50 sin 61.5 = , so that x = 803 yds. If B = 68.5°, then A = 61.5° and we have 700 x sin50 sin18.5 = , so If, on the other hand, B = 111.5°, then A = 18.5° and we have 700 x that x = 289 yds 54. The diagram depicting this situation is essentially the same as #51, with the numbers appropriately changed. Arguing in the same manner, observe that using the sin 45 sin B = . Hence, either B = 53.9° or B = 126.1°. law of Sines yields 700 800
707
Chapter 7
sin 45 sin81.1 = , so that x = 978 yds. 700 x sin 45 sin8.9 If, on the other hand, B = 126.1°, then A = 8.9° and we have = , so that 700 x x = 153 yds. If B = 53.9°, then A = 81.1° and we have
55.
sin35 sin 75 (28 in.)sin 75 = x= ≈ 47 in. 28 in. x sin35
56.
sin33 sin 70 (30 in.)sin 70 = x= ≈ 52 in. 30 in. x sin33
57. We have the following triangle:
Using the Law of Sines, we obtain 23 cm ) sin30 ( sin82 sin30 = b= ≈ 1.2 cm. 23 cm b sin82 58. We have the following triangle:
Using the Law of Sines, we obtain ( 23 cm ) sin 95 ≈ 23 cm. sin82 sin 95 = c= 23 cm c sin82
708
Section 7.1
59. The value of β is incorrect. We see sin β ≈ 1.113 > 1 . Since the range of the sine function is [ −1,1] , there is no angle β such that sin β ≈ 1.113.
60. Only the solution corresponding to the acute angle is found. Should have also provided the solution corresponding to the obtuse value of β
61. False. It applies to all triangles.
62. False. If the angle is acute, it is possible to have a second solution. Also, it is possible for there to be no triangle at all.
63. True. By definition of oblique.
64. True. By definition of oblique.
65. Consider the following diagram:
Note that h = b sin30 = 12 b . So, if h < a < b , then 12 b < a < b .
30
(namely β = 124 ).
β
66. Note that h = b sin 60 = 23 b . So, if h < a < b , then 23 b < a < b .
γ 1 67. Claim: ( a + b ) sin = c cos (α − β ) 2 2 Proof: First, note that the given equality is equivalent to: 1 cos (α − β ) (a + b) = 2 c γ sin 2 We shall start with the right-side, and we will need the following identities: α + β α − β sin α + sin β = 2 sin (1) cos 2 2 γ γ γ sin γ = sin 2 ⋅ = 2 sin cos (2) 2 2 2
709
Chapter 7
Indeed, observe that a + b a b sin α sin β sin α + sin β = + = + = c c c sin γ sin γ sin γ α − β α + β 2 sin cos 2 2 = (by (1)) sin γ α + β α − β 2 sin cos 2 2 (by (2)) = γ γ 2 sin cos 2 2 α +β π γ Now, since α + β + γ = π , it follows that = − . As such, 2 2 2 α + β π γ γ sin = sin − = cos (3). 2 2 2 2 Using (3) in the last line of the above string of equalities then yields α − β α + β α − β 2 cos γ cos α − β 2 sin cos cos 2 2 2 2 = 2 , = γ γ γ γ γ 2 sin cos sin 2 sin cos 2 2 2 2 2
as desired.
68. α − β tan 2 = a−b . Claim: α + β a +b tan 2 Proof: α − β α − β α + β tan sin cos 2 = 2 ⋅ 2 α + β α − β α + β tan cos sin 2 2 2 α β β α α β α β sin cos − sin cos cos cos − sin sin 2 2 2 2⋅ 2 2 2 2 = α β β α α β α β cos cos + sin sin sin cos + sin cos 2 2 2 2 2 2 2 2
710
Section 7.1
Now, multiply the top two binomials together, and the bottom two binomials together, and regroup in the following manner: =1 =1 α α β β β α α β sin cos cos 2 + sin2 − sin cos cos 2 + sin2 2 2 2 2 2 2 2 2 = α α β β β β α α sin cos cos 2 + sin2 + sin cos cos 2 + sin2 2 2 2 2 2 2 2 2 =1
=1
α α β β 1 2 sin α cos α − 2 sin β cos β sin cos − sin cos 2 2 2 2 2 2 2 2= 2 = α α β β 1 α β β α sin cos + sin cos 2 sin cos + 2 sin cos 2 2 2 2 2 2 2 2 2 Now, apply double angle formula, followed by the Law of Sines on the sin β term sin α sin β b sin α sin β = ) to obtain: (i.e., use = a b a b a −b b sin α sin α 1 − sin α − sin α − sin β a = a = a−b . a = = = sin α + sin β sin α + b sin α b a +b a +b sin α 1 + a a a 69. Consider the following diagram:
First, note that γ = 105 . So, sin105 sin30 2 sin105 = c= c sin30 2
30
45
γ 2
= 2 2 sin105 To get the exact value, we apply an addition formula: sin105 = sin 45 cos 60 + sin 60 cos 45 =
( )( ) + ( )( ) = 2 2
1 2
3 2
2 2
Thus, c = 2 2 2 +4 6 = 1 + 3.
711
2+ 6 4
Chapter 7
70. Consider the following diagram:
First, note that γ = 15 . So, sin15 sin 45 6 sin15 = c= c sin 45 6
6
γ
=
6 sin15 2
2
To get the exact value, we apply an addition formula: sin15 = sin 45 cos 30 − sin30 cos 45 =
( )( ) − ( )( ) =
Thus, c =
2 2
3 2
6 6 −4 2 2
2 2
1 2
2
=
6− 2 4
3 2− 6 . 2
71. The following steps constitute the program in this case:
72. The following steps constitute the program in this case:
Program: AYZ :Input “SIDE A =”, A :Input “ANGLE Y =”, Y :Input “ANGLE Z =”, Z :180-Y-Z → X :Asin(Y)/sin(X) → B :Asin(Z)/sin(X) → C :Disp “ANGLE X = “, X :Disp “SIDE B = “, B :Disp “SIDE C =”, C
Program: BXY :Input “SIDE B =”, B :Input “ANGLE X =”, X :Input “ANGLE Y =”, Y :180-X-Y → Z :Bsin(X)/sin(Y) → A :Bsin(Z)/sin(Y) → C :Disp “ANGLE Z = “, Z :Disp “SIDE A = “, A :Disp “SIDE C =”, C
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 10 ANGLE Y = 45 ANGLE Z = 65
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE B = 42.8 ANGLE X = 31.6 ANGLE Y = 82.2
Now, the program will display the following information that solves the triangle: ANGLE X = 70 SIDE B = 7.524873193 SIDE C = 9.64472602
Now, the program will display the following information that solves the triangle: ANGLE Z = 66.2 SIDE A = 22.63602893 SIDE C = 39.5259744
712
Section 7.1
73. The following steps constitute the program in this case:
74. The following steps constitute the program in this case:
Program: ABX :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE X =”, X :sin−1(Bsin(X)/A) → Y :180-Y-X → Z :Asin(Z)/sin(X) → C :Disp “ANGLE Y = “, Y :Disp “ANGLE Z = “, Z :Disp “SIDE C =”, C
Program: BCZ :Input “SIDE B =”, B :Input “SIDE C =”, C :Input “ANGLE Z =”, Z :sin−1(Bsin(Z)/C) → Y :180-Y-Z → X :Csin(X)/sin(Z) → A :Disp “ANGLE Z = “, Z :Disp “ANGLE X = “, X :Disp “SIDE A =”, A
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 22 SIDE B = 17 ANGLE X = 105 Now, the program will display the following information that solves the triangle: ANGLE Y = 48.27925113 ANGLE Z = 26.72074887 SIDE C = 10.24109154
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE B = 16.5 SIDE C = 9.8 ANGLE Z = 79.2 Now, the program will display an error message since there is no triangle in this case.
713
Section 7.2 Solutions -----------------------------------------------------------------------------------1. C
2. C
3. S
4. S
5. S
6. S
7. C
8. S
9. This is SAS, so begin by using Law of Cosines: Step 1: Find b. 2 2 b 2 = a 2 + c 2 − 2ac cos β = ( 4 ) + ( 3 ) − 2 ( 4 )( 3 ) cos (100 ) = 25 − 24 cos (100 ) b ≈ 5.4 ≈ 5 Step 2: Find γ . 3sin100 sin β sin γ sin100 sin γ 3sin100 = = sin γ = γ = sin −1 ≈ 33 b c 5 3 5 5
Step 3: Find α : α ≈ 180 − ( 33 +100 ) = 47
10. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 6 ) + (10 ) − 2 ( 6 ) (10 ) cos ( 80 ) = 136 − 120 cos ( 80 ) c ≈ 10.731 ≈ 11 Step 2: Find α . sin α sin γ sin α sin80 6 sin80 −1 6 sin80 = = sin α = α = sin ≈ 33 a c 6 10.731 10.731 10.731 Step 3: Find β .
β ≈ 180 − ( 33 + 80 ) = 67
11. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = ( 7 ) + ( 2 ) − 2 ( 7 ) ( 2 ) cos (16 ) = 53 − 28cos (16 ) a ≈ 5.107 ≈ 5 Step 2: Find γ . 2 sin16 sin α sin γ sin16 sin γ 2 sin16 = = sin γ = γ = sin −1 ≈6 5.107 2 5.107 a c 5.107 Step 3: Find β .
β ≈ 180 − ( 6 +16 ) = 158
714
Section 7.2
12. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 6 ) + ( 5 ) − 2 ( 6 ) ( 5 ) cos (170 ) = 61 − 60 cos (170 ) c ≈ 10.958 ≈ 11 Step 2: Find α . sin α sin γ sin α sin170 6 sin170 −1 6 sin170 = = sin α = α = sin ≈5 a c 6 10.958 10.958 10.958 Step 3: Find β .
β ≈ 180 − ( 5 +170 ) = 5
13. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = ( 5 ) + ( 5 ) − 2 ( 5 ) ( 5 ) cos ( 20 ) = 50 − 50 cos ( 20 ) a ≈ 1.736 ≈ 2 Step 2: Find β . 5sin 20 sin α sin β sin 20 sin β 5sin 20 = = sin β = β = sin −1 ≈ 80 a b 1.736 5 1.736 1.736 Step 3: Find γ .
γ ≈ 180 − ( 80 + 20 ) = 80
14. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 4.2 ) + ( 7.3 ) − 2 ( 4.2 ) ( 7.3 ) cos ( 25 ) = 70.93 − 61.32 cos ( 25 ) c ≈ 3.9186 ≈ 3.9 Step 2: Find α . 4.2 sin 25 sin α sin γ sin α sin 25 4.2 sin 25 = = sin α = α = sin −1 ≈ 26.9 a c 4.2 3.9186 3.9186 3.9186 Step 3: Find β .
β ≈ 180 − ( 26.9 + 25 ) = 128.1
715
Chapter 7
15. This is SAS, so begin by using Law of Cosines: Step 1: Find b. 2 2 b 2 = a 2 + c 2 − 2ac cos β = ( 9 ) + (12 ) − 2 ( 9 ) (12 ) cos ( 23 ) = 225 − 216 cos ( 23 ) b ≈ 5.1158 ≈ 5 Step 2: Find α . sin β sin α sin 23 sin α 9sin 23 −1 9sin 23 = = sin α = α = sin ≈ 43 b a 5.1158 9 5.1158 5.1158 Step 3: Find γ .
γ ≈ 180 − ( 43 + 23 ) = 114
16. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = ( 6 ) + (13 ) − 2 ( 6 ) (13 ) cos (16 ) = 205 − 156 cos (16 ) a ≈ 7.4191 ≈ 7 Step 2: Find β . 6 sin16 sin α sin β sin16 sin β 6 sin16 = = sin β = β = sin −1 ≈ 13 a b 7.4191 6 7.4191 7.4191 Step 3: Find γ .
γ ≈ 180 − (13 +16 ) = 151
17. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = (12 ) + (16 ) − 2 (12 )(16 ) cos (12 ) c ≈ 4.9 Step 2: Find α . 16 sin12 sin α sin γ sin α sin12 16 sin12 = = sin α = α = sin −1 ≈ 138 a c 16 4.9 4.9 4.9 Step 3: Find β .
β ≈ 180 − (138 +12 ) = 30
716
Section 7.2
18. This is SAS, so begin by using Law of Cosines: Step 1: Find b. 2 2 b 2 = a 2 + c 2 − 2ac cos β = ( 3.6 ) + ( 4.4 ) − 2 ( 3.6 ) ( 4.4 ) cos ( 50 ) b ≈ 3.5 Step 2: Find α . sin β sin α sin50 sin α 3.6 sin50 −1 3.6 sin50 = = sin α = α = sin ≈ 52 b a 3.4 3.6 3.5 3.5 Step 3: Find γ .
γ ≈ 180 − ( 50 + 52 ) = 78
19. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = ( 70 ) + ( 82 ) − 2 ( 70 )( 82 ) cos (170 ) a ≈ 151 Step 2: Find β . 70 sin170 sin α sin β sin170 sin β 70 sin170 = = sin β = β = sin −1 ≈ 4.6 a b 151 70 151 151 Step 3: Find γ .
γ ≈ 180 − (170 + 4.6 ) = 5.4
20. This is SAS, so begin by using Law of Cosines: Step 1: Find b. 2 2 b 2 = a 2 + c 2 − 2ac cos β = ( 20 ) + ( 80 ) − 2 ( 20 ) ( 80 ) cos ( 95 ) b ≈ 84 Step 2: Find α . 80 sin 95 sin β sin α sin 95 sin α 80 sin 95 = = sin α = α = sin −1 ≈ 71 b a 84 80 84 84 Step 3: Find γ .
γ ≈ 180 − ( 95 + 71 ) = 14
717
Chapter 7
21. This is SAS, so begin by using Law of Cosines: Step 1: Find b. 2 2 b 2 = a 2 + c 2 − 2ac cos β = ( 4 ) + ( 8 ) − 2 ( 4 ) ( 8 ) cos ( 60 ) = 80 − 64 cos ( 60 ) b ≈ 6.9282 ≈ 7 Step 2: Find α . sin β sin α sin 60 sin α 4 sin 60 −1 4 sin 60 = = sin α = α = sin ≈ 30 b a 6.9282 4 6.9282 6.9282 Step 3: Find γ .
γ ≈ 180 − ( 30 + 60 ) = 90
22. This is SAS, so begin by using Law of Cosines: Step 1: Find a. a 2 = b 2 + c 2 − 2bc cos α = ( 3 ) + 2
( 18 ) − 2 (3) ( 18 ) cos ( 45 ) = 27 − 6 18 cos ( 45 ) 2
a =3 Step 2: Find β . sin α sin β sin 45 sin β 3sin 45 = = sin β = β = 45 a b 3 3 3 Step 3: Find γ .
γ ≈ 180 − ( 45 + 45 ) = 90
23. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b2 + c 2 − 2bc cos α ( 8 ) = ( 5 ) + ( 6 ) − 2 ( 5 )( 6 ) cos α 64 = 61 − 60 cos α
3 3 so that α = cos −1 − ≈ 92.86 ≈ 93 . 60 60 Step 2: Find either of the remaining two angles using the Law of Sines. sin α sin β sin 92.866 sin β 5sin 92.866 = = sin β = 8 5 8 a b 5sin 92.866 β = sin −1 ≈ 39 8 Step 3: Find the third angle. γ ≈ 180 − ( 93 + 39 ) = 48 Thus, cos α = −
718
Section 7.2
24. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .
c 2 = a 2 + b2 − 2ab cos γ (12 ) = ( 6 ) + ( 9 ) − 2 ( 6 ) ( 9 ) cos γ 144 = 117 − 108cos γ 2
2
2
27 27 so that γ = cos −1 − ≈ 104.477 ≈ 104 . 108 108 Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin104.477 sin β 9sin104.477 = = sin β = c b 12 9 12 9sin104.477 β = sin −1 ≈ 47 12 Step 3: Find the third angle. α ≈ 180 − (104 + 47 ) = 29 Thus, cos γ = −
25. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .
c 2 = a 2 + b2 − 2ab cos γ ( 5 ) = ( 4 ) + ( 4 ) − 2 ( 4 )( 4 ) cos γ 25 = 32 − 32 cos γ 2
2
2
7 7 so that γ = cos −1 ≈ 77.364 ≈ 77 . 32 32 Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin 77.364 sin β 4 sin 77 −1 4 sin 77 = = sin β = β = sin ≈ 51 5 4 5 5 c b Step 3: Find the third angle. α ≈ 180 − ( 77 + 51 ) = 52 Thus, cos γ =
26. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .
c 2 = a 2 + b2 − 2ab cos γ ( 33 ) = (17 ) + ( 20 ) − 2 (17 )( 20 ) cos γ 2
2
2
1089 = 689 − 680 cos γ Thus, cos γ = −
400 400 so that γ = cos −1 − ≈ 126.03 ≈ 126 . 680 680
719
Chapter 7
Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin126.03 sin β 20 sin126.03 = = sin β = c b 33 20 33 20 sin126.03 β = sin −1 ≈ 29 33 Step 3: Find the third angle. α ≈ 180 − (126 + 29 ) = 25 27. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b2 + c 2 − 2bc cos α (14 ) = ( 6 ) + (12 ) − 2 ( 6 ) (12 ) cos α
1 1 so that α = cos −1 − ≈ 96 . 9 9 Step 2: Find either of the remaining two angles using the Law of Sines. 6 sin 96 sin α sin β sin 96 sin β 6 sin 96 = = sin β = β = sin −1 ≈ 25 a b 14 6 14 14 Step 3: Find the third angle. γ ≈ 180 − ( 25 + 96 ) = 59
Thus, cos α = −
28. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b2 + c 2 − 2bc cos α ( 30 ) = ( 41) + (19 ) − 2 ( 41)(19 ) cos α
Thus, cos α ≈ 0.73299 so that α = cos −1 ( 0.73299 ) ≈ 43 . Step 2: Find either of the remaining two angles using the Law of Sines. 19sin 43 sin α sin β sin 43 sin β 19sin 43 = sin β = β = sin −1 = ≈ 25 30 19 30 30 a b Step 3: Find the third angle. γ ≈ 180 − ( 25 + 43 ) = 112
720
Section 7.2
29. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b2 + c 2 − 2bc cos α ( 3.8 ) = (1.2 ) + ( 2.9 ) − 2 (1.2 )( 2.9 ) cos α
Thus, cos α ≈ −0.65948 so that α = cos −1 ( −0.65948 ) ≈ 131 . Step 2: Find either of the remaining two angles using the Law of Sines. sin α sin β sin131 sin β 2.9sin131 −1 2.9sin131 = sin β = β = sin = ≈ 35 3.8 2.9 3.8 3.8 a b Step 3: Find the third angle. γ ≈ 180 − ( 35 +131 ) = 14 30. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b2 + c 2 − 2bc cos α ( 6 ) = (11) + ( 8 ) − 2 (11)( 8 ) cos α
Thus, cos α ≈ 0.84659 so that α = cos −1 ( 0.84659 ) ≈ 32 .
Step 2: Find either of the remaining two angles using the Law of Sines. 8sin32 sin α sin β sin32 sin β 8sin32 = sin β = β = sin −1 = ≈ 45 6 8 6 6 a b Step 3: Find the third angle. γ ≈ 180 − ( 32 + 45 ) = 103 31. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b 2 + c 2 − 2bc cos α ( 8.2 ) = ( 7.1) + ( 6.3 ) − 2 ( 7.1) ( 6.3 ) cos α 67.24 = 90.1 − 89.46 cos α 22.86 22.86 so that α = cos −1 Thus, cos α = ≈ 75.19 ≈ 75 . 89.46 89.46 Step 2: Find either of the remaining two angles using the Law of Sines. sin α sin β sin 75.19 sin β 7.1sin 75.19 = = sin β = 8.2 7.1 8.2 a b 7.1sin 75.19 β = sin −1 ≈ 57 8.2 Step 3: Find the third angle. γ ≈ 180 − ( 75 + 57 ) = 48
721
Chapter 7
32. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is β .
b2 = a 2 + c 2 − 2ac cos β ( 2001) = (1492 ) + (1776 ) − 2 (1492 ) (1776 ) cos β 2
2
2
− 1,376,239 = −5,299,584 cos β 1,376,239 1,376,239 so that β = cos −1 ≈ 74.948 ≈ 75 . 5,299,584 5,299,584 Step 2: Find either of the remaining two angles using the Law of Sines. sin α sin β sin α sin 74.948 1492 sin 74.948 = = sin α = 1492 2001 2001 a b 1492 sin 74.948 α = sin −1 ≈ 46 2001 Step 3: Find the third angle. γ ≈ 180 − ( 75 + 46 ) = 59
Thus, cos β =
33. Since a + b = 9 >10 = c , there can be no triangle in this case. 34. Since a + b = 4.0 > 4.2 = c , there can be no triangle in this case. 35. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .
c 2 = a 2 + b2 − 2ab cos γ (13 ) = (12 ) + ( 5 ) − 2 (12 )( 5 ) cos γ 0 = −120 cos γ 2
2
2
Thus, cos γ = 0 so that γ = cos −1 ( 0 ) = 90 . Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin 90 sin β 5sin 90 −1 5sin 90 = sin β = β = sin = ≈ 23 13 5 13 c b 13 Step 3: Find the third angle. α ≈ 180 − ( 90 + 23 ) = 67
722
Section 7.2
36. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b 2 − 2ab cos γ
( 41 ) = ( 4 ) + (5) − 2 ( 4 )(5) cos γ 0 = −40 cos γ 2
2
2
Thus, cos γ = 0 so that γ = cos −1 ( 0 ) = 90 .
Step 2: Find either of the remaining two angles using the Law of Sines. 5sin 90 sin γ sin β sin 90 sin β 5sin 90 = sin β = β = sin −1 = ≈ 51 c b 5 41 41 41 Step 3: Find the third angle. α ≈ 180 − ( 90 + 51 ) = 39 37. This is AAS, so begin by using Law of Sines: γ = 180 − ( 40 + 35 ) = 105
( 6 ) ( sin35 ) sin α sin β sin35 sin 40 = = b= ≈ 5.354 ≈ 5 6 sin 40 a b b ( 6 ) ( sin105 ) sin α sin γ sin 40 sin105 = = c= ≈9 6 sin 40 a c c
38. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = (19.0 ) + (11.2 ) − 2 (19.0 ) (11.2 ) cos (13.3 ) = 486.44 − 425.6 cos (13.3 )
c ≈ 8.500 ≈ 8.5 Step 2: Find β .
11.2 sin13.3 sin β sin γ sin β sin13.3 11.2 sin13.3 = = sin β = β = sin −1 b c 11.2 8.500 8.500 8.500 ≈ 17.6
Step 3: Find α . α ≈ 180 − (13.3 +17.6 ) = 149.1
723
Chapter 7
39. This is SSA, so begin by using Law of Sines: Step 1: Determine β .
5sin31 sin α sin β sin31 sin β 5sin31 = = sin β = β = sin −1 a b 12 5 12 12 ≈ 12.392 ≈ 12 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of α is 180 − 12 = 168 . In such case, note that β + α > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and α , must sum to 180 ). Step 2: Solve for the triangle. γ ≈ 180 − (12 + 31 ) = 137 sin γ sin α sin137 sin31 12 sin137 = = c= ≈ 16 c a c 12 sin31
40. This is SSA, so begin by using Law of Sines: Step 1: Determine α . 11sin 60 sin α sin γ sin α sin 60 11sin 60 = = sin α = α = sin −1 a c 11 12 12 12
≈ 52.547 ≈ 53 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of α is 180 − 53 = 127 . In such case, note that γ + α > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are γ and α , must sum to 180 ). Step 2: Solve for the triangle. γ ≈ 180 − ( 53 + 60 ) = 67 sin γ sin β sin 60 sin 67 12 sin 67 = = b= ≈ 13 c b 12 b sin 60
41. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is β . b 2 = a 2 + c 2 − 2ac cos β
( 8 ) = ( 7 ) + ( 3 ) − 2 ( 7 ) ( 3 ) cos β 2
2
2
− 2 = −2 21 cos β Thus, cos β =
1 1 so that β = cos −1 ≈ 77.396 ≈ 77 . 21 21
724
Section 7.2
Step 2: Find either of the remaining two angles using the Law of Sines. sin α sin β sin α sin 77.396 7 sin 77.396 = sin α = = a b 7 8 8 7 sin 77.396 ≈ 66 8
α = sin −1 Step 3: Find the third angle. γ ≈ 180 − ( 77 + 66 ) = 37 42. This is AAS, so begin by using Law of Sines: α = 180 − (106 + 43 ) = 31
1( sin106 ) sin α sin β sin31 sin106 = = b= ≈ 1.866 ≈ 2 a b 1 b sin31 1( sin 43 ) sin α sin γ sin31 sin 43 = = c= ≈ 1.32 ≈ 1 a c 1 c sin31
43. This is SSA, so begin by using Law of Sines: Step 1: Determine γ . 2 sin10 sin γ sin β sin γ sin10 2 sin10 = = sin γ = γ = sin −1 c b 2 11 11 11
≈ 1.809 ≈ 2 Note that there is only one triangle in this case since the angle in QII with the same sine as this value of γ is 180 − 2 = 178 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. α ≈ 180 − ( 2 +10 ) = 168
sin α sin β sin168 sin10 11sin168 = = a= ≈ 13 a b a 11 sin10
725
Chapter 7
44. This is SSA, so begin by using Law of Sines: Step 1: Determine γ . 9sin 25 sin α sin γ sin 25 sin γ 9sin 25 = = sin γ = γ = sin −1 ≈ 39 a c 6 9 6 6
This is the solution in QI – label it as γ 1 . The second solution in QII is given by
γ 2 = 180 − γ 1 ≈ 141. Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: γ 1 ≈ 39
β1 ≈ 180 − ( 39 + 25 ) = 116
sin β1 sin γ 1 sin116 sin39 9sin116 = = b1 = ≈ 13 b1 c b1 9 sin39
QII triangle: γ 2 ≈ 141
β 2 ≈ 180 − (141 + 25 ) = 14
sin β 2 sin γ 2 sin14 sin141 9sin14 = = b2 = ≈3 b2 c b2 9 sin141
45. Consider the following diagram:
γ
α Let x = distance between Plane 1 and Plane 2 after 3 hours. In order to determine the value of x, we apply the Law of Cosines as follows: x 2 = 1500 2 + 13052 − 2(1500)(1305) cos150 x = 7,343,514.456 ≈ 2710 mi. Hence, the distance between the two planes after 3 hours is approximately 2,710 mi. .
726
Section 7.2
46. Consider the following diagram:
γ
α
Let x = distance between Plane 1 and Plane 2 after 4 hours. In order to determine the value of x, we apply the Law of Cosines as follows: x 2 = 1600 2 + 1200 2 − 2(1600)(1200) cos120 x = 5,920,000 ≈ 2, 433 mi. Hence, the distance between the two planes after 4 hours is approximately 2, 433 mi. .
47. Consider the following diagram:
γ Since the segment connecting the Pitcher’s mound to Home Base lies on the diagonal of the square (connecting home base to second base), we know that γ = 45 (since diagonals bisect the angles made at their respective vertices). We use the Law of Cosines to find x: x 2 = 90 2 + 60.52 − 2(90)(60.5) cos 45 so that x ≈ 63.7 ft. So, the Pitcher’s mound is approximately 63.7 ft. from Third Base.
727
Chapter 7
48. Consider the following diagram: 14.4
We use the Law of Cosines to find x: x 2 = 182 + 252 − 2(18)(25) cos14.4 so that x ≈ 8.8 ft. 49. Consider the following diagram:
40 125
α
55
Let x = length of the window. First, observe that α ≈ 180 − ( 40 +125 ) = 15 . So,
sin15 sin 40 40 sin15 = x= ≈ 16 ft. . x 40 ft. sin 40
50. Consider the following diagram:
α 40
90
40
5
40
Let x = length of the slide. First, observe that α ≈ 180 − (130 + 5 ) = 45 . So,
sin 45 sin5 6 sin 45 = x= ≈ 49 ft. . x 6 sin5
728
Section 7.2
51. Consider the following diagram:
Let x = distance between the player and the ball. In order to determine the value of x , we apply the Law of Cosines as follows: x 2 = 320 2 + 3952 − 2 ( 320 )( 395 ) cos10 x ≈ 9465.6 ≈ 97.3 feet Hence, the distance between the player had to run is approximately 97.3 feet.
52. Consider the following diagram:
This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle, the angle opposite the 290 ft side. 2902 = 952 + 2002 − 2 ( 95 )( 200 ) cos α 35,075 = −38,000 cos α 35,075 35,075 so that α = cos −1 − ≈ 22.627 ≈ 23 38,000 38,000 Step 2: Find x by using the Law of Sines: 95 sin22.627 sin 22.627 sin x 95 sin22.627 = sin x = x = sin −1 ≈7 290 95 290 290
Thus, cos α = −
Hence, the camera turned approximately 7.
729
Chapter 7
53. Using the Law of Cosines yields c 2 = a 2 + b 2 − 2ab cos c c 2 = ( 25.5 ) + (1.76 ) − 2 ( 25.5 )(1.76 ) cos105 2
2
c 2 ≈ 676.6 c ≈ 26 cm
54. Using the Law of Cosines yields b2 = a 2 + c 2 − 2ac cos B
(1.76 ) = ( 25.5 ) + ( 26 ) − 2 ( 25.5 ) ( 26 ) cos B 2
2
2
0.9979 ≈ cos B B ≈ 3.7
55. Using the Law of Cosines yields a 2 = b 2 + c 2 − 2bc cos α
( 75 ) = (50 ) + ( 50 ) − 2 (50 ) ( 50 ) cos α 2
2
2
−0.125 = cos α
α ≈ 97 Thus, the angle made by the ropes is 97.
56. Using the Law of Cosines yields a 2 = b2 + c 2 − 2bc cos α
(100 ) = ( 75 ) + ( 75 ) − 2 ( 75 ) ( 75 ) cos α 2
2
1 9
= cos α
2
α ≈ 84
Thus, the angle made by the ropes is 84.
57. Using the Law of Cosines yields 2 2 x 2 = ( 35 ) + ( 27 ) − 2 ( 35 ) ( 27 ) cos130 x ≈ 56 ft.
58. Using the Law of Cosines yields 2 2 x 2 = ( 29 ) + ( 36 ) − 2 ( 29 ) ( 36 ) cos135 x ≈ 60 ft.
730
Section 7.2
59. We have the following triangle:
This is a SSS situation. Using the Law of Cosines to find angle B yields b 2 = a 2 + c 2 − 2ac cos B 200 2 = 150 2 + 300 2 − 2(150)(300) cos B 72,500 = 90,000 cos B 72, 500 B = cos −1 ≈ 36 90,000 Next, use the Law of Sines to find angle T: 72,500 sin cos −1 90,000 sinT = 200 150 72,500 150 sin cos −1 90,000 sinT = 200 −1 72,500 150 sin cos 90,000 T = sin −1 ≈ 26 200
731
Chapter 7
60. We have the following triangle:
This is a SSS situation. Using the Law of Cosines to find angle B yields b 2 = a 2 + c 2 − 2ac cos B 60 2 = 100 2 + 50 2 − 2(100)(50) cos B 8,900 = 10,000 cos B 8, 900 ≈ 27 B = cos −1 10,000
61. Should have used the smaller angle β in Step 2.
62. Should have used the larger angle α in Step 1.
63. False. Can use Law of Cosines to solve any such triangle.
64. True. There are infinitely many similar triangles to a given one for which only the angles (and no side) is known.
65. True. By the Law of Cosines, c 2 = a 2 + b2 − 2ab c os 90 = a 2 + b 2
66. False. These only coincide when there is an angle with measure 90.
=0
So, the Pythagorean Theorem is a special case of the Law of Cosines.
732
Section 7.2
67. Let x = length of the longest side of the triangle. The three sides then have length x, 12 x, 34 x . Consider the following diagram:
68. Consider the following triangle:
60
We want to find c such that y = cx . The Law of Cosines yields 2 2 2 ( y ) = ( x ) + ( 14 x ) − 2 ( x ) ( 14 x ) cos 60
α
The Law of Cosines then yields
13 2 y2 = 16 x
( x ) = ( x ) + ( x ) − 2 ( x ) ( x ) cos α 2
1 2
2
3 4
2
1 2
3 4
y = 413 x
0.1875 = −0.75 cos α
α ≈ 104 cos α cos β cos γ a 2 + b 2 + c 2 + + = 69. Claim: a b c 2abc Proof: First, observe that the left-side simplifies to: cos α cos β cos γ bc cos α + ac cos β + ab cos γ (1). + + = a b c abc Next, note that the Law of Cosines yields the following three identities: a 2 − ( b2 + c 2 ) b2 − ( a 2 + c 2 ) c 2 − ( a2 + b2 ) bc cos α = , ac cos β = , ab cos γ = −2 −2 −2 Now, substituting these into the right-side of (1) yields: cos α cos β cos γ bc cos α + ac cos β + ab cos γ + + = a b c abc ( a 2 − b2 − c 2 ) + ( b2 − a 2 − c 2 ) + ( c 2 − a 2 − b2 ) = −2abc 2 2 2 2 a + b2 + c 2 a +b +c =− = −2abc 2abc
733
Chapter 7
70. Claim: α = c cos β + b cos γ . Proof: From the Law of Cosines, we have the following relationships: a 2 = b2 + c 2 − 2bc cos α (1)
b2 = a 2 + c 2 − 2ac cos β
(2)
c 2 = a 2 + b2 − 2ab cos γ
(3)
a + c 2 − b2 (4) Solving equation (2) for c cos β yields: c cos β = 2a −c 2 + a 2 + b 2 Solving equation (3) for b cos γ yields: b cos γ = (5) 2a Now, adding equations (4) and (5) yields: 2
c cos β + b cos γ =
a 2 + c 2 − b 2 −c 2 + a 2 + b 2 a 2 + c 2 −b 2 −c 2 + a 2 + b 2 2a 2 = a, + = = 2a 2a 2a 2a
as desired.
71. Consider the following triangle:
72. Consider the following triangle:
α
80
We are given that x = 32 y . The Law of Cosines yields 2 2 2 ( x ) = ( y ) + ( y ) − 2 ( y ) ( y ) cos α
We are given that x = y + 2 . The Law of Cosines yields 2 2 2 ( x) = ( y) + ( y) −2 ( y ) ( y ) cos80
( y ) = 2 y − 2y cos α 3 2
2
2
2
( y + 2 ) = 2 y2 − 2y2 cos80 2
− = cos α 1 8
α ≈ 97
4 y + 4 = y2 − 2y2 cos80 0.6527y2 − 4 y − 4 = 0 y=
4 ± 16 + 4(0.6527 )( 4 ) 2(0.6527 )
≈ 7.00, −0.8751 So, the sides have approximate lengths 7 in., 7 in., and 9 in.
734
Section 7.2
73. The following steps constitute the program in this case:
74. The following steps constitute the program in this case:
Program: BCX :Input “SIDE B =”, B :Input “SIDE C =”, C :Input “ANGLE X =”, X : (B2+C2-2BCcos(X)) → A :sin−1(Bsin(X)/A) → Y :180-Y-X → Z :Disp “SIDE A = “, A :Disp “ANGLE Y = “, Y :Disp “ANGLE Z =”, Z
Program: BCX :Input “SIDE B =”, B :Input “SIDE C =”, C :Input “ANGLE X =”, X : (B2+C2-2BCcos(X)) → A :sin−1(Bsin(X)/A) → Y :180-Y-X → Z :Disp “SIDE A = “, A :Disp “ANGLE Y = “, Y :Disp “ANGLE Z =”, Z
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE B = 45 SIDE C = 57 ANGLE X = 43 Now, the program will display the following information that solves the triangle: SIDE A = 39.01481143 ANGLE Y = 51.87098421 ANGLE Z = 85.12901579
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE B = 24.5 SIDE C = 31.6 ANGLE X = 81.5 Now, the program will display the following information that solves the triangle: SIDE A = 37.0127263 ANGLE Y = 40.89415545 ANGLE Z = 57.60584455
735
Chapter 7
75. The following steps constitute the program in this case:
76. The following steps constitute the program in this case:
Program: ABC :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “SIDE C =”, C :cos−1((B2+C2-A2)/(2BC)) → X : cos−1((A2+C2-B2)/(2AC)) → Y :180-Y-X → Z :Disp “ANGLE X = “, X :Disp “ANGLE Y = “, Y :Disp “ANGLE Z =”, Z
Program: ABC :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “SIDE C =”, C :cos−1((B2+C2-A2)/(2BC)) → X : cos−1((A2+C2-B2)/(2AC)) → Y :180-Y-X → Z :Disp “ANGLE X = “, X :Disp “ANGLE Y = “, Y :Disp “ANGLE Z =”, Z
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 29.8 SIDE B = 37.6 SIDE C = 53.2 Now, the program will display the following information that solves the triangle: ANGLE X = 32.98051035 ANGLE Y = 43.38013581 ANGLE Z = 103.6393538
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 100 SIDE B = 170 SIDE C = 250 Now, the program will display the following information that solves the triangle: ANGLE X = 16.73494404 ANGLE Y = 29.30810924 ANGLE Z = 133.9569467
736
Section 7.3 Solutions ------------------------------------------------------------------------------------
(
)
1. A =
1 (8) (16 ) sin 60 ≈ 55.4 2
2. A =
1 ( 6 ) 4 3 sin30 ≈ 10.4 2
3. A =
1 (1) 2
( 2 ) sin 45 ≈ 0.5
4. A =
1 2 2 ( 4 ) sin 45 ≈ 4 2
5. A =
1 ( 6 ) (8 ) sin80 ≈ 23.6 2
6. A =
1 ( 9 ) (10 ) sin100 ≈ 44.3 2
7. A =
1 (12 ) (10 ) sin35 ≈ 54 2
8. A =
1 (5 ) ( 6 ) sin 65 ≈ 14 2
9. A =
1 (12 ) ( 6 ) sin 45 ≈ 25 2
10. A =
1 ( 28.3 ) (1.9 ) sin 75 ≈ 26 2
11. A =
1 ( 4 ) ( 7 ) sin 27 ≈ 6.4 2
12. A =
1 ( 6.3) ( 4.8) sin17 ≈ 4.42 2
13. A =
1 (100 ) (150 ) sin36 ≈ 4, 408.4 2
14. A =
1 ( 0.3) ( 0.7 ) sin145 ≈ 0.06 2
15. A =
1 2
( 5 ) (5 5 ) sin50 ≈ 9.6
16. A =
1 2
(
)
( 11) ( 11 ) sin 21 ≈ 2.0
a + b + c 45 = = 22.5 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 97.4
17. First, observe that s =
18. First, observe that s =
19. First, observe that s =
a+b+c 3 = = 1.5 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 0.4 a + b + c 29 = = 14.5 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 29
737
Chapter 7
20. First, observe that s =
a + b + c 100 = 50. So, the area is given by = 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 165
a + b + c 3.4 = 1.7. So, the area is given by = 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 0.39
21. First, observe that s =
22. First, observe that s =
a + b + c 150 = = 75. So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 992
a + b + c 17 + 51 = . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 25.0
23. First, observe that s =
24. First, observe that s =
a + b + c 90 = = 45. So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 180
a + b + c 25 = = 12.5. So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 26.7
25. First, observe that s =
a + b + c 1.5 = = .75. So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 0.0992
26. First, observe that s =
a+b+c = 25.05. So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 111.64
27. First, observe that s =
738
Section 7.3
a+b+c = 196.5. So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 6,885.17
28. First, observe that s =
a+b+c = 24,600. So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 111,632,076
29. First, observe that s =
30. First, observe that s =
a+b+c 2+ 3+ 5 = . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 1.2
31. Observe that the sum of the lengths of the shortest two sides is 155, which does not exceed the length of the third side. Hence, there is no triangle in this case. 32. Observe that the sum of the lengths of the shortest two sides is 22, which does not exceed the length of the third side. Hence, there is no triangle in this case. 1 9 11 33. A = sin 27 ≈ 2.3 2 4 4
35. First, observe that s =
36. First, observe that s =
1 51 55 34. A = sin57 ≈ 174 2 2 4
a + b + c 73 + 83 + 133 14 = = . So, the area is given by 2 2 3 A = s ( s − a )( s − b )( s − c ) ≈ 2.7 a + b + c 256 + 176 + 116 53 So, the area is given by = = 2 2 12 A = s ( s − a ) ( s − b )( s − c ) ≈ 2.1
a+b+c = 1,277 . So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 312,297 sq. nm.
37. First, observe that s =
739
Chapter 7
a+b+c = 1,309.5. So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 328,885 sq. nm.
38. First, observe that s =
a+b+c = 237.5. So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 10,591 sq. ft.
39. First, observe that s =
40. First, we find the area: a+b+c = 42.5. So, the area is given by Observe that s = 2 A = s ( s − a ) ( s − b )( s − c ) ≈ 156.51 sq. ft. Now, since each bag covers 20 sq. ft., we need 8 bags since
156.51 sq. ft. = 7.8, which 20 sq. ft.
must be then rounded up to 8. As such, the total cost is $32 .
41. We need to determine α such that 1000 = sin α =
1000 1000 so that α = sin −1 ≈ 16 . 3600 3600
42. We need to determine α such that 500 = sin α =
1 ( 60 ) (120 ) sin α . This is equivalent to 2
1 ( 40 ) ( 60 ) sin α . This is equivalent to 2
500 500 so that α = sin −1 ≈ 25 . 1200 1200 a+b+c = 44.6 . So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) = ( 44.6 ) ( 22.1)(16.5 )( 6 ) ≈ 312.4 sq. mi.
43. First, observe that s =
44. A =
1 (150 ) ( 60 ) sin 43 ≈ 3,069 sq. ft. 2
740
Section 7.3
45. a. The area is 12 ( 310 ft.) ( 275 ft.) sin 79 ≈ 41,842 sq. ft. b. The total cost is 41,842 * $2.13, which is approximately $89,123. 46. a. The area is 12 ( 475 ft.) ( 310 ft.) sin118 ≈ 65,007 sq. ft. b. The total cost is 65,007 * $1.97, which is approximately $128,064. 47. a+b+c = 262.5 . So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) = 262.5 (17.5 ) (107.5 )(137.5 ) ≈ 8,240 sq. ft.
a. First observe that s =
b. The total cost is ( 8,240 ) ( $3.02), which is approximately $24,885. 48. a+b+c = 410 . So, the area is given by 2 A = s ( s − a ) ( s − b )( s − c ) = 410 ( 85 ) (130 )(195 ) ≈ 29,723 sq. ft.
a. First observe that s =
b. The total cost is ( 29,723 ) ( $3.68), which is approximately $109,381. 49. Let T be a triangle with sides 200 ft., 260 ft., and included angle 65. The area of the parallelogram is equal to 2 × area of T. 1 Note that area of T = ( 200 ) ( 260 ) sin 65 ≈ 23,564 sq. ft. 2 So, the area of the parallelogram is approximately 47,128 sq. ft. 50. Let T be a triangle with sides 250 ft., 300 ft., and included angle 55. The area of the parallelogram is equal to 2 × area of T. 1 Note that area of T = ( 250 ) ( 300 ) sin545 ≈ 30,718 sq. ft. 2 So, the area of the parallelogram is approximately 61,436 sq. ft.
741
Chapter 7
51. First, note that a regular hexagon can be dissected into 6 congruent equilateral 180 ( 6 − 2 ) = 120 , and diagonals bisect the triangles – this is the case since θ = 6 interior angles, thereby making two of the three angles in each of the triangles equal to 60 . As such, area of a regular hexagon = 6 × area of one these triangles. To find the area of one these triangles, we apply Heron’s formula. 3+3+3 3 = 4.5 and so, area A = 4.5 ( 4.5 − 3 ) = 3.897 sq. ft. Observe that s = 2 Thus, the area of the regular hexagon is 6 ( 3.897 sq. ft.) = 23.38 sq. ft. . 52. First, note that a regular decagon can be dissected into 10 congruent triangles. As such, area of a regular decagon = 10 × area of one these triangles. A typical triangle is illustrated below:
To find the area of such a triangle, we first need x. Using the Law of Sines yields sin36 sin 72 5sin 72 = so that x = ≈ 8.09017 in. 5 x sin36 1 Hence, the area A ≈ ( 8.09017 ) ( 8.09017 ) sin36 ≈ 19.236 sq. in. 2 Thus, the area of the regular decagon is 10 (19.236 sq. in.) = 192.36 sq. in. .
53. We identify a = 250, b = 275, c = 295 . Observe that s = area is given by A = s ( s − a ) ( s − b )( s − c ) ≈ 31,913 ft.2
742
a+b+c = 410 . So, the 2
Section 7.3
1
54. We identify a = 12 , b = 83 , c = 23 . Observe that s = 2
+ 38 + 23 11 = . So, the area is 2 16
given by A = s ( s − a ) ( s − b )( s − c ) ≈ 0.09 mi.2
55. A=
56. 1 (12 in.) (15 in.) sin 42 ≈ 60 in.2 2
A=
1 (34 in.) (37 in.) sin15 ≈ 163 in.2 2
57. The semiperimeter is half the a+b+c . perimeter, namely s = 2
58. Area is the square root of s ( s − a )( s − b )( s − c ) .
59. True. The formula works for all triangles.
60. True. The formula works for all triangles.
61. Observe that
1 2
x ( 23 x ) sin 60 = 13 x 2
( )= 3 2
x2 3 6
.
62. Consider the following triangle:
α
β We first find β using the Law of Sines:
( )
3 x 22 sin β sin 45 −1 5 ≈ 25.1041 . = β = sin 3 x x 5x Hence, α = 180 − ( 45 + 25.1041 ) = 109.8959 . As such, the area of the triangle is A = 12 ( x ) ( 53 x ) sin109.8959 = 103 x ( 0.94031) = 0.28x 2 .
743
Chapter 7
63. Consider the following diagram:
α
β
γ
Claim: Area =
a 2 sin β sin γ . 2 sin α
sin α sin β a sin β = x= . a x sin α a 2 sin β sin γ 1 a sin β sin = Hence, the area = ( a ) γ . 2 2 sin α sin α
Proof: First, the Law of Sines gives yields
1 ab sin θ , where θ is the angle between the sides measuring 2 1 a and b with a = b = s (for an isosceles triangle). In such case, A = s 2 sin θ . 2 64. Use the formula A =
( )
π 65. Area of sector = 12 (10 cm ) 80 180 ≈ 69.813 cm 2 2
Area of triangle = 12 (10 cm ) sin80 ≈ 49 cm 2 2
Hence, the area of the shaded region is the difference of these two, namely approximately 21 cm 2 .
( )
π 66. Area of sector = 12 ( 2 cm ) 50 180 ≈ 1.74523 cm 2 2
Area of triangle = 12 ( 2 cm ) sin50 ≈ 1.53 cm 2 Hence, the area of the shaded region is the difference of these two, namely approximately 0.21 cm 2 . 2
744
Section 7.3
67. The following steps constitute the program in this case: Program: ABZ :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Z =”, Z :1/2*ABsin(Z) → W :Disp “AREA W = “, W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 35 SIDE B = 47 ANGLE Z = 68 Now, the program will display the following information that gives the area of the triangle: AREA W = 762.6087204
68. The following steps constitute the program in this case: Program: ABZ :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Z =”, Z :1/2*ABsin(Z) → W :Disp “AREA W = “, W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 1241 SIDE B = 1472 ANGLE Z = 56 Now, the program will display the following information that gives the area of the triangle: AREA W = 757223.0219
69. The following steps constitute the program in this case: Program: ABC :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “SIDE C =”, C :(A+B+C)/2 → S : (S(S-A)(S-B)(S-C)) → W :Disp “AREA W = “, W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 85 SIDE B = 92 SIDE C = 123 Now, the program will display the following information that gives the area of the triangle: AREA W = 3907.492802
70. The following steps constitute the program in this case: Program: ABC :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “SIDE C =”, C :(A+B+C)/2 → S : (S(S-A)(S-B)(S-C)) → W :Disp “AREA W = “, W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 167 SIDE B = 113 SIDE C = 203 Now, the program will display the following information that gives the area of the triangle: AREA W = 65.93319949
745
Section 7.4 Solutions -----------------------------------------------------------------------------------1.
3.
5.
AB = 5 − 2,9 − 7 = 3,2 2 2 AB = ( 3 ) + ( 2 ) = 13 AB = −3 − 4,0 − 1 = −7, −1 2 2 AB = ( −7 ) + ( −1) = 50 = 5 2 AB = −24 − 0,0 − 7 = −24, −7 2 2 AB = ( −24 ) + ( −7 ) = 25
7. Given that u = 3,8 , we have 2 2 u = ( 3 ) + ( 8 ) = 73
tan θ =
8 8 so that θ = tan −1 ≈ 69.4 3 3
2.
4.
AB = 3 − ( −2), − 4 − 3 = 5, −7 2 2 AB = ( 5 ) + ( −7 ) = 74 AB = 2 − ( −1), −5 − ( −1) = 3, −4 2 2 AB = ( 3 ) + ( −4 ) = 5
6.
AB = 4 − ( −2),9 − 1 = 6,8 2 2 AB = ( 6 ) + ( 8 ) = 10
8. Given that u = 4,7 , we have 2 2 u = ( 4 ) + ( 7 ) = 65
tan θ =
7 7 so that θ = tan −1 ≈ 60.3 4 4
9. Given that u = 5, −1 , we have 2 2 u = ( 5 ) + ( −1) = 26
10. Given that u = −6, −2 , we have 2 2 u = ( −6 ) + ( −2 ) = 40 = 2 10
−1 so that 5 −1 θ = tan−1 ≈ 348.7 5
2 so that since the head is in QIII, 6 2 θ = 180 + tan −1 ≈ 198.4 6
tan θ =
tan θ =
746
Section 7.4
11. Given that u = −1, −5 , we have 2 2 | u | = ( −1) + ( −5 ) = 26
12. Given that u = −3, 2 , we have 2 2 | u | = ( −3 ) + ( 2 ) = 13
tan θ =
−5 so that since the head is in −1 −5 QIII, θ = 180° + tan −1 ≈ 259 −1
2 so that since the head is in QII, −3 2 θ = 180 + tan −1 ≈ 146 −3
13. Given that u = −4,1 , we have 2 2 u = ( −4 ) + (1) = 17
14. Given that u = −6,3 , we have 2 2 u = ( −6 ) + ( 3 ) = 45 = 3 5
tan θ =
1 so that since the head is in −4 1 QII, θ = 180 + tan −1 ≈ 166.0 −4
3 so that since the head is in QII, −6 3 θ = 180 + tan−1 ≈ 153.4 −6
15. Given that u = −8,0 , we have 2 2 u = ( −8 ) + ( 0 ) = 8
16. Given that u = 0,7 , we have 2 2 u = ( 0) + (7 ) = 7
Since the vector is on the x-axis pointing in the negative direction, θ = 180.
Since the vector is on the y-axis pointing in the positive direction, θ = 90.
17. Given that u = 0, −5 , we have 2 | u | = 0 2 + ( −5 ) = 5
18. Given that u = 2, 0 , we have | u | = 22 + 02 = 2 Since the vector is on the x-axis pointing in the positive direction, θ = 0.
Since the vector is on the y-axis pointing in the negative direction, θ = 270. 19. Given that u = u =
3,3 , we have
( 3 ) + (3) = 2 3 2
2
3 3 so that θ = tan −1 tan θ = ≈ 60 3 3
tan θ =
tan θ =
20. Given that u = −5, −5 , we have 2 2 u = ( −5 ) + ( −5 ) = 50 = 5 2
−5 = 1 so that since the head is in −5 QIII, θ = 180 + tan −1 (1) ≈ 225 tan θ =
747
Chapter 7
21. Given that u = 54 , 13 , we have 2 2 13 u = ( 54 ) + ( 31 ) = 15
22. Given that u = − 73 , 12 , we have 2 2 u = ( − 73 ) + ( 12 ) = 1437
1 5 tan θ = 3 so that θ = tan −1 ≈ 23 4 12 5
tan θ =
7 so that θ = tan −1 − ≈ 131 − 7 6 1
2
3
23. u + v = −4,3 + 2, −5 = −4 + 2,3 − 5 = −2, −2 . 24. u − v = −4,3 − 2, −5 = −4 − 2,3 − ( −5) = −6,8 25. 3u = 3 −4,3 = −12,9 26. −2u = −2 −4,3 = 8, −6 27. 2u + 4v = 2 −4,3 + 4 2, −5 = −8,6 + 8, −20 = 0, −14 28. 5 ( u + v ) = 5u + 5v = 5 −4,3 + 5 2, −5 = −20,15 + 10, −25 = −10, −10 29. 6 ( u − v ) = 6u − 6v = 6 −4,3 − 6 2, −5 = −24,18 + −12,30 = −36, 48 30. −3 ( u − v ) = −3 −6,8 = 18, −24 31. 10v − 2u − 3v = 7v − 2u = 14, −35 − −8,6 = 22, −41 32. 2u − 3v + 4u = 6u − 3v = 6 −4,3 − 3 2, −5 = −24,18 + −6,15 = −30,33
33. 4 ( u + 2v ) + 3 ( v − 2u ) = 4u + 8v + 3v − 6u = −2u + 11v = 8, −6 + 22, −55 = 30, −61 34. 2 ( 4u − 2v ) − 5 ( 3v − u ) = 8u − 4v −15v + 5u = 13u −19v = −52,39 − 38, −95 = −90,134
748
Section 7.4
35. u = u cos θ , u sin θ = 7 cos 25 , 7 sin 25 ≈ 6.3, 3.0 36. u = u cos θ , u sin θ = 5 cos 75 , 5sin 75 ≈ 1.3, 4.8 37. u = u cos θ , u sin θ = 16 cos100 , 16 sin100 ≈ −2.8, 15.8 38. u = u cos θ , u sin θ = 8cos 200 , 8sin 200 ≈ −7.5, − 2.7 39. u = u cos θ , u sin θ = 4 cos 310 , 4 sin310 ≈ 2.6, − 3.1 40. u = u cos θ , u sin θ = 8cos 225 , 8sin 225 ≈ −5.7, − 5.7 41. u = u cos θ , u sin θ = 9 cos ( 335 ) , 9sin ( 335 ) ≈ 8.2, − 3.8 42. u = u cos θ , u sin θ = 3cos ( 315 ) , 3sin ( 315 ) ≈ 2.1, − 2.1 43. u = u cos θ , u sin θ = 2 cos120 , 2 sin120 ≈ −1, 1.7 44. u = u cos θ , u sin θ = 6 cos330 , 6 sin330 ≈ 5.2, − 3 45. u = u cos θ , u sin θ = 3cos195 , 3sin195 ≈ −2.9, −0.78 46. u = u cos θ , u sin θ = 12 cos 280 , 12 sin 280 ≈ 2.1, −12 v 47. = v v 48. = v
−5, − 12
( −5 ) + ( −12 ) 2
3, 4
(3) + ( 4 ) 2
2
=
2
=
3, 4 5
−5, − 12
13
=
=
−5 −12 , 13 13
3 4 , 5 5
749
Chapter 7
v 49. = v v 50. = v v 51. = v v 52. = v v 53. = v v 54. = v
v 55. = v v 56. = v
v 57. = v
60,11
( 60 ) + (11) 2
2
−7,24
( −7 ) + ( 24 ) 2
( 24 ) + ( −7 ) −10,24
( −10 ) + ( 24 ) 2
( −9 ) + ( −12 ) 40, − 9
( 40 ) + ( −9 ) 2
2
−7 24 , 25 25
24, − 7 = 25
24 −7 , 25 25
25
−10,24
2
−10 24 = , 26 26
−5 12 , 13 13
=
−9 −12 , = 15 15
−3 −4 , 5 5
−9, − 12
15
( −4 3 ) + ( −2 3 ) 2
−2 6,3 6
( −2 6 ) + ( 3 6 )
2
40 −9 , 41 41
2,3 2
=
2 5
−4 3, − 2 3
2
=
40, − 9 = 41
=
( 2 ) + (3 2 )
26
=
2
2,3 2 2
=
=
2
−9, − 12 2
60 11 , 61 61
−7,24
=
2
=
61
=
2
24, − 7 2
60,11
=
2
=
=
2 3 2 , 2 5 2 5
=
−4 3, − 2 3 2 15
−2 6,3 6 78
=
750
=
=
10 3 10 , 10 10
−4 3 −2 3 , 2 15 2 15
−2 6 3 6 , 78 78
=
=
−2 5 − 5 , 5 5
−2 13 3 13 , 13 13
Section 7.4
v 58. = v
6 5, − 4
(6 5 ) + ( −4 ) 2
2
=
6 5, − 4 14
=
6 5 −4 , 14 14
=
3 5 2 ,− 14 7
59. 7i + 3 j
60. −2i + 4 j
61. 5i − 3 j
62. −6i − 2 j
63. −i + 0 j
64. 0i + 2 j
65. 0.8i − 3.6 j
66. − 154 i − 103 j
67. 2i + 0 j
68. 7i − 7 j
69. −5i + 5 j
70. 3i − 4 j
71. 7i + 0 j
72. 0i − 3 j
73. The velocity vector is given by v = 1750 cos 60 ,1750 sin 60 ≈ 875,1516 . So,
the horizontal component is approximately 875 ft/sec. and the vertical component is approximately 1516 ft/sec. 74. The velocity vector is given by v = 2200 cos30 , 2200 sin30 ≈ 1905, 1100 .
So, the horizontal component is approximately 1905 ft./ sec. and the vertical component is approximately 1100 ft./ sec. 75. The force vector is given by 25 cos 20 ,25sin 20 ≈ 23.49,8.55 . So, the
component perpendicular to the bench is approximately 23.49 lb., and the component parallel to the bench is approximately 8.55 lb. 76. The force vector is given by 50 cos 40 , 50 sin 40 ≈ 38.30, 32.14 .
So, the component perpendicular to the bench is approximately 38.30 lbs., and the component parallel to the bench is approximately 32.14 lbs. 630 77. weight = lbs. ≈ 2800.609 lbs. ≈ 2801 lbs. sin13
751
Chapter 7
500 78. weight = lbs. ≈ 1813.97 lbs. ≈ 1814 lbs. sin16 79. Consider the following diagram:
80. Consider the following diagram:
θ
By the Pythagorean Theorem, the actual velocity of the ship is
θ By the Pythagorean Theorem, the actual velocity of the ship is
10 2 + 6 2 ≈ 11.6619 ≈ 11.7 mph .
12 2 + 32 ≈ 12.369 ≈ 12.4 mph .
In order to determine the direction, we use the Law of Sines: sin θ sin 90 = 10 11.6619 10 sin 90 θ = sin −1 ≈ 59.036 11.6619 So the direction is approximately 31 west of due north.
In order to determine the direction, we use the Law of Sines: sin θ sin 90 = 3 12.369 3sin 90 θ = sin −1 ≈ 14.036 12.369 So the direction is approximately 76 west of due north.
752
Section 7.4
81. Consider the following diagram:
82. Consider the following diagram:
60
30
Observe that Plane = 300 cos30 , 300 sin30
Observe that Plane = 400 cos 60 , 400 sin 60
Wind = 40 cos120 , 40 sin120
Wind = 30 cos150 ,30 sin150
Thus, the Resultant = Plane + Wind ≈ 239.81, 184.64 Hence, Airspeed = |Resultant| ≈ 303 mph 184.64 Heading = 90 − tan−1 ≈ 52.41 239.81 (East of due North)
Thus, the Resultant = Plane + Wind ≈ 174.02,361.41 Hence, Airspeed = |Resultant| ≈ 401 mph 361.41 Heading = 90 − tan −1 ≈ 25.71 174.02 (East of due North)
753
Chapter 7
83. We have the following triangle:
β
Using the Law of Cosines yields R 2 = 1752 + 370 2 − 2(175)(370) cos125 R ≈ 491.7348 So, the speed is about 492 mph. To find the heading, we seek β , which is
given by β = 180 − (α + 45 ) . To find α ,
α
we apply the Law of Sines to obtain: sin α sin125 = 370 491.7348 370 sin125 α = sin −1 ≈ 38.0509 491.7348 Thus, β ≈ 97 and so, the heading is about N97E.
84. We have the following triangle:
Using the Law of Cosines yields R 2 = 2052 + 330 2 − 2(205)(330) cos 95 R ≈ 403.38216 So, the speed is about 403 mph. To find the heading, we seek β , which is
α
given by β = 90 − (α +15 ) . To find α ,
β
we apply the Law of Sines to obtain: sin α sin 95 = 330 403.38216 330 sin 95 α = sin −1 ≈ 54.584 403.38216 Thus, β ≈ 20 and so, the heading is about N200E.
754
Section 7.4
85. We have the following triangle:
α
There are technically two triangles. From the Law of Sines, we have sin α1 sin5 = 250 30 250 sin5 α1 = sin −1 ≈ 46.5769 30 We also can have α 2 = 180 − 46.5769 = 133.4231 . Hence, β1 = 180 − ( 5 + 46.5769 ) = 128.4231
β
β 2 = 41.5769 Now, using the Law of Cosines yields R12 = 250 2 + 30 2 − 2(250)(30) cos128.4231 R1 ≈ 270 Similarly, R2 ≈ 228 . So, the two possibilities for the speed of the plane are 228 mph and 270 mph.
86. We have the following triangle:
α
755
β
Chapter 7
There are technically two triangles. From the Law of Sines, we have sin α1 sin8 = 220 40 220 sin8 α1 = sin −1 ≈ 49.947 40 We also can have α 2 = 180 − 49.947 = 130.053 . Hence, β1 = 180 − ( 8 + 49.947 ) = 122.053
β 2 = 41.947 Now, using the Law of Cosines yields R12 = 220 2 + 402 − 2(220)(40) cos122.053 R1 ≈ 243 Similarly, R2 ≈ 192 . So, the two possibilities for the speed of the plane are 192 mph and 243 mph.
87. The force required to hold the box in place is 500 sin30 = 250 lbs.
88. The force required to hold the box in place is 500 sin10 = 87 lbs.
89. vertical component of velocity = 80 sin 40 = 51.4 ft./sec. horizontal component of velocity = 80 cos 40 = 61.3 ft./sec.
90. vertical component of velocity = 100 sin5 = 8.7 ft./sec. horizontal component of velocity = 100 cos5 = 99.6 ft./sec.
91. We need the component form for each of the three vectors involved. Indeed, they are: A = 0, 4 , B = 12,0 , C = 20 cos330 , 20 sin330 ≈ 17.32, − 10 2 As such, A + B + C = 29.32, − 6 . So, A + B + C = 29.32 2 + ( −6 ) ≈ 29.93 yards . 92. Using Exercise 85, we know that A + B + C = 29.32, − 6 . So, the direction angle
−6 is given by tan −1 ≈ −11.57 . 29.32
756
Section 7.4
93. Consider the following diagram:
The given information yields a SAS triangle. So, we use the Law of Cosines: c 2 = a 2 + b 2 − 2ab cos γ = 400 2 + 100 2 − 2(400)(100) cos120
θ
= 170,000 − 80,000 cos120 So, c ≈ 458.26 lbs. Next, we use the Law of Sines to find θ : sin θ sin120 = 100 458.26 100 sin120 θ = sin −1 ≈ 10.9 458.26
94. From Exercise 87, we see that c ≈ 458 lbs. 95. Consider the following diagram:
The given information yields a SAS triangle. So, we use the Law of Cosines: c 2 = a 2 + b 2 − 2ab cos γ = 1000 2 + 500 2 − 2(1000)(500) cos 95 = 1,250,000 − 1,000,000 cos 95 ≈ 1156.3545 ≈ 1156 lbs.
θ
96. Next, we use the Law of Sines to find θ : 500 sin 95 sin θ sin 95 so that θ = sin −1 = ≈ 25.5 500 1156 1156.3545
757
Chapter 7
97. We have the following diagram:
First, we find the magnitude of the resultant vector using the Law of Cosines: 2 Resultant = 900 2 + 750 2 − 2(900)(750) cos155 ≈ 2,596,015.512 So, Resultant ≈ 1611 N . Now, find the measure of the desired angle α using the Law of Sines: sin155 sin α = 1611.215539 750 750 sin155 sin α = 1611.215539 750 sin155 α = sin −1 ≈ 11 1611.215539 Hence, with respect to the bone, add 8 to get that the angle is about 19 .
98. We have the following diagram:
First, we find the magnitude of the resultant vector using the Law of Cosines: 2 Resultant = 8202 + 1,0002 − 2(820)(1,000) cos151 ≈ 3,106,776.32
758
Section 7.4
So, Resultant ≈ 1763 N . Now, find the measure of the desired angle α using the Law of Sines: sin151 sin α = 1762.604981 820 820 sin151 sin α = 1762.604981 −1 820 sin151 α = sin ≈ 13 1762.604981 Hence, with respect to the bone, add 9 to get that the angle is about 22 .
99. Magnitude cannot be negative. Observe that −2, −8 =
( −2 ) + ( −8) = 68 = 2 17 . 2
2
100. Inverse tangent gives an angle whose terminal side is in QI. As such, we must add 180 to obtain the QIII solution. 101. False. i = 1,0 = 1.
102. False. You can move the arrow representative anywhere in the plane.
103. True. Let u = u1 , u2 . Observe that 2 2 u = u1 , u2 = ( u1 ) + ( u2 ) ≥
( u1 ) = u1 = u1 .
104. True. Let u = u1 , u2 . Observe that 2 2 u = u1 , u2 = ( u1 ) + ( u2 ) ≥
( u2 ) = u2 = u2 .
105. Vector (since need two pieces of information to precisely define this quantity). 107.
− a, b =
2
2
106. Scalar (since only need one piece of information to precisely define the quantity).
( −a ) + b 2 = a 2 + b 2 2
b 108. Since −a, b is in QII, the direction angle must be tan −1 − + 180 a in QIV
759
Chapter 7
109. Given that u = 3a, −4a , we have u =
(3a ) + ( −4a ) = 5a and tanθ = − 2
2
4 3
4 so that θ = tan −1 − ≈ 307 . 3 110. Given that u = 34 a, 32 a , we have u = 2
tan θ = 3 = 3
4
( 34 a ) + ( 23 a ) = 12145 a and 2
2
8 8 so that θ = tan −1 ≈ 42 . 9 9
1 111. First, note that u + v = −6,1 . Hence, u + v = 37 and tan θ = − so that 6 1 θ = tan−1 − ≈ 171 . 6 5 112. First, note that u − v = 2,5 . Hence, u − v = 29 and tan θ = so that 2 5 θ = tan −1 ≈ 68 . 2
113. Look ahead to the next section for a formula to use to arrive at 127. 114. Look ahead to the next section for a formula to use to arrive at 87. 115. Enter the following to compute the given quantity: [ [8] [−5] ] + 3[ [−7] [11] ] The output is: [ [−13] [28] ].
116. Enter the following to compute the given quantity: −9( [ [8] [−5] ] −2[ [−7] [11] ] ) The output is: [ [−198] [243] ].
117. First, compute the length of the vector, as follows: (102 + ( −24)2 ) (output 26)
118. First, compute the length of the vector, as follows: (( − 9)2 + ( − 40)2 ) (output 41)
Then, to obtain the unit vector, divide both components of the given vector by this length (26) – to list as a FRACTION, enter the following: 1/26 [ [10] [−24] ] # Frac Output is: [ [5/13] [−12/13] ]
Then, to obtain the unit vector, divide both components of the given vector by this length (41) – to list as a FRACTION, enter the following: 1/41 [ [−9] [−40] ] # Frac Output is: [ [−9/41] [−40/41] ]
760
Section 7.5 Solutions -----------------------------------------------------------------------------------1. 4, −2 ⋅ 3,5 = ( 4 ) ( 3 ) + ( −2 )( 5 ) = 2
2. 7,8 ⋅ 2, −1 = ( 7 ) ( 2 ) + ( 8 )( −1) = 6
3. −5,6 ⋅ 3,2 = ( −5 ) ( 3 ) + ( 6 )( 2 ) = −3
4. 6, −3 ⋅ 2,1 = ( 6 ) ( 2 ) + ( −3 )(1) = 9
5. −1,3 ⋅ 4,10 = ( −1) ( 4 ) + ( 3 )(10 ) = 26
6. 2, −3 ⋅ 3,5 = ( 2 ) ( 3 ) + ( −3 )( 5 ) = −9
7. 1,8 ⋅ −2,6 = (1) ( −2 ) + ( 8 )( 6 ) = 46
8. 10,8 ⋅ 12, −6 = (10 ) (12 ) + ( 8 )( −6 ) = 72
9.
10.
−7, −4 ⋅ −2, −7 = ( −7 )( −2 ) + ( −4 ) ( −7 )
5, −2 ⋅ −1, −1 = ( 5 )( −1) + ( −2 )( −1) = −3
= 42
11. 3, −2 ⋅ 3 3, −1 =
( 3 ) (3 3 ) + ( −2 ) ( −1)
12. 4 2, 7 ⋅ − 2, − 7
(
= 11
13.
5, a ⋅ −3a,2 = ( 5 ) ( −3a ) + ( a )( 2 )
)(
) ( 7 ) ( − 7 ) = −15
= 4 2 − 2 +
14. 4 x, 3 y ⋅ 2 y, −5x = ( 4 x ) ( 2 y ) + ( 3 y )( −5x ) = −7 xy
= −13a
15. 0.8, −0.5 ⋅ 2,6 = ( 0.8 ) ( 2 ) + ( −0.5 )( 6 ) = −1.4
16. −18,3 ⋅ 10, −300 = ( −18 ) (10 ) + ( 3 )( −300 ) = −1080
761
Chapter 7
u v⋅ 17. Use the formula θ = cos with the following computations: u v 2 u v ⋅ = 3, 4 ⋅ 12, −5 = 16 , u = 32 + 4 2 = 5,v = 12 2 + ( −5 ) = 13 −1
16 So, θ = cos −1 ≈ 76 . ( 5 ) (13 ) u v⋅ 18. Use the formula θ = cos with the following computations: u v 2 2 u v ⋅ = −8, 6 ⋅ −15,8 = 168 , u = ( −8 ) + 6 2 = 10,v = ( −15 ) + 82 = 17 −1
168 So, θ = cos −1 ≈ 9 . (10 ) (17 )
u v⋅ 19. Use the formula θ = cos with the following computations: u v 2 2 u v ⋅ = 8,15 ⋅ −4, −3 = −77 , u = 82 + 152 = 17,v = ( −4 ) + ( −3 ) = 5 −1
−77 So, θ = cos −1 ≈ 155 . (17 ) ( 5 )
u v⋅ 20. Use the formula θ = cos with the following computations: u v 2 2 u v ⋅ = 5,12 ⋅ −6, −8 = −126 , u = 52 + 12 2 = 13,v = ( −6 ) + ( −8 ) = 10 −1
−126 So, θ = cos −1 ≈ 166 . (13 ) (10 )
762
Section 7.5
u ⋅ v 21. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −4,3 ⋅ −5, −9 = −7, u = ( −4 ) + ( 3 ) = 5, v = ( −5 ) + ( −9 ) = 106 −1
−7 ≈ 98 . So, θ = cos ( 5 ) 106 −1
(
)
u ⋅ v 22. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 2, −4 ⋅ 4, −1 = 12, u = ( 2 ) + ( −4 ) = 2 5, v = ( 4 ) + ( −1) = 17 −1
12 ≈ 49 . So, θ = cos 2 5 17 −1
(
)(
)
u ⋅ v 23. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −2, −3 ⋅ −3, 4 = −6, u = ( −2 ) + ( −3 ) = 13, v = ( −3 ) + ( 4 ) = 5 −1
−6 ≈ 109 . So, θ = cos ( 5 ) 13 −1
( )
u ⋅ v 24. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 6,5 ⋅ 3, −2 = 8, u = ( 6 ) + ( 5 ) = 61, v = ( 3 ) + ( −2 ) = 13 −1
So, θ = cos −1
8
≈ 73 . 13
( 61 )( )
763
Chapter 7
u ⋅ v 25. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −4,6 ⋅ −6,8 = 72, u = ( −4 ) + ( 6 ) = 2 13, v = ( −6 ) + ( 8 ) = 10 −1
72 ≈ 3 . So, θ = cos 2 13 (10 ) −1
(
)
u ⋅ v 26. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 1,5 ⋅ −3, −2 = −13, u = (1) + ( 5 ) = 26, v = ( −3 ) + ( −2 ) = 13 −1
So, θ = cos −1
−13
= 135 . 13
( 26 )( )
u ⋅ v 27. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −6,2 ⋅ −3,8 = 34, u = ( −6 ) + ( 2 ) = 40, v = ( −3 ) + ( 8 ) = 73 −1
So, θ ≈ cos −1
34
( 40 )(
= 51 . 73
)
u ⋅ v 28. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 7, −2 ⋅ 4,1 = 26, u = ( 7 ) + ( −2 ) = 53, v = ( 4 ) + (1) = 17 −1
So, θ = cos −1
26
( 53 )(
≈ 30 . 17
)
764
Section 7.5
u ⋅ v 29. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −2,2 3 ⋅ − 3,1 = 4 3, u = ( −2 ) + 2 3 = 4, v = − 3 + (1) = 2 −1
(
)
(
)
4 3 So, θ = cos −1 = 30 . ( 4) (2) u ⋅ v 30. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −3 3, −3 ⋅ −2 3,2 = 12, u = −3 3 + ( −3 ) = 6, v = −2 3 + ( 2 ) = 4 −1
(
)
(
12 So, θ = cos −1 = 60 . (6 ) ( 4) u ⋅ v 31. Use the formula θ = cos with the following computations: u v u ⋅ v = −5 3, −5 ⋅ 2, − 2 = 5 2 − 6 , −1
(
)
( −5 3 ) + ( −5) = 10, v = ( 2 ) + ( − 2 ) = 2 5( 2 − 6 ) = 105 . So, θ = cos u =
−1
2
(10 ) ( 2 )
2
2
2
u ⋅ v 32. Use the formula θ = cos with the following computations: u v u ⋅ v = −5, −5 3 ⋅ 2, − 2 = −10 + 5 6, −1
u =
( −5 ) + ( −5 3 ) = 10, v = ( 2 ) + ( − 2 ) = 6 2
2
−10 + 5 6 So, θ = cos = 85 . (10 ) 6 −1
( )
765
2
2
)
Chapter 7
u ⋅ v 33. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 4,6 ⋅ −6, −9 = −78, u = ( 4 ) + ( 6 ) = 2 13, v = ( −6 ) + ( −9 ) = 3 13 −1
−78 = 180 . So, θ = cos 2 13 3 13 −1
(
)(
)
u ⋅ v 34. Use the formula θ = cos with the following computations: u v u ⋅ v = 2,8 ⋅ −12,3 = 0 −1
So, θ = cos −1 ( 0 ) = 90 .
u ⋅ v 35. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −1,6 ⋅ 2, −4 = −26, u = ( −1) + ( 6 ) = 37, v = ( 2 ) + ( −4 ) = 20 −1
So, θ ≈ cos −1
−26
( 37 )(
≈ 163 . 20
)
u ⋅ v 36. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −8, −2 ⋅ 10, −3 = −74, u = ( −8 ) + ( −2 ) = 68, v = (10 ) + ( −3 ) = 109 −1
So, θ ≈ cos −1
−74
( 68 )(
≈ 149 . 109
)
37. Since −6,8 ⋅ −8,6 = 96 ≠ 0, these two vectors are not orthogonal. 38. Since 5, −2 ⋅ −5,2 = −29 ≠ 0, these two vectors are not orthogonal. 39. Since 6, −4 ⋅ −6, −9 = 0, these two vectors are orthogonal.
766
Section 7.5
40. Since 8,3 ⋅ −6,16 = 0, these two vectors are orthogonal. 41. Since 0.8, 4 ⋅ 3, −6 = −21.6 ≠ 0, these two vectors are not orthogonal. 42. Since −7,3 ⋅
1 1 , − = −2 ≠ 0, these two vectors are not orthogonal. 7 3
43. Since 5, −0.4 ⋅ 1.6,20 = 0, these two vectors are orthogonal. 44. Since 12,9 ⋅ 3, −4 = 0, these two vectors are orthogonal. 45. Since
3, 6 ⋅ − 2,1 = 0 , these two vectors are orthogonal.
46. Since
7, − 3 ⋅ 3,7 = 3 7 − 7 3 ≠ 0 , these two vectors are not orthogonal.
47. Since
4 8 1 5 , ⋅ − , = 0 , these two vectors are orthogonal. 3 15 12 24
48. Since
5 6 36 49 6 7 , ⋅ ,− = − ≠ 0 , these two vectors are not orthogonal. 6 7 25 36 5 6
49. W = F d = (100 lbs.) ( 4 ft.) = 400 ft. lbs. . 50. W = F d = (150 lbs.) ( 3.5 ft.) = 525 ft. lbs. 51. Note that 1 ton = 2000 lbs. So, W = F d = ( 4000 lbs.) ( 20 ft.) = 80,000 ft. lbs. 52. Note that 1 ton = 2000 lbs. So, W = F d = ( 5000 lbs.) ( 25 ft.) = 125,000 ft. lbs. 53. Here, F = ( 50 lbs.) cos30 ≈ 43.30 lbs. So, W = F d = ( 43.30 lbs.) ( 30 ft.) = 1299 ft. lbs.
767
Chapter 7
54. Here, F = ( 800 lbs.) cos 20 ≈ 751.75 lbs. So, W = F d = ( 751.75 lbs.) ( 50 ft.) = 37,588 ft. lbs. 55. Here, F = ( 35 lbs.) cos 45 ≈ 24.748 lbs. So, W = F d = ( 24.748 lbs.) ( 6 ft.) = 148 ft. lbs. 56. Here, F = ( 45 lbs.) cos55 ≈ 25.810 lbs. So, W = F d = ( 25.810 lbs.) ( 6 ft.) = 155 ft. lbs. 57. Weight of the car = 2500 lbs. Since we don’t want the car to roll down the hill, we want to apply the correct amount of force that would keep the vertical position of the car the same. That component is ( 2500 lbs.) sin 40 ≈ 1607 lbs. Hence, the force should be 1607 lbs. to keep the car from rolling down the hill.
58. We seek the vertical component of the weight of the car. This is given by ( 40,000 lbs.) sin10 ≈ 6946 lbs.
59. W = F d = ( 40,000 lbs.) sin10 (100 ft.) ≈ 694,593 ft. lbs. 60. Using the same setting as in Exercise 45, we now seek the horizontal component of the weight of the car. This is given by ( 2500 lbs.) cos 40 ≈ 1915 lbs. Hence, W = F d ≈ (1915 lbs.) (120 ft.) = 229,813 ft. lbs. 61. We have the following triangle:
Observe that
u ⋅ v θ = cos u v 2700 = cos −1 3400 3700 −1
(
θ
≈ 40
768
)(
)
Section 7.5
62. We have the following triangle:
Observe that
u ⋅ v
θ = cos −1
u v 2600 −1 = cos 3700 2000
θ
(
)(
)
≈ 144
63. Here, F = ( 70 lbs.) cos 40 So, W = F d = ( ( 70 lbs.) cos 40 ) (100 ft.) = 5,362 ft. lbs. 64. Here, F = ( 65 lbs.) cos 45 So, W = F d = ( ( 65 lbs.) cos 45 ) ( 75 ft.) = 3, 447 ft. lbs. 65. The dot product of two vectors is a scalar, not a vector. Should have summed: ( −3 ) ( 2 ) + ( 2 )( 5 ) = 4
66. The dot product is the sum of the products of corresponding components (first × first, second × second).
67. False. By definition, the dot product of two vectors is a scalar.
68. True. By definition.
69. True. Since u ⋅ v = u v cos θ , if
70. True. If u ⋅ v = u v cos θ = 0 , then assuming that u , v are not the zero vector, then cos θ = 0 , so that θ = 90 .
θ = 90 , then cos 90 = 0 , so that u ⋅v = 0 .
71. 3,7, −5 ⋅ −2, 4,1 = ( 3 ) ( −2 ) + ( 7 )( 4 ) + ( −5 )(1) = 17 72. 1,0, −2,3 ⋅ 5,2,3,1 = (1)( 5 ) + ( 0 ) ( 2 ) + ( −2 )( 3 ) + ( 3 )(1) = 2 73. Let u = a, b , v = c, d . Observe that u ⋅ v = a, b ⋅ c, d = ac + bd = ca + db = c, d ⋅ a, b = v ⋅ u .
769
Chapter 7
74. Let u = a, b . Observe that u ⋅ u = a, b ⋅ a, b = aa + bb = a 2 + b2 =
(
a2 + b2
) = u 2
2
75. Let u = a, b . Observe that 0 ⋅ u = 0,0 ⋅ a, b = 0( a ) + 0( b ) = 0
76. We prove the two equalities separately. Then, we conclude by transitivity that they must all be equal. Claim 1: k ( u ⋅ v ) = ( ku ) ⋅ v k ( u ⋅ v ) = k ( a, b ⋅ c, d ) = k ( ac + bd ) = kac + kbd = ( ka )c + ( kb )d = ka, kb ⋅ c, d = ( k a, b ) ⋅ c, d = ( ku ) ⋅ v Claim 2: ( ku ) ⋅ v = u ⋅ ( kv ) ( ku ) ⋅ v = ( k a, b ) ⋅ c, d = ka, kb ⋅ c, d = ( ka )c + ( kb )d = a( kc ) + b( kd ) = a, b ⋅ ( k c, d ) = u ⋅ ( kv ) 77. The dot product of vectors u and v is a scalar. Since the dot product is only done on two vectors, and there is no longer two vectors available, the expression is undefined. 78.
−2, 3 ⋅ ( 4, 1 + −3, − 5 ) = −2, 3 ⋅ 4, 1 + −2, 3 ⋅ −3, − 5 −2, 3 ⋅ ( 1, − 4 ) = ( −8 + 3 ) + ( 6 − 15 ) −2 − 12 = −5 + ( −9 ) −14 = −14
2 79. First, note that u = a 2 + b 2 , so that u = a 2 + b2 . Also, u ⋅ v = ab + ba = 2ab . 2 As such, u + ( u ⋅ v ) = ( a + b )2 . 80. u ⋅ v = u v cos θ = 2 2 ( 4 ) cos 45 = 8 2
( )= 8
81. u ⋅ v = u v cos θ =
2
(
)
( 10 )( 15 ) cos 30 =
2 2
5 6
770
( 3)
= 152 2
Section 7.5
u ⋅ v a2 −1 1 82. θ = cos = cos −1 = cos ( 2 ) = 60 2 2 a a + 3a u v −1
83. Enter the following two 2 × 1 matrices: −11 15 [ A] = , [B] = 34 −27
84. Enter the following two 2 × 1 matrices: 23 45 [ A] = , [B] = −350 202
Then, compute AT B (using the keystrokes given in the exercise) to obtain the dot product: −1083
Then, compute AT B (using the keystrokes given in the exercise) to obtain the dot product: −69665
85. Let a = 18,0 and b = 18,11 . Note that a ⋅ b (18)(18) + (0)(11) θ = cos −1 = cos −1 ≈ 31.4 182 182 + 112 a b 86. Let a = 0,7,0 and b = 12,7,9 . Note that (0)(12) + (7)(7) + (0)(9) −1 a ⋅ b θ = cos = cos −1 ≈ 65.0 2 2 2 2 a b 7 12 + 7 + 9
771
Chapter 7 Review Solutions --------------------------------------------------------------------------1. Observe that: γ = 180 − (10 + 20 ) = 150
( 4 ) ( sin 20 ) sin α sin β sin10 sin 20 = = b= ≈8 4 sin10 a b b ( 4 ) ( sin150 ) sin α sin γ sin10 sin150 = = c= ≈ 12 4 sin10 a c c
2. Observe that: α = 180 − ( 40 + 60 ) = 80
(10 ) ( sin80 ) sin α sin β sin80 sin 40 = = a= ≈ 15 10 sin 40 a b a (10 ) ( sin 60 ) sin β sin γ sin 40 sin 60 = = c= ≈ 13 10 sin 40 b c c
3. Observe that: γ = 180 − ( 45 + 5 ) = 130
(10 ) ( sin5 ) sin α sin γ sin5 sin130 = = a= ≈ 1.14 ≈ 1 10 sin130 a c a (10 ) ( sin 45 ) sin β sin γ sin 45 sin130 = = b= ≈ 9.23 ≈ 9 10 sin130 b c b
4. Observe that: α = 180 − ( 70 + 60 ) = 50
( 20 ) ( sin 60 ) sin α sin β sin50 sin 60 = = b= ≈ 23 20 sin50 a b b ( 20 ) ( sin 70 ) sin α sin γ sin50 sin 70 = = c= ≈ 25 20 sin50 a c c
772
Chapter 7 Review
5. Observe that: β = 180 − (11 +11 ) = 158
sin α sin γ sin11 sin11 = = a = 11 11 a c a
(11) ( sin158 ) sin β sin γ sin158 sin11 = = b= ≈ 22 11 sin11 b c b
6. Observe that: α = 180 − ( 20 + 50 ) = 110
(8) ( sin110 ) sin α sin β sin110 sin 20 = = a= ≈ 22 8 sin 20 a b a (8) ( sin50 ) sin β sin γ sin 20 sin50 = = c= ≈ 18 8 sin 20 b c c
7. Observe that: β = 180 − ( 45 + 45 ) = 90
( 2 ) ( sin 45 ) sin α sin β sin 45 sin 90 = = a= ≈ 2 2 sin 90 a b a ( 2 ) ( sin 45 ) sin β sin γ sin 90 sin 45 = = c= ≈ 2 2 sin 90 b c c
8. Observe that: γ = 180 − ( 60 + 20 ) = 100
(17 ) ( sin 60 ) sin α sin γ sin 60 sin100 = = a= ≈ 15 17 sin100 a c a (17 ) ( sin 20 ) sin β sin γ sin 20 sin100 = = b= ≈6 17 sin100 b c b
773
Chapter 7
9. Observe that: β = 180 − (12 + 22 ) = 146
( 99 ) ( sin146 ) sin α sin β sin12 sin146 = = b= ≈ 266 a b b 99 sin12 ( 99 ) ( sin 22 ) sin α sin γ sin12 sin 22 = = c= ≈ 178 99 sin12 a c c
10. Observe that: α = 180 − (102 + 27 ) = 51
( 24 ) ( sin102 ) sin α sin β sin51 sin102 = = b= ≈ 30 24 sin51 a b b ( 24 ) ( sin 27 ) sin β sin γ sin51 sin 27 = = c= ≈ 14 24 sin51 b c c
11. Step 1: Determine β . 9sin 20 sin α sin β sin 20 sin β 9sin 20 = = sin β = β = sin −1 ≈ 26 7 9 7 7 a b
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − β1 ≈ 154 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: β1 ≈ 26
γ 1 ≈ 180 − ( 26 + 20 ) = 134
sin β1 sin γ 1 sin 26 sin134 9sin134 = c1 = ≈ 15 = 9 sin 26 b c1 c1
QII triangle: β 2 ≈ 154
γ 2 ≈ 180 − (154 + 20 ) = 6
sin β 2 sin γ 2 sin154 sin 6 9sin 6 = = c2 = ≈2 9 sin154 b c2 c2
774
Chapter 7 Review
12. Step 1: Determine γ . sin γ sin β sin γ sin16 30 sin16 −1 30 sin16 = = sin γ = γ = sin 30 24 24 24 c b
≈ 20.154 ≈ 20 This is the solution in QI – label it as γ 1 . The second solution in QII is given by
γ 2 = 180 − γ 1 ≈ 160 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: γ 1 ≈ 20.154 ≈ 20
α1 ≈ 180 − ( 20.154 +16 ) = 143.84587 ≈ 144
sin α1 sin γ 1 sin143.84587 sin16 24 sin143.84587 = = a1 = ≈ 51.36 ≈ 51 24 sin16 a1 c a1
QII triangle: γ 2 ≈ 160
α 2 ≈ 180 − (160 +16 ) = 4
sin α 2 sin γ 2 sin 4 sin160 30 sin 4 = a2 = ≈6 = 30 sin160 a2 c a2
13. Step 1: Determine γ . 12 sin 24 sin γ sin α sin γ sin 24 12 sin 24 = = sin γ = γ = sin −1 ≈ 29 12 10 10 10 c a
This is the solution in QI – label it as γ 1 . The second solution in QII is given by
γ 2 = 180 − γ 1 ≈ 151 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: γ 1 ≈ 29
α1 ≈ 180 − ( 29 + 24 ) = 127
sin β1 sin α sin127 sin 24 10 sin127 = b1 = ≈ 20 = 10 sin 24 b1 a b1
775
Chapter 7
QII triangle: γ 2 ≈ 151
β 2 ≈ 180 − (151 + 24 ) = 5
sin β 2 sin α sin5 sin 24 10 sin5 = = b2 = ≈2 10 sin 24 b2 a b2
14. Step 1: Determine γ . 116 sin12 sin γ sin β sin γ sin12 116 sin12 = = sin γ = γ = sin −1 ≈ 14 116 100 100 c b 100
This is the solution in QI – label it as γ 1 . The second solution in QII is given by
γ 2 = 180 − γ 1 ≈ 166 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: γ 1 ≈ 14
α1 ≈ 180 − (12 +14 ) = 154
sin α1 sin β sin154 sin12 100 sin154 a = = 1= ≈ 211 a1 b a1 100 sin12
QII triangle: γ 2 ≈ 166
α 2 ≈ 180 − (166 +12 ) = 2
sin α 2 sin β sin 2 sin12 100 sin 2 = = a2 = ≈ 17 a2 b a2 100 sin12
15. First, determine α : sin α sin β sin α sin150 40 sin150 −1 40 sin150 = = sin α = α = sin ≈ 42 40 30 30 30 a b Hence, there is no triangle in this case since α + β > 180 .
776
Chapter 7 Review
16. Step 1: Determine β . sin γ sin β sin165 sin β 2 sin165 −1 2 sin165 = = sin β = β = sin ≈ 10 3 2 3 3 c b Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 − 10 = 170 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. α ≈ 180 − (10 +165 ) = 5
sin γ sin α sin165 sin5 3sin5 = = a= ≈1 3 sin165 c a a
17. Step 1: Determine β . 6 sin10 sin α sin β sin10 sin β 6 sin10 = = sin β = β = sin −1 ≈ 15 4 6 4 4 a b
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − β1 ≈ 165 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: β1 ≈ 15
γ 1 ≈ 180 − (15 +10 ) = 155
sin β1 sin γ 1 sin15 sin155 6 sin155 = = c1 = ≈ 10 b c1 c1 6 sin15
QII triangle: β 2 ≈ 165
γ 2 ≈ 180 − (165 +10 ) = 5
sin β 2 sin γ 2 sin165 sin5 6 sin5 = c2 = ≈2 = 6 sin165 b c2 c2
777
Chapter 7
18. Step 1: Determine α . 37 sin 4 sin α sin γ sin α sin 4 37 sin 4 = = sin α = α = sin −1 ≈6 37 25 25 a c 25 This is the solution in QI – label it as α1 . The second solution in QII is given by
α 2 = 180 − α1 ≈ 174 . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: α1 ≈ 6
β1 ≈ 180 − ( 6 + 4 ) = 170
sin β1 sin γ sin170 sin 4 25sin170 = = b1 = ≈ 62 25 sin 4 b1 c b1
QII triangle: α 2 ≈ 174
β 2 ≈ 180 − (174 + 4 ) = 2
sin β 2 sin γ sin 2 sin 4 25sin 2 = = b2 = ≈ 13 25 sin 4 b2 c b2
19. Consider the following diagram:
20. Consider the following diagram:
sin155 sin10 = 50 ft. y
y=
sin160 sin8 = 40 ft. y
(50 ft.) sin10 ≈ 21 ft.
40 ft.) sin8 ( ≈ 16 ft. y=
sin155 sin155 sin15 = x 50 ft. (50 ft.) sin15 ≈ 31 ft. x= sin155
sin160 sin160 sin12 = x 40 ft. 40 ft.) sin12 ( x= ≈ 24 ft. sin160
778
Chapter 7 Review
21. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 40 ) + ( 60 ) − 2 ( 40 )( 60 ) cos ( 50 ) = 5200 − 4800 cos ( 50 ) c ≈ 45.9849 ≈ 46 Step 2: Find α . sin α sin γ sin α sin50 40 sin50 −1 40 sin50 = = sin α = α = sin ≈ 42 40 45.9849 45.9849 a c 45.9849 Step 3: Find β .
β ≈ 180 − ( 42 + 50 ) = 88
22. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b2 + c 2 − 2bc cos α = (15 ) + (12 ) − 2 (12 ) ( 40 ) cos (140 ) = 369 − 360 cos (140 ) a ≈ 25.392 ≈ 25 Step 2: Find β . 15sin140 sin α sin β sin140 sin β 15sin140 = = sin β = β = sin −1 ≈ 22 25.392 15 25.392 a b 25.392 Step 3: Find γ .
γ ≈ 180 − ( 22 +140 ) = 18
23. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ ( 30 ) = ( 24 ) + ( 25 ) − 2 ( 24 ) ( 25 ) cos γ 2
2
2
900 = 1201 − 1200 cos γ 301 301 Thus, cos γ = so that γ = cos −1 ≈ 75.47 ≈ 75 . 1200 1200 Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin 75.47 sin β 25sin 75.47 −1 25sin 75.47 = = sin β = β = sin 30 25 30 30 c b ≈ 54 Step 3: Find the third angle. α ≈ 180 − ( 54 + 75 ) = 51
779
Chapter 7
24. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ ( 8 ) = ( 6 ) + ( 6 ) − 2 ( 6 ) ( 6 ) cos γ 2
2
2
64 = 72 − 72 cos γ 8 8 so that γ = cos −1 ≈ 83.6206 ≈ 84 . Thus, cos γ = 72 72 Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin83.6206 sin β 6 sin83.6206 = = sin β = 8 6 8 c b −1 6 sin83.6206 β = sin ≈ 48 8 Step 3: Find the third angle. α ≈ 180 − ( 48 + 84 ) = 48 25. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ ( 5 ) = 2
( 11) + ( 14 ) − 2 ( 11)( 14 ) cos γ 2
2
25 = 25 − 2 11 14 cos γ
Thus, cos γ = 0 so that γ = cos −1 ( 0 ) = 90 .
Step 2: Find either of the remaining two angles using the Law of Sines. 14 sin 90 sin γ sin β sin 90 sin β 14 sin 90 = sin β = β = sin −1 = ≈ 48 c b 5 5 5 14 Step 3: Find the third angle. α ≈ 180 − ( 48 + 90 ) = 42
780
Chapter 7 Review
26. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ (122 ) = ( 22 ) + (120 ) − 2 ( 22 )(120 ) cos γ 2
2
2
14,884 = 14,884 − 5280 cos γ Thus, cos γ = 0 so that γ = cos −1 ( 0 ) = 90 . Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin 90 sin β 120 sin 90 −1 120 sin 90 = = sin β = β = sin ≈ 80 122 120 122 122 c b Step 3: Find the third angle. α ≈ 180 − ( 80 + 90 ) = 10
27. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = ( 7 ) + (10 ) − 2 ( 7 ) (10 ) cos (14 ) = 149 −140 cos (14 ) a≈4 Step 2: Find β . sin α sin β sin14 sin β 7 sin14 −1 7 sin14 = = sin β = β = sin ≈ 28 3.627 7 3.627 a b 3.627 Step 3: Find γ .
γ ≈ 180 − ( 28 +14 ) = 138
28. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 6 ) + (12 ) − 2 ( 6 ) (12 ) cos ( 80 ) = 180 −144 cos ( 80 ) c ≈ 12.449 ≈ 12 Step 2: Find β . 12 sin80 sin β sin γ sin β sin80 12 sin80 = = sin β = β = sin −1 ≈ 72 12 12.449 12.449 b c 12.449 Step 3: Find α . α ≈ 180 − ( 72 + 80 ) = 28
781
Chapter 7
29. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = (10 ) + ( 4 ) − 2 (10 ) ( 4 ) cos ( 90 ) = 116 − 80 cos ( 90 ) a ≈ 10.770 ≈ 11 Step 2: Find β . 10 sin 90 sin α sin β sin 90 sin β 10 sin 90 = = sin β = β = sin −1 ≈ 68 10.770 10 10.770 10.770 a b Step 3: Find γ .
γ ≈ 180 − ( 68 + 90 ) = 22
30. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 4 ) + ( 5 ) − 2 ( 4 ) ( 5 ) cos ( 75 ) = 41− 40 cos ( 75 ) c ≈ 5.536 ≈ 6 Step 2: Find β . sin β sin γ sin β sin 75 5sin 75 −1 5sin 75 = = sin β = β = sin ≈ 61 5 5.536 5.536 b c 5.536 Step 3: Find α . α ≈ 180 − ( 61 + 75 ) = 44
31. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ (12 ) = (10 ) + (11) − 2 (10 ) (11) cos γ 2
2
2
144 = 221 − 220 cos γ 77 77 Thus, cos γ = so that γ = cos −1 ≈ 69.512 ≈ 70 . 220 220 Step 2: Find either of the remaining two angles using the Law of Sines. sin γ sin β sin 69.512 sin β 11sin 69.512 −1 11sin 69.512 = = sin β = β = sin 12 11 12 12 c b ≈ 59 Step 3: Find the third angle. α ≈ 180 − ( 70 + 59 ) = 51
782
Chapter 7 Review
32. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ ( 25 ) = ( 22 ) + ( 24 ) − 2 ( 22 ) ( 24 ) cos γ 2
2
2
625 = 1060 − 1056 cos γ 435 435 Thus, cos γ = so that γ = cos −1 ≈ 65.673 ≈ 66 . 1056 1056 Step 2: Find either of the remaining two angles using the Law of Sines. 24 sin 65.673 sin γ sin β sin 65.673 sin β 24 sin 65.673 = = sin β = β = sin −1 25 24 25 25 c b ≈ 61 Step 3: Find the third angle. α ≈ 180 − ( 61 + 66 ) = 53
33. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = (16 ) + (18 ) − 2 (16 )(18 ) cos (100 ) = 580 − 576 cos (100 ) a ≈ 26.077 ≈ 26 Step 2: Find β . 16 sin100 sin α sin β sin100 sin β 16 sin100 = = sin β = β = sin −1 ≈ 37 26.077 16 26.077 a b 26.077 Step 3: Find γ .
γ ≈ 180 − ( 37 +100 ) = 43
34. This is SAS, so begin by using Law of Cosines: Step 1: Find b. 2 2 b 2 = a 2 + c 2 − 2ac cos β = ( 25 ) + ( 25 ) − 2 ( 25 ) ( 25 ) cos ( 9 ) = 1250 −1250 cos ( 9 ) b ≈ 3.923 ≈ 4 Step 2: Find α . 25sin 9 sin β sin γ sin 9 sin γ 25sin 9 = = sin γ = γ = sin −1 ≈ 85.5 3.923 25 3.923 b c 3.923 Step 3: Find γ .
γ ≈ 180 − ( 9 + 85.5 ) = 85.5
783
Chapter 7
35. This is SAS, so begin by using Law of Cosines: Step 1: Find a. 2 2 a 2 = b 2 + c 2 − 2bc cos α = (12 ) + ( 40 ) − 2 (12 )( 4 ) cos (10 ) = 1744 − 960 cos (10 ) a ≈ 28.259 ≈ 28 Step 2: Find β . 12 sin10 sin α sin β sin10 sin β 12 sin10 = = sin β = β = sin −1 ≈4 28.259 12 28.259 28.259 a b Step 3: Find γ .
γ ≈ 180 − ( 4 +10 ) = 166
36. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α . 2 2 2 a 2 = b 2 + c 2 − 2bc cos α ( 26 ) = ( 20 ) + (10 ) − 2 ( 20 ) (10 ) cos α 676 = 500 − 400 cos α
176 176 so that α = cos −1 − ≈ 116.103 ≈ 116 . 400 400 Step 2: Find either of the remaining two angles using the Law of Sines. sin α sin β sin116.103 sin β 20 sin116.103 = = sin β = 26 20 26 a b −1 20 sin116.103 β = sin ≈ 44 26 Thus, cos α = −
Step 3: Find the third angle - γ ≈ 180 − (116 + 44 ) = 20
37. There is no triangle in this case since a + c = 39 > 40 = b . 38. There is no triangle in this case since a + b = 3 > 3 = c .
784
Chapter 7 Review
39. This is SSA, so use Law of Sines: Step 1: Determine β . 4.2 sin15 sin α sin β sin15 sin β 4.2 sin15 = = sin β = β = sin −1 ≈ 10 6.3 4.2 6.3 6.3 a b Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 − 10 = 170 . In such case, note that β + α > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and α , must sum to 180 ). Step 2: Solve for the triangle. γ ≈ 180 − (10 +15 ) = 155
sin γ sin α sin155 sin15 6.3sin155 = c= ≈ 10.3 = 6.3 sin15 c a c
40. This is SSA, so use Law of Sines: Step 1: Determine γ . 6 sin35 sin γ sin β sin γ sin35 6 sin35 = = sin γ = γ = sin −1 ≈ 43 6 5 5 5 c b Note that there is only one triangle in this case since the angle in QII with the same sine as this value of γ is 180 − 43 = 137 . In such case, note that β + γ > 180 , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle. α ≈ 180 − ( 35 + 43 ) = 102
sin α sin β sin102 sin35 5sin102 = = a= ≈9 5 sin 35 a b a
785
Chapter 7
41. Consider the following diagram:
The given information yields a SAS triangle, so we use Law of Cosines: x 2 = 82 + 6 2 − 2(8)(6) cos 50 = 100 − 96 cos 50 ≈ 6.2 mi.
42. Consider the following diagram:
The given information yields a SAS triangle, so we use Law of Cosines: x 2 = 10 2 + 32 − 2(10)(3) cos125 = 109 − 60 cos125 ≈ 12 mi.
43. A =
1 (16 ) (18) sin100 ≈ 141.8 2
44. A =
1 ( 25 ) ( 25 ) sin 9 ≈ 48.89 2
a + b + c 33 = = 16.5 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) = (16.5 ) ( 6.5 )( 5.5 )( 4.5 ) ≈ 51.5
45. First, observe that s =
a + b + c 71 = = 35.5 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) = ( 35.5 )(13.5 ) (11.5 )(10.5 ) ≈ 240.6
46. First, observe that s =
a + b + c 56 = = 28 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) = ( 28 ) ( 2 )( 8 )(18 ) ≈ 89.8
47. First, observe that s =
786
Chapter 7 Review
a + b + c 96 = = 48 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) = ( 48 ) ( 24 )(16 )( 8 ) ≈ 384
48. First, observe that s =
49. A =
1 (12 ) ( 40 ) sin10 ≈ 41.7 2
50. A =
1 ( 21) ( 75 ) sin 60 ≈ 682.0 2
abc with a = b = c = 9 in. and A = 35 in.2 to find r: 4r ( 9 )( 9 )( 9 ) so that 140r = 729 and hence, r = 729 ≈ 5.2 in. 35 = 4r 140
51. Use the formula A =
52. Use the formula A = find r: 54 =
53.
55.
abc with a = 9 in., b = 12 in., c = 15 in. and A = 54 in.2 to 4r
( 9 ) (12 )(15 ) so that 216r = 1620 and hence, r = 7.5 in. 4r
AB = −8 − 4,2 − ( −3) = −12,5 2 2 AB = ( −12 ) + ( 5 ) = 13 AB = 5 − 0,9 − ( −3) = 5,12 2 2 AB = ( 5 ) + (12 ) = 13
54.
56.
AB = 2 − ( −2),8 − 11 = 4, −3 2 2 AB = ( 4 ) + ( −3 ) = 5 AB = 9 − 3, − 3 − ( −11) = 6,8 2 2 AB = ( 6 ) + ( 8 ) = 10
57. Given that u = −10,24 , we have 2 2 u = ( −10 ) + ( 24 ) = 26
58. Given that u = −5, −12 , we have 2 2 u = ( −5 ) + ( −12 ) = 13
24 so that since the head is in −10 24 QII, θ = 180 + tan −1 ≈ 112.6 −10
tan θ =
tan θ =
787
12 12 so that θ = tan −1 ≈ 67.4 5 5
Chapter 7
59. Given that u = 16, −12 , we have 2 2 u = (16 ) + ( −12 ) = 20
60. Given that u = 0,3 , we have 2 2 u = ( 0 ) + (3) = 3
−12 so that 16 −12 θ = tan−1 ≈ 323.1 16
Since the vector is on the y-axis pointing in the positive direction, θ = 90 .
tan θ =
61. 2u + 3v = 2 7, −2 + 3 −4,5 = 14, −4 + −12,15 = 14 − 12, −4 + 15 = 2,11 62. u − v = 7, −2 − −4,5 = 7 − ( −4), −2 − 5 = 11, −7 63. 6u + v = 6 7, −2 + −4,5 = 42, −12 + −4,5 = 42 − 4, −12 + 5 = 38, −7 64. −3 ( u + 2v ) = −3 ( 7, −2 + 2 −4,5 ) = −3 ( 7, −2 + −8,10 ) = −3 −1,8 = 3, −24 65. u = u cos θ , u sin θ = 10 cos 75 , 10 sin 75 ≈ 2.6, 9.7 66. u = u cos θ , u sin θ = 8cos 225 , 8sin 225 ≈ −5.7, − 5.7 67. u = u cos θ , u sin θ = 12 cos105 , 12 sin105 ≈ −3.1, 11.6 68. u = u cos θ , u sin θ = 20 cos15 , 20 sin15 ≈ 19.3, 5.2 v 69. = v v 70. = v 71. 5i + j
6, − 6
( 6 ) + (− 6 ) 2
−11,60
( −11) + ( 60 ) 2
2
2
=
=
6, − 6 2 3 −11,60 61
=
2 2 ,− 2 2
=
−11 60 , 61 61
72. −15i + 2 j
788
Chapter 7 Review
73. Consider the following generic diagram:
θ
Here, u is the plane vector and v is the wind vector. u = u cos θ , u sin θ = 375 cos( −70 ), 375sin( −70 ) ≈ 128.26, −352.38 v = v cos θ , v sin θ = 45 cos(65 ), 45sin(65 ) ≈ 19.02, 40.78 The resultant vector, which is the true course of the plane, is u + v ≈ 147.28, − 311.6 . −311.6 Hence, the actual speed is u + v ≈ 345 mph and θ = tan −1 ( 147.28 ) ≈ −65 , so the actual
heading is 155 clockwise of north.
74. Consider the same generic diagram as in #73. Here, u is the plane vector and v is the wind vector. u = u cos θ , u sin θ = 425 cos( −20 ), 425sin( −20 ) ≈ 399.4, −145.4 v = v cos θ , v sin θ = 25 cos(5 ), 25sin(5 ) ≈ 24.91,2.18 The resultant vector, which is the true course of the plane, is u + v ≈ 424.31, −143.22 . Hence, the actual speed is u + v ≈ 440 mph and 143.22 θ = tan −1 ( −424.31 ) ≈ −19 , so the actual heading is 109 clockwise of north.
75. 6, −3 ⋅ 1, 4 = ( 6 ) (1) + ( −3 )( 4 ) = −6
76.
77. 3,3 ⋅ 3, −6 = ( 3 ) ( 3 ) + ( 3 ) ( −6 ) = −9
78. −2, −8 ⋅ −1,1 = ( −2 ) ( −1) + ( −8 )(1) = −6
79.
80.
−6,5 ⋅ −4,2 = ( −6 ) ( −4 ) + ( 5 )( 2 ) = 34
0,8 ⋅ 1,2 = ( 0 ) (1) + ( 8 )( 2 ) = 16
4, −3 ⋅ −1,0 = ( 4 ) ( −1) + ( −3 )( 0 ) = −4
789
Chapter 7
u ⋅ v 81. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 3, 4 ⋅ −5,12 = 33, u = ( 3 ) + ( 4 ) = 5, v = ( −5 ) + (12 ) = 13 −1
33 So, θ = cos −1 ≈ 59 . ( 5 ) (13 ) u ⋅ v 82. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −4,5 ⋅ 5, −4 = −40, u = ( −4 ) + ( 5 ) = 41, v = ( 5 ) + ( −4 ) = 41 −1
So, θ = cos −1
−40
( 41)(
≈ 167 . 41
)
u ⋅ v 83. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 1, 2 ⋅ −1,3 2 = 5, u = (1) + 2 = 3, v = ( −1) + 3 2 = 19 −1
( )
So, θ = cos −1
5
(
)
≈ 49 . 19
( 3 )( )
u ⋅ v 84. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 7, −24 ⋅ −6,8 = −234, u = ( 7 ) + ( −24 ) = 25, v = ( −6 ) + ( 8 ) = 10 −1
−234 So, θ = cos −1 ≈ 159 . ( 25 ) (10 )
790
Chapter 7 Review
u ⋅ v 85. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 3,5 ⋅ −4, −4 = −32, u = ( 3 ) + ( 5 ) = 34, v = ( −4 ) + ( −4 ) = 32 −1
So, θ = cos −1
−32
( 34 )(
≈ 166 . 32
)
u ⋅ v 86. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −1,6 ⋅ 2, −2 = −14, u = ( −1) + ( 6 ) = 37, v = ( 2 ) + ( −2 ) = 8 −1
So, θ = cos −1
−14
≈ 144 . 8
( 37 )( )
87. Since 8,3 ⋅ −3,12 = 12 ≠ 0 , these two vectors are not orthogonal. 88. Since −6,2 ⋅ 4,12 = 0 , these two vectors are orthogonal. 89. Since 5, −6 ⋅ −12, −10 = 0 , these two vectors are orthogonal. 90. Since 1,1 ⋅ −4, 4 = 0 , these two vectors are orthogonal. 91. Since 0, 4 ⋅ 0, −4 = −16 ≠ 0 , these two vectors are not orthogonal. 92. Since −7,2 ⋅
1 1 ,− = −2 ≠ 0 , these two vectors are not orthogonal. 7 2
93. Since 6z,a − b ⋅ a + b, −6z = 6 za + 6 zb − 6 za + 6bz = 12zb ≠ 0 in general, these two vectors are not orthogonal.
94. Since a − b, −1 ⋅ a + b, a 2 − b2 = ( a − b )( a + b ) − ( a 2 − b2 ) = 0 , these two vectors are orthogonal.
791
Chapter 7
95. W = F d = ( 25 lbs.) cos 25 ( 20 ft.) ≈ 453 ft. lbs. 96. W = F d = ( 30 lbs.) cos 22 ( 25 ft.) ≈ 695 ft. lbs. 97. The following steps constitute the program in this case:
98. The following steps constitute the program in this case:
Program: ACX :Input “SIDE A =”, A :Input “SIDE C =”, C :Input “ANGLE X =”, X :sin−1(Csin(X)/A) → Y :180-Y-X → Z :Asin(Z)/sin(X) → B :Disp “ANGLE Y = “, Y :Disp “ANGLE Z = “, Z :Disp “SIDE B =”, B
Program: ABY :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Y =”, Y :sin−1(Bsin(Y)/A) → X :180-Y-X → Z :Asin(Z)/sin(Y) → C :Disp “ANGLE X = “, X :Disp “ANGLE Z = “, Z :Disp “SIDE C =”, C
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 31.6 SIDE C = 23.9 ANGLE X = 42 Now, the program will display the following information that solves the triangle: ANGLE Y = 30.40327 ANGLE Z = 107.59673 SIDE B = 45.01568
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 137.2 SIDE B = 125.1 ANGLE Y = 54 Now, the program will display the following information that solves the triangle: ANGLE X = 47.53313 ANGLE Z = 78.46687 SIDE C = 166.16441
792
Chapter 7 Review
99. The following steps constitute the program:
100. The following steps constitute the program:
Program: ABZ :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Z =”, Z : (A2+B2-2ABcos(Z)) → C :sin−1(Asin(Z)/C) → Y :180-Y-Z → X :Disp “SIDE C = “, C :Disp “ANGLE Y = “, Y :Disp “ANGLE X =”, X
Program: ABC :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “SIDE C =”, C :cos−1((B2+C2-A2)/(2BC)) → X : cos−1((A2+C2-B2)/(2AC)) → Y :180-Y-X → Z :Disp “ANGLE X = “, X :Disp “ANGLE Y = “, Y :Disp “ANGLE Z =”, Z
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: 33 SIDE A = SIDE B = 29 ANGLE Z = 41.6 Now, the program will display the following information that solves the triangle: SIDE C = ANGLE Y = ANGLE X =
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 3412 SIDE B = 2178 SIDE C = 1576 Now, the program will display the following information that solves the triangle: ANGLE X = ANGLE Y = ANGLE Z =
101. The following steps constitute the program in this case:
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 312 SIDE B = 267 ANGLE Y = 189 Now, the program will display the following information that gives the area of the triangle: AREA W =
Program: ABY :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Y =”, Y :1/2*ABsin(Y) → W :Disp “AREA W = “, W
793
Chapter 7
102. The following steps constitute the program in this case: Program: ABZ :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Z =”, Z :1/2*ABsin(Z) → W :Disp “AREA W = “, W
103. Magnitude = Sum ({25, −60}^ 2) Direction angle = tan −1 ( −60) ÷ (25) The output will be: Magnitude = 65 Direction Angle = −67.38014°
Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 12.7 SIDE B = 29.9 ANGLE Z = 104.8 Now, the program will display the following information that gives the area of the triangle: AREA W =
104. Magnitude = Sum ({−70,10 ∗ 15}^ 2) Direction angle = tan −1 (10 ∗ 15) ÷ (−70) The output will be: Magnitude = 80 Direction Angle = −28.95502°
105. Sum ({14,37}*{9, −26}) /
106. Sum ({−23, −8}*{18, −32}) /
0.76807
−0.17672
(Sum ({14,37}^ 2 ) *Sum ({9, −26}^ 2 ) ) = cos
−1
( Ans ) = 39.81911
So, the angle is approximately 40 .
(Sum ({−23, −8}^ 2 ) *Sum ({18, −32}^ 2 ) ) = cos −1 ( Ans ) = 100.17877
So, the angle is approximately 100 .
794
Chapter 7 Practice Test Solutions ------------------------------------------------------------------1. This is AAS, so we use Law of Sines. Observe that: γ = 180 − ( 30 + 40 ) = 110
(10 ) ( sin30 ) sin α sin β sin30 sin 40 = = a= ≈ 7.8 10 sin 40 a b a (10 ) ( sin110 ) sin β sin γ sin 40 sin110 = = c= ≈ 14.6 10 sin 40 b c c
2. Since only three angles are given, and none of the side lengths is prescribed, there are infinitely many such triangles by similarity. 3. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ . c 2 = a 2 + b2 − 2ab cos γ (12 ) = ( 7 ) + ( 9 ) − 2 ( 7 ) ( 9 ) cos γ 2
2
2
144 = 130 − 126 cos γ 14 14 so that γ = cos −1 − Thus, cos γ = − ≈ 96.4 . 126 126 Step 2: Find either of the remaining two angles using the Law of Sines. 9sin 96.4 sin γ sin β sin 96.4 sin β 9sin 96.4 = = sin β = β = sin −1 ≈ 48.2 c b 12 9 12 12 Step 3: Find the third angle -
α ≈ 180 − ( 96.4 + 48.2 ) = 35.4
4. This is SSA, so we use Law of Sines: Step 1: Determine β . 10 sin 45 sin α sin β sin 45 sin β 10 sin 45 = = sin β = β = sin −1 ≈ 62 a b 8 10 8 8
This is the solution in QI – label it as β1 . The second solution in QII is given by
β 2 = 180 − β1 ≈ 118 . Both are tenable solutions, so we need to solve for two triangles.
795
Chapter 7
Step 2: Solve for both triangles. QI triangle: β1 ≈ 62
γ 1 ≈ 180 − ( 62 + 45 ) = 73
sin α sin γ 1 sin 45 sin 73 8sin 73 = = c1 = ≈ 10.8 8 c1 sin 45 a c1
QII triangle: β 2 ≈ 118
γ 2 ≈ 180 − (118 + 45 ) = 17
sin α sin γ 2 sin 45 sin17 8sin17 = = c2 = ≈ 3.3 8 sin 45 a c2 c2
5. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = (10 ) + (13 ) − 2 (10 )(13 ) cos ( 43 ) ≈ 78.85 c ≈ 8.8 Step 2: Find β .
13sin 43 sin β sin 43 13sin 43 = sin β = β = sin −1 ≈ 87 13 8.8 8.8 8.8 Step 3: Find α . α ≈ 180 − ( 87 + 43 ) = 50
6. This is AAS, so we use Law of Sines. Observe that: α = 180 − ( 26 +13 ) = 141
( 3) ( sin 26 ) sin 26 sin141 = b= ≈ 2.1 3 sin141 b ( 3) ( sin13 ) sin141 sin13 = c= ≈ 1.1 3 sin141 c
7. No triangle
796
Chapter 7 Practice Test
8. This is SSA, so we use Law of Sines: Step 1: Determine γ . 18sin12 sin12 sin γ 18sin12 = sin γ = γ = sin −1 ≈ 11 20 18 20 20
So, α ≈ 180 − (11 +12 ) = 157
9. A =
20 sin157 sin12 sin157 = a= ≈ 38 20 a sin12
1 (10 ) (12 ) sin 72 ≈ 57 2
a + b + c 30 = = 15 . So, the area is given by 2 2 A = s ( s − a ) ( s − b )( s − c ) = (15 ) ( 8 )( 5 )( 3 ) ≈ 34.64
10. First, observe that s =
11. First, observe that s =
a+b+c = 465 . So, the area is given by 2
A = s ( s − a ) ( s − b )( s − c ) =
12. A =
( 465 ) ( 215 )( 90 )(160 ) ≈ 37,943 yd.2
1 ( 20 ) ( 20 ) sin 25 ≈ 85 in.2 2
13. Given that u = −5,12 , we have 2 2 u = ( −5 ) + (12 ) = 13
14. Given that u = −3, −6 , we have 2 2 u = ( −3 ) + ( −6 ) = 3 5
12 so that since the head is in −5 12 QII, θ = 180 + tan −1 ≈ 112.6 −5
tan θ = 2 , so θ = tan −1 ( 2 ) ≈ 63.43 . But, since the head of u is in QIII, we add 180 to conclude that θ = 243 .
tan θ =
15.
v = v
−5,2
( −5 ) + ( 2 ) 2
2
=
−5,2 29
797
=
−5 29 2 29 , 29 29
Chapter 7
16.
v = v
−3, − 4
( −3) + ( −4 ) 2
2
=
−3, − 4 5
=
−3 −4 , 5 5
17. a. 2 −1, 4 − 3 4,1 = −2,8 − 12,3 = −2 − 12,8 − 3 = −14,5 b.
−7, −1 ⋅ 2,2 = ( −7 )( 2 ) + ( −1)( 2 ) = −16
18. Since 1,5 ⋅ −4,1 = 1 ≠ 0 , these two vectors are not orthogonal.
u ⋅ v 19. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = 3, 4 ⋅ −5,12 = 33, u = ( 3 ) + ( 4 ) = 5, v = ( −5 ) + (12 ) = 13 −1
33 So, θ = cos −1 ≈ 59.5 . 5 1 3 ( )( ) u ⋅ v 20. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −2, 4 ⋅ 6,1 = −8, u = ( −2 ) + ( 4 ) = 20, v = ( 6 ) + (1) = 37 −1
−8 So, θ = cos −1 ≈ 107 . 20 37 u ⋅ v 21. Use the formula θ = cos with the following computations: u v 2 2 2 2 u ⋅ v = −4, −4 ⋅ 1,3 = −16, u = ( −4 ) + ( −4 ) = 32, v = (1) + ( 3 ) = 10 −1
−16 So, θ = cos −1 ≈ 153 . 32 10 22. Work = F d = ( 30 cos 60 lbs.) (100 ft.) = 1500 ft. lbs.
798
Chapter 7 Practice Test
23. Boat in still water: u = u cos θ , u sin θ = 18cos50 ,18sin50 Current: v = v cos θ , v sin θ = 10 cos10 ,10 sin10 Hence, actual path of boat is given by u + v ≈ 21.42,15.53 . As such, the actual speed is u + v ≈ 26 mph and since θ = tan −1 ( 15.53 21.42 ) ≈ 36 , the heading is 54 clockwise of north. 24. Here, u is the plane vector and v is the wind vector. u = u cos θ , u sin θ = 300 cos(250 ), 300 sin(250 ) ≈ −102.61, −281.91 v = v cos θ , v sin θ = 45 cos(80 ), 45sin(80 ) ≈ 7.81, 44.32
The resultant vector, which is the true course of the plane, is u + v ≈ −274.10, − 58.29 . Hence, the actual speed is u + v ≈ 280 mph and 58.29 θ = tan −1 ( −−274.10 ) ≈ 12 . Since the head of u + v is in QIII, this corresponds to 192. The corresponding angle measured clockwise from north is 180 + 78 = 258 25. Work = F d = ( 30 cos35 lbs.) ( 50 ft.) = 1229 ft. lbs. 26. Work = F d = ( 3500 cos 40 lbs.) ( 2500 ft.) = 6,702,889 ft. lbs. 27. Let u = u1 , u2 . The magnitude of u is given by
u12 + u22 , which is always non-
negative. Geometrically, since the magnitude represents length, it cannot be negative.
28. We need the component form for each of the three vectors involved. Indeed, they are: A = 0,3 , B = 12,0 , C = 18cos330 , 18sin330 ≈ 15.59, − 9 2 As such, A + B + C = 27.59, − 6 . So, A + B + C = 27.592 + ( −6 ) ≈ 28.23 yards .
−6 Also, the direction angle is given by tan −1 ≈ −12.27 . 27.59 29. The dot product of two vectors is a scalar, as long as the number of components of each vector is the same. 30. The product of a scalor and avector is a vector.
799
Chapter 7 Cumulative Review Solutions ------------------------------------------------------------1. complement: 90 − 79 = 11
supplement: 180 − 79 = 101
2. Impossible since sec θ ∈ ( −∞, −1] ∪ [1, ∞ ) 3. cot θ =
cos θ −2 3 2 5 = 5 =− sin θ 5 3
(
)
1hr . . 4. Using the formula v = st , we obtain s = vt = ( 55 mi hr . ) 36 min. 60 min. = 33 mi.
5. Note that tan θ = −
(
1 3 = − 32 . So, we have 3 2
) (
)
Either sin θ = − 12 and cos θ = 23 or sin θ = 12 and cos θ = − 23 . This occurs when θ =
5π 6
,
11π 6
.
6. Amplitude 8 Period 24ππ = 12 7. The solid graph below is the desired graph:
8. The domain is the set of all real numbers x such that 7x ≠ (2 n+21)π , or x ≠ (2 n+141)π , where n is an integer. This is equivalent to saying the domain is the set of all real numbers for which x ≠ (2 n14+1)π , where n is an integer. The local minimum and maximum values for the guide function y = 3cos(7 x ) are −3 and 3, respectively. So, the desired range is ( −∞, −3] ∪ [3, ∞ ) .
800
Chapter 7 Cumulative Review
9. cos x+1 cos x + sec x 2 + 2 csc 2 x = coscos2 xx−1 + 2 sin x cos x − sec x cos x 2
2 cos 2 x +1 + 2 2 − sin x sin x 2 − cos 2 x − 1 = sin2 x 1 − cos 2 x = sin2 x sin2 x = =1 sin2 x =
10.
x 1+ cos x csc x + cot x sin1 x + cos cos x = cot x = 1 sin x = 1+sincosxx = sec x + 1 sin x cos x + 1 cos x
11. Conditional. For instance, it is false when A = π2 and B = − π2 . 12. Observe that y = sin x cos π4 − cos x sin π4 = sin ( x − π4 ) . The graph is as follows:
801
Chapter 7
13.
cos x (1− 4 sin2 x ) = cos x (1 − 2 sin2 x ) − 2 sin2 x = cos x ( cos 2 x ) − (2 sin x sin x) = cos x cos 2 x − sin x ( 2 sin x cos x ) = cos x cos 2 x − sin x sin 2x = cos(x + 2 x ) = cos3x
1 − cos 2 x 1 − (1 − 2 sin x ) 2 sin 2 x sin x 14. Observe that y = = = = = tan x . The sin 2 x 2 sin x cos x 2 sin x cos x cos x graph is as follows: 2
15.
1 − cos 34π = sin 38π 2
16.
cos(0.2 x ) − cos(0.8x ) = −2 sin ( 0.2 x +2 0.8 x ) sin ( 0.2 x 2−0.8 x )
= −2 sin(0.5x )sin( −0.3x ) = 2 sin(0.5x )sin(0.3x )
17. − π4
1 18. csc −1 (1.0537 ) = sin −1 ( 1.0537 ) = 71.63
802
Chapter 7 Cumulative Review
19. Let x = tan −1 ( 14 ) . Then, tan x = 14 . Consider the following triangle:
20. Observe that cos 2θ = − 12 2θ = 23π , 43π , 83π , 103π θ = π3 , 23π , 43π , 53π
θ So, sec x =
1 1 17 = 4 = . cos x 4 17
21. Factoring yields sec 2 2θ − sec 2θ − 12 = 0 ( sec 2θ − 4 ) ( sec 2θ + 3 ) = 0 . Hence,
sec 2θ = 4 or sec 2θ = −3 . As such, we see that
2θ = cos −1 ( 14 ) , cos −1 ( 14 ) + 360 , 360 − cos −1 ( 14 ) , 720 − cos −1 ( 14 )
cos −1 ( − 13 ) , cos −1 ( − 13 ) + 360 , 360 − cos −1 ( − 13 ) , 720 − cos −1 ( − 13 )
and so
θ ≈ 37.76 , 54.74 , 125.26 , 142.24 , 217.76 , 234.74 , 305.26 , 322.24 . 22. 4 cos 2 x + 4 cos 2 x + 1 = 0
4 cos 2 x + 4 ( 2 cos 2 x − 1) + 1 = 0 12 cos 2 x − 3 = 0 4 cos 2 x − 1 = 0 cos 2 x = 14 cos x = ± 12
x = π3 , 23π , 43π , 53π
803
Chapter 7
23. Observe that: α = 180 − (106.3 + 37.4 ) = 36.3
( 76.1) ( sin106.3 ) sin106.3 sin36.3 = b= ≈ 123.4 m 76.1 sin36.3 b ( 76.1) ( sin37.4 ) sin37.4 sin36.3 = c= ≈ 78.1 m 76.1 sin36.3 c
24. Note that u = 0,900 and v = 750 cos ( −45 ) ,750 sin ( −45 ) . So, u + v = 750 22 ,900 − 750 22 and as such, u + v ≈ 1,525 miles.
( )
( )
25. Orthogonal because 0.5,2 ⋅ −4,1 = 0
804
CHAPTER 8 Section 8.1 Solutions ----------------------------------------------------------------------------------1.
−16 = 16 ⋅ −1 = 4i
3.
−20 = 20 −1 = 2i 5 2 5
2.
i
4
−100 = 100 ⋅ −1 = 10i i
10
4.
i
−24 = 2 6 −1 = 2 6 ⋅i i
5. −4
6. −3
7. 8i
8. 3i 3
9. 3 − 10i
10. 4 − 11i
11. −10 − 12i
12. 7 − ( −5) = 12
13. 3 − 7i − 1− 2i = 2 − 9i
14. 1 + i + 9 − 3i = 10 − 2i
15. 10 −14i
16. −5 + 5i
17. ( 4 − 5i ) − ( 2 − 3i ) = 4 − 5i − 2 + 3i = 2 − 2i
18. ( −2 + i ) − (1 − i ) = −2 + i − 1 + i = −3 + 2i
19. −1 + 2i
20. −1 + 4i
21. 12 − 6i
22. 28 − 24i
23. 96 − 60i
24. −48 − 12i
25. −48 + 27i
26. 15 − 30i
27. −102 + 30i
28. −96 − 36i
29. 6i − 2i 2 = 6i + 2 = 2 + 6i
30. −2i − i 2 = −2i +1 = 1 − 2i
31. 21i + 6i 2 = 21i − 6 = −6 + 21i
32. −40i + 16i 2 = −40i −16 = −16 − 40i
805
Chapter 8 33. (1 − i )(3 + 2i ) = 3 + 2i − 3i − 2i 2
34. ( −3 + 2i )(1 − 3i ) = −3 + 9i + 2i − 6i 2
= 3−i + 2 = 5−i
= −3 + 11i + 6 = 3 + 11i
35. −15 + 21i + 20i + 28 = 13 + 41i
36. −32 + 10i − 16i − 5 = −37 − 6i
37. 42 − 30i + 63i + 45 = 87 + 33i
38. −21 − 14i + 12i − 8 = −29 − 2i
39. −24 + 36i + 12i + 18 = −6 + 48i
40. 16 + 9 = 25
41.
2 9
11 + 98 i − 23 i + 6 = 569 − 18 i
42. − 12 − 13 i + 38 i − 14 = − 34 + 241 i
43. −2i + 34 + 51i + 3 = 37 + 49i
44. 6i + 4 − 9 + 6i = −5 +12i
45. z = 4 − 7i
46. z = 2 − 5i
z z = 4 2 + 7 2 = 65
z z = 2 2 + 52 = 29
47. z = 2 + 3i
48. z = 5 + 3i
z z = 2 2 + 32 = 13
z z = 52 + 32 = 34
49. z = 6 − 4i
50. z = −2 − 7i
z z = 6 2 + 4 2 = 52
z z = 2 2 + 7 2 = 53
51. z = −2 + 6i
52. z = −3 + 9i
z z = 2 2 + 6 2 = 40
z z = 32 + 92 = 90
53. 2 i ⋅ = −2i i i
54. 3 i ⋅ = −3i i i
55. 5 i − ⋅ = 5i i i
56. 7 i − ⋅ = 7i i i
806
Section 8.1 57. 1 (3 + i ) 3 + i 3 + i 3 1 ⋅ = = = + i 2 3 − i (3 + i ) 9 − i 10 10 10
58. 2 7 + i 14 + 2i 7 1 ⋅ = = + i 7 − i 7 + i 49 + 1 25 25
59.
60. 1 4 + 3i 4 + 3i 4 3 ⋅ = = + i 4 − 3i 4 + 3i 16 + 9 25 25
1 ( 3 − 2i ) 3 − 2i 3 − 2i ⋅ = = 3 + 2i ( 3 − 2i ) 9 − 4i 2 9 + 4 =
61.
63.
2 7 − 2i 14 − 4i 14 4 ⋅ = = − i 7 + 2i 7 − 2i 49 + 4 53 53 1 − i (1 − i ) 1 − 2i + i ⋅ = 1 + i (1 − i ) 1− i2 =
65.
3 − 2i 3 2 = − i 13 13 13
2
1 − 2i − 1 −2i = = −i 1 − ( −1) 2
2 + 3i 3 + 5i 6 + 9i + 10i − 15 ⋅ = 3 − 5i 3 + 5i 9 + 25 9 19 =− + i 34 34
67. 4 − 5i ( 7 − 2i ) 28 − 8i − 35i + 10i 2 ⋅ = 7 + 2i ( 7 − 2i ) 49 − 4i 2 28 − 43i − 10 49 + 4 18 − 43i 18 43 = = − i 53 53 53
=
62.
8 1 − 6i 8 − 48i 8 48 ⋅ = = − i 1 + 6i 1 − 6i 1 + 36 37 37
64. 3 − i 3 − i 9 − 6i − 1 8 − 6i ⋅ = = 3+i 3−i 9 +1 10 4 3 = − i 5 5 66. 2 + i 3 + i 6 + 3i + 2i − 1 5 + 5i ⋅ = = 3−i 3+i 9 +1 10 1 1 = + i 2 2 68. 7 + 4i 9 + 3i 63 + 36i + 21i − 12 ⋅ = 9 − 3i 9 + 3i 81 + 9 51 + 57i 17 19 = = + i 90 30 30
807
Chapter 8 69. 8 + 3i 9 + 2i 72 + 27i + 16i − 6 ⋅ = 9 − 2i 9 + 2i 81 + 4 66 43 = + i 85 85
70. 10 − i 12 − 5i 120 − 12i − 50i − 5 ⋅ = 12 + 5i 12 − 5i 144 + 25 115 62 = − i 169 169
71.
72.
i = i ⋅ i = ( i ) ⋅ i = i ( i ) = i ( −1) = −i 15
12
4 3
3
3
2
i 99 = i 96 ⋅ i 3 = ( i 4 ) ⋅ i 3 = i ( i 2 ) = i ( −1) = −i 24
1
73. i 40 = ( i 4 ) = (1) = 1
74. i 18 = i 16 ⋅ i 2 = ( i 4 ) ⋅ i 2 = (1) ⋅ i 2 = −1
75. −3 ⋅ i 56 ⋅ i 2 = −3 (1)( −1) = 3
76. 7i 21 = 7 ⋅ i 20 ⋅ i = 7 (1)( i ) = 7i
77. 2i 36 = 2 ⋅ ( i 4 ) = 2
78. −5i 73 = −5 ⋅ i 72 ⋅ i = −5i
79. 25 − 20i − 4 = 21 − 20i
80. 9 − 30i − 25 = −16 − 30i
81. 4 + 12i − 9 = −5 + 12i
82. 16 − 72i − 81 = −65 − 72i
83.
84.
10
10
9
4
(3 + i ) (3 + i ) = (9 + 6i − 1)(3 + i ) = (8 + 6i )(3 + i ) = 24 + 18i + 8i − 6 = 18 + 26i
(2 + i )2 (2 + i ) = (4 + 4i − 1)(2 + i ) = (3 + 4i )(2 + i ) = 6 + 8i + 3i − 4 = 2 + 11i
2
85.
4
86.
(4 − 3i )2 (4 − 3i ) = (16 − 24i − 9)(4 − 3i ) = (7 − 24i )(4 − 3i ) = 28 − 96i − 21i − 72 = −44 − 117i
(1 − i ) (1 − i ) = (1 − 2i − 1)(1 − i ) = −2i (1 − i ) = −2i − 2 = −2 − 2i 2
87. z = (3 − 6i ) + (5 + 4i ) = 8 − 2i ohms
808
Section 8.1 88.
89. Should have multiplied by the conjugate 4 + i (not 4 − i ). 2 ( 4 + i ) 8 + 2i 8 + 2i ⋅ = = 4 − i ( 4 + i ) 16 − i 2 17
1 1 1 = + z 3 − 6i 5 + 4i 1 3 + 6i 1 5 − 4i = ⋅ + ⋅ 3 − 6i 3 + 6i 5 + 4i 5 − 4i 3 + 6i 5 − 4i = + 45 41 1 5 2 4 = + + − i 15 41 15 41 116 22 = + i 615 615
=
91. True.
90. i 2 = −1 not + 1. 10 − 7i − 12i 2 = 10 − 7i − 12 = 22 − 7i 92. True.
93. True.
94. False.
95.
x + 2 x + 1 = ( x + 1) = ( x + i ) ( x − i ) 4
2
2
2
2
8 2 + i 17 17
96. 2
( x + 9 ) = ( ( x + 3i )( x − 3i )) 2
2
= ( x + 3i )2 ( x − 3i )2
97. 41− 38i
98. −352 − 936i
2 11 99. 125 + 125 i
100.
809
7 625
24 − 625 i
2
Section 8.2 Solutions -----------------------------------------------------------------------------------1. – 8. The points listed in Exercises 1 – 8 are all plotted on the axes below:
9. The head of the vector is the complex number.
10. The head of the vector is the complex number. 8
4
7
3
6
2
5 4
1
3
−4
−3
−2
−1
1
2
3
4
2
5
−1
1
−4
−3
−2
−1
1
2
3
4
5
1
12. The head of the vector is the complex number.
11. The head of the vector is the complex number. 1
−4
−3
−2
−1
1 1
−1
2
3
4
5
−7
−6
−5
−4
−3
−2
−1
1 −1
−2 −3
−2 −3
810
Section 8.2 13. In order to express 1− i in polar form, observe that x = 1, y = −1, so that the point is in QIV. Now,
(1) + ( −1) = 2 2
r = x2 + y2 =
2
y −1 π = = −1, so that θ = tan −1 ( −1) = −45 or − . x 1 4 7π . Since 0 ≤ θ < 2π , we must use the reference angle 315 or 4 tan θ =
7π 7π Hence, 1 − i = 2 cos + i sin = 2 cos ( 315 ) + i sin ( 315 ) . 4 4 14. In order to express 2 + 2i in polar form, observe that x = 2, y = 2, so that the point is in QI. Now,
(2) + (2) = 2 2 2
r = x2 + y2 =
tan θ =
2
y 2 π = = 1, so that θ = tan −1 (1) = 45 or . x 2 4
π π Hence, 2 + 2i = 2 2 cos + i sin = 2 2 cos ( 45 ) + i sin ( 45 ) . 4 4 15. In order to express 1 + 3 i in polar form, observe that x = 1, y = 3, so that the point is in QI. Now, r = x2 + y2 =
(1) + ( 3 ) = 2
y 3 tan θ = = = 3, so that θ = tan −1 x 1
2
2
3 π 2 3 = tan 1 = 60 or 3 . 2
( )
−1
π π Hence, 1 + 3 i = 2 cos + i sin = 2 cos ( 60 ) + i sin ( 60 ) . 3 3
811
Chapter 8
16. In order to express −3 − 3 i in polar form, observe that x = −3, y = − 3, so that the point is in QIII. Now, r = x2 + y2 = tan θ =
( −3 ) + ( − 3 ) = 2 3 2
2
y − 3 1 π 7π 1 = = =π + = , so that θ = tan −1 or 210. x 6 6 −3 3 3
1 (Remember to add π to tan −1 since the angle has terminal side in QIII.) 3 7π 7π Hence, −3 − 3 i = 2 3 cos + i sin = 2 3 cos ( 210 ) + i sin ( 210 ) . 6 6 17. In order to express −4 + 4i in polar form, observe that x = −4, y = 4, so that the point is in QII. Now, r = x2 + y2 =
( −4 ) + ( 4 ) = 4 2 2
2
y 4 π 3π or 135. = = −1, so that θ = π + tan −1 ( −1) = π − = x −4 4 4 −1 (Remember to add π to tan ( −1) since the angle has terminal side in QII.) tan θ =
3π 3π Hence, −4 + 4i = 4 2 cos + i sin = 4 2 cos (135 ) + i sin (135 ) . 4 4 18. In order to express 5 − 5i in polar form, observe that x = 5, y = − 5, so that the point is in QIV. Now, r = x2 + y2 =
( 5 ) + ( − 5 ) = 10 2
2
y − 5 π = = −1, so that θ = tan −1 ( −1) = −45 or − . x 4 5 7π . Since 0 ≤ θ < 2π , we must use the reference angle 315 or 4
tan θ =
Hence,
7π 7π 5 − 5i = 10 cos + i sin = 10 cos ( 315 ) + i sin ( 315 ) . 4 4
812
Section 8.2
19. In order to express point is in QIV. Now,
3 − 3i in polar form, observe that x = 3, y = −3, so that the
( 3 ) + ( −3) = 2 3 π y −3 = − 3, so that θ = tan ( − 3 ) = −60 or − . tan θ = = x 3 3 2
r = x 2 + y2 =
2
−1
Since 0 ≤ θ < 2π , we must use the reference angle 300 or Hence,
5π . 3
5π 5π 3 − 3i = 2 3 cos + i sin = 2 3 cos ( 300 ) + i sin ( 300 ) . 3 3
20. In order to express − 3 + i in polar form, observe that x = − 3, y = 1, so that the point is in QII. Now, r = x2 + y2 =
tan θ =
( − 3 ) + (1) = 2 2
2
y −3 1 π 5π 1 = =− =π − = , so that θ = π + tan −1 − or 150. x 6 6 3 3 3
1 (Remember to add π to tan −1 − since the angle has terminal side in QII.) 3 5π 5π Hence, − 3 + i = 2 cos + i sin = 2 cos (150 ) + i sin (150 ) . 6 6 21. In order to express 3 + 0 i in polar form, observe that x = 3, y = 0, so that the point is on the positive x-axis. Now, r = x2 + y2 =
(3 ) + ( 0 ) = 3 and θ = 0 or 0. 2
2
Hence, 3 + 0i = 3 cos ( 0 ) + i sin ( 0 ) = 3 cos ( 0 ) + i sin ( 0 ) . 22. In order to express −2 + 0i in polar form, observe that x = −2, y = 0, so that the point is on the negative x-axis. Now, r = x2 + y2 =
( −2 ) + ( 0 ) = 2 and θ = π or 180. 2
2
Hence, −2 + 0i = 2 cos (π ) + i sin (π ) = 2 cos (180 ) + i sin (180 ) .
813
Chapter 8
23. In order to express 2 3 − 2i in polar form, observe that x = 2 3, y = −2, so that the point is in QIV. Now,
( 2 3 ) + ( −2 ) = 4 2
r = x2 + y2 =
tan θ =
2
y −2 1 1 11π = =− , so that θ = tan −1 − = 6 . x 2 3 3 3
11π 11π Hence, 2 3 − 2i = 4 cos + i sin . 6 6 24. In order to express −8 − 8 3 i in polar form, observe that x = −8, y = −8 3, so that the point is in QIII. Now, r = x 2 + y2 =
( −8) + ( −8 3 ) = 16 2
2
y −8 3 π 4π = = 3, so that θ = tan −1 3 = π + = . x −8 3 3 (Remember to add π to tan −1 3 since the angle has terminal side in QIII.)
( )
tan θ =
( )
4π 4π Hence, −8 − 8 3 i = 16 cos + i sin . 3 3 25. In order to express − 12 + 23 i in polar form, observe that x = − 12 , y = 23 , so that the point is in QII. Now,
( − 12 ) + ( 23 ) = 1 2
r = x2 + y2 =
tan θ =
2
3 y π 2π = 2 1 = − 3, so that θ = π + tan −1 − 3 = π − = . x −2 3 3
(
(
)
)
(Remember to add π to tan −1 − 3 since the angle has terminal side in QII.) 2π 2π Hence, − 12 + 23 i = cos + i sin . 3 3
814
Section 8.2 26. In order to express − 53 − 53 i in polar form, observe that x = − 53 , y = − 53 , so that the point is in QIII. Now, r = x 2 + y2 = tan θ =
( − 53 ) + ( − 53 ) = 5 32 2
2
π 5π y − 53 = 5 = 1, so that θ = tan −1 (1) = π + = . x −3 4 4
(Remember to add π to tan −1 (1) since the angle has terminal side in QIII.) 5π 5π Hence, − 53 − 53 i = 5 3 2 cos + i sin . 4 4 27. In order to express 83 − 18 i in polar form, observe that x = 83 , y = − 18 , so that the point is in QIV. Now,
( ) + (− ) =
r = x 2 + y2 = tan θ =
Hence,
3 8
3 8
2
1 2 8
1 4
y − 18 1 1 11π , so that θ = tan −1 − = 3 =− = 6 . x 3 3 8
11π 11π − 81 i = 14 cos + i sin . 6 6
28. In order to express 167 + 167 i in polar form, observe that . x = 167 , y = 167 , so that the point is in QI. Now,
( 167 ) + ( 167 ) = 7162 2
r = x2 + y2 = tan θ =
2
π y 167 = 7 = 1, so that θ = tan −1 (1) = 45 or . 4 x 16
π π Hence, 167 + 167 i = 7162 cos + i sin . 4 4 29. In order to express 3 − 7i in polar form, observe that x = 3, y = −7, so that the point is in QIV. Now, r = x 2 + y2 =
tan θ =
(3 ) + ( −7 ) = 58 2
2
y −7 7 = , so that θ = tan −1 − ≈ −66.80. x 3 3
815
Chapter 8 Since 0 ≤ θ < 360 , we must use the reference angle 293.2. Hence, 3 − 7i ≈ 58 cos ( 293.2 ) + i sin ( 293.2 ) . 30. In order to express 2 + 3i in polar form, observe that x = 2, y = 3, so that the point is in QI. Now, r = x2 + y2 =
tan θ =
( 2 ) + ( 3) = 13 2
2
y 3 3 = , so that θ = tan −1 ≈ 56.3. x 2 2
Hence, 2 + 3i ≈ 13 cos ( 56.3 ) + i sin ( 56.3 ) . 31. In order to express −6 + 5i in polar form, observe that x = −6, y = 5, so that the point is in QII. Now, r = x 2 + y2 =
( −6 ) + ( 5 ) = 61 2
2
y 5 5 = , so that θ = 180 + tan −1 − ≈ 140.2. x −6 6 5 (Remember to add 180 to tan −1 − since the angle has terminal side in QII.) 6 tan θ =
Hence, −6 + 5i ≈ 61 cos (140.2 ) + i sin (140.2 ) . 32. In order to express −4 − 3 i in polar form, observe that x = −4, y = −3, so that the point is in QIII. Now, r = x2 + y2 =
( −4 ) + ( −3) = 5 2
2
y 3 3 = , so that θ = 180 + tan −1 ≈ 216.9. x 4 4 3 (Remember to add 180 to tan −1 since the angle has terminal side in QIII.) 4
tan θ =
Hence, −4 − 3i ≈ 5 cos ( 216.9 ) + i sin ( 216.9 ) .
816
Section 8.2 33. In order to express −5 + 12 i in polar form, observe that x = −5, y = 12, so that the point is in QII. Now,
( −5 ) + (12 ) = 13 2
r = x2 + y2 =
2
y −12 12 = , so that θ = 180 + tan −1 − ≈ 112.6. 5 x 5 12 (Remember to add 180 to tan −1 − since the angle has terminal side in QII.) 5
tan θ =
Hence, −5 + 12 i ≈ 13 cos (112.6 ) + i sin (112.6 ) . 34. In order to express 24 + 7 i in polar form, observe that x = 24, y = 7, so that the point is in QI. Now,
( 24 ) + ( 7 ) = 25 2
r = x 2 + y2 =
tan θ =
2
y 7 7 = , so that θ = tan −1 ≈ 16.3. x 24 24
Hence, 24 + 7i ≈ 25 cos (16.3 ) + i sin (16.3 ) . 35. In order to express 8 − 6i in polar form, observe that x = 8, y = −6, so that the point is in QIV. Now,
(8 ) + ( −6 ) = 10 2
r = x 2 + y2 =
2
y −6 6 = , so that θ = tan−1 − ≈ −36.87. x 8 8 Since 0 ≤ θ < 360 , we must use the reference angle 323.1.
tan θ =
Hence, 8 − 6i ≈ 10 cos ( 323.1 ) + i sin ( 323.1 ) . 36. In order to express −3 + 4 i in polar form, observe that x = −3, y = 4, so that the point is in QII. Now, r = x2 + y2 =
tan θ =
( −3 ) + ( 4 ) = 5 2
2
y 4 4 = , so that θ = 180 + tan−1 − ≈ 126.9. x −3 3
817
Chapter 8 4 (Remember to add 180 to tan −1 − since the angle has terminal side in QII.) 3 Hence, −3 + 4 i ≈ 5 cos (126.9 ) + i sin (126.9 ) . 37. In order to express −6 + 2 i in polar form, observe that x = −6, y = 2, so that the point is in QII. Now, r = x2 + y2 =
( −6 ) + ( 2 ) = 2 10 2
2
y −2 1 = , so that θ = π + tan −1 − ≈ 2.82 rad. x 6 3 1 (Remember to add 180 to tan −1 − since the angle has terminal side in QII.) 3
tan θ =
Hence, −6 + 2 i ≈ 2 10 cos ( 2.82 ) + i sin ( 2.82 ) . 38. In order to express −5 − 8 i in polar form, observe that x = −5, y = −8, so that the point is in QIII. Now, r = x2 + y2 =
( −5 ) + ( −8) = 89 2
2
y 8 8 = , so that θ = 180 + tan −1 ≈ 4.15 rad. x 5 5 8 (Remember to add π to tan −1 since the angle has terminal side in QIII.) 5
tan θ =
Hence, −5 − 8i ≈ 89 cos ( 4.15 ) + i sin ( 4.15 ) . 39. In order to express 35 − 10i in polar form, observe that x = 35, y = −10, so that the point is in QIV. Now, r = x2 + y2 =
tan θ =
(35 ) + (10 ) = 5 53 2
2
y −10 10 = , so that θ = tan −1 − ≈ 6.00 rad. x 35 35
Hence, 35 − 10i ≈ 5 53 cos ( 6.00 ) + i sin ( 6.00 ) .
818
Section 8.2 40. In order to express 20 + i in polar form, observe that x = 20, y = 1, so that the point is in QI. Now,
( 20 ) + (1) = 401 2
r = x2 + y2 =
tan θ =
2
y 1 1 = , so that θ = tan −1 ≈ 0.05. x 20 20
Hence, 20 + i ≈ 401 cos ( 0.05 ) + i sin ( 0.05 ) . 41. In order to express 34 + 53 i in polar form, observe that x = 34 , y = 53 , so that the point is in QI. Now, r = x2 + y2 = tan θ =
( 34 ) + ( 53 ) ≈ 1.83 2
2
y 53 20 = 3 , so that θ = tan −1 ≈ 1.15 rad. x 4 9
Hence, 34 + 53 i ≈ 1.83 cos (1.15 ) + i sin (1.15 ) . 42. In order to express − 125 + 65 i in polar form, observe that x = − 125 , y = 65 , so that the point is in QII. Now, r = x2 + y2 = tan θ =
( − 125 ) + ( 65 ) ≈ 2.54 2
2
5 y 25 = 612 , so that θ = 180 + tan −1 − ≈ 2.81 rad. x −5 72
25 (Remember to add π to tan −1 − since the angle has terminal side in QII.) 72 Hence, − 125 + 65 i ≈ 2.54 cos ( 2.81) + i sin ( 2.81) . 43. In order to express − 252 − 134 i in polar form, observe that x = − 252 , y = − 134 , so that the point is in QIII. Now, r = x 2 + y2 = tan θ =
( − 252 ) + ( − 134 ) ≈ 12.9 2
2
y 134 26 = 25 , so that θ = π + tan −1 ≈ 3.40 rad. x 2 100
26 (Remember to add π to tan −1 since the angle has terminal side in QIII.) 100 Hence, − 252 − 134 i ≈ 12.9 cos ( 3.40 ) + i sin ( 3.40 ) .
819
Chapter 8 27 27 44. In order to express 13 − 113 i in polar form, observe that x = 13 , y = − 113 , so that the point is in QIV. Now,
r = x2 + y2 = tan θ =
( 1327 ) + ( − 113 ) ≈ 2.09 2
2
y − 113 39 = 27 , so that θ = tan −1 − ≈ 6.15 rad. x 297 13
27 − 113 i ≈ 2.09 cos ( 6.15 ) + i sin ( 6.15 ) . Hence, 13
45. 5 cos (180 ) + i sin (180 ) = 5 [ −1 + 0i ] = −5
2 2 46. 2 cos (135 ) + i sin (135 ) = 2 − + i = − 2 + 2i 2 2
2 2 i = 47. 2 cos ( 315 ) + i sin ( 315 ) = 2 − 2 2
2 − 2i
48. 3 cos ( 270 ) + i sin ( 270 ) = 3 [ 0 − i ] = −3i
1 3 49. −4 cos ( 60 ) + i sin ( 60 ) = −4 + i = −2 − 2 3i 2 2 3 1 50. −4 cos ( 210 ) + i sin ( 210 ) = −4 − − i = 2 3 + 2i 2 2 51.
3 1 3 3 i 3 cos (150 ) + i sin (150 ) = 3 − + i = − + 2 2 2 2
52.
3 1 3 3 i 3 cos ( 330 ) + i sin ( 330 ) = 3 − i = − 2 2 2 2
53.
2 2 π π 2 cos + i sin = 2 i = 1+ i + 2 4 4 2
820
Section 8.2
3 1 5π 5π 54. 2 cos i sin 2 + = − + i = − 3 + i 6 6 2 2
π π 5 5 3 55. 5 cos + i sin = + i 3 3 2 2 57.
3 7π 7π 3 3 3 + i sin − i cos = − 2 6 6 4 4
5π 5π + i sin = 5 − 5i 3 59. 10 cos 3 3
7π 7π 56. 8 cos + i sin = 4 2 − 4i 2 4 4 58.
9 3π 3π 9 + i sin = 0 − i cos 5 5 2 2
60. 15 ( cos π + i sin π ) = −15 + 0i
61. 5 cos ( 295 ) + i sin ( 295 ) ≈ 2.1131 − 4.5315i 62. 4 cos ( 35 ) + i sin ( 35 ) ≈ 3.2766 + 2.2943i 63. 3 cos (100 ) + i sin (100 ) ≈ −0.5209 + 2.9544i 64. 6 cos ( 250 ) + i sin ( 250 ) ≈ −2.0521 − 5.6382i 65. −7 cos (140 ) + i sin (140 ) ≈ 5.3623 − 4.4995i 66. −5 cos ( 320 ) + i sin ( 320 ) ≈ −3.8302 + 3.2139i
11π 11π 67. 3 cos + i sin ≈ −2.8978 + 0.7765i 12 12 4π 4π 68. 2 cos + i sin = −0.4450 + 1.9499i 7 7 3π 3π 69. −2 cos + i sin ≈ 0.6180 − 1.9021i 5 5
821
Chapter 8
15π 15π 70. −4 cos + i sin ≈ 1.6617 + 3.6385i 11 11
π π 71. 6 cos + i sin ≈ 5.54 + 2.30i 8 8 73. Consider the following diagram:
5π 5π 72. 14 cos + i sin ≈ 3.62 + 13.5i 12 12 Force A = 100 ( cos 0 + i sin 0 ) = 100
Force B = 120 ( cos30 + i sin30 )
3 1 = 120 + i = 60 3 + i 2 2 Resultant Force = Force A + Force B = 100 + 60 3 + i = 100 + 60 3 + 60i
(
(
)
) (
)
So, the magnitude of the resultant force =
(100 + 60 3 ) + 60 ≈ 212.6 lbs. 2
74. Consider the following diagram:
2
Force A = 40 ( cos 0 + i sin 0 ) = 40
Force B = 50 ( cos 45 + i sin 45 )
2 2 = 50 + i = 25 2 + 2i 2 2 Resultant Force = Force A + Force B = 40 + 25 2 + 2i = 40 + 25 2 + 25 2i
(
(
) (
)
)
So, the magnitude of the resultant force =
( 40 + 25 2 ) + ( 25 2 ) ≈ 83.2 lbs. 2
822
2
Section 8.2 75. Consider the following diagram:
Force A = 80 ( cos 0 + i sin 0 ) = 80
Force B = 150 ( cos30 + i sin30 )
3 1 = 150 + i = 75 3 + i 2 2 Resultant Force = Force A + Force B = 80 + 75 3 + i = 80 + 75 3 + 75i
(
(
) (
)
)
So, the resultant angle = tan θ =
75 80 + 75 3
75 so that θ = tan −1 ≈ 19.7 . 80 + 75 3
76. Consider the following diagram:
Force A = 20 ( cos 0 + i sin 0 ) = 20
Force B = 60 ( cos 60 + i sin 60 )
1 3 = 60 + i = 30 1 + 3i 2 2 Resultant Force = Force A + Force B = 20 + 30 1 + 3i = 50 + 30 3i
(
(
)
)
So, the resultant angle = tan θ =
30 3 50
30 3 so that θ = tan−1 ≈ 46.1 . 50 77. Plane: u = 300 ( cos165 + i sin165 ) Wind: v = 30 ( cos 270 + i sin 270 ) Resultant: u + v ≈ −289.78 + 47.65i So, u + v ≈ 294 mph and θ = tan−1 ( −47.65 289.78 ) ≈ 171 . So,
u + v ≈ 294 ( cos171 + i sin171 ) .
78. Plane: u = 150 ( cos10 + i sin10 ) Wind: v = 20 ( cos 90 + i sin 90 ) Resultant: u + v ≈ 147.72 + 46.05i So, u + v ≈ 155 mph and 46.05 θ = tan−1 ( 147.72 ) ≈ 17.3. So,
u + v ≈ 155 ( cos17.3 + i sin17.3 ) .
823
Chapter 8 79. Boat: u = 15 ( cos140 + i sin140 ) Current: v = 5 ( cos180 + i sin180 ) Resultant: u + v ≈ −16.49 + 9.64i So, u + v ≈ 19.10 mph and
80. Boat: u = 22 ( cos310 + i sin310 ) Current: v = 9 ( cos 270 + i sin 270 ) Resultant: u + v ≈ 14.14 − 25.85i So, u + v ≈ 29.5 mph and
θ = tan−1 ( −9.64 16.49 ) ≈ −30.31 + 180 ≈ 150 . So, u + v ≈ 19.10 ( cos150 + i sin150 ) .
25.85 θ = tan−1 ( −14.14 ) ≈ −61.32 , but we use 299 as the angle with corresponding positive measure. So, u + v ≈ 29.5 ( cos 299 + i sin 299 ) .
81. The point is in QIII, not QI. As 8 such, you should add 180 to tan −1 . 3
82. The point is in QII, not QIV. As such, 8 you should add 180 to tan −1 − . 3
83. True. All points that lie along the 84. True. All points that lie along the y-axis x-axis are of the form a + 0i. So, they are of the form 0 + bi. So, they must be must be real numbers. (purely) imaginary numbers. 85. True. Let z = a + bi . Then, z = a − bi . Observe that z = a + b 2
2
and z = a 2 + ( −b ) = a 2 + b 2 . 2
86. False. Consider, for example, z = i. Then, z = −i. Note that the argument of z is 90 , while the argument of z is 270.
87. The argument of z = a, where a is a positive real number, is 0 since it lies on the positive x-axis.
88. The argument of z = bi , where b is a positive real number, is 90 since it lies on the positive y-axis.
89. The modulus of z = bi is
90. The modulus of z = a is
b = b. 2
824
a2 = a .
Section 8.2 91. In order to express a − 2a i , a > 0, in polar form, observe that x = a, y = −2a, so that the point is in QIV. Now, r = x2 + y2 =
( a ) + ( −2 a ) = a 5 2
2
y −2a = = −2, so that θ = tan −1 ( −2 ) ≈ −63.43. x a Since 0 ≤ θ < 360 , we must use the reference angle 296.6. tan θ =
Hence, a − 2ai ≈ a 5 cos ( 296.6 ) + i sin ( 296.6 ) . 92. In order to express −3a − 4a i , a > 0, in polar form, observe that x = −3a, y = −4a, so that the point is in QIII. Now, r = x2 + y2 =
( −3a ) + ( −4a ) = 5a 2
2
y −4 a 4 4 = = , so that θ = 180 + tan −1 ≈ 233.1. x −3a 3 3 4 (Remember to add 180 to tan −1 since the angle has terminal side in QIII.) 3 tan θ =
Hence, −3a − 4ai ≈ 5a cos ( 233.1 ) + i sin ( 233.1 ) . 93. We use the addition formulae to simplify first:
( ) + ( )( ) = sin = sin ( − ) = sin cos − cos sin = ( ) ( ) − ( ) ( ) = π π So, 4 cos + i sin = ( 6 + 2 ) + ( 6 − 2 ) i 12 12 cos 12π = cos ( π3 − π4 ) = cos π3 cos π4 + sin π3 sin π4 = ( 21 ) π
π
π
π
π
π
π
12
3
4
3
4
3
4
3 2
2 2
3 2
2 2
1 2
2 2
2 2
94. We use the half-angle formulae to simplify first:
cos 58π = cos sin 58π = sin
( ) 5π 4
2
( ) 5π 4
2
=−
=
1 + cos 54π 1 − 22 2− 2 =− =− 2 2 2
1 − cos 54π 1 + 22 2+ 2 = = 2 2 2
5π 5π + i sin = −2 2 − 2 + 2 2 + 2 i. So, 4 cos 8 8
825
2+ 6 4 6− 2 4
Chapter 8 95. Observe that 3i ( 2 + 4i )( 3 − 2i ) = 3i ( 6 + 12i − 4i + 8 ) = 3i (14 + 8i ) = −24 + 42i.
So, r = ( −24)2 + 422 = 6 65 and θ = tan −1 ( −4224 ) ≈ −1.05 + π ≈ 2.09 (since in QII). Hence, the result is 6 65 ( cos 2.09 + i sin 2.09 ) . 96. Observe that −2i 3 (1 + 4i )( 2 − 5i ) = 2i (1 + 4i )( 2 − 5i ) = 2i ( 22 + 3i ) = −6 + 44i .
So, r = ( −6)2 + 442 = 2 493 and θ = tan −1 ( − 446 ) ≈ −1.43 + π ≈ 1.71 (since in QII). Hence, the result is 2 493 ( cos1.71 + i sin1.71) . 97. Let z = 1 + 2i . Observe that z 0 = (1 + 2i ) = 1 + 0i 0
−10
z1 = 1 + 2i z 2 = (1 + 2i ) = −3 + 4i 2
10
20
30
40
−10
z 3 = (1 + 2i ) = −11 − 2i 3
−20
z = (1 + 2i ) = −7 − 24i 4
4
−30
z 5 = (1 + 2i ) = 41 − 38i 5
−40
98. Let z = −1 + i . Observe that
4
z 0 = ( −1 + i ) = 1 + 0i
3
z 1 = −1 + i
2
0
z 2 = ( −1 + i ) = 0 − 2i 2
1
z 3 = ( −1 + i ) = 2 + 2i 3
−4
−3
−2
−1
1
−1
z 4 = ( −1 + i ) = −4 + 0i 4
−2
z 5 = ( −1 + i ) = 4 − 4i 5
−3
−4
826
2
3
4
5
Section 8.2 99. From the calculator, we obtain: abs (1 + i ) ≈ 1.41421 (actually =
2)
angle (1 + i ) = 45
100. From the calculator, we obtain: abs (1 − i ) ≈ 1.41421 (actually =
2)
angle (1 − i ) = 315
So, 1 + i ≈ 1.4142 ( cos 45 + i sin 45 ) .
So, 1 − i ≈ 1.4142 ( cos315 + i sin315 ) .
101. In order to express 2 + i in polar form, observe that x = 2, y = 1 , so that the point is in QI. Now, r = x2 + y2 =
tan θ =
( 2 ) + (1) = 5 2
2
y 1 1 = , so that θ = tan −1 ≈ 26.57. x 2 2
Hence, 2 + i ≈ 5 cos ( 26.57 ) + i sin ( 26.57 ) . 2 2 3 2 3 2 + i = + i 102. 3 ( cos 45 + i sin 45 ) = 3 2 2 2 2
827
Section 8.3 Solutions -----------------------------------------------------------------------------------1. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 4 )( 3 ) cos ( 40 + 80 ) + i sin ( 40 + 80 )
1 3 i = −6 + 6 3i = 12 cos (120 ) + i sin (120 ) = 12 − + 2 2 2. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 2 )( 5 ) cos (100 + 50 ) + i sin (100 + 50 )
3 1 = 10 cos (150 ) + i sin (150 ) = 10 − + i = −5 3 + 5i 2 2 3. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 4 )( 2 ) cos ( 80 + 145 ) + i sin ( 80 + 145 )
2 2 = 8 cos ( 225 ) + i sin ( 225 ) = 8 − − i = −4 2 − 4 2i 2 2 4. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 3 )( 4 ) cos (130 + 170 ) + i sin (130 + 170 )
1 3 = 12 cos ( 300 ) + i sin ( 300 ) = 12 − i = 6 − 6 3i 2 2 5. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 = ( 2 )( 4 ) cos (10 + 80 ) + i sin (10 + 80 ) = 8 cos ( 90 ) + i sin ( 90 ) = 8 [ 0 + i ] = 0 + 8i
828
Section 8.3
6. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 = ( 3 )( 5 ) cos (190 + 80 ) + i sin (190 + 80 ) = 15 cos ( 270 ) + i sin ( 270 ) = 15 [ 0 − i ] = 0 − 15i
7. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 6 )( 8 ) cos ( 20 + 10 ) + i sin ( 20 + 10 ) = 48 cos ( 30 ) + i sin ( 30 ) ≈ 24 3 + 24i 8. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 5 )( 2 ) cos ( 200 + 40 ) + i sin ( 200 + 40 ) = 10 cos ( 240 ) + i sin ( 240 ) = 10 − 12 − 23 i = −5 − 5 3i 9. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 = ( 12 ) ( 29 ) cos ( 280 + 50 ) + i sin ( 280 + 50 )
= 19 cos ( 330 ) + i sin ( 330 ) = 19 23 − 12 i = 183 − 181 i
10. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 65 )( 125 ) cos (15 + 195 ) + i sin (15 + 195 ) = 2 cos ( 210 ) + i sin ( 210 ) = 2 − 23 − 12 i = − 3 − i 11. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 =
( 3 )( 27 ) cos 12π + π6 + i sin 12π + π6
2 2 9 2 9 2 π π i = i = 9 cos + i sin = 9 + + 2 2 2 2 4 4
829
Chapter 8
12. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 =
( 5 )( 5 ) cos 15π + 415π + i sin 15π + 415π
1 3 5 5 3 π π i = + i = 5 cos + i sin = 5 + 2 3 3 2 2 2 13. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields 3π π 3π π z1z2 = ( 4 )( 3 ) cos + + i sin + 8 8 8 8 π π = 12 cos + i sin = 12 [ 0 + i ] = 0 + 12i 2 2
14. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
2π π 2π π + + i sin + z1z2 = ( 6 )( 5 ) cos 9 9 9 9 1 3 π π = 30 cos + i sin = 30 + i = 15 + 15 3i 2 2 3 3 15. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields 4π π 4π π + + i sin + z1z2 = ( 9 )(1) cos 3 3 3 3 5π 5π 3 9 3 9 1 = 9 cos + i sin = 9 2 − 2 i = 2 − 2 i 3 3
16. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
3π 7π 3π 7π + + z1z2 = (13 )( 4 ) cos + i sin 10 10 10 10 = 52 cos (π ) + i sin (π ) = 52 [ −1 + 0i ] = −52 + 0i
830
Section 8.3
17. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields 7π π 7π π + + i sin + z1z2 = ( 3 )( 5 ) cos 12 4 12 4 10π 10π 3 15 3 15 1 = 15 cos + i sin = 15 − 2 + 2 i = − 2 + 2 i 12 12
18. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields π 3π π 3π z1z2 = (18 )( 2 ) cos + + i sin + 3 2 3 2 11π 11π 3 1 = 36 cos + i sin = 36 2 − 2 i = 18 3 − 18i 6 6
19. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 =
( )( ) cos π2 + π4 + i sin π2 + π4 7 2
7 4
3π 3π = 78 cos + i sin = 78 − 22 + 22 i = − 7162 + 7162 i 4 4
20. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 =
( ) ( ) cos 87π + 67π + i sin 87π + 67π 3 3
2 5
= 156 cos ( 2π ) + i sin ( 2π ) = 156 [1 + 0i ] = 156 + 0i
21. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 6 = cos (100 − 40 ) + i sin (100 − 40 ) z2 2
1 3 3 3 3 = 3 cos ( 60 ) + i sin ( 60 ) = 3 + i = + i 2 2 2 2
831
Chapter 8
22. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 8 = cos ( 80 − 35 ) + i sin ( 80 − 35 ) z2 2 2 2 = 4 cos ( 45 ) + i sin ( 45 ) = 4 + i = 2 2 + 2 2i 2 2
23. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 10 = cos ( 200 − 65 ) + i sin ( 200 − 65 ) z2 5 2 2 = 2 cos (135 ) + i sin (135 ) = 2 − + i = − 2 + 2i 2 2
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2 z1 4 = cos ( 280 − 55 ) + i sin ( 280 − 55 ) z2 4
24. Using the formula
= cos ( 225 ) + i sin ( 225 ) = −
25. Using the formula
2 2 − i 2 2
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 12 cos ( 350 − 80 ) + i sin ( 350 − 80 ) = z2 3 = 2 cos ( 270 ) + i sin ( 270 ) = 0 − 2i
832
Section 8.3
26. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 40 cos (110 − 20 ) + i sin (110 − 20 ) = z2 10 = 2 cos ( 90 ) + i sin ( 90 ) = 0 + 2i
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2 z1 2 = cos ( 213 − 33 ) + i sin ( 213 − 33 ) z2 4
27. Using the formula
= 12 cos (180 ) + i sin (180 ) = − 12 + 0i 28. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 12 = cos ( 315 − 15 ) + i sin ( 315 − 15 ) z2 3 = 4 cos ( 300 ) + i sin ( 300 ) = 4 12 − 23 i = 2 − 2 3i
29. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 53 = cos ( 295 − 55 ) + i sin ( 295 − 55 ) z2 104 = 32 cos ( 240 ) + i sin ( 240 ) = 32 − 21 − 23 i = − 34 − 3 43 i
30. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 23 = cos ( 355 − 235 ) + i sin ( 355 − 235 ) z2 89 = 34 cos (120 ) + i sin (120 ) = 34 − 21 + 23 i = − 38 + 3 8 3 i
833
Chapter 8
31. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 9 5π π 5π π = cos − + i sin − z2 3 12 12 12 12
1 3 3 3 3 π π i = 3 cos + i sin = 3 + i = + 2 3 3 2 2 2 32. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 8 5π 3π 5π 3π = cos − − + i sin 8 8 z2 4 8 8
2 2 π π = 2 cos + i sin = 2 + i = 2 2 4 4 33. Using the formula
2 + 2i
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 45 22π 2π 22π 2π = − − cos + i sin 9 z2 15 15 15 15
1 3 5 5 3 4π 4π = 5 cos i = − − i + i sin = 5 − − 2 2 2 2 3 3 34. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 22 11π 5π 11π 5π π π = − − cos + i sin = 2 cos + i sin z2 11 3 18 18 18 18 3
1 3 = 2 + i = 1 + 3i 2 2
834
Section 8.3
35. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 25 13π 7π 13π 7π = − − cos + i sin 9 9 z2 5 9 9
1 3 5 5 3 6π 6π = 5 cos i = − + i + i sin = 5 − + 2 2 9 9 2 2 36. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 30 3π π 3π π = cos − + i sin − z2 15 2 6 2 6
1 3 8π 8π = 2 cos + i sin = 2 − − i = −1 − 3i 2 2 6 6 37. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 12 17π π 17π π = 3 cos − + i sin − z2 8 12 4 12 4
2 3 2 14π 14π 4 3 1 = 34 cos − i = − − i + i sin = 3 − 2 2 3 3 12 12 38. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 125 61π 31π 61π 31π = 1 cos − − + i sin 36 36 z2 8 36 36
5 3 5 30π 30π 10 3 1 = 103 cos + i = − + i + i sin = 3 − 2 2 3 3 36 36
835
Chapter 8
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
39. Using the formula
z1 152 35π 5π 35π 5π = 25 cos − − + i sin z2 18 18 18 18 4
3 3 3 3 30π 30π 6 1 i = − i = 65 cos + i sin = 5 − 5 18 18 2 2 5 z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
40. Using the formula
z1 127 13π π 13π π = 14 cos − + i sin − z2 3 12 12 12 12 1 = 18 cos (π ) + i sin (π ) = 18 [ −1 + 0i ] = − + 0i 8
41. In order to express ( −1 + i ) in rectangular form, follow these steps: 5
Step 1: Write −1 + i in polar form. Since x = −1, y = 1 , the point is in QII. So, r = x2 + y2 =
( −1) + (1) = 2 2
2
π 3π y 1 = = −1, so that θ = π + tan −1 ( −1) = π − = . x −1 4 4 (Remember to add π to tan −1 ( −1) since the angle has terminal side in QII.) tan θ =
3π 3π Hence, −1 + i = 2 cos + i sin . 4 4 Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( −1 + i ) = ( 2 ) cos 5 ⋅ 5
5
2 3π 2 3π − i = 4 − 4i + i sin 5 ⋅ = 4 2 4 2 4 2
836
Section 8.3
42. In order to express (1 − i ) in rectangular form, follow these steps: 4
Step 1: Write 1 − i in polar form. Since x = 1, y = −1, the point is in QIV. So,
(1) + ( −1) = 2 2
r = x2 + y2 =
2
π y −1 = = −1, so that θ = tan −1 ( −1) = − . 4 x 1 7π Since 0 ≤ θ < 2π , we must use the reference angle . 4 7π 7π Hence, 1 − i = 2 cos + i sin . 4 4 tan θ =
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
(1 − i ) = ( 2 ) cos 4 ⋅ 4
4
7π 7π + i sin 4 ⋅ = 4 cos ( 7π ) + i sin ( 7π ) 4 4
= 4 [ −1 + 0i ] = −4 + 0i
(
43. In order to express − 3 + i
) in rectangular form, follow these steps: 6
Step 1: Write − 3 + i in polar form. Since x = − 3, y = 1, the point is in QII. So, r = x 2 + y 2 = tan θ =
( − 3 ) + (1) = 2 2
2
y −3 1 π 5π 1 , so that θ = π + tan −1 − . =π − = = =− x 6 6 3 3 3
1 (Remember to add π to tan −1 − since the angle has terminal side in QII.) 3 5π 5π Hence, − 3 + i = 2 cos + i sin . 6 6 Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( − 3 + i ) = ( 2 ) cos 6 ⋅ 56π + i sin 6 ⋅ 56π = 64 cos (5π ) + i sin (5π ) 6
6
= 64 [ −1 + 0i ] = −64 + 0i
837
Chapter 8
44. In order to express
( 3 − i ) in rectangular form, follow these steps: 8
3 − i in polar form.
Step 1: Write
Since x = 3, y = −1 , the point is in QIV. So, r = x 2 + y 2 =
( 3 ) + ( −1) = 2 2
2
y −1 π 1 =− . = , so that θ = tan −1 − x 6 3 3 11π Since 0 ≤ θ < 2π , we must use the reference angle . 6 11π 11π Hence, 3 − i = 2 cos + i sin . 6 6 tan θ =
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( 3 − i ) = 2 cos 8 ⋅ 116π + i sin 8 ⋅ 116π 8
8
88π 44π 2 = = 14π + π and so, 6 3 3 2 1 2 88π 2 88π cos sin = cos 14π + π = cos π = − = sin 14π + π 3 3 2 6 3 6
In order to simplify this further, observe that
3 2 = sin π = 3 2
Thus,
( 3 − i ) = 2 − 12 + 23 i = −128 + 128 3i . 8
8
838
Section 8.3
(
45. In order to express 1 − 3i
) in rectangular form, follow these steps: 4
Step 1: Write 1 − 3i in polar form. Since x = 1, y = − 3 , the point is in QIV. So, r = x2 + y2 =
(1) + ( − 3 ) = 2 2
2
3 π , so that θ = tan −1 − = − . 3 2 1 5π Since 0 ≤ θ < 2π , we must use the reference angle . 3 5π 5π Hence, 1 − 3i = 2 cos + i sin . 3 3 r = 1, θ =
π
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
(1 − 3i ) = 2 cos 4 ⋅ 53π + i sin 4 ⋅ 53π 4
4
20π 2 = 6π + π and so, 3 3 2 1 20π 2 cos = cos 6π + π = cos π = − 2 3 3 3
In order to simplify this further, observe that
2 3 20π 2 sin = sin 6π + π = sin π = 3 3 3 2 4 1 3 i = −8 + 8 3i . Thus, 1 − 3i = 2 4 − + 2 2
(
)
839
Chapter 8
(
46. In order to express −1 + 3i
) in rectangular form, follow these steps: 5
Step 1: Write −1 + 3i in polar form. Since x = −1, y = 3 , the point is in QII. So, r = x 2 + y2 =
( −1) + ( 3 ) = 2 2
2
y − 3 π 2π = , so that θ = π + tan −1 − 3 = π − = . x 1 3 3 (Remember to add π to tan −1 − 3 since the angle has terminal side in QII.)
(
tan θ =
(
)
)
2π 2π Hence, −1 + 3i = 2 cos + i sin . 3 3 Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( −1 + 3i ) = ( 2 ) cos 5 ⋅ 23π + i sin 5 ⋅ 23π = 32 cos 103π + i sin 103π 5
5
1 3 i = −16 − 16 3i = 32 − − 2 2 47. In order to express ( 4 − 4i ) in rectangular form, follow these steps: 8
Step 1: Write 4 − 4i in polar form. Since x = 4, y = −4 , the point is in QIV. So, r = x2 + y2 =
( 4 ) + ( −4 ) = 4 2 2
2
y −4 π = = −1 , so that θ = tan −1 ( −1) = − . 4 x 4 7π Since 0 ≤ θ < 2π , we must use the reference angle . 4 7π 7π Hence, 4 − 4i = 4 2 cos + i sin . 4 4 tan θ =
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( 4 − 4i ) = ( 4 2 ) cos 8 ⋅ 8
8
7π 7π + i sin 8 ⋅ = 1,048,576 cos (14π ) + i sin (14π ) 4 4
840
Section 8.3
In order to simplify this further, observe that 14π = 7 ( 2π ) and so, cos (14π ) = 1 sin (14π ) = 0
Thus, ( 4 − 4i ) = 1,048,576 [1 + 0i ] = 1,048,576 + 0i . 8
48. In order to express ( −3 + 3i ) in rectangular form, follow these steps: 10
Step 1: Write −3 + 3i in polar form. Since x = −3, y = 3 , the point is in QII. So, r = x 2 + y2 =
( −3) + (3 ) = 3 2 2
2
y 3 π 3π = = −1 , so that θ = π + tan −1 ( −1) = π − = . x −3 4 4 (Remember to add π to tan −1 ( −1) since the angle has terminal side in QII.) tan θ =
3π 3π Hence, −3 + 3i = 3 2 cos + i sin . 4 4 Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
3π 3π 15π 15π + i sin 10 ⋅ = 1,889,568 cos + i sin 4 4 2 2 15π 3 = 6π + π and so, In order to simplify this further, observe that 2 2 3 15π 3 cos = cos 6π + π = cos π = 0 2 2 2 3 15π 3 sin = sin 6π + π = sin π = −1 2 2 2
( −3 + 3i ) = (3 2 ) cos 10 ⋅ 10
10
Thus, ( −3 + 3i ) = 1,889,568 [ 0 − i ] = 0 − 1,889,568i 10
(
49. In order to express 4 3 + 4i
) in rectangular form, follow these steps: 7
Step 1: Write 4 3 + 4i in polar form. Since x = 4 3, y = 4 , the point is in QI. So, r = x 2 + y2 = tan θ =
(4 3 ) + (4) = 8 2
2
y 4 1 1 π = = , so that θ = tan −1 = . x 4 3 3 3 6
841
Chapter 8
π π Hence, 4 3 + 4i = 8 cos + i sin . 6 6 Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( 4 3 + 4i ) = 8 cos 7 ⋅ π6 + i sin 7 ⋅ π6 = 2,097,152 − 23 − 21 i 7
7
= −1,048,576 3 − 1,048,576i
(
50. In order to express −5 + 5 3i
) in rectangular form, follow these steps: 7
Step 1: Write −5 + 5 3i in polar form. Since x = −5, y = 5 3 , the point is in QII. So, r = x2 + y2 =
( −5 ) + (5 3 ) = 10 2
2
y 5 3 π 2π = = − 3 , so that θ = π + tan −1 − 3 = π − = . 3 3 x −5 (Remember to add π to tan −1 − 3 since the angle has terminal side in QII.)
(
tan θ =
(
)
)
2π 2π Hence, −5 + 5 3i = 10 cos + i sin . 3 3 Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( −5 + 5 3i ) = 10 cos 7 ⋅ 23π + i sin 7 ⋅ 23π 7
7
14π 2 = 4π + π and so, 3 3 2 1 14π 2 cos = cos 4π + π = cos π = − 2 3 3 3
In order to simplify this further, observe that
2 3 14π 2 sin = sin 4π + π = sin π = 3 3 3 2 7 1 3 i = −5,000,000 + 5,000,000 3i Thus, −5 + 5 3i = 107 − + 2 2
(
)
842
Section 8.3
51. Given z = 2 − 2 3i , in order to compute z 2 , follow these steps: Step 1: Write 2 − 2 3i in polar form. 1
Since x = 2, y = −2 3 , the point is in QIV. So, r = x2 + y2 =
( 2 ) + ( −2 3 ) = 4 2
2
y −2 3 = = − 3 , so that θ = tan −1 − 3 = −60 . x 2 Since 0 ≤ θ < 360 , we must use the reference angle 300 . Hence, z = 2 − 2 3i = 4 cos ( 300 ) + i sin ( 300 ) .
(
tan θ =
)
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
n
1
n
300 360 k 300 360 k z = 4 cos + + + i sin , k = 0, 1 2 2 2 2 1
2
= 2 cos (150 ) + i sin (150 ) , 2 cos ( 330 ) + i sin ( 330 ) Step 3: Plot the roots in the complex plane:
843
Chapter 8
52. Given z = 2 + 2 3i , in order to compute z 2 , follow these steps: Step 1: Write 2 + 2 3i in polar form. 1
Since x = 2, y = 2 3 , the point is in QI. So, r = x 2 + y2 =
(2) + (2 3 ) = 4 2
2
y 2 3 = = 3 , so that θ = tan −1 x 2 Hence, z = 2 + 2 3i = 4 cos ( 60 ) + i sin ( 60 ) . tan θ =
( 3 ) = 60 .
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
n
1
n
60 360 k 60 360 k z = 4 cos + + + i sin , k = 0, 1 2 2 2 2 1
2
= 2 cos ( 30 ) + i sin ( 30 ) , 2 cos ( 210 ) + i sin ( 210 ) Step 3: Plot the roots in the complex plane:
844
Section 8.3
53. Given z = 18 − 18i , in order to compute z 2 , follow these steps: Step 1: Write z = 18 − 18i in polar form. 1
Since x = 18, y = − 18 , the point is in QIV. So, r = x2 + y2 =
( 18 ) + ( − 18 ) = 6 2
2
y − 18 = = −1 , so that θ = tan −1 ( −1) = −45 . x 18 Since 0 ≤ θ < 360 , we must use the reference angle 315 . Hence, z = 18 − 18i = 6 cos ( 315 ) + i sin ( 315 ) . θ 360 k θ 360 k 1 1 i sin + Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n 315 360 k 315 360 k 1 z 2 = 6 cos i sin + + + , k = 0, 1 2 2 2 2 tan θ =
6 cos (157.5 ) + i sin (157.5 ) , 6 cos ( 337.5 ) + i sin ( 337.5 ) Step 3: Plot the roots in the complex plane: =
845
Chapter 8
54. Given z = − 2 + 2i , in order to compute z 2 , follow these steps: Step 1: Write z = − 2 + 2i in polar form. Since x = − 2, y = 2 , the point is in QII. So, 1
r = x2 + y2 =
(− 2 ) + ( 2 ) = 2 2
2
y 2 = = −1 , so that θ = 180 + tan −1 ( −1) = 135 . x − 2 (Remember to add 180 to tan −1 ( −1) since the angle has terminal side in QII.)
tan θ =
Hence, z = − 2 + 2i = 2 cos (135 ) + i sin (135 ) .
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
n
1
n
135 360 k 135 360 k z = 2 cos + + + i sin , k = 0, 1 2 2 2 2 1
2
2 cos ( 67.5 ) + i sin ( 67.5 ) , 2 cos ( 247.5 ) + i sin ( 247.5 ) Step 3: Plot the roots in the complex plane: =
846
Section 8.3
55. Given z = 4 + 4 3i , in order to compute z 3 , follow these steps: Step 1: Write 4 + 4 3i in polar form. 1
Since x = 4, y = 4 3 , the point is in QI. So, r = x 2 + y2 =
(4) + (4 3 ) = 8 2
2
y 4 3 = = 3 , so that θ = tan −1 x 4 Hence, z = 4 + 4 3i = 8 cos ( 60 ) + i sin ( 60 ) . tan θ =
( 3 ) = 60 .
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
n
1
n
60 360 k 60 360 k z = 8 cos + + + i sin , k = 0, 1, 2 3 3 3 3 1 3
3
= 2 cos ( 20 ) + i sin ( 20 ) , 2 cos (140 ) + i sin (140 ) , 2 cos ( 260 ) + i sin ( 260 ) Step 3: Plot the roots in the complex plane:
847
Chapter 8
27 27 3 1 + i , in order to compute z 3 , follow these steps: 2 2 27 27 3 i in polar form. Step 1: Write z = − + 2 2 27 27 3 , the point is in QII. So, Since x = − , y = 2 2 56. Given z = −
2
2 27 27 3 r = x + y = − + = 27 2 2 2
2
27 3 y 2 = − 3 , so that θ = 180 + tan −1 − 3 = 120 . = 27 x − 2 (Remember to add 180 to tan −1 − 3 since the angle has terminal side in QII.)
(
tan θ =
(
)
)
27 27 3 + i = 27 cos (120 ) + i sin (120 ) . 2 2 θ 360 k θ 360 k 1 1 n n Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n Hence, z = −
120 360 k 120 360 k + + z = 27 cos + i sin , k = 0, 1, 2 3 3 3 3 1 3
3
= 3 cos ( 40 ) + i sin ( 40 ) , 3 cos (160 ) + i sin (160 ) , 3 cos ( 280 ) + i sin ( 280 ) Step 3: Plot the roots in the complex plane:
848
Section 8.3
57. Given z = 3 − i , in order to compute z 3 , follow these steps: Step 1: Write z = 3 − i in polar form. 1
Since x = 3, y = −1 , the point is in QIV. So, r = x 2 + y2 =
tan θ =
( 3 ) + ( −1) = 2 2
2
y −1 −1 = −30 . = , so that θ = tan −1 x 3 3
Since 0 ≤ θ < 360 , we must use the reference angle 330 . Hence, z = 3 − i = 2 cos ( 330 ) + i sin ( 330 ) . θ 360 k θ 360 k 1 1 + i sin Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n 330 360 k 330 360 k 1 sin + + + z 3 = 3 2 cos i , k = 0, 1, 2 3 3 3 3 =
3
2 cos (110 ) + i sin (110 ) , 3 2 cos ( 230 ) + i sin ( 230 ) ,
3
2 cos ( 350 ) + i sin ( 350 )
Step 3: Plot the roots in the complex plane:
849
Chapter 8
58. Given z = 4 2 + 4 2i , in order to compute z 3 , follow these steps: Step 1: Write 4 2 + 4 2i in polar form. Since x = 4 2, y = 4 2 , the point is in QI. So, 1
r = x2 + y2 =
tan θ =
(4 2 ) + (4 2 ) = 8 2
2
y 4 2 = = 1 , so that θ = tan −1 (1) = 45 . x 4 2
Hence, z = 4 2 + 4 2i = 8 cos ( 45 ) + i sin ( 45 ) .
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
n
1
n
45 360 k 45 360 k z = 8 cos + + + i sin , k = 0, 1, 2 3 3 3 3 1 3
3
= 2 cos (15 ) + i sin (15 ) , 2 cos (135 ) + i sin (135 ) , 2 cos ( 255 ) + i sin ( 255 ) Step 3: Plot the roots in the complex plane:
850
Section 8.3
59. Given z = 8 2 − 8 2i , in order to compute z 4 , follow these steps: Step 1: Write z = 8 2 − 8 2i in polar form. Since x = 8 2, y = −8 2 , the point is in QIV. So, 1
r = x 2 + y2 =
(8 2 ) + ( −8 2 ) = 16 2
2
y −8 2 = = −1 , so that θ = tan −1 ( −1) = −45 . x 8 2 Since 0 ≤ θ < 360 , we must use the reference angle 315 . Hence, z = 8 2 − 8 2i = 16 cos ( 315 ) + i sin ( 315 ) . θ 360 k θ 360 k 1 1 n n Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 315 360 k 315 360 k 1 + + + z 4 = 4 16 cos i sin , k = 0, 1, 2, 3 4 4 4 4
tan θ =
=
2 cos ( 78.75 ) + i sin ( 78.75 ) , 2 cos (168.75 ) + i sin (168.75 ) , 2 cos ( 258.75 ) + i sin ( 258.75 ) , 2 cos ( 348.75 ) + i sin ( 348.75 )
Step 3: Plot the roots in the complex plane:
851
Chapter 8
60. Given z = − 128 + 128i , in order to compute z 4 , follow these steps: Step 1: Write z = − 128 + 128i in polar form. Since x = − 128, y = 128 , the point is in QII. So, 1
r = x 2 + y2 =
( − 128 ) + ( 128 ) = 16 2
2
y 128 = = −1 , so that θ = 180 + tan −1 ( −1) = 135 . x − 128 (Remember to add 180 to tan −1 ( −1) since the angle has terminal side in QII.)
tan θ =
Hence, z = − 128 + 128i = 16 cos (135 ) + i sin (135 ) .
θ 360 k θ 360 k 1 1 i sin + Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n 135 360 k 135 360 k 1 + + + z 4 = 4 16 cos i sin , k = 0, 1, 2, 3 4 4 4 4 =
2 cos ( 33.75 ) + i sin ( 33.75 ) , 2 cos (123.75 ) + i sin (123.75 ) , 2 cos ( 213.75 ) + i sin ( 213.75 ) , 2 cos ( 303.75 ) + i sin ( 303.75 )
Step 3: Plot the roots in the complex plane:
852
Section 8.3
61. Given z = 10 3 − 10i , in order to compute z 4 , follow these steps: Step 1: Write z = z = 10 3 − 10i in polar form. Since x = 10 3, y = −10 , the point is in QIV. So, 1
r = x 2 + y2 =
tan θ =
(10 3 ) + ( −10 ) = 20 2
2
y −10 −1 −1 = = , so that θ = tan −1 = −30 . x 10 3 3 3
Since 0 ≤ θ < 360 , we must use the reference angle 330 . Hence, z = 10 3 − 10i = 20 cos ( 330 ) + i sin ( 330 ) . θ 360 k θ 360 k 1 1 i sin + Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n 330 360 k 330 360 k 1 z 4 = 4 20 cos + + i + sin , k = 0, 1, 2, 3 4 4 4 4 = 2 5 cos82.5 + i sin82.5 , 2 5 cos172.5 + i sin172.5 , 2 5 cos 262.5 + i sin 262.5 , 2 5 cos352.5 + i sin352.5 Step 3: Plot the roots in the complex plane: 4 3 2 1
−4
−3
−2
−1
1 −1 −2 −3 −4 5
853
2
3
4
5
Chapter 8
62. Given z = −5 − 5 3i , in order to compute z 4 , follow these steps: Step 1: Write z = −5 − 5 3i in polar form. Since x = −5, y = −5 3 , the point is in QIII. So, 1
r = x 2 + y2 =
( −5 ) + ( −5 3 ) = 10 2
2
y −5 3 = = 3 , so that θ = tan −1 x −5 Hence, z = −5 − 5 3i = 10 cos ( 240 ) + i sin ( 240 ) . tan θ =
( 3 ) = 240 .
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
1
n
n
240 360 k 240 360 k z = 10 cos + + + i sin , k = 0, 1, 2, 3 4 4 4 4 1
4
4
= 4 10 cos 60 + i sin 60 , 4 10 cos150 + i sin150 , 4
10 cos 240 + i sin 240 , 4 10 cos330 + i sin330
Step 3: Plot the roots in the complex plane 4 3 2 1
−4
−3
−2
−1
1 −1 −2 −3 −4 5
854
2
3
4
5
Section 8.3
63. We seek all complex numbers x such that x 4 − 16 = 0 . Observe that this equation can be written equivalently as: ( x2 − 4 )( x2 + 4 ) = 0
( x − 2 )( x + 2 ) ( x2 + 4 ) = 0
The solutions are therefore x = ±2, ± 2i . (Note: We could have alternatively used the formula for complex roots, namely θ 360 k θ 360 k 1 1 z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n However, the left-side readily factored in a way that is easy to see what all of the solutions were. As such, factoring was the more expedient route for this problem.) 64. We seek all complex numbers x such that x3 − 8 = 0 , or equivalently x3 = 8 . While it is clear that x = 2 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = 8 in polar form. Since x = 8, y = 0 , the point is on the positive x-axis. So, r = 8, θ = 0 .
Hence, 8 = 8 cos ( 0 ) + i sin ( 0 ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 0 2π k 0 2π k 8 3 = 3 8 cos + + i sin + , k = 0, 1, 2 3 3 3 3
2π 2π 4π 4π = 2 cos ( 0 ) + i sin ( 0 ) , 2 cos + i sin , 2 cos + i sin 3 3 3 3
= 2, − 1 + 3i , − 1 − 3i Step 3: Hence, the complex solutions of x3 − 8 = 0 are 2, − 1 + 3i , − 1 − 3i .
855
Chapter 8
65. We seek all complex numbers x such that x3 + 8 = 0 , or equivalently x3 = −8 . While it is clear that x = −2 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −8 in polar form. Since x = −8, y = 0 , the point is on the negative x-axis. So, r = 8, θ = π .
Hence, −8 = 8 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 π 2π k π 2π k ( −8) 3 = 3 8 cos + + i sin + , k = 0, 1, 2 3 3 3 3
π π 5π 5π = 2 cos + i sin , 2 cos (π ) + i sin (π ) , 2 cos + i sin 3 3 3 3
= −2, 1 − 3i , 1 + 3i Step 3: Hence, the complex solutions of x3 + 8 = 0 are −2, 1 − 3i , 1 + 3i . 66. We seek all complex numbers x such that x3 + 1 = 0 , or equivalently x3 = −1 . While it is clear that x = −1 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −1 in polar form. Since x = −1, y = 0 , the point is on the negative x-axis. So, r = 1, θ = π .
Hence, −1 = 1 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
856
Section 8.3
π
3
( −1) = 3 1 cos 1
3
+
2π k π 2π k + i sin + , k = 0, 1, 2 3 3 3
π π 5π 5π = 1 cos + i sin , 1 cos (π ) + i sin (π ) , 1 cos + i sin 3 3 3 3
= −1,
1 3 1 3 − i, + i 2 2 2 2
Step 3: Hence, the complex solutions of x3 + 1 = 0 are −1,
1 3 1 3 − i, + i . 2 2 2 2
67. We seek all complex numbers x such that x 4 + 16 = 0 , or equivalently x 4 = −16 . We shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −16 in polar form. Since x = −16, y = 0 , the point is on the negative x-axis. So, r = 16, θ = π .
Hence, −16 = 16 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 π 2π k π 2π k ( −16 ) 4 = 4 16 cos + + i sin + , k = 0, 1, 2, 3 4 4 4 4
π π 3π 3π = 2 cos + i sin , 2 cos + i sin , 4 4 4 4 5π 5π 7π 7π 2 cos + i sin , 2 cos + i sin 4 4 4 4
= 2 + 2i , − 2 + 2i , − 2 − 2i ,
2 − 2i
Step 3: Hence, the complex solutions of x + 16 = 0 are 4
2 + 2i , − 2 + 2i , − 2 − 2i ,
857
2 − 2i .
Chapter 8
68. We seek all complex numbers x such that x 6 + 1 = 0 , or equivalently x 6 = −1 . We shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −1 in polar form. Since x = −1, y = 0 , the point is on the negative x-axis. So, r = 1, θ = π .
Hence, −1 = 1 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
π
6
( −1) = 6 1 cos 1
6
+
2π k π 2π k + i sin + , k = 0, 1, 2, 3, 4, 5 6 6 6
π π π π 5π 5π = 1 cos + i sin , 1 cos + i sin , 1 cos + i sin 6 6 2 2 6 6 7π 7π 3π 3π 11π 11π 1 cos + i sin , 1 cos + i sin , 1 cos + i sin 6 6 2 2 6 6 3 1 3 1 3 1 3 1 + i, i, − + i, − − i, − i, − i 2 2 2 2 2 2 2 2 Step 3: Hence, the complex solutions of x6 + 1 = 0 are =
3 1 3 1 3 1 3 1 + i, i, − + i, − − i, − i, − i . 2 2 2 2 2 2 2 2
69. We seek all complex numbers x such that x6 − 1 = 0 , or equivalently x6 = 1 . While it is clear that x = 1 is a solution by inspection, the remaining complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = 1 in polar form. Since x = 1, y = 0 , the point is on the positive x-axis. So, r = 1, θ = 0 .
Hence, 1 = 1 cos ( 0 ) + i sin ( 0 ) .
858
Section 8.3
1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
(1) = 6 1 cos 0 + 1
6
2π k 2π k + i sin 0 + , k = 0, 1, 2, 3, 4, 5 6 6
π π 2π 2π = 1 cos ( 0 ) + i sin ( 0 ) , 1 cos + i sin , 1 cos + i sin 3 3 3 3 4π 4π 5π 5π 1 cos (π ) + i sin (π ) , 1 cos + i sin , 1 cos + i sin 3 3 3 3 1 3 1 3 1 3 1 3 + i, − + i , − 1, − − i, − i 2 2 2 2 2 2 2 2 Step 3: Hence, the complex solutions of x 6 − 1 = 0 are = 1,
1,
1 3 1 3 1 3 1 3 + i, − + i , − 1, − − i, − i . 2 2 2 2 2 2 2 2
70. We seek all complex numbers x such that 4 x 2 + 1 = 0 . Observe that this equation can be written equivalently as: 1 1 1 x2 = − so that x = ± − = ± i 4 4 2 1 1 The solutions are therefore x = − i , i . 2 2 (Note: We could have alternatively used the formula for complex roots, namely θ 360 k θ 360 k 1 1 + z n = r n cos + i sin + , k = 0, 1, ... , n − 1 . n n n n However, the left-side readily factored in a way that is easy to see what all of the solutions were. As such, factoring was the more expedient route for this problem.)
71. We seek all complex numbers x such that x 2 + i = 0 . Observe that this equation can be written equivalently as: x 2 = − i so that x = ± −i Now, in order to express these solutions in rectangular form, we proceed as follows: Step 1: Write z = −i in polar form. 3π . Since x = 0, y = −1 , the point is on the negative y-axis. So, r = 1, θ = 2
859
Chapter 8
3π 3π Hence, −i = 1 cos + i sin . 2 2 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
( −i )
1
2
3π 3π 2π k 2π k 2 2 = 1 cos + + i sin + , k = 0, 1 2 2 2 2 2 2 2 2 3π 3π 7π 7π = 1 cos + i sin , 1 cos + i, − i + i sin = − 2 2 2 2 4 4 4 4
The solutions are therefore x = −
2 2 2 2 + i, − i . 2 2 2 2
72. We seek all complex numbers x such that x 2 − i = 0 . Observe that this equation can be written equivalently as: x 2 = i so that x = ± i Now, in order to express these solutions in rectangular form, we proceed as follows: Step 1: Write z = i in polar form.
Since x = 0, y = 1 , the point is on the positive y-axis. So, r = 1, θ =
π
2
.
π π Hence, i = 1 cos + i sin . 2 2 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
(i )
1
2
π π 2π k 2π k 2 2 = 1 cos + + i sin + , k = 0, 1 2 2 2 2 2 2 2 2 π π 5π 5π = 1 cos + i sin , 1 cos + i, − − i + i sin = 2 2 2 4 4 4 4 2
The solutions are therefore x =
2 2 2 2 + i, − − i . 2 2 2 2
860
Section 8.3
73. We seek all complex numbers x such that x3 + 8i = 0 , or equivalently x3 = −8i . We shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −8i in polar form. Since x = 0, y = −8 , the point is on the negative y-axis. So, r = 8, θ = 32π .
Hence, −8i = 8 cos ( 32π ) + i sin ( 32π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 3π 2π k 3π 2π k + ( −8i ) 3 = 3 8 cos + + i sin , k = 0, 1, 2 3 3 6 6
= 2i , − 3 − i , 3 − i
Step 3: Hence, the complex solutions of x3 + 8i = 0 are 2i , − 3 − i , 3 − i . 74. We seek all complex numbers x such that x3 − 8i = 0 , or equivalently x3 = 8i . We shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = 8i in polar form. Since x = 0, y = 8 , the point is on the positive y-axis. So, r = 8, θ = π2 .
Hence, 8i = 8 cos ( π2 ) + i sin ( π2 ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 π 2π k π 2π k (8i ) 3 = 3 8 cos + + i sin + , k = 0, 1, 2 3 3 6 6
= −2i , − 3 + i , 3 + i
Step 3: Hence, the complex solutions of x3 − 8i = 0 are −2i , − 3 + i , 3 + i .
861
Chapter 8
2 2 1 − i , in order to compute z 5 , follow these steps: 2 2 2 2 − i in polar form. Step 1: Write z = − 2 2 2 2 Since x = − , y=− , the point is in QIII. So, 2 2 75. Given z = −
2
2
2 2 r = x + y = − + − = 1 2 2 2
2
2 y − 2 = = 1 , so that θ = 180 + tan −1 (1) = 225 . x − 2 2 −1 (Remember to add 180 to tan (1) since the angle has terminal side in QIII.) tan θ =
2 2 − i = 1 cos ( 225 ) + i sin ( 225 ) . 2 2 θ 360 k θ 360 k 1 1 + i sin Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n
Hence, z = −
225 360 k 225 360 k 1 + + + z 5 = 5 1 cos i sin , k = 0, 1, 2, 3, 4 5 5 5 5 cos ( 45 ) + i sin ( 45 ) , cos (117 ) + i sin (117 ) , cos (189 ) + i sin (189 ) , = cos ( 261 ) + i sin ( 261 ) , cos ( 333 ) + i sin ( 333 ) Step 3: Plot the roots in the complex plane and connect consecutive roots to form a regular pentagon:
862
Section 8.3
76. Given z = 16i , in order to compute z 4 , follow these steps: Step 1: Write z = 16i in polar form. Since x = 0, y = 16 , the point is on the positive y-axis. So, r = 16, θ = 90 1
Hence, z = 16i = 16 cos ( 90 ) + i sin ( 90 ) .
θ 360 k θ 360 k 1 1 Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 90 360 k 90 360 k 1 + + + z 4 = 4 16 cos i sin , k = 0, 1, 2, 3 4 4 4 4
=
2 cos ( 22.5 ) + i sin ( 22.5 ) , 2 cos (112.5 ) + i sin (112.5 ) , 2 cos ( 202.5 ) + i sin ( 202.5 ) , 2 cos ( 292.5 ) + i sin ( 292.5 )
Step 3: Plot the roots in the complex plane and connect consecutive roots to form a square:
−
863
Chapter 8
1 3 1 − i , in order to compute z 6 , follow these steps: 2 2 1 3 i in polar form. Step 1: Write z = − 2 2 1 3 Since x = , y = − , the point is in QIV. So, 2 2 77. Given z =
2
2 3 1 r = x + y = + − =1 2 2 2
tan θ =
2
3 y − 2 = = − 3 , so that θ = −60 , so we use 300 . 1 x 2
1 3 − i = 1 cos ( 300 ) + i sin ( 300 ) . 2 2 θ 360 k θ 360 k 1 1 + i sin Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n
Hence, z =
300 360 k 300 360 k 1 + + + sin z 6 = 6 1 cos i , k = 0, 1, 2, 3, 4, 5 6 6 6 6
= cos ( 50 + 60 k ) + i sin ( 50 + 60 k ) , k = 0, 1, 2, 3, 4, 5 Step 3: Plot the roots in the complex plane and connect consecutive roots to form a regular hexagon:
864
Section 8.3
78. Given z = 2i , in order to compute z 8 , follow these steps: Step 1: Write z = 2i in polar form. Since x = 0, y = 2 , the point is on the positive y-axis. So, r = 2, θ = 90 . 1
Hence, z = 2i = 2 cos ( 90 ) + i sin ( 90 ) .
θ 360 k θ 360 k 1 1 Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 90 360 k 90 360 k 1 z 8 = 8 2 cos + + i + sin , k = 0, 1, 2, 3, 4, 5, 6, 7 8 8 8 8 90 90 = 8 2 cos + 45 k + i sin + 45 k , k = 0, 1, 2, 3, 4, 5, 6, 7 8 8 Step 3: Plot the roots in the complex plane and connect consecutive roots to form a regular octagon:
79. Reversed order of angles being subtracted.
80. Should use the product formula, namely z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) ,
to multiply. 81. Should use DeMoivre’s formula. In 82. Only found one of the five solutions. 6 general, ( a + b ) ≠ a 6 + b6 .
865
Chapter 8
83. True. From the product formula, we see that z1z2 = rr 1 2 1 2 cos (θ1 + θ 2 ) + 1 2 sin (θ1 + θ 2 ) cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) = rr rr i ,
which is of the form of a complex number. 84. True. From the quotient formula, we see that z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) . z2 r2 85. We proceed as follows: Step 1: Write z = n + n 3i in polar form. Let x = n, y = n 3 . Observe
(
r = n2 + n 3
) = 2n, θ = tan ( ) = π3 . 3
−1
n 3 n
π π Hence, z = n + n 3i = 2 n cos + i sin . 3 3 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 2 π 2π k π 2π k n + n 3i = 2 n cos + + i sin + , k = 0, 1 2 2 6 6
(
)
π π 7π 7π = 2 n cos + i sin , 2 n cos + i sin 6 6 6 6 =
6n 2n 6n 2n + − i, − i 2 2 2 2
866
Section 8.3
86. We proceed as follows: Step 1: Write z = n − n 3i in polar form. Let x = n, y = −n 3 . Observe
(
r = n2 + −n 3
) = 2n, θ = tan ( ) = 53π . 3
−1
−n 3 n
5π 5π Hence, z = n − n 3i = 2 n cos + i sin . 3 3 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
(
n − n 3i
) = 2n cos 56π + 2π2k + i sin 56π + 2π2k , k = 0, 1 1
2
5π 5π 11π 11π = 2n cos + i sin , 2 n cos + i sin 6 6 6 6 = −
6n 2n 6n 2n + i, − i 2 2 2 2
i θ1 87. Observe that z1z2 = ( re )( r2ei θ2 ) = rr1 2e ( 1 2 ) = rr1 2 cos (θ1 + θ2 ) + i sin (θ1 + θ2 ) 1 i θ +θ
88. Observe that
iθ
z1 r1e 1 r1 i (θ1 −θ2 ) r1 = = e = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) z2 r2 eiθ2 r2 r2
89. Observe that z n = ( re i θ ) = r n ei nθ = r n cos ( n θ ) + i sin ( n θ ) n
1 1 1 1 n θ θ i θ 90. Observe that z n = ( rei θ ) = r n e ( n ) = r n cos + i sin . Hence, by n n 1 θ 2 kπ θ 2kπ periodicity (period = 2π ), we have wk = r n cos + , + i sin + n n n n n where k = 0, ... , n − 1
867
Chapter 8
3 1 1 − i , in order to compute z 5 , follow these steps: 2 2 3 1 − i in polar form. Step 1: Write z = 2 2 3 1 Since x = , y = − , the point is in QIV. Also, we have r = 1 and 2 2 1 3 11π y , so that θ = . tan θ = sin θ = − and cos θ = 6 x 2 2 11π 11π Hence, z = cos . + i sin 6 6 1 1 θ 2 kπ θ 2 kπ Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 91. Given z =
11π 2kπ 11π 2kπ 1 + + z 5 = cos + i sin , k = 0, 1, 2, 3, 4 5 5 6(5) 6(5) Step 3: Plot the roots in the complex plane:
868
Section 8.3
2 2 1 + i , in order to compute z 4 , follow these steps: 2 2 2 2 + i in polar form. Step 1: Write z = − 2 2 2 2 Since x = − , y= , the point is in QII. Also, we have r = 1 and 2 2 y 3π tan θ = = −1 , so that θ = . x 4 3π 3π Hence, z = cos + i sin . 4 4 92. Given z = −
1 1 θ 2 kπ θ 2 kπ Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 3π 3π 1 2 kπ 2 kπ + + z 4 = cos + i sin , k = 0, 1, 2, 3 4 4 4(4) 4(4) Step 3: Plot the roots in the complex plane:
869
Chapter 8
1 3 1 93. Given z = − − i , in order to compute z 6 , follow these steps: 2 2 1 3 i in polar form. Step 1: Write z = − − 2 2 1 3 Since x = − , y = − , the point is in QIII. Also, we have r = 1 and 2 2 3 1 4π y . tan θ = sin θ = − and cos θ = − , so that θ = 3 x 2 2 4π 4π Hence, z = cos . + i sin 3 3 1 1 θ 2 kπ θ 2 kπ Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 4π 2kπ 4π 2 kπ + + z 6 = cos + i sin , k = 0, 1, 2, 3, 4, 5 6 6 3(6) 3(6) Step 3: Plot the roots in the complex plane:
870
Section 8.3
94. Given z = −4 + 4i , in order to compute z 5 , follow these steps: Step 1: Write z = −4 + 4i in polar form. Since x = −4, y = 4 , the point is in QII. Also, we have y 3π . r = ( −4)2 + 42 = 4 2 and tan θ = = −1 , so that θ = x 4 3π 3π Hence, z = 4 2 cos + i sin . 4 4 1
1 1 θ 2 kπ θ 2 kπ Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 3π 2 kπ 3π 2 kπ 1 5 + + z 5 = 4 2 cos + i sin , k = 0, 1, 2, 3, 4 5 5 4(5) 4(5) Step 3: Plot the roots in the complex plane:
(
)
871
Section 8.4 Solutions -----------------------------------------------------------------------------------1. – 12. The following graph shows the points in Exercises 1 – 12:
#11 #12
13. In order to convert the point 2,2 3 (expressed in rectangular
(
)
coordinates) to polar coordinates, first observe that x = 2, y = 2 3, so that the point is in QI. Hence, r=
14. In order to convert the point ( 3, −3 ) (expressed in rectangular coordinates) to polar coordinates, first observe that x = 3, y = −3, so that the point is in QIV. Hence, r=
(2) + (2 3 ) = 4 2
2
tan θ =
y 2 3 = 3 so that tan θ = = x 2
θ = tan −1
π
( 3) = 3 .
So, the given point can be expressed in π polar coordinates as 4, . 3
( 3 ) + ( −3 ) = 3 2 2
2
y −3 = = −1 so that x 3
π θ = tan−1 ( −1) = − .
4 Since 0 ≤ θ < 2π , we use the reference 7π angle θ = . 4 So, the given point can be expressed in 7π polar coordinates as 3 2, . 4
872
Section 8.4
15. In order to convert the point −1, − 3 (expressed in rectangular
16. In order to convert the point 6,6 3 (expressed in rectangular
coordinates) to polar coordinates, first observe that x = −1, y = − 3, so that the point is in QIII. Hence,
coordinates) to polar coordinates, first observe that x = 6, y = 6 3, so that the point is in QI. Hence,
(
)
r=
(
)
( −1) + ( − 3 ) = 2
r=
( 6 ) + ( 6 3 ) = 12
tan θ =
y 6 3 = = 3 so that x 6
2
2
y − 3 = = 3 so that x −1 4π θ = π + tan −1 3 = . 3 (Remember to add π to tan −1 3 since tan θ =
( )
( )
the point is in QIII.) So, the given point can be expressed in
2
2
θ = tan−1
( 3 ) = π3 .
So, the given point can be expressed in π polar coordinates as 12, . 3
4π polar coordinates as 2, . 3 17. In order to convert the point ( −4, 4 ) (expressed in rectangular coordinates) to polar coordinates, first observe that x = −4, y = 4, so that the point is in QII. Hence, r=
( −4 ) + ( 4 ) = 4 2 2
2
y −4 = = −1 so that x 4 3π θ = π + tan −1 ( −1) = . 4 −1 (Remember to add π to tan ( −1) since the point is in QII.) So, the given point can be expressed in
tan θ =
(
18. In order to convert the point 0, 2
)
(expressed in rectangular coordinates) to polar coordinates, first observe that x = 0, y = 2, so that the point is on the positive y-axis. Hence, r=
π 2 ( 0 ) + ( 2 ) = 2, θ = 2
2 So, the given point can be expressed in
π polar coordinates as 2, . 2
3π polar coordinates as 4 2, . 4
873
Chapter 8 19. In order to convert the point ( 3,0 )
(expressed in rectangular coordinates) to polar coordinates, first observe that x = 3, y = 0, so that the point is on the positive x-axis. Hence, r=
20. In order to convert the point ( −7, −7 ) (expressed in rectangular coordinates) to polar coordinates, first observe that x = −7, y = −7, so that the point is in QIII. Hence,
2 2 (3) + ( 0 ) = 3, θ = 0
So, the given point can be expressed in polar coordinates as ( 3,0 ) .
r=
( −7 ) + ( −7 ) = 7 2 2
2
y −7 = = 1 so that x −7 5π θ = π + tan −1 (1) = . 4 −1 (Remember to add π to tan (1) since the point is in QIII.) So, the given point can be expressed in
tan θ =
5π polar coordinates as 7 2, . 4 21. In order to convert the point − 3, −1 (expressed in rectangular
22. In order to convert the point 2 3, −2 (expressed in rectangular
coordinates) to polar coordinates, first observe that x = − 3, y = −1, so that the point is in QIII. Hence,
coordinates) to polar coordinates, first observe that x = 2 3, y = −2, so that the point is in QIV. Hence,
(
)
r=
( − 3 ) + ( −1) = 2
tan θ =
y −1 1 = = so that x − 3 3
2
(
2
1 7π . = 3 6
)
r= tan θ =
( 2 3 ) + ( −2 ) = 4 2
2
y −2 1 = =− so that x 2 3 3
1 (Remember to add π to tan −1 3 since the point is in QIII.) So, the given point can be expressed in
1 π =−6. 3 Since 0 ≤ θ < 2π , we use the reference 11π angle θ = . 6 So, the given point can be expressed in
7π polar coordinates as 2, . 6
11π polar coordinates as 4, . 6
θ = π + tan −1
θ = tan −1 −
874
Section 8.4
23. In order to convert the point 3 3 3 (expressed in rectangular − , 2 2 coordinates) to polar coordinates, first 3 3 3 observe that x = − , y = , so that the 2 2 point is in QII. Hence, 2
2 3 3 3 r = − + = 9 =3 2 2
24. In order to convert the point ( 0, −2 )
(expressed in rectangular coordinates) to polar coordinates, first observe that x = 0, y = −2 , so that the point is on the negative y-axis. Hence, 3π 2 2 . r = ( 0 ) + ( −2 ) = 2,θ = 2 So, the given point can be expressed in 3π polar coordinates as 2, . 2
3 3 y 2 = − 3 so that tan θ = = −3 x 2 2π . θ = tan1 − 3 = 3 So, the given point can be expressed in
(
)
2π polar coordinates as 3, . 3 25. In order to convert the point 5π 4, (expressed in polar coordinates) 3 to rectangular coordinates, first observe 5π that r = 4, θ = . Hence, 3 5π 1 x = r cos θ = 4 cos = 4 = 2 3 2 3 5π = − y = r sin θ = 4 sin 4 2 = −2 3 3 So, the given point can be expressed in
(
)
rectangular coordinates as 2, − 2 3 .
3π 26. In order to convert the point 2, 4 (expressed in polar coordinates) to rectangular coordinates, first observe 3π . Hence, that r = 2, θ = 4 2 3π x = r cos θ = 2 cos = 2 − = − 2 4 2 2 3π y = r sin θ = 2 sin = 2 = 2 4 2 So, the given point can be expressed in
(
)
rectangular coordinates as − 2, 2 .
875
Chapter 8
27. In order to convert the point 5π −1, (expressed in polar 6 coordinates) to rectangular coordinates, 5π first observe that r = −1, θ = . Hence, 6 3 3 5π x = r cos θ = − cos = = − − 6 2 2
28. In order to convert the point 7π −2, (expressed in polar 4 coordinates) to rectangular coordinates, 7π first observe that r = −2, θ = . Hence, 4 2 7π x = r cos θ = −2 cos = − 2 = −2 4 2
1 5π 1 y = r sin θ = − sin = − = − 2 6 2 So, the given point can be expressed in
2 7π y = r sin θ = −2 sin = 2 = −2 − 2 4
3 1 rectangular coordinates as , − . 2 2
So, the given point can be expressed in
29. In order to convert the point 11π 0, (expressed in polar coordinates) 6 to rectangular coordinates, first observe 11π that r = 0, θ = . Hence, 6 11π x = r cos θ = 0 cos =0 6 11π y = r sin θ = 0 sin =0 6 So, the given point can be expressed in rectangular coordinates as ( 0,0 ) .
30. In order to convert the point ( 6,0 ) (expressed in polar coordinates) to rectangular coordinates, first observe that r = 6, θ = 0 . Hence, x = r cos θ = 6 cos ( 0 ) = 6 (1) = 6
(
)
rectangular coordinates as − 2, 2 .
y = r sin θ = 6 sin ( 0 ) = 6 ( 0 ) = 0 So, the given point can be expressed in rectangular coordinates as ( 6,0 ) .
876
Section 8.4
31. In order to convert the point π −6, (expressed in polar coordinates) 2 to rectangular coordinates, first observe
that r = −6, θ =
π
2
. Hence,
π x = r cos θ = −6 cos = 0 2 π y = r sin θ = −6 sin = −6 2 So, the given point can be expressed in rectangular coordinates as ( 0, −6 ) . π 33. In order to convert the point 8, 3 (expressed in polar coordinates) to rectangular coordinates, first observe that r = 8, θ =
π
3
. Hence,
π x = r cos θ = 8cos = 4 3 π y = r sin θ = 8sin = 4 3 3 So, the given point can be expressed in
(
)
rectangular coordinates as 4, 4 3 .
32. In order to convert the point 3π −4, (expressed in polar coordinates) 2 to rectangular coordinates, first observe 3π . Hence, that r = −4, θ = 2 3π x = r cos θ = −4 cos = 0 2 3π y = r sin θ = −4 sin = 4 2 So, the given point can be expressed in rectangular coordinates as ( 0, 4 ) . 34. In order to convert the point 4π 10, (expressed in polar coordinates) 3 to rectangular coordinates, first observe 4π that r = 10, θ = . Hence, 3 4π x = r cos θ = 10 cos = −5 3 4π y = r sin θ = 10 sin = −5 3 3 So, the given point can be expressed in
(
)
rectangular coordinates as −5, −5 3 .
877
Chapter 8
35. In order to convert the point 5π −6, (expressed in polar 4 coordinates) to rectangular coordinates, 5π first observe that r = −6,θ = . Hence, 4 − 2 5π = −6 x = r cos θ = −6 cos = 3 2 4 2 y = r sin θ = −6 sin
5π 2 = −6 − = 3 2. 4 2
So, the given point can be expressed in
36. In order to convert the point 7π 5, (expressed in polar coordinates) 6 to rectangular coordinates, first observe 7π that r = 5,θ = . Hence, 6 − 3 7π 5 3 = 5 x = r cos θ = 5 cos = − 6 2 2 7π 5 1 = 5 − = − . 6 2 2 So, the given point can be expressed in
y = r sin θ = 5sin
rectangular coordinates as 3 2,3 2 .
5 3 5 rectangular coordinates as − , − . 2 2
37. In order to convert the point ( 4, 45° )
38. In order to convert the point (1, 60° )
(expressed in polar coordinates) to rectangular coordinates, first observe that r = 4,θ = 45° . Hence,
(expressed in polar coordinates) to rectangular coordinates, first observe that r = 1,θ = 60° . Hence, 1 x = r cos θ = 1cos 60° = 2 3 . y = r sin θ = 1sin 60° = 2 So, the given point can be expressed in
(
)
2 x = r cos θ = 4 cos 45° = 4 = 2 2 2 2 y = r sin θ = 4 sin 45° = 4 = 2 2. 2 So, the given point can be expressed in
(
)
rectangular coordinates as 2 2,2 2 .
1 3 rectangular coordinates as , . 2 2
878
Section 8.4
39. In order to convert the point ( 2, 240 ) (expressed in polar
40. In order to convert the point ( −3, 150 ) (expressed in polar
coordinates) to rectangular coordinates, first observe that r = 2, θ = 240 . Hence, 1 x = r cos θ = 2 cos ( 240 ) = 2 − = −1 2
coordinates) to rectangular coordinates, first observe that r = −3, θ = 150 . Hence, 3 x = r cos θ = −3cos (150 ) = −3 − 2
3 y = r sin θ = 2 sin ( 240 ) = 2 − = − 3 2 So, the given point can be expressed in rectangular coordinates as −1, − 3 .
(
3 3 2 3 1 y = r sin θ = −3sin (150 ) = −3 = − 2 2 So, the given point can be expressed in =
)
3 3 3 rectangular coordinates as , − . 2 2
41. In order to convert the point ( −1, 135 ) (expressed in polar
42. In order to convert the point (5, 315 ) (expressed in polar coordinates)
coordinates) to rectangular coordinates, first observe that r = −1, θ = 135. Hence, 2 2 x = r cos θ = − cos (135 ) = − − = 2 2
to rectangular coordinates, first observe that r = 5, θ = 315 . Hence, 2 5 2 x = r cos θ = 5 cos ( 315 ) = 5 = 2 2
2 2 y = r sin θ = − sin (135 ) = − = − 2 2 So, the given point can be expressed in
2 5 2 y = r sin θ = 5sin ( 315 ) = 5 − = − 2 2 So, the given point can be expressed in rectangular coordinates as
2 2 rectangular coordinates as ,− . 2 2
5 2 5 2 ,− . 2 2
879
Chapter 8
43. In order to convert the point ( 52 , 330 ) (expressed in polar
44. In order to convert the point ( 34 , 225 ) (expressed in polar
coordinates) to rectangular coordinates, first observe that r = 52 , θ = 330. Hence,
coordinates) to rectangular coordinates, first observe that r = 34 , θ = 225. Hence,
y = r sin θ = 52 sin ( 330 ) = − 54
y = r sin θ = 34 sin ( 225 ) = − 3 8 2
x = r cos θ = 52 cos ( 330 ) = 5 4 3
So, the given point can be expressed in rectangular coordinates as
(
5 3 4
x = r cos θ = 34 cos ( 225 ) = − 3 8 2
So, the given point can be expressed in
)
, − 54 .
rectangular coordinates as
(−
3 2 8
)
, − 3 82 .
45. In order to convert the point (3, 180 ) (expressed in polar
46. In order to convert the point (5, 270 ) (expressed in polar coordinates)
coordinates) to rectangular coordinates, first observe that r = 3, θ = 180. Hence,
to rectangular coordinates, first observe that r = 5, θ = 270. Hence,
y = r sin θ = 3sin (180 ) = 0
y = r sin θ = 5sin ( 270 ) = −5
x = r cos θ = 3cos (180 ) = −3
x = r cos θ = 5 cos ( 270 ) = 0
So, the given point can be expressed in rectangular coordinates as ( −3,0 ) .
So, the given point can be expressed in rectangular coordinates as ( 0, −5 ) .
47. In order to convert the point ( 7, 315 ) (expressed in polar
48. In order to convert the point ( 2, 120 ) (expressed in polar coordinates)
coordinates) to rectangular coordinates, first observe that r = 7, θ = 315 . Hence,
to rectangular coordinates, first observe that r = 2, θ = 120 . Hence, x = r cos θ = 2 cos (120 ) = −1
x = r cos θ = 7 cos ( 315 ) =
7 2 2 7 2 y = r sin θ = 7 sin ( 315 ) = − 2 So, the given point can be expressed in rectangular coordinates as
y = r sin θ = 2 sin (120 ) = 3
So, the given point can be expressed in
(
)
rectangular coordinates as −1, 3 .
7 2 7 2 ,− . 2 2
880
Section 8.4
49. d
50. b
51. a
53. The graph of r = 5 is a circle centered at the origin with radius 5 since for any given angle, the same distance from the origin is used to plot the point.
52. c
54. The graph of r = −3 is a circle centered at the origin with radius 3 since for any given angle, the same distance from the origin is used to plot the point. 4 3 2 1
−4
−3
−2
−1
1
2
3
4
5
−1 −2 −3 −4 −5
55. The graph θ = 4π
is a line through 3 the origin which makes an angle of 4π 3 with the positive x-axis.
56. The graph θ = − π
is a line through 3 the origin which makes an angle of − π 3 with the positive x-axis.
1
−4
−3
−2
−1
1 −1 −2 −3 −4 5
881
Chapter 8 57. In order to graph r = 2 cos θ , consider the following table of points:
θ
r = 2 cos θ
0
2
π
0
2
π 3π
−2 2
2π
0 2
The graph is as follows:
58. In order to graph r = 3sin θ , consider the following table of points:
( r,θ ) ( 2,0 )
( 0,π 2 )
θ
r = 3sin θ
0
0
π
3
2
π
( −2,π )
( 0,3π 2 )
3π
0 2
2π
( 2,2π )
−3 0
The graph is as follows:
882
( r,θ ) ( 0,0 )
(3,π 2 ) ( 0, π )
( −3, 3π 2 ) ( 0,2π )
Section 8.4 59. In order to graph r = 4 sin 2θ , consider the following table of points:
θ
r = 4 sin 2θ
0
0
π
4
π
4 2
3π
4
π 5π 3π 7π
0 −4 0
4 2 4
2π
4 0 −4 0
The graph is as follows:
60. In order to graph r = 5 cos 2θ , consider the following table of points:
( r,θ ) ( 0,0 )
( 4,π 4 ) ( 0,π 2 )
θ
r = 5 cos 2θ
0
5
π
0
π
( −4,3π 4 )
4 2
3π
4
π
( 0, π )
( 4, 5π 4 ) ( 0, 3π 2 ) ( −4, 7π 4 )
5π 3π 7π
0 5
4 2 4
2π
( 0,2π )
−5
0 −5 0 5
The graph is as follows:
883
( r,θ ) ( 5,0 )
( 0,π 4 ) ( −5,π 2 ) ( 0,3π 4 ) (5, π )
( 0, 5π 4 ) ( −5, 3π 2 ) ( 0, 7π 4 ) (5,2π )
Chapter 8 61. In order to graph r = 3sin3θ , consider the following table of points:
θ
r = 3sin3θ
0
0
π
3
6
2π
π
6 2
4π 5π
6 6
π 7π 8π 3π
0 −3 0 3 0
6 6 2
10π 11π 2π
6 6
−3 0 3 0 −3 0
The graph is as follows:
62. In order to graph r = 4 cos 3θ , consider the following table of points:
( r,θ ) ( 0,0 )
(3,π 6 ) ( 0, 2π 6 ) ( −3,π 2 ) ( 0, 4π 6 ) (3, 5π 6 )
θ
r = 4 cos 3θ
0
4
π
0
6
2π
π
6 2
4π 5π
6 6
π
( 0, π )
( −3, 7π 6 ) ( 0, 8π 6 ) (3, 3π 2 ) ( 0,10π 6 ) ( −3,11π 6 )
7π 8π 3π
0 4 0 −4
6 6 2
10π 11π
( 0,2π )
−4
2π
6 6
0 4 0 −4 0 4
The graph is as follows:
884
( r,θ ) ( 4,0 )
( 0,π 6 )
( −4, 2π 6 ) ( 0,π 2 ) ( 4, 4π 6 ) ( 0, 5π 6 ) ( −4, π )
( 0, 7π 6 ) ( 4, 8π 6 ) ( 0,3π 2 )
( −4,10π 6 ) ( 0,11π 6 ) ( 4,2π )
Section 8.4 63. In order to graph r 2 = 9 cos 2θ , consider the following table of points:
θ
r 2 = 9 cos 2θ
r
0
9
±3
π
0
0
−9 0
No points 0
9
±3
0
0
−9 0
No points 0
9
±3
π
4 2
3π
4
π 5π 3π 7π
4 2 4
2π
The graph is as follows:
64. In order to graph r 2 = 16 sin 2θ , consider the following table of points:
( r,θ ) ( ±3,0 )
( 0,π 4 )
θ
r 2 = 16 sin 2θ
r
0
0
0
π
16
±4
0
0
−16 0
No points 0
16
±4
0
0
−16
No points 0
π
( 0, 3π 4 )
4 2
3π
4
π
( ±3, π )
( 0, 5π 4 )
5π 3π
( 0, 7π 4 )
7π
4 2 4
2π
( ±3,2π )
0
The graph is as follows:
885
( r,θ ) ( 0,0 )
( ±4,π 4 ) ( 0,π 2 ) ( 0, π )
( ±4,5π 4 ) ( 0, 3π 2 ) ( 0,2π )
Chapter 8 65. In order to graph r = −2 cos θ , consider the following table of points:
θ
r = −2 cos θ
0
−2
π
0
2
π 3π
2 2
2π
0 −2
66. In order to graph r = −3sin3θ , consider the following table of points:
( r,θ ) ( −2,0 )
( 0,π 2 ) ( 2, π )
θ
r = −3sin3θ
0
0
π
−3
6
2π
( 0, 3π 2 )
π
( −2,2π )
6 2
4π 5π
The graph is as follows:
6 6
π 7π 8π 3π
0 3 0 −3 0
6 6 2
10π 11π 2π
6 6
3 0 −3 0 3 0
The graph is as follows:
886
( r,θ ) ( 0,0 )
( −3,π 6 ) ( 0, 2π 6 ) (3,π 2 ) ( 0, 4π 6 ) ( −3, 5π 6 ) ( 0, π )
(3, 7π 6 ) ( 0,8π 6 ) ( −3, 3π 2 ) ( 0,10π 6 ) (3,11π 6 ) ( 0,2π )
Section 8.4 67. In order to graph r = 4θ , consider the following table of points:
θ
r = 4θ
0
0
π
2π
2
π 3π
4π 2
2π
6π 8π
( r,θ ) ( 0,0 )
( 2π ,π 2 ) ( 4π , π )
(6π , 3π 2 ) (8π , 2π )
68. In order to graph r = −2θ , consider the following table of points:
θ
r = −2θ
0
0
π
−π
2
π 3π
−2π 2
2π
−3π −4π
r = −3 + 2 cos θ
0
−1
π
−3
2
π 3π
−5 2
2π
−3 −1
The graph is as follows:
( r,θ ) ( 0,0 )
( −π ,π 2 ) ( −2π , π )
( −3π , 3π 2 ) ( −4π , 2π )
69. In order to graph r = −3 + 2 cos θ , consider the following table of points:
θ
The graph is as follows:
The graph is as follows:
( r,θ ) ( −1,0 )
( −3,π 2 ) ( −5, π )
( −3, 3π 2 ) ( −1,2π )
887
Chapter 8 70. In order to graph r = 2 + 3sin θ , consider the following table of points:
θ
r = 2 + 3sin θ
0
2
π
5
2
π 3π
2π
( r,θ ) ( 2,0 )
(5,π 2 ) ( 2, π )
2 2
The graph is as follows:
−1
( −1,3π 2 ) ( 2,2π )
2
71. Using x = r cos θ , y = r sin θ , we see that the equation becomes r ( sin θ + 2 cos θ ) = 1
72. Using x = r cos θ , y = r sin θ , we see that the equation becomes r ( sin θ − 3cos θ ) = 2
r sin θ + 2r cos θ = 1
r sin θ − 3r cos θ = 2
y + 2x = 1
y − 3x = 2
y = −2 x + 1 The graph of this curve is a line.
y = 3x + 2 The graph of this curve is a line.
73. Using x = r cos θ , y = r sin θ , we see that the equation becomes r 2 cos 2 θ − 2r cos θ + r 2 sin 2 θ = 8
74. Using x = r cos θ , y = r sin θ , we see that the equation becomes r 2 cos 2 θ − r sin θ = −2
x 2 − 2x + y2 = 8 x2 − 2x + 1 + y2 = 8 + 1 ( x − 1)2 + y 2 = 9 The graph of this curve is a circle. 75. Using x = r cos θ , y = r sin θ , we see that the equation becomes r 2 sin2 θ + 2r cos θ = 3
x 2 − y = −2 y = x2 + 2 The graph of this curve is a parabola. 76. Using x = r cos θ , y = r sin θ , we see that the equation becomes r 2 cos 2 θ + r sin θ = 3
y2 + 2x − 3 = 0 x = 12 (3 − y2 ) The graph is a parabola.
x2 + y = 3 y = −x 2 + 3 The graph is a parabola.
888
Section 8.4
77. The graph of r =
0.587 (1 + 0.967 ) is 1 − 0.967 cos θ
as follows:
78. The graph of r =
as follows:
79. Consider the following table of values for the graphs of r = θ and r= θ : θ r =θ r= θ
0
0
0
π
π
π
2
π 3π
2
π 2
2π
2
2
π 3π
2π
2π
3π
2
Notice that the graph of r = θ (dotted) is more tightly wound than is the graph of r = θ (solid).
889
29.62 (1 + 0.249 ) is 1 − 0.249 cos θ
Chapter 8
80. Consider the following table of values for the graphs of r = θ and 4 r =θ 3 : 4 θ r =θ r =θ 3 0 0 0
π
2
π
2
π
π
3π
3π
2
2π
( 2) π
(π )
4
4
3
(3π 2 )
2
2π
( 2π )
3
4
4
3
3
Notice that the graph of r = θ 3 (dotted) is more loosely wound than is the graph of r = θ (solid). 4
81. Consider the following table of values for the graphs of r 2 = 4 cos 2θ and r 2 = 14 cos 2θ :
θ
r 2 = 4 cos 2θ
r 2 = 14 cos 2θ
0
4 −4
1
π
2 π 3π 2 2π
4 −4 4
4
− 14 1
4
− 14 1
4
Notice that the graph of r 2 = 14 cos 2θ (solid) is much closer to the origin than is the graph of r 2 = 4 cos 2θ (dotted).
890
Section 8.4 82. The graphs of r 2 = 4 cos 2θ and r 2 = 4 cos ( 2θ + 2 ) are given to the right.
Notice that the graph of r 2 = 4 cos ( 2θ + 2 ) (dotted) is simply a rotation of the graph of r 2 = 4 cos 2θ (solid). The same set of points is generated, but starting/ending at different positions on the graph.
83. In order to graph r = 2 + 2 sin θ , consider the following table of points:
θ
r = 2 + 2 sin θ
0
2
π
4
2
π 3π
2 2
2π
0 2
( r, θ ) ( 2,0 )
( 4,π 2 ) ( 2, π )
( 0, 3π 2 ) ( 2,2π )
84. In order to graph r = −4 − 4 sin θ , consider the following table of points:
θ
r = −4 − 4 sin θ
0
−4
π
−8
2
π 3π
−4 2
2π
0 −4
The graph is as follows:
The graph is as follows:
( r, θ ) ( −4,0 )
( −8,π 2 ) ( −4, π )
( 0, 3π 2 ) ( −4,2π )
891
Chapter 8
85. a) – c) All three graphs generate the same set of points, as seen below:
86. We must choose L = 12, and choose a such that r 2 = cos( aθ ) makes 8 complete cycles in the given interval. Using the observations in Exercise 77, we see that a must be 8, so that the equation
is r 2 = 12 cos ( 8θ ) . 87. The point is in QIII, so needed to add π to the angle. 88. The point is in QII, so needed to add π to the angle. 89. True. By definition.
Note that all three graphs are figure eights. Extending the domain in b results in twice as fast movement, while doing so in c results in movement that is four times as fast.
90. False. The equation of a cardioid is either r = a + a cos θ or r = a + a sin θ , while the more general limacon has equation either r = a + b cos θ or r = a + b sin θ . So, simply choose a ≠ b to get a limacon that is not a cardioid.
91. Observe that if x = a , then a r cos θ = a , so that r = , which cos θ represents a vertical line.
92. Observe that if y = b , then b r sin θ = b , so that r = , which sin θ represents a horizontal line.
93. The point ( a,θ ) can also be represented as ( −a,θ ± 180 ) . This is because the
point ( −a,θ ) is diametrically opposed (through the origin) to the given point. As
such, we need to rotate from it 180 clockwise or counterclockwise to arrive at the given point.
892
Section 8.4 94. In order to convert the point ( −a, b )
(expressed in rectangular coordinates) to polar coordinates, first observe that x = −a, y = b , so that the point is in QII. Hence, r=
( −a ) + ( b ) = a 2 + b 2 2
2
y b b = = − so that x −a a b θ = 180 + tan −1 − . a So, the given point can be expressed in polar coordinates as tan θ =
2 b 2 −1 a + b ,180 + tan − a . 96. We solve the following equation: 1 − 2 sin2 θ = 1 − 6 cos 2 θ tan2 θ = 3 tan θ = ± 3
θ = π3 , 23π , 43π , 53π The corresponding r-values for all of these values of θ are −1/2. So, the points of intersection are ( − 12 , π3 ) , ( − 12 , 23π ) , ( − 12 , 43π ) , ( − 12 , 53π ) .
95. We solve the following equation: 1 − 2 sin 2θ = 1 − 2 cos 2θ sin 2θ = cos 2θ 2θ = π4 , 54π , 94π , 134π
θ = π8 , 58π , 98π , 138π The corresponding r-values are: θ = π8 , 98π : r = 1 − 2 sin ( 2 ⋅ π8 ) = 1 − 2
θ = 58π , 138π : r = 1 − 2 sin ( 2 ⋅ 58π ) = 1 + 2 So, the points of intersection are 1 − 2, π8 , 1 − 2, 98π , 1 + 2, 58π ,
( )( (1 + 2, ) .
)(
13π 8
).
97. The tests for symmetry in polar coordinates are: x-axis symmetry: r( −θ ) = r(θ ), for all θ y-axis symmetry: −r( −θ ) = r(θ ), for all θ origin: −r(θ ) = r(θ ), for all θ Applying these to the given curve, we notice that −r = sin ( 3 ( −θ ) ) −r = sin ( −3θ )
−r = − sin ( 3θ ) r = sin ( 3θ ) . Hence, it possesses y-axis symmetry, but none of the others.
98. We apply each of the tests in #97 to the given curve: 2 ( −r ) = cos ( 4θ ) r 2 = cos ( 4θ ) symmetry to the origin.
r 2 = cos ( 4 ( −θ ) ) r 2 = cos ( −4θ ) r 2 = cos ( 4θ ) symmetry to the x-axis
( −r ) = cos ( 4 ( −θ ) ) r 2 = cos ( −4θ ) r 2 = cos ( 4θ ) symmetry to the y-axis 2
893
Chapter 8
θ 99. The graph of r = cos is as 2 follows:
3θ 100. The graph of r = 2 cos is as 2 follows:
The inner loop is generated beginning π 3π . with θ = and ending with 2 2
The petal in the 1st quadrant is generated
101. The graph of r = 1 + 3cos θ is as follows:
102. The graphs of r = 1 + sin(2θ ) (solid) and r = 1 − cos(2θ ) (dashed) are as follows:
beginning with θ = 0 and ending with
The very tip of the inner loop begins with θ = π2 , then it crosses the origin (the first time) at θ = cos −1 ( − 13 ) , winds around,
and eventually ends with θ = 32π .
π
3
.
Solving 1 + sin(2θ ) = 1 − cos(2θ ) , we see that sin(2θ ) + cos(2θ ) = 0 , which is satisfied when 2θ is an integer multiple of π4 that has terminal side in QII or QIV. As such, we see that 3π 7π 11π 15π 2θ = , , , 4 4 4 4 3π 7π 11π 15π . and hence, θ = , , , 8 8 8 8
894
Section 8.5 Solutions -----------------------------------------------------------------------------------1. In order to graph the curve defined parametrically by x = t + 1, y = t , t ≥ 0 , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since t = x − 1 , substituting this into the expression for y yields y = x − 1, x ≥ 1 .
The graph is as follows:
2. In order to graph the curve defined parametrically by x = 3t, y = t 2 − 1, t in [ 0, 4 ] ,
The graph is as follows:
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since t = x , 3 substituting this into the expression for y yields 2
1 x y = − 1 = x 2 − 1, x in [ 0,12 ] . 9 3 3. In order to graph the curve defined parametrically by x = −3t, y = t 2 + 1, t in [ 0, 4 ] ,
The graph is as follows:
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since t = − x , 3 substituting this into the expression for y yields 2
1 x y = − + 1 = x 2 + 1, x in [ −12,0 ] . 9 3
895
Chapter 8
4. In order to graph the curve defined parametrically by x = t 2 − 1, y = t 2 + 1, t in [ −3,3] ,
The graph is as follows:
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since t 2 = x + 1 , substituting this into the expression for y yields y = ( x + 1) + 1 = x + 2, x in [ −1, 8] . 5. In order to graph the curve defined parametrically by x = t 2 , y = t3 , t in [ −2,2 ] , it is easiest to be guided by a sequence of tabulated points, as follows:
t −2 −1 0 1 2
x = t2 4 1 0 1 4
The graph is as follows:
y = t3 −8 −1 0 1 8
6. In order to graph the curve defined parametrically by x = t3 + 1, y = t3 − 1, t in [ −2,2 ] ,
The graph is as follows:
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since t3 = x − 1 , substituting this into the expression for y yields y = ( x − 1) − 1 = x − 2, x in [ −7, 9] .
896
Section 8.5
7. In order to graph the curve defined parametrically by x = t , y = t, t in [ 0,10 ] , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, substituting y = t into the expression for x yields x = y , y in [ 0,10 ] .
The graph is as follows:
8. In order to graph the curve defined parametrically by
The graph is as follows:
x = t, y = t 2 + 1, t in [ 0,10 ] ,
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, substituting x = t into the expression for y yields y = x 2 + 1, x in [ 0,10 ] .
9. In order to graph the curve defined parametrically by 2 3 x = ( t + 1) , y = ( t + 2 ) , t in [ 0,1] ,
The graph is as follows:
it is easiest to be guided by a sequence of tabulated points, as follows: t
x = ( t + 1)
0 0.25 0.50 0.75 1.00
1 1.5625 2.25 3.0625 4
2
y = (t + 2 )
3
8 11.391 15.63 20.80 27
897
Chapter 8
10. In order to graph the curve defined parametrically by 3 2 x = ( t − 1) , y = ( t − 2 ) , t in [ 0, 4 ] ,
The graph is as follows:
it is easiest to be guided by a sequence of tabulated points, as follows: t
x = ( t − 1)
y = (t − 2 )
0 1 2 3 4
−1 0 1 8 27
4 1 0 1 4
3
2
11. In order to graph the curve defined parametrically by x = 3sin t, y = 2 cos t, t in [ 0,2π ] , consider the following sequence of tabulated points:
t 0 π
2
π
3π
2
2π
x = 3sin t 0 3 0 −3 0
The graph is as follows:
y = 2 cos t 2 0 −2 0 2
Eliminating the parameter reveals that the graph is an ellipse. Indeed, observe that x 2 = 9sin 2 t, y 2 = 4 cos 2 t so that x2 y2 = sin2 t, = cos 2 t . 9 4 x2 y2 + = 1. Summing then yields 9 4
898
Section 8.5
12. In order to graph the curve defined parametrically by x = cos 2t, y = sin t, t in [ 0,2π ] , consider the following sequence of tabulated points: t x = cos 2t y = sint 0 1 0 π −1 1 2 π 1 0 3π −1 −1 2 2π 1 0 Notice that the curve retraces itself within this interval. Eliminating the parameter reveals that the path that the graph takes is a portion of a parabola opening to the right. Indeed, observe that x = cos 2t = 2 sin 2 t − 1 = 2 y 2 − 1 .
The graph is as follows:
13. In order to graph the curve defined parametrically by x = sin t + 1, y = cos t − 2, t in [ 0,2π ] , consider the following sequence of tabulated points: t x = sin t + 1 y = cos t − 2
The graph is as follows:
0
1 −1 2 −2 2 π 1 −3 3π 0 −2 2 2π 1 −1 Eliminating the parameter reveals that the graph is a circle with radius 1 centered at (1,−2). Indeed, observe that x − 1 = sin t, y + 2 = cos t π
( x − 1) = sin2 t, ( y + 2 ) = cos2 t 2
2
so that
( x − 1) + ( y + 2 ) = 1. 2
2
899
Chapter 8
14. In order to graph the curve defined The graph is as follows: parametrically by x = tan t, y = 1, t in − π , π , 4 4 observe that since the y-coordinate is always 1, the path is a horizontal line segment starting at the point (−1, 1) and ending at the point (1, 1).
15. In order to graph the curve defined parametrically by x = 1, y = sin t, t in [ −2π , 2π ] ,
The graph is as follows:
observe that since the x-coordinate is always 1, the path is a vertical line segment starting and ending (after retracing its path) at the point (1, 0). The retracing occurs simply because of the periodicity of the sine curve.
16. In order to graph the curve defined parametrically by x = sin t, y = 2, t in [ 0,2π ] ,
The graph is as follows:
observe that since the y-coordinate is always 2, the path is a horizontal line segment starting at the point (0, 2), goes to (1,2) and comes back through (0, 2) and visits (−1, 2), and then finally ends at (0,2).
900
Section 8.5
17. In order to graph the curve defined parametrically by x = sin2 t, y = cos 2 t, t in [ 0,2π ] ,
The graph is as follows:
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, simply adding the two equations yields x + y = 1 , 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 , which is equivalent to y = 1 − x , 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 .
18. In order to graph the curve defined parametrically by x = 2 sin2 t, y = 2 cos 2 t, t in [ 0,2π ] ,
The graph is as follows:
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that x y = sin2 t, = cos 2 t, t in [ 0,2π ] 2 2 so that adding these two equations yields x y + = 1 , 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 , 2 2 which is equivalent to y = 2 − x , 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 .
901
Chapter 8
19. In order to graph the curve defined parametrically by x = 2 sin3t, y = 3cos 2t, t in [ 0,2π ] ,
The graph is as follows:
consider the following sequence of tabulated points: t
x = 2 sin3t
0
0 2
π
6
π π
4 3
π
2
2π
3
3π
4
5π
6
π
2 0 −2 0 2 2 0
y = 3cos 2t 3 3
2
0 − 32 −3 − 32 0 3
2
3
The pattern replicates for t in [π ,2π ] . 20. In order to graph the curve defined parametrically by x = 4 cos 2t, y = t, t in [ 0,2π ] ,
The graph is as follows:
consider the following sequence of tabulated points: t 0 π π
4 2
3π
4
π
5π 3π 7π
4 2 4
2π
x = 4 cos 2t 4 0 −4 0 4 0 −4 0 4
y=t 0 π π
4 2
3π
4
π
5π 3π 7π
4 2 4
2π
902
Section 8.5
21. In order to express the parametrically-defined curve given by 1 x = , y = t 2 as a rectangular equation, t we eliminate the parameter t in the following sequence of steps: 2
1 1 1 t = so that y = = 2 . x x x So, the equation is y =
So, the equation is y = x + 2 .
1 . x2
23. In order to express the parametrically-defined curve given by x = 5t, y = 4t + 2 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: x 4 x t = so that y = 4 + 2 = x + 2. 5 5 5
So, the equation is y =
22. In order to express the parametricallydefined curve given by x = t 2 − 1, y = t 2 + 1 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: t 2 = x + 1 so that y = ( x + 1) + 1 = x + 2 .
4 x+2. 5
24. In order to express the parametricallydefined curve given by x = 2t, y = −5t − 1 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: x 5 x t = so that y = −5 − 1 = − x − 1. 2 2 2 5 So, the equation is y = − x − 1. 2
25. In order to express the parametrically-defined curve given by x = t + 2, y = t 2 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: 2 t = x − 2 so that y = ( x − 2 ) .
26. In order to express the parametricallydefined curve given by x = t − 2, y = t3 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: 3 t = x + 2 so that y = ( x + 2 ) .
So, the equation is y = ( x − 2 ) .
So, the equation is y = ( x + 2)3 .
2
903
Chapter 8
27. In order to express the parametrically-defined curve given by x = 2t, y = 2t − 3 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: t=
x x so that y = 2 − 3 = x − 3. 2 2
28. In order to express the parametricallydefined curve given by x = t , y = 9t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: y y y t = so that x = = 3x = y . 9 9 3
So, the equation is y = x − 3 .
So, the equation is 3x = y .
29. In order to express the parametrically-defined curve given by x = t3 + 1, y = t3 − 1 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: t3 = x − 1 so that y = ( x − 1) − 1 = x − 2 .
30. In order to express the parametricallydefined curve given by x = 3t, y = t 2 − 1 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: 2
x2 x x −1 . t = so that y = − 1 = 9 3 3
So, the equation is y = x − 2 .
1 So, the equation is y = x 2 − 1 . 9 31. In order to express the parametrically-defined curve given by x = t, y = t 2 + 1 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: t = x so that y = x 2 + 1 .
So, the equation is y = x 2 + 1 .
32. In order to express the parametricallydefined curve given by x = sin2 t, y = cos 2 t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: x + y = sin2 t + cos 2 t = 1
So, the equation is x + y = 1 .
904
Section 8.5
33. In order to express the parametrically-defined curve given by x = 2 sin2 t, y = 2 cos 2 t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: x y + = sin2 t + cos 2 t = 1 so that 2 2 x+y=2
So, the equation is x + y = 2 . 35. In order to express the parametrically-defined curve given by x = 4 ( t 2 + 1) , y = 1 − t 2 as a rectangular
equation, we eliminate the parameter t in the following sequence of steps: x x Note that = t 2 + 1 so that t 2 = − 1 . 4 4 x x Hence, y = 1 − − 1 = 2 − . 4 4 Simplifying, we see that the equation is x + 4y = 8 . 37. In order to express the parametrically-defined curve given by x = 2t, y = 2 sin t cos t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: Note that x = 2t t = x2 . Hence, y = 2 sin x2 cos x2 = sin(2 ⋅ x2 ) = sin x .
We see that the equation is y = sin x .
34. In order to express the parametricallydefined curve given by x = sec 2 t, y = tan2 t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: sin2 t 1 − cos 2 t 1 = = y = tan2 t = −1 2 2 cos t cos t cos 2 t = sec 2 t − 1 = x − 1 So, the equation is y = x − 1 . 36. In order to express the parametricallydefined curve given by x = t − 1, y = t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: Note that x 2 = t − 1 so that t = x 2 + 1 .
Hence, y = x 2 + 1 . Squaring both sides and simplifying, we see that the equation is y 2 − x 2 = 1 .
38. In order to express the parametricallyt defined curve given by x = , y = 2 tan t as 2 a rectangular equation, we eliminate the parameter t in the following sequence of steps: t Note that x = t = 2x . Hence, 2 y = 2 tan 2x .
We see that the equation is y = 2 tan 2x .
905
Chapter 8
39. In order to express the parametrically-defined curve given by x = t , y = t + 2 as a rectangular equation, we eliminate the parameter t in the following sequence of steps:
We see that the equation is y = x 2 + 2 .
40. In order to express the parametricallyt defined curve given by x = t + 1, y = − 1 4 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: Note that x = t + 1, so that t = x 2 − 1 .
Hence, y =
x2 −1 x2 − 5 . We see that −1 = 4 4
x2 − 5 . the equation is y = 4 41. In order to express the parametrically-defined curve given by x = sin t, y = sin t as a rectangular equation, we eliminate the parameter t by simply noting that y = x . Thus, the
equation is y = x .
equation is y = −x 2 .
43. Assuming that the initial height h = 0, we need to find t such that y = −16t 2 + ( 400 sin 45 ) t + 0 = 0 .
We solve this equation as follows: −16t 2 + ( 400 sin 45 ) t + 0 = 0 −16t 2 + 200 2 t = 0
(
42. In order to express the parametricallydefined curve given by x = cos t, y = − cos 2 t as a rectangular equation, we eliminate the parameter t by simply noting that y = −x 2 . Thus, the
)
t −16t + 200 2 = 0 t = 0,
44. We know from Exercise 37 that the time traveled is 17.7 seconds. So, the horizontal distance traveled is x (17.7 ) = ( 400 cos 45 ) (17.7 ) ≈ 5,000 ft.
Also, the maximum altitude must occur at 17.7 t≈ = 8.85 sec. The height at this time 2 is given by y(8.85) = −16(8.85)2 + ( 400 sin 45 ) (8.85) ≈ 1250 ft.
200 2 ≈ 17.7 sec. 16 So, it hits the ground approximately 17.7 seconds later.
906
Section 8.5
45. First, we convert all measurements to feet per second: 105 mi. 1 hr. 5280 ft. ft. = 154 sec. 1 hr. 3600 sec. 1 mi. Now, from the given information, we know that the parametric equations are: x = (154 cos 20 ) t
y = −16t 2 + (154 sin 20 ) t + 3
We need to determine the time t0 corresponding to x = 420 ft. Then, we will compare y ( t0 ) to 15; if y ( t0 ) < 15 , then it doesn’t clear the fence. Observe that 420 ≈ 2.9023 sec. 420 = (154 cos 20 ) t0 so that t0 = 154 cos 20 Then, we have y ( 2.9023 ) ≈ 21.0938 , so it does clear the fence. 46. We use the same information as in Exercise 39. This time, we need to determine t0 such that y ( t0 ) = 0 . Then, we will compute x ( t0 ) = total horizontal distance traveled
t y 0 = maximum height 2 Observe that solving −16t 2 + (154 sin 20 ) t + 3 = 0 using the quadratic formula yields t=
−154 sin 20 ±
(154 sin 20 ) − 4( −16)(3) 2
2( −16) −52.6711 ± 54.4632 ≈ ≈ −0.056 or 3.3479 −32
Now, we have x ( 3.3479 ) ≈ 484.5 ft. = total horizontal distance traveled
y (1.67395 ) ≈ 46.3 ft. = maximum height
907
Chapter 8
47. From the given information, we know that the parametric equations are: x = ( 700 cos 60 ) t
y = −16t 2 + ( 700 sin 60 ) t + 0
We need to determine t0 such that y ( t0 ) = 0 . Then, we will compute x ( t0 ) = total horizontal distance traveled
t y 0 = maximum height 2 Observe that solving −16t 2 + ( 700 sin 60 ) t = 0 yields −16t 2 + ( 700 sin 60 ) t + 0 = 0
t ( −16t + 700 sin 60 ) = 0 t = 0,
700 sin 60 ≈ 37.89 sec. 16
Now, we have x ( 37.89 ) ≈ 13,261 ft. = total horizontal distance traveled y (18.94 ) ≈ 5742 ft. = maximum height
48. From the given information, we know that the parametric equations are: x = ( 2000 cos 60 ) t
y = −16t 2 + ( 2000 sin 60 ) t + 0
We need to determine t0 such that y ( t0 ) = 0 . Then, we will compute x ( t0 ) = total horizontal distance traveled
t y 0 = maximum height 2 Observe that solving −16t 2 + ( 2000 sin 60 ) t = 0 yields −16t 2 + ( 2000 sin 60 ) t + 0 = 0 t ( −16t + 2000 sin 60 ) = 0
t = 0,
908
2000 sin 60 ≈ 108.25 sec. 16
Section 8.5
Now, we have x (108.25 ) ≈ 108,253 ft. = total horizontal distance traveled y ( 54.127 ) = 46,875 ft. = maximum height
49. From the given information, we know that the parametric equations are: x = ( 4000 cos30 ) t
y = −16t 2 + ( 4000 sin30 ) t + 20
We need to determine t0 such that y ( t0 ) = 0 .
Observe that solving −16t 2 + ( 4000 sin30 ) t + 20 = 0 using the quadratic formula yields t=
−4000 sin30 ±
( 4000 sin30 ) − 4( −16)(20) 2
2( −16)
−2000 ± 2000 2 + 1280 ≈ −0.001 or 125 −32 So, it stays in flight for 125 seconds. ≈
50. From the given information, we know that the parametric equations are: x = ( 5000 cos 40 ) t
y = −16t 2 + ( 5000 sin 40 ) t + 20
We need to determine t0 such that y ( t0 ) = 0 . Then, we calculate x ( t0 ) . If this is larger than 2(5,280) = 10,560 ft., then it can hit the target. Observe that solving −16t 2 + ( 5000 sin 40 ) t + 20 = 0 using the quadratic formula yields t=
−5000 sin 40 ±
(5000 sin 40 ) − 4( −16)(20) 2
2( −16)
≈ −0.006 or 200.877 sec.
So, x(200.877) = ( 5000 cos 40 ) (200.877) ≈ 769, 405 ft. Since this is significantly larger than 10,560 ft., it can definitely hit a target 2 miles away.
909
Chapter 8
51. The parametric equations are x = (100 cos35 ) t
52. The parametric equations are x = (150 cos55 ) t
The graph is as follows:
The graph is as follows:
53. The points are as follows:
54. In general, multiply the arguments of the trigonometric functions by coefficients larger than 1 in order to increase speed. Note that A+B represents the length of the arm when fully extended.
y = −16t 2 + (100 sin35 ) t
t 0
π
2 π 3π 2 2π
x(t) A+B 0
y(t) 0 A+B
−A−B 0
0 −A−B
A+B
0
55. Since both sin(10t ) and cos(10t ) have period 210π , we conclude that one revolution occurs in 210π ≈ 0.63 seconds.
y = −16t 2 + (150 sin55 ) t
56. We want to sweep out the curve in the opposite direction, so we must negate all time inputs. This results in the parametric equations x = sin( −10t ), y = cos( −10t ) .
910
Section 8.5
57. We simply evaluate the equations t t x = −100 sin , y = 75 cos 4 4 at times t = 10, 20, 30: t = 10: x(10) ≈ −60, y(10) ≈ −60 t = 20: x(20) ≈ 96, y(20) ≈ 21 t = 30: x(30) ≈ −94, y(30) ≈ 26 So, the desired points are: ( −60, −60), (96,21), ( −94,26)
58. One period occurs in 2π = 8π ≈ 25.1 sec of time. 1
4
59. The first racer completes one lap in 2π = 8π seconds, while the second racer 1 4
completes one lap in 21π3 = 6π seconds. So, the second racer is going faster.
60. Computing the given equations at the integer values 1, 2, etc., we observe that Racer 1 completes 3 laps in 24π seconds, while Racer 2 completes 4 laps in the same amount of time. So, Racer 2 laps Racer 1 in 24π ≈ 75 seconds.
61. The original domain must be t ≥ 0 . Therefore, only the portion of the parabola where y ≥ 0 is part of the plane curve.
62. The original domain must be t ≥ 0 . Therefore, only the portion of the parabola where x ≥ 0 is part of the plane curve.
63. False. They simply constitute a set of points in the plane.
65. Observe that the parametric equations x = nt, y = sin t can be described by the equation y = sin ( xn ) .
66. Observe that the parametric equations t x = , y = cot t n can be described by the equation y = cot( nx ). These are just cotangent graphs with period
These are just sine curves with period 2π n
64. True. By definition of parametric equation (viewed as how points are generated by action of a third parameter).
π n
911
.
Chapter 8
67. Given the parametric equations x = t, y = 1− t, t ≥ 0 , observe that since t = x 2 , we have
68. Given the parametric equations x = ln t, y = t, t > 0 ,
we see that x = ln y , so that y = e x .
y = 1 − x 2 . Now, since x = t is
always ≥ 0 and y = 1 − x 2 is also always ≥ 0 , this generates a quarter circle in QI. 69. We eliminate the parameter in the parametric equations x = tan t, y = sec t . Observe that sin2 t 1−sin2 t y2 − x 2 = cos12 t − cos = 1, 2 = cos2 t t
which is the equation of a vertical hyperbola.
70. We eliminate the parameter in the parametric equations x = et , y = −t. Observe that t = ln x, so that y = − ln x. This is the equation of a natural log curve reflected over the x-axis.
71. Consider the curve defined parametrically by: { x = a sin t − sin at, y = a cos t + cos at
We have the following graphs:
Graph for a = 2
Graph for a = 3
Graph for a = 4
Notice that as the value of a increases, the distance between the vertices and the origin gets larger, and the number of distinct vertices (and corresponding congruent arcs) increases (and = a, when a is a positive integer).
912
Section 8.5 72. Consider the curve defined by: { x = a cos t − b cos at, y = a sin t + sin at
We have the following graphs:
Graph for a = 3, b = 1
Graph for a = 4, b = 2
Graph for a = 6, b = 2
In each case, one cycle of the curve is completed in 2π units of time. 73. Consider the curve defined by: { x = cos at, y = sin bt We have the following graphs:
Graph for a = 2, b = 4
Graph for a = 4, b = 2
Graph for a = 1, b = 3
Graph for a = 3, b = 1
In each case, one cycle of the curve is traced out in 2π units of time.
913
Chapter 8
74. First, we consider the curve defined by the following parametric equations: { x = a sin at − sin t, y = a cos at − cos t
Graph for a = 3
Graph for a = 2
Now, in comparison, we consider the curve defined by the following parametric equations: { x = a sin at − sin t, y = a cos at + cos t
Graph for a = 2
Graph for a = 3
All curves are generated in 2π units of time.
914
Chapter 8 Review Solutions --------------------------------------------------------------------------1. i 39 = i 4(9)+3 = i 4(9)i 3 = ( i 4 ) i 3 = (1) ( −i ) = −i 9
9
2. i 100 = i 4(25) = ( i 4 ) = (1) = 1 25
3.
25
−49 = −1 49 = 7i
4.
−50 = −1 50 = 5i 2
5. ( 2 − 3i ) + ( 4 + 6i ) = ( 2 + 4 ) + ( −3 + 6 ) i = 6 + 3i 6. ( −3 − i ) − ( 5 − 2i ) = ( −3 − 5 ) + ( −1 + 2 ) i = −8 + i 7. (1 + 4i )( −6 − 2i ) = −6 − 2i − 24i − 8i 2 = −6 − 26i + 8 = 2 − 26i 8. ( 2 + 5i )( −3 + 2i ) = −6 − 15i + 4i − 10 = −16 − 11i 9.
3 − 6i 2 − i 6 − 12i − 3i − 6 15i ⋅ = =− = −3i 2 +i 2 −i 4 +1 5
10.
4 −i 4 − i 2 − 3i 8 − 12i − 2i + 3i 2 5 − 14i 5 14 = ⋅ = = = − i 2 2 + 3i 2 + 3i 2 − 3i 4 − 9i 13 13 13
11.
8 − 12i 8 − 12i i 12 + 8i = ⋅ = = −3 − 2i 4i 4i i −4
12.
1 − 2i 1 − 2i 1 − 2i 1 − 2i − 2i − 4 3 4 = ⋅ = = − − i 1 + 2i 1 + 2i 1 − 2i 1+ 4 5 5
13. i ( 2 + 3i ) + 4 ( −3 + 2i ) = 2i + 3i 2 − 12 + 8i = −15 + 10i 14. −3 ( 2 − 3i ) + i ( −6 + i ) = −6 + 9i − 6i + i 2 = −7 + 3i
915
Chapter 8
15. The complex solutions to x 2 − 4 x + 9 = 0 are x=
4±
( −4 ) − 4 (1)( 9 ) 4 ± −20 4 ± 2i 5 = 2 ± 5i . = = 2 (1) 2 2 2
16. The complex solutions to 2x 2 + 2 x + 11 = 0 are x=
−2 ± 2 2 − 4 ( 2 )(11) 2 (2)
=
−2 ± −84 −2 ± 2i 21 1 21 = = − ± i . 4 4 2 2
17. The complex solutions to 5x 2 + 6x + 5 = 0 are x=
−6 ± 6 2 − 4 ( 5 )( 5 ) 2 (5 )
=
−6 ± −64 −6 ± 8i 3 4 = = − ± i . 10 10 5 5
18. The complex solutions to x 2 + 4 x + 5 = 0 are x=
19. 20.
−4 ± 4 2 − 4 (1)( 5 ) 2 (1)
=
−4 ± −4 −4 ± 2i = = −2 ± i . 2 2
−25 −9 = ( 5i )( 3i ) = 15i 2 = −15
( 2 + −100 )( −3 + −4 ) = ( 2 + 10i )( −3 + 2i ) = −6 + 4i − 30i + 20i = −26 − 26i 2
21. The head of the vector is the complex number.
22. The head of the vector is the complex number. 2 1
1
−4
−3
−2
−1
−8
1
−7
−6
−5
−4
−3
−2
−1
1 −1
−1 −2
−2 −3 −4 5
916
Chapter 8 Review
23. & 24. The complex numbers –6 + 2i and 5i are plotted below:
25. In order to express 2 − 2i in polar form, observe that x = 2, y = − 2 , so that the point is in QIV. Now,
( 2 ) + (− 2 ) = 2 2
r = x2 + y2 =
2
y − 2 π = = −1 , so that θ = tan −1 ( −1) = −45 or − . x 4 2 7π Since 0 ≤ θ < 2π , we must use the reference angle 315 or . 4 tan θ =
Hence,
7π 7π 2 − 2i = 2 cos + i sin = 2 cos ( 315 ) + i sin ( 315 ) . 4 4
26. In order to express point is in QI. Now,
3 + i in polar form, observe that x = 3, y = 1 , so that the r = x2 + y2 =
tan θ =
Hence,
( 3 ) + (1) = 2 2
2
y 3 1 1 π = = , so that θ = tan −1 = or 30 . x 3 3 3 6
π π 3 + i = 2 cos + i sin = 2 cos ( 30 ) + i sin ( 30 ) . 6 6
917
Chapter 8 27. In order to express 0 − 8 i in polar form, observe that x = 0, y = −8 , so that the point is on the negative y-axis. Now, r = x 2 + y2 =
28. In order to express −8 − 8i in polar form, observe that x = −8, y = −8 , so that the point is in QIII. Now,
( 0 ) + ( −8) = 8 and 2
r = x2 + y2 =
2
( −8 ) + ( −8 ) = 8 2 2
2
y −8 = = 1 , so that x −8 θ = 180 + tan −1 (1) = 225 .
θ = 270 .
tan θ =
Hence, 0 − 8i = 8 cos ( 270 ) + i sin ( 270 ) .
(Remember to add 180 to tan −1 (1) since the point is in QIII.) Hence, −8 − 8i = 8 2 cos ( 225 ) + i sin ( 225 ) .
29. In order to express −60 + 11i in polar form, observe that x = −60, y = 11, so that the point is in QII. Now, r = x2 + y2 =
30. In order to express 9 − 40 i in polar form, observe that x = 9, y = −40 , so that the point is in QIV. Now,
( −60 ) + (11) = 61 2
2
r = x 2 + y2 =
y −11 = , so that x 60 −11 θ = 180 + tan−1 ≈ 169.6 . 60 −11 (Remember to add 180 to tan −1 60 since the angle has terminal side in QII.) Hence,
( 9 ) + ( −40 ) = 41 2
2
y −40 = , so that x 9 −40 θ = tan−1 ≈ −77.3196 . 9 Since 0 ≤ θ < 360 , we must use the reference angle 282.7 . Hence,
tan θ =
tan θ =
−60 + 11i ≈ 61 cos (169.6 ) + i sin (169.6 ) .
918
9 − 40i ≈ 41 cos ( 282.7 ) + i sin ( 282.7 ) .
Chapter 8 Review
31. In order to express 15 + 8i in polar form, observe that x = 15, y = 8 , so that the point is in QI. Now, r = x 2 + y2 =
32. In order to express −10 − 24i in polar form, observe that x = −10, y = −24 , so that the point is in QIII. Now,
(15 ) + (8 ) = 17 2
2
y 8 = , so that x 15 8 θ = tan −1 ≈ 28.1 . 15
tan θ =
Hence, 15 + 8i ≈ 17 cos ( 28.1 ) + i sin ( 28.1 ) .
r = x2 + y2 =
( −10 ) + ( −24 ) = 26 2
2
y 24 = , so that x 10 24 θ = 180 + tan−1 ≈ 247.4 . 10 24 (Remember to add 180 to tan −1 10 since the angle has terminal side in QIII.) Hence, tan θ =
−10 − 24i ≈ 26 cos ( 247.4 ) + i sin ( 247.4 ) . 33.
1 3 6 cos ( 300 ) + i sin ( 300 ) = 6 − i 2 2
34.
3 1 − i 4 cos ( 210 ) + i sin ( 210 ) = 4 − 2 2
= 3 − 3 3i 35. 2 cos (135 ) + i sin (135 )
2 2 i = −1 + i = 2 − + 2 2
= −2 3 − 2i 36.
3 1 + i 4 cos (150 ) + i sin (150 ) = 4 − 2 2 = −2 3 + 2i
37. 4 cos ( 200 ) + i sin ( 200 )
38. 3 cos ( 350 ) + i sin ( 350 )
≈ −3.7588 − 1.3681i
≈ 2.9544 − 0.5209i
39. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 = ( 3 )( 4 ) cos ( 200 + 70 ) + i sin ( 200 + 70 ) = 12 cos ( 270 ) + i sin ( 270 ) = 12 [ 0 − i ] = −12i
919
Chapter 8 40. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 3 )( 4 ) cos ( 20 + 220 ) + i sin ( 20 + 220 )
1 3 = 12 cos ( 240 ) + i sin ( 240 ) = 12 − − i = −6 − 6 3i 2 2 41. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields z1z2 = ( 7 )( 3 ) cos (100 + 140 ) + i sin (100 + 140 )
1 3 21 21 3 = 21 cos ( 240 ) + i sin ( 240 ) = 21 − − i = − − i 2 2 2 2 42. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = (1)( 4 ) cos ( 290 + 40 ) + i sin ( 290 + 40 )
3 1 = 4 cos ( 330 ) + i sin ( 330 ) = 4 − i = 2 3 − 2i 2 2 43. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 6 cos ( 200 − 50 ) + i sin ( 200 − 50 ) = z2 6 = cos (150 ) + i sin (150 ) = −
3 1 + i 2 2
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2 z1 18 = cos (190 − 100 ) + i sin (190 − 100 ) z2 2
44. Using the formula
= 9 cos ( 90 ) + i sin ( 90 ) = 9i
920
Chapter 8 Review
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2 z1 24 cos ( 290 − 110 ) + i sin ( 290 − 110 ) = 4 z2
45. Using the formula
= 6 cos (180 ) + i sin (180 ) = −6 46. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
z1 200 cos ( 93 − 48 ) + i sin ( 93 − 48 ) = z2 2
2 2 = 10 cos ( 45 ) + i sin ( 45 ) = 10 + i = 5 2 + 5 2i 2 2 47. In order to express ( 3 + 3i ) in rectangular form, follow these steps: Step 1: Write 3 + 3i in polar form. Since x = 3, y = 3 , the point is in QI. So, 4
r = x2 + y2 =
(3) + (3) = 3 2 2
2
y 3 = = 1 , so that θ = tan −1 (1) = 45 . x 3 Hence, 3 + 3i = 3 2 cos ( 45 ) + i sin ( 45 ) . tan θ =
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
(3 + 3i ) = (3 2 ) cos ( 4 ⋅ 45 ) + i sin ( 4 ⋅ 45 ) = 324 cos (180 ) + i sin (180 ) 4
4
= 324 [ −1 + 0i ] = −324
921
Chapter 8
(
48. In order to express 3 + 3i
) in rectangular form, follow these steps: 4
Step 1: Write 3 + 3i in polar form. Since x = 3, y = 3, the point is in QI. So,
(3) + ( 3 ) = 2 3
tan θ =
2
2
r = x2 + y2 =
y 3 1 1 , so that θ = tan −1 = = = 30 . 3 x 3 3
Hence, 3 + 3i = 2 3 cos ( 30 ) + i sin ( 30 ) .
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
(3 + 3i ) = ( 2 3 ) cos ( 4 ⋅ 30 ) + i sin ( 4 ⋅ 30 ) = 144 cos (120 ) + i sin (120 ) 4
4
1 3 i = −72 + 72 3i = 144 − + 2 2
(
49. In order to express 1 + 3i
) in rectangular form, follow these steps: 5
Step 1: Write 1 + 3i in polar form. Since x = 1, y = 3 , the point is in QI. So, r = x2 + y2 =
(1) + ( 3 ) = 2 2
y 3 = , so that θ = tan −1 x 1 Hence, 1 + 3i = 2 cos ( 60 ) + i sin ( 60 ) . tan θ =
2
( 3 ) = 60.
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
(1 + 3i ) = ( 2 ) cos (5 ⋅ 60 ) + i sin (5 ⋅ 60 ) = 32 cos (300 ) + i sin (300 ) 5
5
1 3 = 32 − i = 16 − 16 3i 2 2
922
Chapter 8 Review
50. In order to express ( −2 − 2i ) in rectangular form, follow these steps: 7
Step 1: Write −2 − 2i in polar form. Since x = −2, y = −2 , the point is in QIII. So, r = x2 + y2 =
( −2 ) + ( −2 ) = 2 2 2
2
y −2 = = 1, so that θ = 180 + tan −1 (1) = 225. x −2 (Remember to add 180 to tan−1 (1) since the angle has terminal side in QIII.) tan θ =
Hence, −2 − 2i = 2 2 cos ( 225 ) + i sin ( 225 ) .
Step 2: Apply DeMoivre’s theorem z n = r n cos ( n θ ) + i sin ( n θ ) .
( −2 − 2i ) = ( 2 2 ) cos ( 7 ⋅ 225 ) + i sin ( 7 ⋅ 225 ) = 1024 2 − 7
7
2 2 + i 2 2
= −1024 + 1024i 51. Given z = 2 + 2 3i , in order to compute z 2 , follow these steps: Step 1: Write 2 + 2 3i in polar form. Since x = 2, y = 2 3, the point is in QI. So, 1
r = x 2 + y2 =
(2) + (2 3 ) = 4 2
2
y 2 3 = = 3, so that θ = tan −1 x 2 Hence, z = 2 + 2 3i = 4 cos ( 60 ) + i sin ( 60 ) . tan θ =
( 3 ) = 60.
θ 360 k θ 360 k 1 1 + Step 2: Now, apply z n = r n cos + i sin + , k = 0, 1, ... , n − 1: n n n n 60 360 k 60 360 k 1 + + + z 2 = 4 cos i sin , k = 0, 1 2 2 2 2
= 2 cos ( 30 ) + i sin ( 30 ) , 2 cos ( 210 ) + i sin ( 210 )
923
Chapter 8
Step 3: Plot the roots in the complex plane:
52. Given z = −8 + 8 3i , in order to compute z 4 , follow these steps: 1
Step 1: Write z = −8 + 8 3i in polar form. Since x = −8, y = 8 3, the point is in QII. So, r = x2 + y2 =
( −8) + (8 3 ) = 16 2
2
(
)
y 8 3 = = − 3, so that θ = 180 + tan −1 − 3 = 120. x −8 (Remember to add 180 to tan −1 − 3 since the angle has terminal side in QII.) tan θ =
(
)
Hence, z = −8 + 8 3i = 16 cos (120 ) + i sin (120 ) . θ 360 k θ 360 k 1 1 + Step 2: Now, apply z n = r n cos + sin i + , k = 0, 1, ... , n − 1: n n n n 120 360 k 120 360 k 1 z 4 = 4 16 cos + + + i sin , k = 0, 1, 2, 3 4 4 4 4 =
2 cos ( 30 ) + i sin ( 30 ) , 2 cos (120 ) + i sin (120 ) , 2 cos ( 210 ) + i sin ( 210 ) , 2 cos ( 300 ) + i sin ( 300 )
924
Chapter 8 Review
Step 3: Plot the roots in the complex plane:
53. Given z = −256 + 0i , in order to compute z 4 , follow these steps: Step 1: Write z = −256 + 0i in polar form. Since x = −256, y = 0 , the point is on the negative x-axis. So, 1
r = 256, θ = 180
Hence, z = −256 + 0i = 256 cos (180 ) + i sin (180 ) .
θ 360 k θ 360 k 1 1 i sin + Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n 180 360 k 180 360 k 1 sin + + + z 4 = 4 256 cos i , k = 0, 1, 2, 3 4 4 4 4 =
4 cos ( 45 ) + i sin ( 45 ) , 4 cos (135 ) + i sin (135 ) , 4 cos ( 225 ) + i sin ( 225 ) , 4 cos ( 315 ) + i sin ( 315 )
925
Chapter 8
Step 3: Plot the roots in the complex plane:
54. Given z = 0 − 18i , in order to compute z 2 , follow these steps: Step 1: Write z = 0 − 18i in polar form. Since x = 0, y = −18 , the point is on the negative y-axis. So, r = 18, θ = 270 1
Hence, z = 0 − 18i = 18 cos ( 270 ) + i sin ( 270 ) .
θ 360 k θ 360 k Step 2: Now, apply z = r cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1
n
1
n
270 360 k 270 360 k 1 sin z 2 = 18 cos + + i + , k = 0, 1 2 2 2 2 = 3 2 cos (135 ) + i sin (135 ) , 3 2 cos ( 315 ) + i sin ( 315 )
926
Chapter 8 Review
Step 3: Plot the roots in the complex plane:
55. We seek all complex numbers x such that x3 + 216 = 0 , or equivalently x3 = −216 . While it is clear that x = −6 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −216 in polar form. Since x = −216, y = 0 , the point is on the negative x-axis. So, r = 216, θ = π .
Hence, −216 = 216 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 π 2π k π 2π k ( −216 ) 3 = 3 216 cos + + i sin + , k = 0, 1, 2 3 3 3 3
π π 5π 5π = 6 cos + i sin , 6 cos (π ) + i sin (π ) , 6 cos + i sin 3 3 3 3 = 3 + 3 3i , − 6, 3 − 3 3i Step 3: Hence, the complex solutions of x3 + 216 = 0 are 3 + 3 3i , − 6, 3 − 3 3i .
927
Chapter 8
56. We seek all complex numbers x such that x 4 − 1 = 0 . Observe that this equation can be written equivalently as: ( x2 − 1)( x2 + 1) = 0
( x − 1)( x + 1) ( x 2 + 1) = 0
The solutions are therefore x = ±1, ± i . (Note: We could have alternatively used the formula for complex roots, namely θ 360 k θ 360 k 1 1 z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n However, the left-side readily factored in a way that is easy to see what all of the solutions were. As such, factoring was the more expedient route for this problem.) 57. We seek all complex numbers x such that x 4 + 1 = 0 , or equivalently x 4 = −1 . We shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −1 in polar form. Since x = −1, y = 0 , the point is on the negative x-axis. So, r = 1, θ = π .
Hence, −1 = 1 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n
π
4
( −1) = 4 1 cos 1
4
+
2π k π 2π k + i sin + , k = 0, 1, 2, 3 4 4 4
π π 3π 3π = cos + i sin , cos + i sin , 4 4 4 4 5π 5π 7π 7π cos 4 + i sin 4 , cos 4 + i sin 4 2 2 2 2 2 2 2 2 + + i, − i, − − i, − i 2 2 2 2 2 2 2 2 Step 3: Hence, the complex solutions of x 4 + 1 = 0 are =
2 2 2 2 2 2 2 2 + i, − + i, − − i, − i . 2 2 2 2 2 2 2 2
928
Chapter 8 Review
58. We seek all complex numbers x such that x3 − 125 = 0 , or equivalently x3 = 125 . While it is clear that x = 5 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = 125 in polar form. Since x = 125, y = 0 , the point is on the positive x-axis. So, r = 125, θ = 0 .
Hence, 125 = 125 cos ( 0 ) + i sin ( 0 ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 0 2π k 0 2π k 125 3 = 3 125 cos + + i sin + , k = 0, 1, 2 3 3 3 3
2π 2π 4π 4π = 5 cos ( 0 ) + i sin ( 0 ) , 5 cos + i sin , 5 cos + i sin 3 3 3 3 5 5 3 5 5 3 i, − − i = 5, − + 2 2 2 2 Step 3: Hence, the complex solutions of x3 − 125 = 0 are 5 5 3 5 5 3 5, − + i, − − i . 2 2 2 2
929
Chapter 8 59. In order to convert the point ( −2,2 )
(expressed in rectangular coordinates) to polar coordinates, first observe that x = −2, y = 2 , so that the point is in QII. Hence, r=
( −2 ) + ( 2 ) = 2 2 2
2
y −2 = = −1 so that x 2 3π . θ = π + tan −1 ( −1) = 4 (Remember to add π to tan −1 ( −1) since the point is in QII.) So, the point can be expressed in polar
60. In order to convert the point 4, −4 3 (expressed in rectangular
(
coordinates) to polar coordinates, first observe that x = 4, y = −4 3 , so that the point is in QIV. Hence, r=
tan θ =
3π coordinates as 2 2, , plotted 4 below:
)
tan θ =
( 4 ) + ( −4 3 ) = 8 2
2
y −4 3 = = − 3 so that x 4
(
)
θ = tan −1 − 3 = −
π
. 3 Since 0 ≤ θ < 2π , we use the reference 5π . angle θ = 3 So, the point can be expressed in polar 5π coordinates as 8, , plotted below: 3
930
Chapter 8 Review
61. In order to convert the point −5 3, −5 (expressed in rectangular
62. In order to convert the point 3, 3 (expressed in rectangular
coordinates) to polar coordinates, first observe that x = −5 3, y = −5 , so that the point is in QIII. Hence,
coordinates) to polar coordinates, first observe that x = 3, y = 3 , so that the point is in QI. Hence,
(
)
(
)
r=
( −5 3 ) + ( −5 ) = 10
r=
tan θ =
y −5 1 = = so that x −5 3 3
tan θ =
2
2
1 7π . = 3 6
θ = π + tan −1
1 (Remember to add π to tan −1 3 since the point is in QIII.) So, the given point can be expressed in
( 3) + ( 3) = 6 2
2
y 3 = = 1 so that x 3
θ = tan −1 (1) =
π
. 4 So, the given point can be expressed in
π polar coordinates as 6, , which is 4 plotted below:
7π polar coordinates as 10, , which is 6 plotted below:
931
Chapter 8 63. In order to convert the point ( 0, −2 )
64. In order to convert the point (11,0 )
(expressed in rectangular coordinates) to polar coordinates, first observe that x = 0, y = −2 , so that the point is on the negative y-axis. Hence, 3π 2 2 r = ( 0 ) + ( −2 ) = 2, θ = 2 So, the given point can be expressed in
(expressed in rectangular coordinates) to polar coordinates, first observe that x = 11, y = 0 , so that the point is on the positive x-axis. Hence,
3π polar coordinates as 2, , which is 2 plotted below:
r=
2 2 (11) + ( 0 ) = 11, θ = 0
So, the given point can be expressed in polar coordinates as (11,0 ) , which is plotted below:
5π 65. In order to convert the point −3, (expressed in polar coordinates) to 3 5π . Hence, rectangular coordinates, first observe that r = −3, θ = 3 3 5π 1 x = r cos θ = −3cos = −3 = − 2 3 2 3 3 3 5π y = r sin θ = −3sin = = −3 − 2 3 2 3 3 3 So, the given point can be expressed in rectangular coordinates as − , . 2 2
932
Chapter 8 Review
5π 66. In order to convert the point 4, (expressed in polar coordinates) to 4 5π . Hence, rectangular coordinates, first observe that r = 4, θ = 4 2 5π x = r cos θ = 4 cos = −2 2 = 4 − 4 2 2 5π y = r sin θ = 4 sin = −2 2 = 4 − 4 2 So, the given point can be expressed in rectangular coordinates as
( −2 2, − 2 2 ) .
π 67. In order to convert the point 2, (expressed in polar coordinates) to 3 rectangular coordinates, first observe that r = 2, θ =
π
. Hence, 3 π 1 x = r cos θ = 2 cos = 2 = 1 3 2 3 π y = r sin θ = 2 sin = 2 = 3 2 3
(
)
So, the given point can be expressed in rectangular coordinates as 1, 3 . 7π 68. In order to convert the point 6, (expressed in polar coordinates) to 6 7π . Hence, rectangular coordinates, first observe that r = 6, θ = 6 3 7π x = r cos θ = 6 cos = −3 3 = 6 − 6 2 7π 1 y = r sin θ = 6 sin = 6 − = −3 6 2 So, the given point can be expressed in rectangular coordinates as
933
( −3 3, − 3) .
Chapter 8 4π 69. In order to convert the point 1, (expressed in polar coordinates) to 3 4π . Hence, rectangular coordinates, first observe that r = 1, θ = 3 1 4π x = r cos θ = 1cos =− 2 3 3 4π y = r sin θ = 1sin =− 2 3 1 3 So, the given point can be expressed in rectangular coordinates as − , − . 2 2
7π 70. In order to convert the point −3, (expressed in polar coordinates) to 4 7π . Hence, rectangular coordinates, first observe that r = −3, θ = 4 2 3 2 7π x = r cos θ = −3cos = − = −3 2 4 2 2 3 2 7π y = r sin θ = −3sin = = −3 − 2 2 4 3 2 3 2 So, the given point can be expressed in rectangular coordinates as − , . 2 2
934
Chapter 8 Review
71. In order to graph r = 4 cos 2θ , consider the following table of points:
θ
r = 4 cos 2θ
0
4
π
0
π
4 2
3π
4
π 5π 3π 7π
−4 0 4
4 2 4
2π
0 −4 0 4
72. In order to graph r = sin3θ , consider the following table of points:
( r,θ ) ( 4,0 )
( 0,π 4 ) ( −4,π 2 ) ( 0,3π 4 )
θ
r = sin3θ
0
0
π
1
6
2π
π
( 4, π )
6 2
4π
( 0, 5π 4 ) ( −4,3π 2 ) ( 0, 7π 4 )
5π
6 6
π 7π
( 4,2π )
8π 3π
The graph is as follows:
0 −1 0 1 0
6 6 2
10π 11π 2π
6 6
−1 0 1 0 −1 0
The graph is as follows:
935
( r,θ ) ( 0,0 )
(1,π 6 )
( 0, 2π 6 ) ( −1,π 2 ) ( 0, 4π 6 ) (1,5π 6 ) ( 0,π )
( −1, 7π 6 ) ( 0, 8π 6 ) (1,3π 2 ) ( 0,10π 6 ) ( −1,11π 6 ) ( 0,2π )
Chapter 8 73. In order to graph r = −θ , consider the following table of points:
θ
r = −θ
0
0
π
−π
2 π
3π
2 −π
2
− 3π
2π
−2π
2
The graph is as follows:
74. In order to graph r = 4 − 3sin θ , consider the following table of points:
( r,θ ) ( 0,0 )
( − π 2 ,π 2 )
θ
r = 4 − 3sin θ
0
4
π
1
2 π
( −π , π )
( − 3π 2 , 3π 2 )
3π
( −2π , 2π )
2
2π
4 7 4
The graph is as follows:
936
( r,θ ) ( 4,0 )
(1,π 2 ) ( 4,π )
(7,3π 2 ) ( 4,2π )
Chapter 8 Review
The graph is as follows:
75. In order to graph the curve defined parametrically by x = sin t, y = 4 cos t, t in [ −π , π ] ,
consider the following sequence of tabulated points: t x = sint y = 4 cost −π 0 −4 π −1 0 − 2 0 0 4 π 1 0 2 π 0 −4 Eliminating the parameter reveals that the graph is an ellipse. Indeed, observe that 2
y x = sin t, = cos 2 t 4 y2 so that summing then yields x 2 + = 1. 16 2
2
The graph is as follows:
76. In order to graph the curve defined parametrically by x = 5sin2 t, y = 2 cos 2 t, t in [ −π , π ] ,
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, x y = sin2 t, = cos 2 t 5 2 so that adding the two equations yields x y + = 1 , 0 ≤ x ≤ 5, 0 ≤ y ≤ 2 , 5 2 which is a line segment.
937
Chapter 8
The graph is as follows:
77. In order to graph the curve defined parametrically by x = 4 − t 2 , y = t 2 , t in [ −3,3] ,
it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since t 2 = y , substituting this into the expression for x yields x = 4 − y x in [ −5, 4 ] .
The graph is as follows:
78. In order to graph the curve defined parametrically by x = t + 3, y = 4, t in [ −4, 4 ] ,
observe that since the y-coordinate is always 4, the path is a horizontal line segment starting at the point (−1, 4) and ends at (7, 4).
79. In order to express the parametrically-defined curve given by x = 4 − t 2 , y = t as a rectangular equation, we eliminate the parameter t by simply substituting y = t into
the equation for x. So, the equation is x = 4 − y2 . 80. In order to express the parametrically-defined curve given by x = 5sin2 t, y = 2 cos 2 t as a rectangular equation, we eliminate the parameter t. y x = cos 2 t , so that adding the two equations yields Indeed, observe that = sin2 t, 2 5 x y + = 1 , which is equivalent to 2 x + 5 y = 10 . 5 2
938
Chapter 8 Review
81. In order to express the parametrically-defined curve given by x = 2 tan2 t, y = 4 sec 2 t as a rectangular equation, we eliminate the parameter t in the following sequence of steps: sin2 t 1 − cos 2 t = x = 2 tan2 t = 2 2 2 2 cos t cos t 1 = 2 − 2 = 2 ( sec 2 t ) − 2 2 t cos 2 1 = ( 4 sec 2 t ) − 2 = y − 2 4 2 So, the equation is y = 2 x + 4 .
So, the equation is y = x − 9 . 83. 2,868 − 6,100i 84. 0.0028 + 0.0096i
85. From the calculator, we obtain: − 23 − 11i = 12
(
82. In order to express the parametrically-defined curve given by x = 3t 2 + 4, y = 3t 2 − 5 as a rectangular equation, we eliminate the parameter t in the following sequence of steps: 3t 2 = x − 4 so that y = ( x − 4) − 5 = x − 9 .
86. From the calculator, we obtain: 11 + 23i = 12
)
(
angle − 23 − 11i ≈ 66 , so we use 246
)
angle 11 + 23i ≈ 78
So, − 23 − 11i ≈ 12 ( cos 246 + i sin 246 ) .
So, 11 + 23i ≈ 12 ( cos 78 + i sin 78 ) .
87. Given z = −8 + 8 3 i , in order to compute z 4 , follow these steps: 1
Step 1: Write z = −8 + 8 3 i in polar form. Since x = −8, y = 8 3 , the point is in QII. So, using the calculator yields r = 16, θ = 120
Hence, z = −8 + 8 3 i = 16 cos (120 ) + i sin (120 ) .
θ 360 k θ 360 k 1 1 + i sin Step 2: Now, apply z n = r n cos + + , k = 0, 1, ... , n − 1 : n n n n 120 360 k 120 360 k 1 + + + z 4 = 4 16 cos i sin , k = 0, 1, 2, 3 4 4 4 4
939
Chapter 8
Step 3: Plot the roots in the complex plane and connect consecutive roots to form a square:
88. Given z = 8 3 + 8i , in order to compute z 4 , follow these steps: 1
Step 1: Write z = 8 3 + 8i in polar form. Since x = 8 3, y = 8 , the point is in QI. So, using the calculator yields r = 16, θ = 30
Hence, z = −8 + 8 3 i = 16 cos ( 30 ) + i sin ( 30 ) .
θ 360 k θ 360 k 1 1 + Step 2: Now, apply z n = r n cos + i sin + , k = 0, 1, ... , n − 1 : n n n n 30 360 k 30 360 k 1 + + + z 4 = 4 16 cos i sin , k = 0, 1, 2, 3 4 4 4 4
940
Chapter 8 Review
Step 3: Plot the roots in the complex plane and connect consecutive roots to form a square:
89. Consider the following graph of r = 1 − 2 sin3θ :
90. Consider the following graph of r = 1 + 2 cos 3θ :
Note that the curve self-intersects at the origin, multiple times. This occurs when r = 0. To find these angles for which this is true, we solve the following equation: 0 = 1 − 2 sin3θ
Note that the curve self-intersects at the origin, multiple times. This occurs when r = 0. To find these angles for which this is true, we solve the following equation: 0 = 1 + 2 cos 3θ cos 3θ = 12
sin3θ = 12 3θ = π6 , 56π , 136π , 176π , 256π , 296π
3θ = π3 , 53π , 73π , 113π , 133π , 173π
θ = 18π , 518π , 1318π , 1718π , 2518π , 2918π
θ = π9 , 59π , 79π , 119π , 139π , 179π
941
Chapter 8
91. First, we consider the curve defined by the following parametric equations: x = a cos at + b sin bt, y = a sin at + b cos bt
Graph for a = 2, b = 3
Graph for a = 3, b = 2
All curves are generated in 2π units of time. 92. First, we consider the curve defined by the following parametric equations: x = a sin at − b cos bt, y = a cos at − b sin bt
Graph for a = 1, b = 2
Graph for a = 2, b = 1
All curves are generated in 2π units of time.
942
Chapter 8 Practice Test Solutions ------------------------------------------------------------------1. Observe that
2. The complex number –3 – 5i is plotted below:
i 20 = i 4(5) = i 4(5) = ( i ) = (1) = 1, 4 5
5
i 13 = i 4(3)+1 = i 4(3)i 1 = ( i 4 ) i 1 = (1) (i ) = i 3
3
i 17 = i 4( 4 )+1 = i 4( 4 )i1 = ( i 4 ) i 1 = (1) (i ) = i , 4
4
i 1000 = i 4(250 ) = i 4(250 ) = ( i 4 )
250
= (1)
250
=1
So, i 20 − i 13 + i 17 − i 1000 = 1 − i + i − 1 = 0
3. 2 − 4i 2 − 4i i 4 + 2i 4 2 = ⋅ = = − − i 3i 3i i −3 3 3
4. 8 − i 8 − i 3 − i 24 − 3i − 8i − 1 = ⋅ = 3+i 3+i 3−i 9 +1 23 − 11i 23 11 = = − i 10 10 10
5. In order to express 3 2 (1 − i ) in polar
6. First, note that 2 6 3 2 − 6i = 6 12 − 12i = 12 3 − 12i.
form, observe that x = 3 2, y = −3 2 , so that the point is in QIV. Now, r = x 2 + y2 =
(
3 2
) ( 2
+ −3 2
)
2
(
So, r=
=6
y −3 2 = = −1 , so that x 3 2 θ = tan−1 ( −1) = −45 .
tan θ =
)
(12 3 ) + ( −12 ) = 24 and θ = tan ( ) = . 2
2
−1
−12 12 3
11π 6
Hence, 24 ( cos 116π + i sin 116π ) .
Since 0 ≤ θ < 2π , we must use the reference angle 315 . Hence,
3 2 (1 − i ) = 6 cos ( 315 ) + i sin ( 315 ) .
943
Chapter 8
7. The modulus is r=
( −3 ) + (3 3 ) 2
2
8. The modulus of –5 – 5i is
= 6, and the
( )
argument is θ = tan −1 3−33 = 23π since in QII.
= 5 2 . Since tan θ =
( −5 ) + ( −5 ) 2
2
−5 = 1 , the argument −5
is θ = 180 + tan −1 (1) = 225 .
9. Using the formula z1z2 = rr 1 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) yields
z1z2 = ( 4 )( 2 ) cos ( 75 + 15 ) + i sin ( 75 + 15 ) = 8 cos ( 90 ) + i sin ( 90 ) = 8 [ 0 + i ] = 8i
10. Using the formula
z1 r1 = cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) yields z2 r2
1 z1 4 3 = cos ( 75 − 15 ) + i sin ( 75 − 15 ) = 2 cos ( 60 ) + i sin ( 60 ) = 2 + i z2 2 2 2 = 1 + 3i 11. Given that z = 16 cos120 + i sin120 , we apply DeMoivre’s theorem
z n = r n cos ( n θ ) + i sin ( n θ ) to compute z 4 . Indeed, observe that
1 3 4 i = 32,768 −1 + 3i z 4 = (16 ) cos ( 4 ⋅120 ) + i sin ( 4 ⋅120 ) = 65,536 − + 2 2 12. Given that z = 16 cos120 + i sin120 , we apply
θ 360 k θ 360 k 1 1 + z n = r n cos + i sin + , k = 0, 1, ... , n − 1 , n n n n 1
to compute z 4 . Indeed, observe that 120 360 k 120 360 k 1 z 4 = 4 16 cos + + i + sin , k = 0, 1, 2, 3 4 4 4 4 = 2 cos ( 30 ) + i sin (30 ) , 2 cos (120 ) + i sin (120 ) , 2 cos ( 210 ) + i sin ( 210 ) , 2 cos ( 300 ) + i sin ( 300 ) , =
3 + i , − 1 + 3i , − 3 − i , 1 − 3i
944
Chapter 8 Practice Test
13. The point ( a,θ ) can also be represented as ( −a,θ ± 180 ) . This is because the
point ( −a,θ ) is diametrically opposed (through the origin) to the given point. As
such, we need to rotate from it 180 clockwise or counterclockwise to arrive at the given point. 14. In order to convert the point ( 3, 210 ) (expressed in polar coordinates) to
rectangular coordinates, first observe that r = 3, θ = 210. Hence, 3 3 3 x = r cosθ = 3cos(210 ) = 3 − = − 2 2 3 1 y = r sin θ = 3sin(210 ) = 3 − = − 2 2
3 3 3 So, the given point can be expressed in rectangular coordinates as − , − 2 2
15. In order to convert the point ( −3, −4 ) (expressed in rectangular coordinates) to
polar coordinates, first observe that x = −3, y = −4 , so that the point is in QIII. Hence, r=
( −3) + ( −4 ) = 5 2
2
y −4 4 so that θ = 180 + tan −1 ≈ 233. = x −3 3 4 (Remember to add 180 to tan −1 since the point is in QIII.) 3 tan θ =
So, the given point can be expressed in polar coordinates as ( 5,233 ) .
945
Chapter 8 16. In order to graph r = 6 sin 2θ , consider the following table of points:
θ
r = 6 sin 2θ
0
0
π
6
π
4 2
3π
4
π 5π
2
7π
−6 0
4
3π
0
4
2π
6 0 −6 0
( r,θ ) ( 0,0 )
(6,π 4 ) ( 0,π 2 )
( −6,3π 4 ) ( 0,π )
(6,5π 4 ) ( 0, 3π 2 ) ( −6, 7π 4 ) ( 0,2π )
17. In order to graph r = 1 + cos3θ , consider the following table of points:
θ
r = 1 + cos3θ
0
2
π
1
6
2π
π
6 2
4π
6
0 1 2
The graph is as follows:
( r,θ ) ( 2,0 )
(1,π 6 )
( 0, 2π 6 ) (1,π 2 ) ( 2, 4π 6 )
946
The graph is as follows:
Chapter 8 Practice Test
18. The graph is a spiral, as shown below:
19. In order to graph r 2 = 9 cos 2θ , consider the following table of points:
θ
r 2 = 9 cos 2θ
r
0
9
±3
π
0
0
−9 0
No points 0
9
±3
0
0
−9 0
No points 0
9
±3
π
4 2
3π
4
π 5π 3π 7π
4 2 4
2π
The graph is as follows:
( r,θ ) ( ±3,0 )
( 0,π 4 ) ( 0,3π 4 ) ( ±3, π )
( 0, 5π 4 ) ( 0, 7π 4 ) ( ±3,2π )
20. Given the parametric equations x = 1 − t , y = t , t in [ 0,1] ,
observe that since t = y 2 , we have x = 1 − y 2 . Now, since y = t is always ≥ 0 and x = 1 − y 2 is also always ≥ 0 , this generates a quarter circle in QI.
947
Chapter 8
21. From the given information, we know that the parametric equations are: x = (120 cos 45 ) t
y = −16t 2 + (120 sin 45 ) t + 0
We need to determine t0 such that y ( t0 ) = 0 . Then, we will compute x ( t0 ) = total horizontal distance traveled
Observe that solving −16t 2 + (120 sin 45 ) t = 0 yields −16t 2 + (120 sin 45 ) t = 0 t ( −16t + 120 sin 45 ) = 0
120 sin 45 ≈ 5.3 sec. 16 Now, we conclude that x ( 5.3 ) = 450 ft. = total horizontal distance traveled t = 0,
22. Note that A is 150 ( cos 0 + i sin 0 ) ,
23. Plane: u = 400 ( cos 205 + i sin 205 ) Wind: v = 24 ( cos 270 + i sin 270 )
Hence,
Resultant: u + v = 400 cos 205 , 400 sin 205 − 24 So, u + v ≈ 411 mph and
B is 250 ( cos 45 + i sin 45 ) .
A + B = 150 + 250
( ) , 250 ( ) . 2 2
2 2
The magnitude (or force) is A + B ≈ 372 and the direction is 2 θ = tan −1 150125 ≈ 16.6 , +125 2
(
)
− 24 θ = tan −1 ( 400sin205 ≈ 28 . So, 400 cos205 )
u + v ≈ 411( cos 208 + i sin 208 ) .
so 73.4 off north. 24. The arrow position is 50 cos 60 ,50 sin 60 . So, the horizontal ground distance
the arrow travels is 50 cos 60 = 50 ( 12 ) = 25 ft. 25. Evaluate the given equations at t=10: x(10) ≈ 96, y(10) ≈ 21. So, the desired position is (96, 21).
948
Chapter 8 Practice Test
26. We seek all complex numbers x such that x3 + 8 = 0 , or equivalently x3 = −8 . While it is clear that x = −2 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = −8 in polar form. Since x = −8, y = 0 , the point is on the negative x-axis. So, r = 8, θ = π .
Hence, −8 = 8 cos (π ) + i sin (π ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 π 2π k π 2π k ( −8) 3 = 3 8 cos + + i sin + , k = 0, 1, 2 3 3 3 3
π π 5π 5π = 2 cos + i sin , 2 cos (π ) + i sin (π ) , 2 cos + i sin 3 3 3 3 = −2, 1 − 3i , 1 + 3i Step 3: Hence, the complex solutions of x3 + 8 = 0 are −2, 1 − 3i , 1 + 3i . The triangle is as follows:
4 3 2 1
−4
−3
−2
−1
1 −1 −2 −3 −4 5
949
2
3
4
5
Chapter 8 Cumulative Test Solutions -----------------------------------------------------------1. The Pythagorean theorem yields a 2 + 52 = 132 , so that a = 12.
2. Note that tan θ = 201 = 0.05 . So, the grade is 5%.
3. Consider the following triangle:
π 4. 140 180 = 79π
( )
5. Using the formula s = rθ d
( )
( ) yields π
180
s = ( 900 km ) (15 ) 180π = 75π km
−
θ
6. First, note that π v = r ω = 10sec. ( 4.2 in.) . Since 5 minutes
(
)
corresponds to 300 seconds, we have π s = vt = 10sec. ( 4.2 in.)( 300 sec.) ≈ 396 in.
(
)
= 33 ft. Observe that sec θ = cos1 θ = − 157 = − 115757 . 11
7. Amplitude = 0.009 cm and frequency = 1/period = 400 hertz (since period 2π 1 800π = 400 ).
8. Some information about the graph is as follows: Period 2 Amplitude 3 Phase shift 1 to the right Vertical shift down 2 units
The graph is to the right:
950
Chapter 8 Cumulative Test
9. Note that the asymptotes occur when sin ( 14 x ) = 0 , which occurs when x = 0, 4π . The graph is to the right:
10. Conditional, because cot 2 x − csc 2 x = cot 2 x − (1 + cot 2 x )
11. tan ( 715π − 215π ) = tan ( π3 )
= −1 ≠ 1.
12. Consider the following triangle:
Observe that cos x = − 53 , sin x = 54 . So, 1 1 sec 2x = = 2 cos 2 x cos x − sin2 x 1 25 = = − 2 2 7 ( − 35 ) − ( 54 )
951
Chapter 8
13. 2
1 1 1 1 2 x 2 x sec ( 2 ) − csc ( 2 ) = − = − cos ( x ) sin ( x ) ± 1 + cos x ± 1 − cos x 2 2 2 2 2 (1 − cos x ) − 2 (1 + cos x ) 2 2 = − = 1 + cos x 1 − cos x (1 + cos x )(1 − cos x ) 2
=
2
2
−4 cos x −4 cos x cos x 1 = = −4 = −4 cot x csc x 2 2 1 − cos x sin x sin x sin x 15. −1.56
14. cos A + cos B 2 cos ( A+2 B ) cos ( A−2 B ) = sin A + sin B 2 sin ( A+2 B ) cos ( A−2 B )
= cot ( A+2 B )
16. Let x = sec −1
( ) . Then, sec x = 29 2
so that cos x = 229 . Consider the following triangle:
29 2
,
17. Note that 5tan θ + 7 = 0 implies that tan θ = − 75 and so, θ = tan −1 ( − 75 ) .
This is approximately −54.46 . We need the positive measures of the angles in QII and QIV. These are 180 − 54.46 = 125.54 and 360 − 54.46 = 305.54 .
18. Observe that sin 2θ + 13 sin θ = 0 3 ( 2 sin θ cos θ )
We conclude that tan x = 52 .
+ sin θ = 0 sin θ ( 6 cos θ + 1) = 0 So, sin θ = 0, 6 cos θ + 1 = 0 . The solutions of the first equation are θ = 0 , 180 . The solution of the second equation is θ = cos −1 ( − 61 ) ≈ 95.59 . We also need the QIII angle, which is 360 − 95.59 = 260.41 .
952
Chapter 8 Cumulative Test
19. No triangle 20. This is SAS, so begin by using Law of Cosines: Step 1: Find c. 2 2 c 2 = a 2 + b 2 − 2ab cos γ = ( 8 ) + (13 ) − 2 ( 8 )(13 ) cos (19 ) c≈6 Step 2: Find α . 13sin19 sin α sin19 13sin19 = sin α = α = sin −1 ≈ 45 13 6 6 6 Step 3: Find β . β ≈ 180 (19 + 45 ) = 116
21. A = 12 23.
( 13 )( 11) sin31 ≈ 3.1
22. 15 cos110 ,15sin110 ≈ −5.1,14.1
4 + i 2 + 3i 8 + 2i + 12i − 3 5 14 ⋅ = = + i 2 − 3i 2 + 3i 4+9 13 13
24. We seek all complex numbers x such that x3 − 27 = 0 , or equivalently x3 = 27 . We shall apply the approach for computing complex roots involving the formula 1 1 θ 2π k θ 2π k z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 . n n n n To this end, we follow these steps. Step 1: Write z = 27 in polar form. Since x = 27, y = 0 , the point is on the positive x-axis. So, r = 27, θ = 0 .
Hence, 27 = 27 cos ( 0 ) + i sin ( 0 ) . 1 1 θ 2π k θ 2π k Step 2: Now, apply z n = r n cos + + i sin + , k = 0, 1, ... , n − 1 : n n n n 1 0 2π k 0 2π k 27 3 = 3 27 cos + + i sin + , k = 0, 1, 2 3 3 3 3
2π 2π 4π 4π = 3 cos ( 0 ) + i sin ( 0 ) , 3 cos + i sin , 3 cos + i sin 3 3 3 3 = 3, − 32 + 3 2 3 i , − 23 − 3 2 3 i Step 3: Hence, the complex solutions of x3 − 27 = 0 are 3, − 32 + 3 2 3 i , − 32 − 3 2 3 i .
953
Chapter 8 25. The graph of θ = − π4 is a line, as seen below:
954