KT Classroom: Trigonometry Graphs and Equations

MATHEMATICS Learner’s Study and Revision Guide for Grade 12 SOLUTION OF TRIGONOMETRIC EQUATIONS AND INEQUALITIES

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Preparation for the Mathematics examination brought to you by Kagiso Trust

Contents Unit 17 Solving trigonometric equations and inequalities graphically

3

Solving trigonometric equations analytically – Type 1

5

Other inequalities that can be solved graphically

6

Solving trigonometric equations analytically – Type 2

7

Solving trigonometric equations analytically – Type 3

9

Answers to exercises

10

Refining techniques for sketching trigonometric functions

17

Examination questions with solution hints and answers

21

More questions from past examination papers

26

Answers

32

How to use this revision and study guide 1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

Solution of trigonometric equations and inequalities analytically and graphically

REVISION UNIT 17: TRIGONOMETRIC EQUATIONS AND INEQUALITIES SOLVING TRIGONOMETRIC EQUATIONS AND INEQUALITIES GRAPHICALLY Example 1: Below are shown the graphs of f ( x ) = cos

[

]

(

)

x and g ( x ) = sin x − 30o for 2

x ∈ − 180o ; 180o . y

1

g

B

f

A x -180

-160

-140

-120

-100

-80

-60

-40

-20

20

40

60

80

100

120

140

160

180

-1

1. Where the graphs meet is where f ( x ) = g (x ) The graphs meet at the point A where x = −160 o and at the point B where x = 80 o .

So the solution of the equation

f ( x ) = g (x ) or cos

[

(

)

x = sin x − 30o 2

]

for values of x ∈ − 180o ; 180o is x = −160 o and x = 80 o .

2. To find out the values of x for which f ( x ) > g (x ) , look at the graphs and see in what interval or intervals the graph of f is above the graph of g .

Answer is − 160 o < x < 80 o .

3

Preparation for the Mathematics examination brought to you by Kagiso Trust

Example 2: Below are shown again the graphs of f ( x ) = cos

[

]

(

)

x and g ( x ) = sin x − 30o for 2

x ∈ − 180o ; 180o . y

1

g

B

f

A x -180

-160

-140

-120

-100

-80

-60

-40

-20

20

40

60

80

100

120

140

160

180

-1

1. For what values of x is

f (x ) < 0? g (x )

The answer will be given by values of x for which f ( x ) and g ( x ) have opposite signs (one negative and the other positive).

(

)

Looking at the graph the answer is given by x ∈ − 150o ; 30o .

3. At what values of x is: a)

b)

c)

f ( x ) − g ( x ) = 1?

Answer : x =

f ( x ) − g ( x ) = 1,5 ?

Answer: x =

f ( x ) − g ( x ) = −1?

Answer: x =

Solution of trigonometric equations and inequalities analytically and graphically

SOLVING TRIGONOMETRIC EQUATIONS ANALYTICALLY TYPE 1: REPLACING tan θ by

sin θ whenever tan θ appears in cos θ

the equation to be solved

Example 3: Solve sin 2 x =

1 tan x for values of − 90 o ≤ x ≤ 180 o .

2 1 tan x 2

Solution:

sin 2 x =

1 sin x

2 sin x cos x = .

2 cos x

4 sin x cos 2 x = sin x

4 sin x cos 2 x − sin x = 0

sin x 4 cos 2 x − 1 = 0

Either

sin x = 0

or 4cos 2 x ‐ 1 = 0

Therefore

(

)

x = 0 o or 180 o

1

2 x = −60 o ; 60 o ; or 120 o

Cos x = ±

or

Verifying the above analytical solutions by means of graphs The graphs of f ( x ) = sin 2 x and g ( x ) =

1 tanx are drawn below on the same set of axes. The solution 2

to the equation is then given by the x ‐coordinates of the points of intersection of the two graphs indicated by the arrows in the diagram below.

y

g

1

↓ -80

-70

↑ -60

-50

-40

-30

-20

-10

↓ 10

20

30

40

-1

5

f

50

60

x

70

80

90

100

110

↑ 120

130

140

150

160

170

↑

Preparation for the Mathematics examination brought to you by Kagiso Trust

ONCE MORE ON SOLVING TRIGONOMETRIC INEQUALITIES As shown at the beginning of this unit, we can use graphs to solve not only trigonometric equations as in Example 1 but also trigonometric inequalities as in Example 2. Let’s use the above example to explain further how you would solve trigonometric inequalities.

For example, we can use the graphs to determine the values of x for which

g (x ) > f (x ) or, in other words

1 tan x > sin 2 x 2

All you need to do is to look at the graphs to find the range of values of x for which the y values of the graph of g ( x ) are greater than the y values of the graph of f ( x ) .

Simply explained, this means that look for the range of x values for which the graph of g ( x ) is above

the graph of f ( x ) .

Answer: − 60 o < x < 0 o or 60 o < x < 90 o ; or 120 o < x < 180 o

OTHER INEQUALITIES THAT CAN BE SOLVED GRAPHICALLY By looking at the graphs of say any f ( x ) and any g ( x ) drawn on the same set of axes, we can also solve inequalities in the form:

f (x ) < 0 g (x )

or

f (x ).g (x ) > 0

In the first case (which we looked at in Example 1), we would look for a range of x values for which

f ( x ) and g ( x ) have opposite signs. That is, a range of x values in which the graphs of f ( x ) and g ( x ) lie on opposite sides of the x ‐axis.

In the second case, we would look for a range of x for which both f ( x ) and g ( x ) are positive or

negative. That is, a range of x values in which the graphs of f ( x ) and g ( x ) lie on the same side of the

x ‐axis.

GENERAL SOLUTIONS General solutions of trigonometric equations covered in Unit 16 are applied in the sections that follow.

Commit them to memory or learn to derive them by repeatedly practising to write them down because they are not included on the information sheet that will be given to you in the examination. They are referred to in the next section.

Solution of trigonometric equations and inequalities analytically and graphically

(

TYPE 2: USING sin θ = cos 90o − θ

)

or

(

cosθ = sin 90o − θ

)

Suppose you are solving the equation:

sin x = sin α

Obviously in this case the two sides could not be equal unless

x = α

Because sine is positive in the first and second quadrants we must in fact have

x = α in the first quadrant and x = 180 o − α in the second quadrant

Furthermore, sine has a period of 360 o ; so these solutions repeat themselves every 360 o .

Hence the general solutions are:

x = α + n.360 o and x = 180 o − α + n.360 o where n ∈ Ζ

Does this solution ring a bell?

It is in fact the general solution you met in Unit 16 with calc ∠ replaced by α .

So too application of the general solutions to:

cos x = cosα

will result in the solution:

x = ±α + n.360 o

and application of the general formula to:

tan x = tan α

will result in the solution:

x = a + n.180 o

We now show how the above together with the property of complementary angles in the box at the top of this page is used to solve equations.

(

)

7

Preparation for the Mathematics examination brought to you by Kagiso Trust

Example 4: Solve the equation sin x = cos 2 x

You can proceed as in either column 1 or column 2 below:

sin x = cos 2 x

sin x = sin 90o − 2 x

The general solution is thus:

x = 90o − 2 x + n.360o

and

x = 180o − 90o − 2 x + n.360o

Example 5: Solve the equation sin x = cos x − 45o

(

sin x = cos 2 x cos 2 x = sin x cos 2 x = cos 90o − x

The general solution is thus:

2 x = 90o − x + n.360o

and

2 x = − 90o − x + n.360o

)

(

(

)

(

(

)

(

)

)

(

)

)

You can proceed as in either column 1 or column 2 below:

o sin x = cos x − 45o sin x = cos x − 45

cos x − 45o = sin x sin x = sin 90o − x − 45o cos x − 45o = cos 90o − x

The general solution is thus: The general solution is thus:

x = 90o − x − 45o + n.360o x − 45o = 90o − x + n.360o and and x = 180o − 90o − x − 45o + n.360o x − 45o = − 90o − x + n.360o

Exercise 17.1: Use both answers in the above examples to find solutions for values of x ∈ − 270 o ;180 o

(

)

{

{

(

(

)}

{

( (

)}

(

)}

( ) )

(

) (

(

) (

)

(

)

)

)

[

]

Exercise 17.2: Do rough sketches to show how the above equations can be solved graphically. Where in the sketches do you find the solutions to the equations?

Exercise 17.3: Looking at the sketches, for what values of x is a) sin x ≥ cos 2 x

(

)

b) cos x − 45 o > sin x

Solution of trigonometric equations and inequalities analytically and graphically

TYPE 3: USING DOUBLE ANGLE FORMULAE

Use of a double angle formula will apply if the equation includes a trigonometric ratio of a single angle and another of a double angle. For example, x and 2 x . The idea is to reduce all the ratios to one of a single angle. Factorization or use of the quadratic formula will usually follow substitution using a double angle formula.

Example 6: Solve 2 cos x = sin 2 x by factorizing.

2 cos x = sin 2 x 2 cos x = 2 sin x cos x

cos x = sin x cos x

cos x − sin x cos x = 0 which you can now factorise : cos x(1 − sin x ) = 0

Either cos x = 0 or 1 − sin x = 0

Solution:

Exercise 17.4: Complete the above example by finding the solutions for − 270 o ≤ x ≤ 180 o .

Exercise 17.5: Do rough sketches to show how the equation in above example can be solved graphically.

Exercise 17.6: Looking at your rough sketches, for what values of x is 2 cos x ≥ sin 2 x ?

Exercise 17.7: Looking at your rough sketches, for what values of x is 2 cos x sin 2 x < 0 ?

Example 7: Solve sin x = cos 2 x for − 270 o ≤ x ≤ 180 o using the quadratic formula.

Solution Step 1: sin x = cos 2 x

Step 2: sin x = 1 − 2 sin 2 x

Step 3: 2 sin 2 x + sin x − 1 = 0

Thus

sin x =

Explanation The third step is a quadratic equation in sin x with

a = 2, b = 1 and c = −1

Substitute for a, b and c

or solve by factorization.

− b ± b 2 − 4ac

2a

Exercise 17.8: Complete the above example by finding the solutions for − 270 o ≤ x ≤ 180 o .

Exercise 17.9: Instead of using the quadratic formula, use factorization to solve the above problem.

Exercise 17.10: Verify the solutions of Exercises 17.8 and 17.9 graphically.

9

Preparation for the Mathematics examination brought to you by Kagiso Trust

ANSWERS EXERCISE 17.1

Solution to Example 6: The solutions to the first example are fully worked out in order to show you how to work them out.

Bear in mind that we are looking for solutions in the range [−270 o ;180 o ] .

First answer in the first column:

x = 90 o − 2 x + n.360 o

(

)

x + 2 x = 90 + n.360o 3x = 90o + n.360o x = 30o + n.120o o

n = −3 : n = −2 : n = −1 : n = 0 : n = 1 : n = 2 :

x = 30o − 360o = −330o x = 30o − 240o = −210o x = 30o − 120o = −90o x = 30o x = 30o + 120o = 150o x = 30o + 240o = 270o

Simplifies to

n = −1 :

x = −90 − n.360

Reject this answer. Why?

Reject this answer. Why?

Second answer in the first column:

x = 180 o − 90 o − 2 x + n.360 o

(

o

x = −90o + 360o x = −90o

)

o

No need to complete the answer. Why?

n = 0 :

Is there need to complete this? Why? n = 1 :

Combining the two sets of answers and writing them in ascending (or descending) order we have all the solutions in the interval [−270 o ;180 o ] :

Answers: x = −210 o ; − 90 o ; 30 o ; 150 o

NOTE: Using the second column would yield the same answers. Give yourself practice by trying this.

Solution to Example 7: Working out the general solutions for n ∈ {−,−,−2;−1;0;1;2;−;−;−;} , the answers are given by: x = −112,5o ; 67,5o

Solution of trigonometric equations and inequalities analytically and graphically

EXERCISE 17.2 Sketch of Example 4 y

1

sin x

↓

↓ A

-270

-240

-210

-180

-150

-120

-90

↓

↓

D

C

-60

-30

30

60

90

120

150

x 180

cos 2x -1

B

Graphical solutions are given by the x ‐coordinates of the points A, B, C and D which are the points of intersection of the two graphs. The arrows indicate the solutions.

Sketch of Example 5 y

1

sin x

B

↓ -270

-240

-210

-180

-150

-120

↓ -90

-60

-30

30

60

x 90

120

150

180

cos (x-45) A -1

Similarly, the solutions which are indicated by the arrows are given by the x ‐coordinates of the points of intersection A and B of the two graphs give.

11

Preparation for the Mathematics examination brought to you by Kagiso Trust

EXERCISE 17.3 Pay particular attention to whether the relationship in the question is ≥ , that is “greater than and equal to”; or simply >, that is “greater than”. The type of relation indicates how you write the solution as explained below.

Solution to Example 4: Inspect the graphs for values of x for which the graph of sin x is above the graph of cos 2 x .

These values are x ∈ − 270 0 ; − 210 o and x ∈ 30 o ; 150 o

We use the square brackets [ …. ] because of ≥ in the question. This means that x takes the value that is against the square bracket [ .

Using set notation the two sets can be combined by using the symbol ∪ which stands for “union”:

x ∈ − 270 o ; − 210 o ∪ 30 o ; 150 o

Solution to Example 5: We must inspect the graphs looking out for values of x for which the graph of cos x − 45o lies above the graph of sin x .

These values are x ∈ 30 o ; 15 .

Note that in this case we do not use the square brackets because of > in the questions. The bracket ( against a value means that x does not take that value.

[

]

[

] [

[

]

]

(

(

)

)

Solution of trigonometric equations and inequalities analytically and graphically

EXERCISE 17.4 cos x = 0

1 − sin x = 0 sin x = 1

Solution on calculator or calc ∠ = 90

Applying the general solution:

o x = ±calc∠ + n.360 o

Solution on calculator or calc ∠ = 90

Applying the general solution:

o

x = calc∠ + n.360 or x = (180 o − calc∠) + n.360 o

n = −1 : x = ±90o − 1× 360o o o o o o o x = 90 + n.360 x = +90 − 360 or - 90 − 360 o o o o o x = −270 √ or − 450 Reject. Why? or x = (180 − 90 ) + n.360

n = 0: x = ±90o √ that is accept. Why? which both reduce to

o

x = ±90 + n.360

o

o

x = 90 + n.360

n = −1 : x = −270o √ Why? o

n = 0:

o

x = 90o √ Why?

Putting together all the answers we thus get: x = −270 o ; − 90 o ; 90∗

EXERCISE 17.5

y

2

↓ -270

↓ -240

-210

-180

-150

-120

-90

↓

1

-60

-30

30

60

90

x 120

150

180

sin 2x -1

2cos x

-2

The solutions are indicated by the arrows and confirm those found in Exercise 17.4 above.

13

Preparation for the Mathematics examination brought to you by Kagiso Trust

EXERCISE 17.6

y

←

The ordina te s of 2cos x a re gre a te r tha n those of sin 2x in this ra nge 2

→

1

x -240

-210

-180

-150

-120

-90

-60

-30

30

sin 2x

60

90

120

150

180

-1

2cos x -2

We must inspect the graphs and look for where the graph of 2 cos x lies above the graph of sin 2 x .

In other words, where are the ordinates or the y coordinates of 2 cos x greater than or equal to those of sin 2 x ?

The answer is indicated by the arrows, that is x ∈ − 90 o ; 90 o

Why do we use “close” brackets in this case?

EXERCISE 17.7

[

]

y

←

The ordinates have opposite signs in this range

→ 2

1

x -240

-210

-180

-150

-120

sin 2x

-90

-60

-30

30

60

90

120

150

-1

2cos x -2

In this case we inspect the graphs to observe a range in which the graphs or their ordinates have opposite signs.

This range is indicated by the arrows in the above diagram.

The answer is thus: x ∈ − 180 o ; 0 o

Why do we use “open” brackets in this case?

(

)

180

Solution of trigonometric equations and inequalities analytically and graphically

EXERCISE 17.8

Finding the solutions by using the quadratic formula:

sin x =

−1± 1+ 8 −1± 3 − 4 2 = = or 4 4 4 4

sin x = −1 or 0,5

Calculator answers are: x = −90 or x = 30

Inserting these solutions into the two general solutions:

o o o First, x = −90 : x = −90 + n.360 or x = 180 o − − 90 o + n.360 o = 270 o + n.360 o

o o Determine the solutions in the interval − 270 ≤ x ≤ 180 :

o

o

{

n = 1:

x = −90 + 1 × 360o = 270o Reject. Why?

(

)}

x = 270o + 1× 360o Reject. Why?

n = 0 : x = −90 Accept. Why?

x = 270o Reject. Why?

o

n = −1 : x = −90 + (− 1) × 360 Reject. Why? o

x = 270 o + (− 1)× 360 o = −90 o Accept. Why?

o

You can extend the integer values of n in both negative and positive directions to see if you will uncover more solution in the interval − 270 ≤ x ≤ 180 . Do try.

o

o

(

)

Second, x = 30 : x = 30 + n.360 or x = 180 o − 30 o + n.360 o = 150 o + n.360 o

o

o

o

n = −1 : x = 30 o + (− 1)× 360 o Reject. Why?

x = 150o − 360o = −210o Accept. Why?

n = 0 : x = 30 Accept. Why? o

x = 150o Accept. Why?

Again, you can try more integer values of n to see if there are any more solutions in the given interval.

Combining all the solutions and writing them in ascending order of magnitude, the answer is:

x ∈ − 210 o ; − 90 o ; 30 o ; 150 o

Compare these solutions with the graphical solutions illustrated on previous page.

{

}

15

Preparation for the Mathematics examination brought to you by Kagiso Trust

EXERCISE 17.9

Finding the solutions by means of factorization:

sin x = cos 2 x

sin x = 1 − 2 sin x

2 2 sin x + sin x − 1 = 0

(2 sin x − 1)(sin x + 1) = 0

So that either 2 sin x − 1 = 0 or sin x + 1 = 0

2

sin x =

1 = 0,5 2

or

sin x = −1

Leading as we have seen in the previous Exercise to the solutions already found, namely:

x ∈ − 210 o ; − 90 o ; 30 o ; 150 o

EXERCISE 17.10

{

}

Finding the solutions graphically

Below is a graphical illustration of the solution of sin x = cos 2x which confirms the solutions obtained by the alternative methods used in Exercises 17.8 and 17.9:

y

1

sin x

↓

↓ A

-270

-240

-210

-180

-150

-120

-90

↓

↓

D

C

-60

-30

30

60

90

120

150

x 180

cos 2x

B

The solutions are seen to be x ∈ − 210 o ;

{

-1

− 90 o ; 30 o ;

}

150 o

Solution of trigonometric equations and inequalities analytically and graphically

REFINING TECHNIQUES FOR SKETCHING TRIGONOMETRIC FUNCTIONS We sketched trigonometric functions in UNIT 7 and in the preceding section. Let’s use a few more examples to show in greater detail what to do in order to obtain sketches of the curves of trigonometric functions.

[

]

Example 8: Sketch the graph of f ( x ) = sin 3x for the domain x ∈ − 90 o ; 180 o .

Characteristics

Parent function sin x

Period

360 o

x − intercepts at which y = 0

x = 0 o + k .180 o

y = f ( x ) = sin 3x 360 o = 120 o 3 x is replaced by 3x 3 x = 0 o + k .180 o x = 0 o + k .60 o

k = −3, x = −180 o k = −2, x = −120 o k = −1, x = −60 o k = 0, x = 0 o k = 1, x = 60 o k = 2, x = 120 o k = 3, x = 180 o k = 4, x = 240 o

Coordinates that fall within the domain:

− 120 o ; 0 o

( ) (− 60 ; 0 ) (0 ; 0 ) (60 ; 0 ) (120 ; 0 ) (180 ; 0 ) o

o

o

o

o

o

o

o

o

o

y = intercepts at

y = sin 0 o = 1

y = sin 3 × 0 o = sin 0 o = 0

(0 o ; 0)

which x = 0 o Maximum value at y = 1

x = 90 o + k .360 o

x is replaced by 3x 3 x = 90 o + k .360 o

Coordinates that fall within the domain:

− 90 o ; 1

Minimum value y = −1

k = −1, x = −90 o k = 0, x = 30 o k = 1, x = 150 o k = 2, x = 270 o x is replaced by 3x 3 x = −90 o + k .360 o

x = −90 o + k .360 o

k = −1, k = 0, k = 1, k = 2, 17

Coordinates

x = −150 o x = −30 o x = 90 o x = 210 o

(

)

(30 ; 1) (150 ; 1) o

o

Coordinates that fall within the domain:

o − 30 ; - 1

(

)

(90 ; - 1) o

Preparation for the Mathematics examination brought to you by Kagiso Trust

What a lot of working! It had to be at this stage because we are making explanations. But with much practice on your part you will not have to do so much working out and writing down. On many occasions you are required to state or indicate on your sketch the coordinates of the intercepts and turning points so you must know how to derive them. In practice though these can be more easily derived from your knowledge of the parent function instead of getting bogged down by using the general formulae.

What you next do is to plot the coordinates on your axes: y

1

x -120

-105

-90

-75

-60

-45

-30

-15

15

30

45

60

75

90

105

120

135

150

165

180

75

90

105

120

135

150

165

180

-1

Now join the points you see with a smooth curve (not straight lines!): y

1

x -120

-105

-90

-75

-60

-45

-30

-15

15

30

45

60

-1

The first sketch above is for purposes of showing that first you must plot the points. Otherwise, all you need to show when solving this question is the last sketch.

What is the amplitude of f ( x ) = sin 3x ?

Solution of trigonometric equations and inequalities analytically and graphically

(

[

)

]

Example 9: Sketch the graph of f ( x ) = cos x + 30 o for the domain x ∈ − 180 o ; 240 o .

Characteristics

(

)

Parent function cos x

y = f (x ) = cos x + 30 o

Coordinates

360 o x = (2k + 1).90 o

360 o

Coordinates that fall within the domain:

− 120 o ; 0 o

Period x − intercepts at which y = 0

x is replaced by x + 30 o x + 30 o = (2k + 1).90 o x = 2k .90 o + 60 o

k = −1, k = 0, y = intercepts at

( (

which x = 0

cos x = 1 x = 0 o + k .360 o

) )

Coordinates that fall within the domain:

− 30 o ; 1

(

k = 0, x = −30 o k = 1, x = 330 o

cos x = −1 x = 180 o + k .360 o

(60 ; 0 ) o

(0; 0,87)

Minimum value y = −1

)

o

y = cos x + 30 o = cos 0 o +30 o = cos 30 o 3 = = 0,87 2 x is replaced by x + 30 o x + 30 o = 0 o + k .360 o x = −30 o + k .360 o

o

Maximum value at y = 1

(

x = −120 o x = 60 o

x is replaced by x + 30 o x + 30 o = 180 o + k .360 o x = 150 o + k .360 o

)

Coordinates that fall within the domain:

150 o ; - 1

(

k = 0, x = 150 o

)

Plot the points and join them with a smooth curve to get: y

1

x -180

-165

-150

-135

-120

-105

-90

-75

-60

-45

-30

-15

15

-1

What is the amplitude of f ( x ) = cos x + 30 o ?

(

)

19

30

45

60

75

90

105

120

135

150

165

180

195

210

225

240

Preparation for the Mathematics examination brought to you by Kagiso Trust

(

[

)

]

Example 10: Sketch the graph of f ( x ) = tan x − 60 o for the domain x ∈ − 135 o ; 180 o .

(

)

Characteristics

Parent function tan x

y = f ( x ) = tan x − 60 o

Coordinates

Period x − intercepts at which y = 0

180 o x = k .180 o

180 o

Coordinates that fall within the domain:

− 120 o ; 0

x is replaced by x − 60 o x − 60 o = k .180 o x = 60 o + k .180 o

(

k = −1, x = −120 o k = 0, x = 60 o

(

)

(60 ; 0) o

)

y = intercepts

y = tan 0 o = 0

tan 0 o + 60 o = tan 60 o = 1,73

(0; 1,73)

at which x = 0 Asymptotes

x = (2k + 1).90 o

x is replaced by x − 60 o x − 60 o = (2k + 1).90 o

Asymptotes at

x = −30 o

x = 150 o

o

x = (2k + 1).90 o + 60 o k = −1, x = −30 o k = 0, x = 150 o

Plot the points and join them with a smooth curve to get: y 9 8 7 6 5 4 3 2 1

x -130

-120 -110

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

180

-1 -2 -3 -4 -5 -6 -7 -8 -9

We will now apply all of the above to solving trigonometric equations and inequalities.

Solution of trigonometric equations and inequalities analytically and graphically

PAPER 2 QUESTION 6

DoE/ADDITIONAL EXEMPLAR 2008

PAPER 2 QUESTION8

DoE/NOVEMBER 2008

PAPER 2 QUESTION 7

DoE/PREPARATORY EXAMINATION 2008

(

)

Given the following functions: f ( x ) = sin x + 30 o and g ( x ) = cos 2 x

[

]

7.1

Sketch the functions for x ∈ − 90 o ;180 o on DIAGRAM SHEET 1.

Clearly show the intercepts with axes as well as the turning points.

7.2

Calculate values of x for f ( x ) = g ( x ) for x ∈ − 90 o ;90 o .

7.3

Use your graph to determine the value(s) of x for which f ( x ) ≥ g ( x ) for x ∈ 0 o ;180 o (2)

7.4

State the value(s) of x for which g ( x ) is negative and x is increasing for x ∈ − 90 o ;180 o .

[

(8)

(8)

]

[

21

[

]

]

(2)[20]

Preparation for the Mathematics examination brought to you by Kagiso Trust

PAPER 2 QUESTION 6

Number Hints and answers 6.1

Look at each function.

State its amplitude.

State its period, and its x ‐ and y ‐ intercepts.

State its turning points.

Sketch each graphs.

Check your answer with a friend.

DoE/ADDITIONAL EXEMPLAR 2008

Work out the solutions in the boxes below DIAGRAM SHEET 1

6.2

6.3

Some of the trigonometric equations you will have met in Grades 11 & 12 can be solved like algebraic equations in one unknown.

To get a single unknown we must transform the equation so that it only has sines or cosines, and furthermore, simplified to sines or cosines of a single angle x rather than a double angle 2 x .

sin x as a sine of a single x is already in simplified form. So tackle cos 2 x reducing this to an expression in terms of a sine of single x by using an appropriate formula.

Do this and you will end up with a quadratic equation in sin x . Solve for sin x by using the quadratic formula. Or, first let sin x = k .

Answer: x = 21,5 o or x = 158,5 o Look at your two graphs drawn on the same axes and find out where along the x ‐axis the difference between the two graphs is equal to 3 units.

If you could not see where the graphs differ by 3 units, use the answer to see this and this will show you how to work out such a question in future.

Answer: x = 90 o

Solution of trigonometric equations and inequalities analytically and graphically

PAPER 2 QUESTION 8

Number Hints and answers

8.1 Have either a cosine or a sine on both sides of the equation by using the cosine of an angle is the same as the sine of the complement of the angle, and vis a verse.

First find the general solutions and then find the solutions within − 90 o ;180 o

Answer: x = −67,5 o ;22,5 o ;112,5 o

[

8.2

DoE/NOVEMBER 2008

Work out the solutions in the boxes below

]

or x = −45 o ;135 o Sketch strictly within the given domain; otherwise you will lose a mark or two.

Use a different colour pen for each curve so you can see each clearly when you sketch them.

Use broken horizontal lines to indicate the maximum and minimum of each function.

The sine curve is easy to sketch in because its period is 360 o . Mark the points where it crosses the x ‐axis, attains its maximum and minimum values.

Do likewise for the cosine curve dividing by 3 the points where its parent curve cos x crosses the x ‐axis and attains its maximum and minimum values within the domain − 90 o ;180 o .

DIAGRAM SHEET 2

[

8.3

Look at the graphs and see for which values of x within − 90 o ;180 o the graph of f ( x ) is below the graph of g ( x ) .

[

]

23

Answer:

]

Preparation for the Mathematics examination brought to you by Kagiso Trust

PAPER 2 QUESTION 7 Number 7.1

DoE/PREPARATORY EXAMINATION 2008

Hints and answers Work out the solutions in the boxes below First do sketches on rough paper if this approach suits you better and enables you to see the shape of its curve together with the location of its turning points and intercepts on the axes.

On rough paper make a quick sketch of sin x for x ∈ − 90 o ;180 o . Then shift the graph by

[

]

o

30 to the left. Mark its intercepts with the axes and its turning points.

Similarly on rough paper make a quick sketch of cos x for x ∈ − 180 o ;360 o taking into consideration that the period is to be halved when sketching cos 2 x .

Now halve the markings on the x ‐axis to give the x ‐intercepts of cos 2 x for x ∈ − 90 o ;180 o .

On the diagram given to you, now sketch the graphs as neatly and correctly as possible, superimposing one on the other: DIAGRAM SHEET 1

Answers:

f (x ) g (x ) o x ‐intercepts − 30 ;0 − 45 o ;0

[

[

]

( ) (150 ;0) o

(

)

(45 ;0) (135 ;0) (0 ;1) o

o

y ‐ intercepts

]

(0 ;0.5) o

o

Solution of trigonometric equations and inequalities analytically and graphically

Number

Hints and answers Turning points 60 o ;1

(

)

Work out the solutions in the boxes below

(0 ;1) o

(− 90 ;−1) (90 ;−1) (180 ;1) o

o

o

7.2

You are solving the equation

sin x + 30 o = cos2 x in the interval

( ) x ∈ [− 90 ;90 ] . o

o

Use the fact that the sin of an angle is equal to the cosine of the complement of the angle, or vice a versa. This way you will have either a sin or a cosine on both sides of the equation enabling you to equate the angles. Remember to apply the general solutions and only thereafter restrict the solutions to x ∈ − 90 o ;90 o .

Answers: x = −60 o ;20 o Inspect your sketches in 7.1 to see where the graph of f ( x ) is above the

[

7.3

]

graph of g ( x ) .

Note: answer is sought in interval x ∈ 0 o ;180 o .

For what range of values is f ( x )

[

above g ( x ) ? This will give you the answer to the question.

Use your solutions in 7.2 to mark the end points of the range of values of x for which f ≥ g .

Answer: x ∈ 20 o ;140 o Answer:

You write the answer.

[

7.4

]

]

25

Preparation for the Mathematics examination brought to you by Kagiso Trust

MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008

Paper 2

Solution of trigonometric equations and inequalities analytically and graphically

Feb – March 2008

Paper 2

DIAGRAM SHEET 2

27

Preparation for the Mathematics examination brought to you by Kagiso Trust

Feb – March 2009

DIAGRAM SHEET 1

Paper 2

Solution of trigonometric equations and inequalities analytically and graphically

November 2009 (Unused paper)

DIAGRAM SHEET 1 This sheet is the one shown above.

29

Paper 2

Preparation for the Mathematics examination brought to you by Kagiso Trust

November 2009(1)

DIAGRAM SHEET 4

Paper 2

Solution of trigonometric equations and inequalities analytically and graphically

Feb – March 2010

Paper 2

DIAGRAM SHEET 3

31

Preparation for the Mathematics examination brought to you by Kagiso Trust

ANSWERS November 2009(1) 12.1 Sketch:

Exemplar 2008 8.1 A(−160 o ;0,174)

(

)

B 80 o ;0,766 8.2 − 160 < x < 80 o

Feb/March 2009 8.1 Sketch:

o

12.2 Points A and B shown on the graph.

12.3 x = 90 o or x = −30 o 12.4 x = 30 o and x = 210 o 12.5 x ∈ − 90 o ;−60 o U 120 o ;270 o

Feb/March 2010 11.1 x ∈ 180 o ;210 o ;330 o ;360 o

11.2 Sketch:

8.2 x = 0 o or 180 o or x = 60 o ; − 60 o or 120 o

[

8.3 − 60 o < x < 0 o or 60 o < x < 90 o or 120 o < x < 180 o

{

1 9 Maximum value = 3 1 Minimum value = 4

]

) (

}

November 2009 (Unused paper) 8.1 a = 2 8.2 Sketch:

8.3 2 8.4 Reading from the graph, 14,5 o lies to the right of the point of intersection.

So θ < 14,5 o .

11.3 180 o ≤ x ≤ 210 o or 330 o ≤ x ≤ 360 o