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CHAPTER

TRIGONOMETRIC FUNCTIONS:

5

UNIT CIRCLE APPROACH

LEARNING OUTCOMES By the end of this chapter, you should be able to

 Use reference number to find terminal point on the unit circle.  Determine trigonometric function using properties of unit circle.  Evaluate trigonometric function using definition of trigonometric functions.  Sketch and transform graphs of sine, cosine, tangent, secant and cosecant.

5.1

UNIT CIRCLE This section is about properties of circle radius 1 unit and centred at the origin.

The definition of unit circle is, Definition: Unit Circle The unit circle is the circle in

xy 

plane centred at origin and has one unit

radius. The equation for unit circle is x2  y2  1

Example 1: Verify that the point

 2 6 5 P  ,  7 7 

is on the unit circle.

Solution We must verify that P satisfies equation of unit circle,

Since

and 2 6 x 7

5 y 7

, substitute into

x2  y2  1

x2  y2  1

Hence, 2

 2 6   5  2 24 25       1    7 7 49 49  

1

.

.

Example 2: Find

y

coordinate if the point

 5  P , y   6 

is on the unit circle in

Quadrant IV. Solution Since point P is on unit circle, substitute

into equation of unit circle 5 x 6

x2  y2  1

.

2

 5 2    6   y 1   25 11  36 36 11 y 6 y2  1 

Since the point is in Quadrant IV,

y

coordinate must be negative, thus

.

y

11 6

 Terminal Points on The Unit Circle 2

Terminal point is a point of

measured in radians.

t

P ( x, y )

determined by the real number

which is

t

is the length of the arc intercepted by the angle,



in

radians.

The terminal point

P ( x, y )

usually obtained when we travel start from point

and move along the unit circle in the direction of counter clockwise

clockwise

P ( x, y )

 t  0

1,0

 t  0

or

. Figure (a) and (b) show the relationship of terminal point

and . t

Terminal point

P(1,0)

counter clockwise

when travel

t   0

.

Figure (a) Terminal point

P(1,0)

counter clockwise

when travel

 t  0 2

.

Figure (b) Note: Different values of

t

can determine the same terminal point.

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Example 3: Find the terminal points on the unit circle determined by given real number . t (a)

(b)

2

 2

(c) 

(d)

3 2

3

Solution From the graph, (a) terminal point is

1,0

(b) terminal point is

1,0

(c) terminal point is

(d) 4

 0,1

terminal point

  1,0

We can configure other terminal points using

Figure (c)

 t 4

.

P lies on line

yx

.

Since P is the intersection of line yx

and unit circle

x2  y2  1

x2  x2  1 2x 2  1 1 x2  2 1 2 x  2 2 y

2 2

P is in Quadrant I, thus terminal point

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 2 2  P ,  2 2  

.

,

We can use same method to find other terminal points. Table 1 shows the special values of . t

t

Terminal point

0

1,0

 6

 3 1    2 ,2  

 4

 2 2    2 , 2   

 3

1 3  ,  2 2   

 2

 0,1 Table 1

 The Reference Number Definition: Reference Number The reference number denoted as

associated with a real number, t is the

t shortest distance along the unit circle between terminal point and the

x

-axis.

Let’s observe this graph.

We can determine the location of terminal point based on real number t

The reference number, is in

t between terminal point and axis

x Steps to find terminal point

P (  a,b)

6

using reference number,

1) Find reference number, . t 2) Find terminal point determined by . t 3) Choose the sign of terminal point

P (  a,b)

according to the quadrant in

which terminal point, lies. t

Example 4: Find the terminal points of given real number, t. (a)

(b)

(c)

7 6

5 4

(d)

4  3



41 4

Solution (a) The reference number,

.

The terminal point,

Terminal point of

negative and

y

(from Table 1).

is located at Quadrant III, where

is negative.

Therefore, terminal point of

(b) The reference number,

is

.

7

.

is

The terminal point,

(from Table 1).

Terminal point of

negative and

y

is located at Quadrant II, where

is positive.

Therefore, terminal point of

(c) The reference number,

is

Terminal point of

y

.

.

The terminal point,

negative and

is

(from Table 1).

is located at Quadrant III, where

is

is negative.

Therefore, terminal point of

(d) The reference number,

is

.

The terminal point,

Terminal point of

.

(from Table 1).

is located at Quadrant IV, where

8

is

positive

and

y

is negative.

Therefore, terminal point of

is

.

EXERCISE 1.1 1. (a) In your own words, explain what is unit circle. (b) State the equation of unit circle.

2. (a) What is reference number, ? t (b) Let the reference number be

 t 3

for terminal point

P  x, y 

lies in Quadrant IV. 3. Verify the given point whether it is on the unit circle. (a)  5 11   ,   6 6    4. Given the real number, t.

Find, (i) reference number for each t. (ii) terminal point of for each t.

P  x, y  (a)

3 4

(b)

7  6

(c)

(d)

11 3

9

31 6

(e)



29 6

, find P if it

5.2

TRIGONOMETRIC FUNCTIONS OF REAL NUMBERS In this section, we will see how the properties of unit circle can be used to define the trigonometric functions. Before that, what are trigonometric functions?

Definition: Trigonometric Functions Let be the terminal point on the unit circle which determined by any P ( x, y ) real number,

t

and we define sin t  y

csc t 

cos t  x

tan t 

1 y denominator

1 sec t  x

y x

cot t 

 0.WHY?

x y

Let’s create table of special values of the trigonometric functions using above definition and terminal points in Figure 1.

t

sin t

cos t

tan t

csc t

sec t

cot t

0

0

1

0

-

1

-

 6

1 2

3 2

3 3

2

2 3 3

3

 4

2 2

2 2

1

2

2

1

 3

3 2

1 2

3

2 3 3

2

3 3

 2

1

0

-

1

-

0

y

  1 

3  ; ,  3  2 2  

  2 2  ; , 4  2 2 

  3 1  ; , 6  2 2 

x

Figure 1

Table 1 Example 1: Find all six trigonometric functions determined by real number, . t

10

a)

b)

 t 4

t

 6

Solution a) Terminal point of

 t 4

is

 2 2    2 , 2   

.

Then all six trigonometric functions are, sin t  y  csc t 

1  y

cot t 

x  y

2 2 1

cos t  x 

2/2 2/2 2/2

b) Terminal point of



2 2

2 2

y 2/2  1 x 2/2 1 1 2 sec t     2 x 2/2 2 tan t 

 2

1

 t 6

is

 3 1    2 , 2  

.

Then all six trigonometric functions are, sin t  y 

1 2

cos t  x 

csc t 

1 1  2 y 1/ 2

cot t 

x 3/2   3 y 1/ 2

3 2

tan t 

y 1/ 2 1 3    x 3 3/2 3

sec t 

1 1 2 2 3    x 3 3/2 3

Steps to evaluate the trigonometric functions for any real number . t 1. Find the reference number, . t 2. Determine the sign of the trigonometric function of

t

by locating which

quadrant that terminal point lies.

Quadrant I

: All functions are positive

Quadrant II Quadrant III

: All functions are negative except for sine 11 : All functions are negative except for tangent

Quadrant IV

: All functions are negative except for cosine

Sine

All

Tangent

Cosine

3. Evaluate trigonometric function using reference number and add (obtained in step 2).

Example 2: Evaluate the trigonometric functions. a) b) c) 5  7       5  sin   cos  tan     4   6   3  d)

  sec    3

e)

d)

 2  csc   3 

 25  cot    2 

Solution a) Reference number,

For real number,

The sign for

.

the terminal point lies on Quadrant III.

is negative.

Evaluate:

b) Reference number,

For real number,

.

.

the terminal point lies on Quadrant II.

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

sign

The sign for

is negative.

Evaluate:

c) Reference number,

For real number,

The sign for

.

.

the terminal point lies on Quadrant IV.

is positive.

Evaluate:

d) Reference number,

For real number,

The sign for

.

.

the terminal point lies on Quadrant IV.

is the reciprocal of cosine. Cosine is negative in

Quadrant IV. Evaluate:

.

13

e) Reference number,

 t 3

For real number,

The sign for

.

the terminal point lies on Quadrant II.

 2  csc   3 

is the reciprocal of sine. Sine is positive in

Quadrant II. Evaluate:

  2 3  2  2 3 csc    csc  3 3 3  3 

f) Reference number,

For real number,

Evaluate

 t 2

25 t 2

.

.

the terminal point is

 25  x 0 cot    0  2  y 1

 0,1

.

.

 Even and Odd Properties.

Let think of

the relationship of the

trigonometric

functions of

t

and

t

. We can

see that,

sin( t )   y   sin t cos( t ) the x even-odd cos t From this, we can complete properties for other functions.

Properties : Even and Odd Trigonometric Functions 14

Odd Functions sin( t )   sin t

Even Functions cos( t )  cos t

tan( t )   tan t

sec( t )  sec t

csc( t )   csc t cot( t )   cot t Example 3: Use even-odd properties to find the value of the trigonometric functions. a)   sin     2

b)

c)

  cos    4

  csc    6

d)

  sec    3

Solution a) Sine is an odd function. Therefore,  3   3  sin      sin    (1)  1  2   2  b) Cosine is an even function. Therefore, 2     cos    cos   2  4 4 c) Cosecant is an odd function. Therefore, 1 1     csc     csc      2 sin   / 6  1/ 2  6 6 d) Secant is an even function. Therefore, 1 1     sec    sec    2  3  3  cos  / 3 1 / 2  Fundamental identities We use fundamental identities to relate any trigonometric function to any other.

If we know the value of any trigonometric functions at , then we can use it to t find other trigonometric functions at . t

15

The fundamental identities are as follows.

Fundamental Identities Phytagorean Identities

Reciprocal Identities

sin 2 t  cos 2 t  1

csc t 

1 sin t

tan 2 t  1  sec 2 t

sec t 

1 cos t

1  cot 2 t  csc 2 t

tan t 

sin t cos t

cot t 

1 cos t  tan t sin t

Example 4: Evaluate all trigonometric functions of given. a) tan t  2

 /2t 

b)

terminal point of

4 csc t  3

c) sin t  d)

1 3

12 cos t  13

t

t

by using the information

is in Quadrant I

sec t  0 terminal point of

t

is in Quadrant IV

Solution a) Using phtagorean identity

.

  2 2  1  sec 2 t sec 2 t  5 sec t   5 1  5 cos t 1 cos t   5 16

Since

t

is in Quadrant II, cosine is negative. Similar with secant.

sec t   5 , cos t  

1 5

Using phytagorean identity

sin 2 t  cos 2 t  1

2

 1  sin t      1 5  1 4 sin 2 t  1   5 5 4 2 5 2 5 sin t     5 5 5 5 2

Using reciprocal identity

csc t 

cot t 

1 2/ 5

2/ 5

and cot t 

cos t sin t

5 2



 1/ 5

1 csc t  sin t



1 5



5 1  2 2

Answers: tan t  2

,

, , 1 5 sec t   5 , cos t   cot t   sin t  2 5 csc t  2 5 2 5 1

,

 Trigonometric Graphs In this section, you will learn how to sketch the graph of sine, cosine, tangent, cotangent, secant and cosecant.

Graph

Properties

Graph

Name

(a=1,k=1)

17

Equation: Sine

y  a sin kx Period:

2 / k

Amplitude:

a

Equation: Cosine

y  a cos kx Period:

2 / k

Amplitude:

a

Equation: Tangent

y  a tan kx Period:

 /k

Equation: Cotangent

y  a cot kx Period:

Cosecant

 /k

Equation: y  a csc kx Period:

2 / k

18

Equation: Secant

y  a sec kx Period:

2 / k

Notes:

k

is a constant and

k 0

 Transformation of Graphs The focus in this section is what happened to the sine, cosine and tangent curve if we shift them to the right or to the left. Let’s see some properties that will helps us to sketch those curves.

Graph’s

Properties

Example of Shifted Graph

Name Shift to the right

 /6 Equation: Sine Period:

Period: 2π

y  a sin k ( x  b)

2 / k

Amplitude:

Phase shift: Interval:

  Amplitude=2 y  2 sin  x  

a



b

 b, b   2 / k  

19

6

Equation: Cosine

Period:

y  a cos k ( x  b)

2 / k

Amplitude:

Phase shift: Interval:

  y  5 cos 3x   7 

b

 b, b   2 / k  

Equation: Tangent Period:

a

y  a tan k ( x  b)

 /k

Phase shift:

b y  tan  x   / 4 

Equation: Cotangent Period:

y  a cot k ( x  b)

 /k

Phase shift:

b y  5 cot  3 x   / 2 

EXERCISE 5.2

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