EXAMPLE 1 Evaluate trigonometric expressions

Find the exact value of (a) cos 165° and (b) tan .π12

a. cos 165° 12

= cos (330°)

= – 1 + cos 330°2

= –1 +

2

32

= – 2 + 2

3

b. tan π12

12

π6

= tan ( )

=1 – cos π

6

sinπ6

= 1 – 3

212

3= 2 –

EXAMPLE 2 Evaluate trigonometric expressions

Given cos a = with < a < 2π, find (a) sin 2a and

(b) sin .

513

3π2

a2

SOLUTION

a. Using a Pythagorean identity gives sin a 1213

= – .

sin 2a = 2 sin a cos a 1213

513

= 2(– )( ) 120169

= –

b. Because is in Quadrant II,a2

a2

sin is positive.

a2

sin = 1 – cos a2

= 1 –

2

513 = 4

13= 2 13

13

EXAMPLE 3 Standardized Test Practice

SOLUTION

sin 21 – cos 2

Use double-angle formulas.=2 sin cos

1 – (1 – 2 sin2)

Simplify denominator.=2 sin cos

2 sin2

Divide out common factor 2 sin .=cos

sinUse cotangent identity.= cot

ANSWER The correct answer is B.

GUIDED PRACTICE for Examples 1, 2, and 3

Find the exact value of the expression.

1. tan π8 ANSWER

2 – 1

2. sin 5π8 ANSWER

22 +2

3. cos 15° ANSWER

32 +2

GUIDED PRACTICE for Examples 1, 2, and 3

Find the exact value of the expression.

4. π2

Given sin a = with 0 < a < , find cos 2a and tan .a2

22

ANSWER

2 – 1

3π25. Given cos a = – with π < a < , find sin 2a and sin .a

235

ANSWER

2425’

2 55

–

GUIDED PRACTICE for Examples 1, 2, and 3

Simplify the expression.

6. cos 2 sin + cos ANSWER

cos – sin

7. tan 2xtan x

ANSWER

21 – tan2 x

8. sin 2x tan x2 ANSWER

2 cos x(1 – cos x)