EXAMPLE 1

1. Find the exact value of (a)cos165° and (b) tan . 3 3 π π π 6 12 b. tan = tan ( ) = cos(330°) 12 1 – cos 1 + cos 330° = = – 1 1 2 sin 2 2 3 π 3 π 1 2 6 6 2 2 1 + 1 – = – = 2 2 + = – 2 = 2 – EXAMPLE 1 Evaluate trigonometric expressions a. cos165°

2. Given cos a = with < a < 2π, find (a)sin 2a and (b)sin. 5 12 13 13 3π 2 = – . a a a a 2 2 2 2 12 5 120 =2(– )( ) = – 13 13 169 b. Because is in Quadrant II, sinis positive. 5 1 – 1 – cos a 4 13 13 = 2 = sin = = 13 13 2 2 EXAMPLE 2 Evaluate trigonometric expressions SOLUTION a. Using a Pythagorean identity gives sin a sin 2a =2sin acos a

3. sin 2q 2 sin q cos q = 1 – (1 – 2 sin2q ) 1 – cos 2q 2 sin q cos q = 2 sin2q cos q = sinq ANSWER The correct answer is B. EXAMPLE 3 Standardized Test Practice SOLUTION Use double-angle formulas. Simplify denominator. Divide out common factor 2 sinq. = cotq Use cotangent identity.

4. 3 2 – 1 2 2. sin 1. tan 5π ANSWER ANSWER 8 2 + π 2 8 ANSWER 2 + 2 for Examples 1, 2, and 3 GUIDED PRACTICE Find the exact value of the expression. 3. cos 15°

5. 5 2 – 1 4. Givensina = with 0 < a < , findcos 2a andtan . 3π ANSWER 2 3 π a a 2 2 2 2 2 5 24 5. Givencosa = – withπ < a < , findsin 2a andsin . 25’ for Examples 1, 2, and 3 GUIDED PRACTICE Find the exact value of the expression. ANSWER 2 – 5

6. cos 2 q 6. 8. sin 2x tan sin q+ cos q ANSWER ANSWER cos q– sin q 2 cosx(1 – cosx) x 2 tan 2x 7. tan x ANSWER 2 1 – tan2 x for Examples 1, 2, and 3 GUIDED PRACTICE Simplify the expression.