1. EXAMPLE 1 Evaluate trigonometric expressions Find the exact value of (a) cos 165° and (b) tan. π 12 a. cos 165° 1 2 = cos (330°) = – 1 + cos 330° 2 = – 1 + 2 3 2 = – 2 + 2 3 b. tan π 12 1 2 π 6 = tan ( ) = 1 – cos π 6 sin π 6 = 1 – 3 2 1 2 3= 2 –

2. EXAMPLE 2 Evaluate trigonometric expressions Given cos a = with < a < 2π, find (a) sin 2a and (b) sin. 5 13 3π3π 2 a 2 SOLUTION a. Using a Pythagorean identity gives sin a 12 13 = –. sin 2a = 2 sin a cos a 12 13 5 = 2(– )( ) 120 169 = – b. Because is in Quadrant II, a 2 a 2 sin is positive. a 2 sin = 1 – cos a 2 = 1 – 2 5 13 = 4 = 2 13

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

4. GUIDED PRACTICE for Examples 1, 2, and 3 Find the exact value of the expression. 1. tan π 8 ANSWER 2 – 1 2. sin 5π5π 8 ANSWER 22 + 2 3. cos 15° ANSWER 32 + 2

5. GUIDED PRACTICE for Examples 1, 2, and 3 Find the exact value of the expression. 4. π 2 Given sin a = with 0 < a <, find cos 2a and tan. a 2 2 2 ANSWER 2 – 1 3π3π 2 5. Given cos a = – with π < a <, find sin 2a and sin. a 2 3 5 ANSWER 24 25’ 2 5 5 –

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