EXAMPLE 1 Evaluate trigonometric expressions Find the exact
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EXAMPLE 1 Evaluate trigonometric expressions Find the exact value of (a) cos 165° and (b) tan π. 12 a. cos 165° = cos 1 (330°) b. tan π = tan 1 ( π ) 2 6 12 2 π 1 – cos 6 = = – 1 + cos 330° π 2 sin 6 =– 1+ 3 2 2 = =– 2+ 2 =2– 3 3 1– 3 2 1 2
EXAMPLE 2 Evaluate trigonometric expressions Given cos a = 5 with 3π < a < 2π, find (a) sin 2 a and 2 13 a (b) sin. 2 SOLUTION a. Using a Pythagorean identity gives sin a = – 12. 13 sin 2 a = 2 sin a cos a = 2(– 12 )( 5 ) = – 120 169 13 13 b. Because a is in Quadrant II, sin a is positive. 2 2 5 1 – cos a = 13 = 4 = 2 13 sin a = 2 13 13 2 2
EXAMPLE 3 Standardized Test Practice SOLUTION sin 2 q 2 sin q cos q Use double-angle formulas. = 2 1 – (1 – 2 sin q ) 1 – cos 2 q 2 sin q cos q Simplify denominator. = 2 2 sin q cos q Divide out common factor 2 sin q. = sinq Use cotangent identity. = cot q ANSWER The correct answer is B.
GUIDED PRACTICE for Examples 1, 2, and 3 Find the exact value of the expression. 1. tan π 8 ANSWER 2– 1 2. sin 5π 8 ANSWER 2+ 2 2 3. cos 15° ANSWER 2+ 3 2
GUIDED PRACTICE for Examples 1, 2, and 3 Find the exact value of the expression. a 4. Given sin a = 2 with 0 < a < π , find cos 2 a and tan. 2 2 2 ANSWER 2– 1 5. Given cos a = – 3π 3 with π < a < , find sin 2 a and sin a. 2 5 2 ANSWER 24 – 2 5 25’ 5
GUIDED PRACTICE for Examples 1, 2, and 3 Simplify the expression. 6. cos 2 q sin q + cos q ANSWER cos q – sin q 7. tan 2 x tan x ANSWER 2 1 – tan 2 x 8. sin 2 x tan x 2 ANSWER 2 cos x(1 – cos x)
