712 CHAPTER 7 Trigonometric Identities and Equations We find sin 2A using sin1A + B2 = sin A cos B + cos A sin B, with B = A. sin 2 A = sin1A + A2 2A = A + A = sin A cos A + cos A sin A Sine sum identity sin 2 A =2 sin A cos A Add. Using the identity for tan1A + B2, we find tan 2A. tan 2 A = tan1A + A2 2A = A + A = tan A + tan A 1 - tan A tan A Tangent sum identity tan 2 A = 2 tan A 1 −tan2 A Simplify. LOOKING AHEAD TO CALCULUS The identities cos 2 A = 1 - 2 sin2 A and cos 2 A = 2 cos2 A - 1 can be rewritten as sin2 A = 1 2 11 - cos 2 A2 and cos2 A = 1 2 11 + cos 2 A2. These identities are used to integrate the functions ƒ1A2 = sin2 A and g1A2 = cos2 A. NOTE In general, for a trigonometric function ƒ, ƒ12 A2 ≠2ƒ1A2. Double-Angle Identities cos 2 A =cos2 A −sin2 A cos 2 A =1 −2 sin2 A cos 2 A =2 cos2 A −1 sin 2 A =2 sin A cos A tan 2 A = 2 tan A 1 −tan2 A EXAMPLE 1 Finding Function Values of 2U Given Information about U Given cos u = 3 5 and sin u 60, find sin 2u, cos 2u, and tan 2u. SOLUTION To find sin 2u, we must first find the value of sin u. sin2 u + cos2 u = 1 Pythagorean identity sin2 u + a 3 5b 2 = 1 cos u = 3 5 sin2 u = 16 25 A 3 5B 2 = 9 25 ; Subtract 9 25 . sin u = - 4 5 Now use the double-angle identity for sine. sin 2u = 2 sin u cos u = 2a- 4 5ba 3 5b = - 24 25 sin u = - 4 5 and cos u = 3 5 Now we find cos 2u, using the first of the double-angle identities for cosine. cos 2u = cos2 u - sin2 u = 9 25 - 16 25 = - 7 25 Take square roots. Choose the negative square root because sin u 60. Pay attention to signs here. Any of the three forms may be used. cos u = 3 5 and A 3 5B 2 = 9 25 ; sin u = - 4 5 and A - 4 5B 2 = 16 25
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