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Double-Angle, Half-Angle, And Sum-Product Identities

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  Section 7.3 Double-Angle, Half-Angle, and Sum-ProductIdentities Double-Angle Formulas EXAMPLE: If cos x  =  − 23 and  x  is in quadrant II, find cos2 x  and sin2 x. Solution: Using one of the double-angle formulas for cosine, we getcos2 x  = 2cos 2 x − 1 = 2  − 23  2 − 1 = 89  − 1 =  − 19To use the formula sin2 x  = 2sin x cos x,  we need to find sin x  first. We havesin x  = √  1 − cos 2 x  =   1 −  − 23  2 = √  53where we have used the positive square root because sin x  is positive in quadrant II. Thussin2 x  = 2sin x cos x  = 2  √  53  − 23  =  − 4 √  59EXAMPLES:(a) Write cos3 x  in terms of cos x. (b) Prove the identity sin3 x sin x cos x  = 4cos x − sec x. 1  EXAMPLES:(a) Write cos3 x  in terms of cos x. Solution: We havecos3 x  = cos(2 x  +  x )= cos2 x cos x − sin2 x sin x = (2cos 2 x − 1)cos x − (2sin x cos x )sin x = 2cos 3 x − cos x − 2sin 2 x cos x = 2cos 3 x − cos x − 2cos x (1 − cos 2 x )= 2cos 3 x − cos x − 2cos x  + 2cos 3 x = 4cos 3 x − 3cos x (b) Prove the identity sin3 x sin x cos x  = 4cos x − sec x. Solution: We havesin3 x sin x cos x  = sin( x  + 2 x )sin x cos x = sin x cos2 x  + cos x sin2 x sin x cos x = sin x (2cos 2 x − 1) + cos x (2sin x cos x )sin x cos x = sin x (2cos 2 x − 1)sin x cos x  + cos x (2sin x cos x )sin x cos x = 2cos 2 x − 1cos x  + 2cos x = 2cos x −  1cos x  + 2cos x = 4cos x − sec x 2  Half-Angle Formulas EXAMPLE: Express sin 2 x cos 2 x  in terms of the first power of cosine.Solution: We use the formulas for lowering powers repeatedly.sin 2 x cos 2 x  =  1 − cos2 x 2  1 + cos2 x 2  = (1 − cos2 x )(1 + cos2 x )2 · 2= 1 − cos 2 2 x 4 = 14  −  14 cos 2 2 x  = 14  −  14  1 + cos4 x 2  = 14  −  1 + cos4 x 8 = 14  −  18  −  cos4 x 8 = 18  −  18 cos4 x  = 18(1 − cos4 x )Another way to obtain this identity is to use the double-angle formula for sine in the formsin x cos x  = 12 sin2 x . Thussin 2 x cos 2 x  = 14 sin 2 2 x  = 14  1 − cos4 x 2  = 18(1 − cos4 x )EXAMPLE: Find the exact value of sin22 . 5 ◦ .3  EXAMPLE: Find the exact value of sin22 . 5 ◦ .Solution: Since 22 . 5 ◦ is half of 45 ◦ , we use the half-angle formula for sine with  u  = 45 ◦ . Wechoose the + sign because 22 . 5 ◦ is in the first quadrant.sin 45 ◦ 2 =   1 − cos45 ◦ 2 =   1 −√  2 / 22 =   2 −√  24 = 12   2 −√  2EXAMPLE: Find the exact value of sin15 ◦ .Solution: Since 15 ◦ is half of 30 ◦ , we use the half-angle formula for sine with  u  = 30 ◦ . Wechoose the + sign because 15 ◦ is in the first quadrant.sin 30 ◦ 2 =   1 − cos30 ◦ 2 =   1 −√  3 / 22 =   2 −√  34 = 12   2 −√  3Note that 12   2 −√  3 can also be rewritten as √  6 −√  24  .  Indeed,12   2 −√  3 = √  6 −√  24 ⇑ 2   2 −√  3 = √  6 −√  2 ⇑  2   2 −√  3  2 =  √  6 −√  2  2 ⇑ 4  2 −√  3  =  √  6  2 − 2 √  6 √  2 +  √  2  2 ⇑ 8 − 4 √  3 = 6 − 2 √  12 + 2 ⇑ 8 − 4 √  3 = 8 − 2 √  12 ⇑ 8 − 4 √  3 = 8 − 2 √  4 √  3EXAMPLE: Find tan( u/ 2) if sin u  = 25 and  u  is in quadrant II.Solution: To use the half-angle formulas for tangent, we first need to find cos u.  Since cosine isnegative in quadrant II, we havecos u  =  −   1 − sin 2 u  =  −   1 −  25  2 =  −√  215Thustan  u 2 = 1 − cos u sin u  = 1 + √  21 / 52 / 5 = 5 + √  2124
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