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EXAMPLE 10.50Prove that cos' A cos 3A +sin' Asin 3A--cos 2A |
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Answer» LHS = cos^3A cos3A + sin^3A sin3A = cos^3A (4cos^3A - 3cosA) + sin^3A (3sinA - 4sin^3A) (using identities of cos3A and sin3A) = 4cos^6A - 3cos^4A + 3sin^4A - 4sin^6A = 4 (cos^6A -sin^6A) + 3 (sin^4A -cos^4A) = 4 ( (cos^2A)^3- (sin^2A)^3) + 3 ( (sin^2A)^2- (cos^2A)^2) = 4 [ (cos^2A -sin^2A) (cos4A +cos^2Asin^2A +sin^4A)] + 3 [(sin^2A+cos^2A)(sin^2A - cos^2A)] using a^3- b^3= (a-b)(a^2+ab+b^2) and a^2-b^2=(a-b)(a+b) = 4 [ cos2A ((sin^2A+cos^2A)^2- cos^2Asin^2A) ] + 3 [ 1 (-cos2A)]= 4 [cos2A(1 - cos^2Asin^2A )] - 3 cos2A= 4cos2A - 4cos2A cos^2Asin^2A - 3cos2A = cos2A- 4cos2Acos^2Asin^2A = cos2A (1 - 4cos^2Asin^2A )= cos2A (1 - sin^2A) (Since2sinAcosA=sin2A)= cos2A cos^22A= cos^32A= RHS |
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