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Question

Q27


27. If a cos3 θ + 3a sin2 θ cos θ = m and a sin3 θ + 3a sin θ cos2 θ = n, prove that m + n2/3 + m - n2/3 = 2a2/3

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Solution

Dear student

Given:m=acos3θ+3asin2θ cosθn=asin3θ+3asinθ cos2θConsider, LHSm+n23+m-n23=acos3θ+3asin2θ cosθ+asin3θ+3asinθ cos2θ23+acos3θ+3asin2θ cosθ-asin3θ-3asinθ cos2θ23=a23cos3θ+sin3θ+3cosθ sinθcosθ+sinθ23+a23cos3θ-sin3θ-3cosθ sinθcosθ-sinθ23=a23cosθ+sinθ323+a23cosθ-sinθ323=a23cos2θ+sin2θ+2sinθcosθ+cos2θ+sin2θ-2sinθcosθ=a23×2×cos2θ+sin2θ=2a23=RHSIdentity used:1) cos2x+sin2x=1
Regards

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