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Question

Evaluate
cos4θ+sin4θ=12cos2θ.sin2θ

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Solution

R.H.S=12cos2θ.sin2θ

=[sin2θ+cos2θ]22cos2θ.sin2θ

=sin4θ+cos4θ+2cos2θ.sin2θ2cos2θ.sin2θ

=sin4θ+cos4θ

Hence proved

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