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Question

Show that sin4θcos4θ=12cos2θ

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Solution

LHS =sin4θcos4θ

=(sin2θ)2(cos2θ)2 [a2b2=(ab)(a+b)]

=(sin2θ+cos2θ)(sin2θcos2θ)

=sin2θcos2θ [sin2θ+cos2θ=1]

=(cos2θsin2θ) [cos2θ=cos2θsin2=2cos2θ1]

=cos2θ

=(2cos2θ1)

=12cos2θ=R.H.S.


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