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Q(24) Prove the following 

\small \cos4x = 1 - 8\sin^{2}x\cos^{2}x

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We know that 
                 cos2x=1-2\sin^{2}x
We use this in our problem 
 cos 4x = cos 2(2x)
            =  1-2\sin^{2}2x
             = 1-2(2\sin x \cos x)^{2}                                       (\because \sin2x = 2\sin x \cos x)
             = 1-8\sin^{2}x\cos^{2}x = R.H.S.

Posted by

Gautam harsolia

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