Question

Prove that cos 4x = 1 – 8sin^2 x cos^2 x

Answer 1

Use the identities,

$\cos 2\theta=1-2\sin^2 \theta\\ \sin 2\theta=2\sin \theta \cos \theta$

Now,

$LHS=\cos 4x=1-2\sin ^2 2x\\ LHS=\cos 4x=1-2(2\sin x\cos x)^2\\ LHS=\cos 4x=1-8\sin^2 x\cos^2 x=RHS$

Answer 2

Answer:

you can solve  

cos4x as

cos2(2x) =1-2sin2(2x)

            =1-2(2sinx . cosx)2                          { sin2x = 2sinx. cosx}

            =1-2(4sin2x.cos2x)

           =1-8sin2x.cos2x

LHS=RHS