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Question no. 4
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Solution: you can just replace x by alpha according to your question

Explanation : Observe that there are n terms of cos, and, in 2n,

the no. of 2 is

also n. So, we utilise each 2 with every term of cos.

We will also use the identity :sin2θ=2sinθcosθ. Thus,

The L.H.S.=(2cosx)(2cos2x)(2cos4x)(2cos8x)...(2cos2n−1x)

But, x=π/2n+1⇒2nx+x=π,or,2nx=π−x, so that,

sin(2nx)=sin(π−x)=sinx. Therefore,

The L.H.S.=sinx/sinx=1=The R.H.S

Hence proved.

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