**Solution:** you can just replace x by alpha according to your question

**Explanation :** Observe that there are n terms of cos, and, in 2^{n},

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)...(2cos2^{n−1}x)

⋮

⋮

But, x=π/2^{n}+1⇒2^{n}x+x=π,or,2^{n}x=π−x, so that,

sin(2^{n}x)=sin(π−x)=sinx. Therefore,

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

Hence proved.