Question

Question #5176d

Answer 1

See the Proof in Explanation Section.

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 `: sin2theta=2sinthetacostheta`. Thus,

The`L.H.S.=(2cosx)(2cos2x)(2cos4x)(2cos8x)...(2cos2^(n-1)x)`

`={(2sinxcosx)(2cos2x)(2cos4x)...........(2cos2^(n-1)x)}/sinx`

`={(sin2x)(2cos2x)(2cos4x).........(2cos2^(n-1)x)}/sinx`

`={(2sin2xcos2x)(2cos4x).......(2cos2^(n-1)x)}/sinx`

`={(sin4x)(2cos4x)..........(2cos2^(n-1)x)}/sinx`

`={(2sin4xcos4x)..........(2cos2^(n-1)x)}/sinx`
`vdots`
`vdots`
`vdots`
`=sin(2*2^(n-1)x)/sinx`

`=sin(2^nx)/sinx`

But, `x=pi/(2^n+1) rArr 2^nx+x=pi, or, 2^nx=pi-x`, so that,

`sin(2^nx)=sin(pi-x)=sinx`. Therefore,

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

Hence, the Proof. Enjoy Maths.!