Question #5176d

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Answer 1 · 2016-08-11T19:29:26

See the Proof in Explanation Section.

Observe that there are $n$ $n$ terms of $cos$ $cos$ , and, in $2^{n}$ $2^n$ , the no. of $2$ $2$ is

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

We will also use the identity $: sin 2 θ = 2 sin θ cos θ$ $: sin2theta=2sinthetacostheta$ . 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.!