# Question #5176d

Aug 11, 2016

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 $: \sin 2 \theta = 2 \sin \theta \cos \theta$. Thus,

The$L . H . S . = \left(2 \cos x\right) \left(2 \cos 2 x\right) \left(2 \cos 4 x\right) \left(2 \cos 8 x\right) \ldots \left(2 \cos {2}^{n - 1} x\right)$

$= \frac{\left(2 \sin x \cos x\right) \left(2 \cos 2 x\right) \left(2 \cos 4 x\right) \ldots \ldots \ldots . . \left(2 \cos {2}^{n - 1} x\right)}{\sin} x$

$= \frac{\left(\sin 2 x\right) \left(2 \cos 2 x\right) \left(2 \cos 4 x\right) \ldots \ldots \ldots \left(2 \cos {2}^{n - 1} x\right)}{\sin} x$

$= \frac{\left(2 \sin 2 x \cos 2 x\right) \left(2 \cos 4 x\right) \ldots \ldots . \left(2 \cos {2}^{n - 1} x\right)}{\sin} x$

$= \frac{\left(\sin 4 x\right) \left(2 \cos 4 x\right) \ldots \ldots \ldots . \left(2 \cos {2}^{n - 1} x\right)}{\sin} x$

$= \frac{\left(2 \sin 4 x \cos 4 x\right) \ldots \ldots \ldots . \left(2 \cos {2}^{n - 1} x\right)}{\sin} x$
$\vdots$
$\vdots$
$\vdots$
$= \sin \frac{2 \cdot {2}^{n - 1} x}{\sin} x$

$= \sin \frac{{2}^{n} x}{\sin} x$

But, $x = \frac{\pi}{{2}^{n} + 1} \Rightarrow {2}^{n} x + x = \pi , \mathmr{and} , {2}^{n} x = \pi - x$, so that,

$\sin \left({2}^{n} x\right) = \sin \left(\pi - x\right) = \sin x$. Therefore,

The $L . H . S . = \sin \frac{x}{\sin} x = 1 = T h e R . H . S$

Hence, the Proof. Enjoy Maths.!