Question

Given that $\cos \frac{x}{2}.\cos \frac{x}{4}.\cos \frac{x}{8}....=\frac{\sin x}{x}$, prove
$\frac{1}{2^2}{\sec }^2\frac{x}{2}+\frac{1}{2^4}{\sec }^2\frac{x}{4}+....={\csc }^{2}x-\frac{1}{x^2}$

Answer 1

Given that,

$\cos \frac{x}{2}.\cos \frac{x}{4}.\cos \frac{x}{8}.......=\frac{\sin x}{x}......\left(1\right)$

Prove that,

$\frac{1}{2^2}{\sec }^2\frac{x}{2}+\frac{1}{2^4}{\sec }^2\frac{x}{4}+.......={\csc }^{2}x-\frac{1}{x^2}$

Proof:-

By equation (1) to and we get,

$\cos \frac{x}{2}.\cos \frac{x}{4}.\cos \frac{x}{8}.......=\frac{\sin x}{x}$

Using summation and log both side and we get,

$\sum _{k=1}^{\infty }\log \cos \left(\frac{x}{2^k}\right)=\log \sin x-\log x$

On differentiating this equation with respect to x and we get,

$\sum _{k=1}^{\infty }-\frac{1}{2^k}\tan \left(\frac{x}{2^k}\right)=\cot x-\frac{1}{x}$

Again differentiating and we get,

$\sum _{k=1}^{\infty }-\frac{1}{2^{2k}}{\sec }^2\left(\frac{x}{2^k}\right)=-{\csc }^{2}x+\frac{1}{x^2}$

And solving summation,

$\Rightarrow -(\frac{1}{2^2}{\sec }^2\frac{x}{2}+\frac{1}{2^4}{\sec }^2\frac{x}{4}+.......)=-({\csc }^{2}x-\frac{1}{x^2})$

$\Rightarrow \frac{1}{2^2}{\sec }^2\frac{x}{2}+\frac{1}{2^4}{\sec }^2\frac{x}{4}+.......={\csc }^{2}x-\frac{1}{x^2}$

Hence proved.