Question

If $cos\frac{x}{2}cos\frac{x}{4}cos\frac{x}{8}.....=\frac{sinz}{x}$ then the sum $\frac{1}{2^2}se{c}^2\frac{x}{2}+\frac{1}{2^1}se{c}^2\frac{x}{4}+...$ is -

• A. $co{s}^{2}x$
• B. $co{s}^{2}x+\frac{1}{z}$
• C. $cose{c}^{2}x-\frac{1}{x^2}$
• D. $co{s}^{2}x+x$

Answer 1

The correct option is C $cose{c}^{2}x-\frac{1}{x^2}$

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

${\sum }_{k=1}^{\infty }\cos \frac{x}{2^k}=\frac{\sin x}{x}$

Take $\log$ on both sides

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

Diff. both sides w.r.t.x

$\frac{1}{{\sum }_{k=1}^{\infty }\cos \frac{x}{2^k}}{\sum }_{k=1}^{\infty }\sin \frac{x}{2^k}.{\sum }_{k=-1}^{\infty }\frac{1}{2^k}=\frac{1}{\sin x}\cos x-\frac{1}{x}$

$\Rightarrow {\sum }_{k=1}^{\infty }\frac{-1}{2^k}\tan \frac{x}{2^k}=\cot x-\frac{1}{x}$

Diff. both sides w.r.t.

${\sum }_{k=1}^{\infty }\frac{-1}{2^k}{\sec }^2\frac{x}{2^k}.\frac{1}{2^k}=-{\csc }^{2}x+\frac{1}{2}$

${\sum }_{k=1}^{\infty }{\sec }^2\frac{x}{2^k}={\csc }^{2}x-\frac{1}{x^2}$

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