Question

$\cos \frac{x}{2}\times \cos \frac{x}{2^2}\times \cos \frac{x}{2^3}..........\times \cos \frac{x}{2^n}$$A\to \infty$$\frac{\sin 2^{n}A}{2^n\sin A}$

Answer 1

$\begin{array}{cc}\cos \frac{x}{2}\cdot \left(0\right)\frac{x}{2^2}\cdot \cos \frac{x}{2^3}.......\cos \frac{x}{}{2}^{n}n\to \frac{\infty \sin 2^{n}A}{2^n\sin A}& \\ now,& \\ \cos \frac{x}{2}& \\ =\frac{1}{2}\left(\sin \frac{x}{2}\cdot \cos \frac{x}{2}\right)& \frac{1}{2}\sin \frac{x}{2}=\frac{\sin x}{2\sin \frac{x}{2}}\\ \cos \frac{x}{2^2}& \\ =\frac{1}{2}\sin \frac{x}{2^2}\cos \frac{x}{2^2}& \sin \frac{x}{2^2}=\frac{\sin \frac{x}{2}}{2^2\sin }\\ Similarly& \\ \cos \frac{x}{2^n}=\frac{1}{2}\cos \frac{x}{2^n}\sin \frac{x}{2^n}& \sin \frac{x}{2^n}=\frac{\sin 2\frac{x}{2}}{2^n\sin \frac{n}{2}n}\\ \cos \frac{n}{2}\cdot \cos \frac{n}{2}\cos \frac{n}{2^3}...\cos \frac{n}{2^n}A\to & \\ =\frac{\sin 2^{n}A}{2^n\sin x}& \end{array}$