Question

$\cos x+\cos 3x+......+\cos (2n-1)x=\frac{\sin 2^{n}x}{2\sin x},x\ne k\pi$.

• A. True
• B. False

Answer 1

The correct option is A True

Given

$\cos x+\cos 3x+\cdots +\cos nx=\frac{\sin 2^{n}x}{2\sin x}$

LHS

$\cos x+\cos 3x+\cdots +\cos nx$

Multipying and dividing by $2\sin x$

$\frac{2\sin x\cos x}{2\sin x}+\cos 3x+\cdots +\cos nx$

$\frac{\sin 2x}{2\sin x}+\cos 3x+\cdots +\cos nx$

$\frac{\sin 2x+2\sin x\cos 3x}{2\sin x}+\cdots +\cos nx$

$\frac{\sin 2x+\sin 4x-\sin 2x}{2\sin x}+\cdots +\cos nx$

$\frac{\sin 4x}{2\sin x}+\cos 5x+\cdots +\cos nx$

$⟹\frac{\sin 2^{n-1}x}{2\sin x}+\cos nx$

$⟹\frac{\sin 2^{n-1}x+2\sin x\cos nx}{2\sin x}$

$⟹\frac{\sin 2^{n-1}x+\sin 2^{n}x-\sin 2^{n-1}x}{2\sin x}$

$⟹\frac{\sin 2^{n}x}{\sin x}$

RHS

Hence proved