Question

Prove that =$\frac{\sin 5x-2sin3x+\sin x}{\cos 5x-\cos x}=\tan x$

Answer 1

Formula:

$\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$\cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$LHS=\frac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}$

$=\frac{(\sin 5x-\sin 3x)-(\sin 3x-\sin x)}{\cos 5x-\cos x}$

$=\frac{2\cos 4x\sin x-2\cos 2x\sin x}{-2\sin 3x\sin 2x}$

$=\frac{2\sin x(\cos 4x-\cos 2x)}{-2\sin 3x\sin 2x}$

$=\frac{2\sin x(-2\sin 3x\sin x)}{-2\sin 3x\left(2\sin x\cos x\right)}$

$=\frac{\sin x}{\cos x}$

$=\tan x=RHS$

$LHS=RHS$