Question

$\frac{\sin 7x+6\sin 5x+17\sin 3x+12\sin x}{\sin 6x+5\sin 4x+12\sin 2x}$ equals:

• (A) $\sin x$
• (B) $2\sin x$
• (C) $\cos x$
• (D) $2\cos x$

Answer 1

The Correct Option is D

Explanation:
Given,$\frac{\sin 7x+6\sin 5x+17\sin 3x+12\sin x}{\sin 6x+5\sin 4x+12\sin 2x}$$=\frac{\sin 7\mathrm{x}+\sin 5\mathrm{x}+5\sin 5\mathrm{x}+5\sin 3\mathrm{x}+12\sin 3\mathrm{x}+12\sin \mathrm{x}}{\sin 6\mathrm{x}+5\sin 4\mathrm{x}+12\sin 2\mathrm{x}}$By using, $\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$Therefore,$=\frac{(2\sin (\frac{7x+5x}{2})\cos (\frac{7x-5x}{2})+5\times (2\sin (\frac{5x+3x}{2})\cos (\frac{5x-3x}{2}))+12\times (2\sin (\frac{3x+x}{2})\cos (\frac{3x-x}{2}))}{(\sin 6x+5\sin 4x+12\sin 2x)}$$=\frac{2\sin 6x\cos x+10\sin 4x\cos x+24\sin 2x\cos x}{(\sin 6x+5\sin 4x+12\sin 2x)}$$=\frac{2\cos x(\sin 6x+5\sin 4x+12\sin 2x)}{(\sin 6x+5\sin 4x+12\sin 2x)}$$=2\cos x$Hence, the correct option is (D).