Question

If $0<x<\pi$, then $\frac{\sin 8x+7\sin 6x+18\sin 4x+12\sin 2x}{\sin 7x+6\sin 5x+12\sin 3x}$ equal to

• A. 2 sin x
• B. sin x
• C. sin 2x
• D. 2 cos x
• E. cos x

Answer 1

Answer: (D)

We have, $0<x<\pi$
$\frac{\sin 8x+7\sin 6x+18\sin 4x+12\sin 2x}{\sin 7x+6\sin 5x+12\sin 3x}$
$\frac{\left[\left(\sin 8x+\sin 6x\right)+6\left(\sin 6x+\sin 4x\right)+12\left(\sin 4x+\sin 2x\right)\right]}{\sin 7x+6\sin 5x+12\sin 3x}$
$\frac{\left[2\sin \left(\frac{8x+6x}{2}\right)\cos \left(\frac{8x-6x}{2}\right)+6\cdot 2\sin \frac{\left(6x+4x\right)}{2}\cos \frac{\left(6x-4x\right)}{2}+12\cdot 2\sin \frac{\left(4x+2x\right)}{2}\cos \frac{\left(4x-2x\right)}{2}\right]}{\sin 7x+6\sin 5x+12\sin 3x}$
$\frac{\left[2\sin 7x\cos x+12\sin 5x\cos x+24\sin 3x\cos x\right]}{\sin 7x+6\sin 5x+12\sin 3x}$
$=\frac{2\cos x[\sin 7x+6\sin 5x+12\sin 3x]}{\sin 7x+6\sin 5x+12\sin 3x}0$
$=2\cos x$