Question

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