Question:

$\frac{\sin 5 x}{\sin x}$ is equal to

(a) $16 \cos ^{4} x-12 \cos ^{2} x+1$

(b) $16 \cos ^{4} x+12 \cos ^{2} x+1$

(c) $16 \cos ^{4} x-12 \cos ^{2} x-1$

(d) $16 \cos ^{4} x+12 \cos ^{2} x-1$

Solution:

(a) $16 \cos ^{4} x-12 \cos ^{2} x+1$

To find : $\frac{\sin 5 x}{\sin x}$

Now,

$\sin 5 x=\sin (3 x+2 x)$

$=\sin 3 x \cos 2 x+\cos 3 x \sin 2 x$

$=\left(3 \sin x-4 \sin ^{3} x\right)\left(1-2 \sin ^{2} x\right)+\left(4 \cos ^{3} x-3 \cos x\right)(2 \sin x \cos x)$

$=\left(3 \sin x-6 \sin ^{3} x-4 \sin ^{3} x+8 \sin ^{5} x\right)+2 \sin x \cos ^{2} x\left(4 \cos ^{2} x-3\right)$

$=\left(3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x\right)+2 \sin x\left(1-\sin ^{2} x\right)\left[4\left(1-\sin ^{2} x\right)-3\right]$

$=\left(3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x\right)+\left(2 \sin x-2 \sin ^{3} x\right)\left(4-4 \sin ^{2} x-3\right)$

$=\left(3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x\right)+\left(2 \sin x-8 \sin ^{3} x-2 \sin ^{3} x+8 \sin ^{5} x\right)$

$=5 \sin x-20 \sin ^{3} x+16 \sin ^{5} x$

$\therefore \frac{\sin 5 x}{\sin x}=\frac{5 \sin x-20 \sin ^{3} x+16 \sin ^{5} x}{\sin x}$

$=5-20 \sin ^{2} x+16 \sin ^{4} x$

$=5-20\left(1-\cos ^{2} x\right)+16\left(1-\cos ^{2} x\right)^{2}$

$=5-20+20 \cos ^{2} x+16\left(1+\cos ^{4} x-2 \cos ^{2} x\right)$

$=5-20+20 \cos ^{2} x+16+16 \cos ^{4} x-32 \cos ^{2} x$

$=16 \cos ^{4} x-12 \cos ^{2} x+1$