$\sin 5 x=5 \cos ^{4} x \sin x-10 \cos ^{2} x \sin ^{3} x+\sin ^{5} x$
LHS $=\sin 5 \mathrm{x}$
$=\sin (3 x+2 x)$
$=\sin 3 x \times \cos 2 x+\cos 3 x \times \sin 2 x$
$=\left(3 \sin x-4 \sin ^{3} x\right)\left(2 \cos ^{2} x-1\right)+\left(4 \cos ^{3} x-3 \cos x\right) \times 2 \sin x \cos x$
$=-3 \sin x+4 \sin ^{3} x+6 \sin x \cos ^{2} x-8 \sin ^{3} x \cos ^{2} x+8 \sin x \cos ^{4} x-6 \sin x \cos ^{2} x$
$=8 \sin x \cos ^{4} x-8 \sin ^{3} x \cos ^{2} x-3 \sin x+4 \sin ^{3} x$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-3 \sin x+3 \sin x \cos ^{4} x+4 \sin ^{3} x+2 \sin ^{3} x \cos ^{2} x$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-3 \sin x\left(1-\cos ^{4} x\right)+2 \sin ^{3} x\left(2+\cos ^{2} x\right)$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-3 \sin x\left(1-\cos ^{2} x\right)\left(1+\cos ^{2} x\right)+2 \sin ^{3} x\left(2+\cos ^{2} x\right)$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-3 \sin ^{3} x\left(1+\cos ^{2} x\right)+2 \sin ^{3} x\left(2+\cos ^{2} x\right)$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-\sin ^{3} x\left[3\left(1+\cos ^{2} x\right)-2\left(2+\cos ^{2} x\right)\right]$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-\sin ^{3} x\left[3+3 \cos ^{2} x-4-2 \cos ^{2} x\right]$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-\sin ^{3} x\left[\cos ^{2} x-1\right]$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x-\sin ^{3} x \times\left(-\sin ^{2} x\right)$
$=5 \sin x \cos ^{4} x-10 \sin ^{3} x \cos ^{2} x+\sin ^{5} x$
$=5 \cos ^{4} x \sin x-10 \cos ^{2} x \sin ^{3} x+\sin ^{5} x$
= RHS
Hence proved.