Prove that:

LHS $=\sin 5 x$

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

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

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

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

$=3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x+\left\{8 \sin x\left(1-\sin ^{2} x\right)^{2}-6 \sin x\left(1-\sin ^{2} x\right)\right\}$

$=3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x+\left\{8 \sin x\left(1-2 \sin ^{2} x+\sin ^{4} x\right)-6 \sin x+6 \sin ^{3} x\right\}$

$=3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x+8 \sin x-16 \sin ^{3} x+8 \sin ^{5} x-6 \sin x+6 \sin ^{3} x$

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

= RHS

Hence proved.