Prove that:

Solution:

We know,

$\cos 3 x=4 \cos ^{3} x-3 \cos x$

$\Rightarrow \cos ^{3} x=\frac{\cos 3 x+3 \cos x}{4} \ldots$ (i)

Also,

$\sin 3 x=3 \sin x-4 \sin ^{3} x$

$\Rightarrow \sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4} \ldots$ (ii)

Now,

LHS $=\cos ^{3} x \sin 3 x+\sin ^{3} x \cos 3 x$

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

[Using (i) and (ii)]

$=\frac{1}{4}[3(\sin 3 x \cos x+\sin x \cos 3 x)+(\cos 3 x \sin 3 x-\sin 3 x \cos 3 x)]$

$=\frac{1}{4}[3 \sin (3 x+x)+0]$

$=\frac{3}{4} \sin 4 x$

$=\mathrm{RHS}$

Hence proved.