Prove that: $\cos ^{3} x \sin 3 x+\sin ^{3} x \cos 3 x=\frac{3}{4} \sin 4 x$
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.
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