Solve:
$sin 3 x = 3 sin x - 4 {sin}^{3} x$

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Solution

We know that,
$sin (α + β) = sin α cos β + cos α sin β$
$cos (2 α) = {cos}^{2} α - {sin}^{2} α = 2 {cos}^{2} α - 1 = 1 - 2 {sin}^{2} α$
then,
$L . H . S$ = $sin (3 x) = sin (2 x + x) = sin (2 x) cos x + cos (2 x) sin x =$
$= (2 sin x . cos x .) . cos x + (1 - 2 {sin}^{2} x) sin x$
$\begin{matrix} = 2 sin x . {cos}^{2} x + sin x - 2 {sin}^{3} x = 2 sin x (1 - {sin}^{2} x) + sin x - 2 {sin}^{3} x = 2 sin x - 2 {sin}^{3} x + sin x - 2 {sin}^{3} x \end{matrix}$
$= 3 sin x - 4 {sin}^{3} x$
$R . H . S$ proved

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