# Prove the following identity?

## I know this is a double angle question, but not sure how to prove it! If you could show me the steps, that would be great! $\sin \left(3 x\right) = 3 \sin x - 4 {\sin}^{3} x$

Mar 14, 2018

#### Explanation:

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

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

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

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

= $\sin x - 2 {\sin}^{3} x + 2 \sin x {\cos}^{2} x$

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

= $\sin x - 2 {\sin}^{3} x + 2 \sin x - 2 {\sin}^{3} x$

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

Mar 14, 2018

See Explanation.

#### Explanation:

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

Hope you got it.

Mar 14, 2018

Kindly go through a Proof in the Explanation.

#### Explanation:

$\sin 3 x = \underline{\sin 3 x - \sin x} + \sin x$,

$= 2 \cos \left(\frac{3 x + x}{2}\right) \sin \left(\frac{3 x - x}{2}\right) + \sin x$,

$= 2 \cos 2 x \sin x + \sin x$,

$= 2 \left(1 - 2 {\sin}^{2} x\right) \sin x + \sin x$,

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

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