# Question #7cfc8

Oct 5, 2016

Proof below

#### Explanation:

First we will find the expansion of $\sin \left(3 x\right)$ separately (this will use the expansion of trig functions formulae):
$\sin \left(3 x\right) = \sin \left(2 x + x\right)$
$= \sin 2 x \cos x + \cos 2 x \sin x$
$= 2 \sin x \cos x \cdot \cos x + \left({\cos}^{2} x - {\sin}^{2} x\right) \sin x$
$= 2 \sin x {\cos}^{2} x + \sin x {\cos}^{2} x - {\sin}^{3} x$
$= 3 \sin x {\cos}^{2} x - {\sin}^{3} x$
$= 3 \sin x \left(1 - {\sin}^{2} x\right) - {\sin}^{3} x$
$= 3 \sin x - 3 {\sin}^{3} x - {\sin}^{3} x$
$= 3 \sin x - 4 {\sin}^{3} x$

Now to solve the original question:
$\frac{\sin 3 x}{\sin x} = \frac{3 \sin x - 4 {\sin}^{3} x}{\sin} x$
$= 3 - 4 {\sin}^{2} x$
$= 3 - 4 \left(1 - {\cos}^{2} x\right)$
$= 3 - 4 + 4 {\cos}^{2} x$
$= 4 {\cos}^{2} x - 1$
$= 4 {\cos}^{2} x - 2 + 1$
$= 2 \left(2 {\cos}^{2} x - 1\right) + 1$
$= 2 \left(\cos 2 x\right) + 1$