Trigonometry and computations - what to do? I want to prove the following equality:
$$\frac{\sin3x}{\sin x}\cdot\frac{\sin(2n+1)x}{\sin x}=\frac{\sin(2n-1)x}{\sin x}+\frac{\sin(2n+1)x}{\sin x}+\frac{\sin(2n+3)x}{\sin x}$$
I don't know which theorem I have to use to come to the given result. Can someone help me with this computation? Thanks.

HINT:
Using $\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2,$
$\sin(2n-1)x+\sin(2n+3)x=2\sin(2n+1)x\cos 2x$
As $\cos2x=1-2\sin^2x,$  taking out the common factor $\frac{\sin(2n+1)x}{\sin x}$ from the RHS, we get  $$2\cos2x+1 =2(1-2\sin^2x)+1=3-4\sin^2x=\frac{3\sin x-4\sin^3x}{\sin^2x}=\frac{\sin3x}{\sin^2x}$$
