TITLE: Verify the following trigonometric identity.
QUESTION [5 upvotes]: Verify the following trig identity: $$\sin(3\theta)-\sin\theta = 2\cos(2\theta)\sin\theta$$

Here is my work so far. 
$\sin(3\theta)-\sin\theta = 2\cos(2\theta)\sin\theta$
LHS:$$\sin(\theta+2\theta)-\sin\theta$$
$$\sin\theta \cos(2\theta)+\sin(2\theta)\cos\theta-\sin\theta$$
$$\sin\theta \cos(2\theta)+(2\sin\theta \cos\theta)\cos\theta-\sin\theta$$
Where do I go from here? I think I should leave the first term in the line above as is, and try and manipulate the second two terms to equal $\sin\theta \cos(2\theta)$, then the LHS will add together to equal $2\cos(2\theta)\sin\theta$, and the identity will be verified. How do you suggest I get there?
Any hints or advice would be appreciated. 
EDIT:
I ended up verifying this identity using the identity frank000 mentioned in comments. Thanks to everyone for the input, it was all very helpful.

REPLY [4 votes]: You're almost done! In the second and third term, factor the $\sin \theta$  and you'll get $2 \cos^2 \theta -1$ which you might recognize as $\cos 2\theta$.

REPLY [2 votes]: $\sin(3θ)−\sinθ=2\cos(2θ)\sinθ$
$\sinθ\cos(2θ)+\sin(2θ)\cosθ−sinθ=2(\cos^2θ-\sin^2θ)\sinθ$
$\sinθ(\cos^2θ-\sin^2θ)+(2\sinθ\cosθ)\cosθ−\sinθ=2(\cos^2θ-\sin^2θ)\sinθ$
$\cos^2θ-\sin^2θ−1=-2\sin^2θ$
$-\sin^2θ-\sin^2θ=-2\sin^2θ$
$-2\sin^2θ=-2\sin^2θ$