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.
I ended up verifying this identity using the identity frank000 mentioned in comments. Thanks to everyone for the input, it was all very helpful.