I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\ &= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\ &= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\ &= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\ &= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\ &= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\ &= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\ &= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\ &= 2\sin2\theta\cos\theta.}$$