[Math] Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$

Question

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 $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + (\cos^2\theta – \sin^2\theta)\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta – \sin^2\theta + 1)\\
&= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\
&= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\
&= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= 2\sin2\theta\cos\theta.}$$

Answer 1

Use the trigonometric identity about $\sin A + \sin B= 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$.