Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ 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 $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\
&= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\
&= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\
&= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\ 
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= 2\sin2\theta\cos\theta.}$$ 
 A: Use the trigonometric identity about $\sin A + \sin B= 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$.
A: Use the sum of angles identities for the $\sin$ function:- 
$$\sin(3\theta)=\sin(2\theta+\theta)=\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta)$$
$$\sin(\theta)=\sin(2\theta-\theta)=\sin(2\theta)\cos(\theta)-\cos(2\theta)\sin(\theta)$$
Adding both equations results in
$$\sin(3\theta)+\sin(\theta)=2\sin(2\theta)\cos(\theta)$$
A: Yes, you can differentiate both sides and check whether they are equal: $$\eqalign{
\dfrac{\mathrm d}{\mathrm d\theta}\big[\sin3\theta+\sin\theta\big]&\overset?=\dfrac{\mathrm d}{\mathrm d\theta}\big[2\sin2\theta\cos\theta\big] \\
3\cos3\theta+\cos\theta&\overset?= 2\big[\tfrac12( \cos\theta+3\cos3\theta)\big]\tag{$\overset{\rm Chain}{\underset{\sf rule}{}}\overset{\,+}{}\overset{\rm Prod.}{\underset{\sf rule}{}}$}\\
3\cos3\theta+\cos\theta&\overset?= \cos\theta+3\cos3\theta. \quad\checkmark
}$$
So you can conclude that they're the same. 
A: $$\sin(3\theta)+\sin(\theta)=\Im[e^{3i\theta}]+\Im[e^{i\theta}]=\Im[e^{3i\theta}+e^{i\theta}]=$$
$$=\Im[e^{2i\theta}(e^{i\theta}+e^{-i\theta})]=2\cos(\theta)\Im[e^{2i\theta}]=2\cos(\theta)\sin(2\theta)\ .$$
