TITLE: prove trig equivalence
QUESTION [5 upvotes]: $$\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
I want to use the the $e^{ix}$ identities but I'm not sure if it can be done that way, let alone how to do it.
Any tips would be appreciated.

REPLY [2 votes]: As requested, we will prove the identity of interest through direct use Euler's Identity, 
$$e^{iz}=\cos z+i\sin z$$
From this we have $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$ and $\cos z=\frac{e^{iz}+e^{-iz}}{2}$.  Then, we can write
$$\begin{align}
2\sin\left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)&=2\left(\frac{e^{i(x+y)/2}-e^{-i(x+y)/2}}{2i}\right)\left(\frac{e^{i(x-y)/2}+e^{-i(x-y)/2}}{2}\right)\\\\
&=\frac{e^{ix}+e^{iy}-e^{-iy}-e^{-ix}}{2i}\\\\
&=\frac{e^{ix}-e^{-ix}}{2i}+\frac{e^{iy}-e^{-iy}}{2i}\\\\
&=\sin x+\sin y 
\end{align}$$
as was to be shown!