prove trig equivalence $$\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
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.

\begin{align}
\sin u \cos v + \cos u \sin v & = \sin(u+v) \\
\sin u \cos v - \cos u \sin v & = \sin(u-v)
\end{align}
Apply the above in the case where $u = \dfrac{x+y} 2$ and $v=\dfrac{x-y} 2$:
\begin{align}
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin \left( \frac{x+y}{2} + \frac{x-y}{2} \right) \\[10pt]
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin \left( \frac{x+y}{2} - \frac{x-y}{2} \right) 
\end{align}
Or in other words:
\begin{align}
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin x \\[10pt]
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin y 
\end{align}
Add left sides and add right sides:
$$
 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = \sin x + \sin y.
$$
