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.
 A: HINT
It may be easier to use
$$
\sin (a \pm b) = \sin a \cos b \pm \cos a \sin b \\
\cos (a \pm b) = -(\cos^2 a \pm \sin^2 a)
$$
A: From $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$, get:
$$\require{cancel}\begin{align}
\sin x+\sin y&=\sin\left(\frac{x+y}{2}+\frac{x-y}{2}\right)+\sin\left(\frac{x+y}{2}-\frac{x-y}{2}\right) \\[2ex]
&=\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)+\cancel{\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)}\\&\qquad +\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)-\cancel{\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)} \\[2ex]
&=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)
\end{align}$$
A: 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!
A: \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:
$$
 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = \sin x + \sin y.
$$
A: Hint:
You really should know the $3$ linearisation formulae:
\begin{alignat*}{2}
&2\cos a\cos b&&=\cos(a-b)+\cos(a+b)\\
&2\sin a\sin b&&=\cos(a-b)-\cos(a+b)\\
&2\sin a\cos b&&=\sin(a-b)+\sin(a+b)
\end{alignat*}
Set $\;a-b=x,\enspace a+b=y$, and solve for $a$ and $b$.
