Prove the sum to product formulas with complex numbers First of all, I don't know if it's possible but I'm just wondering about it. I'm pretty standard when it comes to complex numbers knowledge. So my goal: $$\sin x+\sin y=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ so I thought I'd start from the right hand side. Recall the complex defn of sine and cosine: $$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$ and $$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$ Applying both we get: $$2\frac{e^{\frac{xi+yi}{2}}-e^{\frac{-xi+yi}{2}}}{2i}\frac{e^{\frac{ix+iy}{2}}+e^{\frac{-ix+yi}{2}}}{2}$$ but I don't really know what to do now. I've tried expanding the numerator and simplifying the 2's out but I'm pretty stuck. Any hints?

HINT
Let use instead

*

*$\sin \alpha = \frac{e^{i\alpha}-e^{-i\alpha}}{2i}$

*$\cos \beta = \frac{e^{i\beta}+e^{-i\beta}}{2}$
with $\alpha=\frac{x+y}2$ and $\beta=\frac{x-y}2\in \mathbb R$  and then by your way we find out
$$2\sin \alpha\cos \beta  =2\frac{e^{i\alpha}-e^{-i\alpha}}{2i}\frac{e^{i\beta}+e^{-i\beta}}{2}$$
form which you can easily conclude grouping the right terms after multiplication.
