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?
Prove the sum to product formulas with complex numbers
complex numbersproof-writingtrigonometry
Best Answer
HINT
Let use instead
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.