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

• $$\sin \alpha = \frac{e^{i\alpha}-e^{-i\alpha}}{2i}$$
• $$\cos \beta = \frac{e^{i\beta}+e^{-i\beta}}{2}$$

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.

• umm, I simplified to $\frac{e^{i2\alpha}-e^{-i2\beta}}{2i}$ but I don't see how that's equal to $\sin x+\sin y$ Aug 29 '21 at 21:27
• @Acyex Don't you get the following? $$\frac{e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)}+e^{i(\alpha-\beta)}-e^{-i(\alpha-\beta)}}{2i}$$
– user
Aug 29 '21 at 21:35
• OH wait a second let me check Aug 29 '21 at 21:37
• my bad, I now see it, I copied the problem wrongly, thanks! Btw now I will be trying the product to sum formula proof with complex numbers, I'll likely ask a new question if I don't get it, anyways just wanted to let it know if you wanted to help whenever I post it and if I do Aug 29 '21 at 21:39
• @Acyex Ah ok sorry I didn't noticed that! Then you are looking for the same formula on the complex?
– user
Aug 29 '21 at 21:41