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?



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.

  • $\begingroup$ umm, I simplified to $\frac{e^{i2\alpha}-e^{-i2\beta}}{2i}$ but I don't see how that's equal to $\sin x+\sin y$ $\endgroup$
    – Acyex
    Aug 29 '21 at 21:27
  • $\begingroup$ @Acyex Don't you get the following? $$\frac{e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)}+e^{i(\alpha-\beta)}-e^{-i(\alpha-\beta)}}{2i}$$ $\endgroup$
    – user
    Aug 29 '21 at 21:35
  • $\begingroup$ OH wait a second let me check $\endgroup$
    – Acyex
    Aug 29 '21 at 21:37
  • $\begingroup$ 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 $\endgroup$
    – Acyex
    Aug 29 '21 at 21:39
  • $\begingroup$ @Acyex Ah ok sorry I didn't noticed that! Then you are looking for the same formula on the complex? $\endgroup$
    – user
    Aug 29 '21 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.