Proving the sum to product formula with complex numbers starting from the left hand side

I've just asked a similar question today Prove the sum to product formulas with complex numbers but I was wondering what if we we wanted to start with the left hand side instead of the right hand side? Because at the end I want to be able to think, to derive this formula without knowing the outcome, the right hand side, so I'm wondering about how to do it starting from the left hand side. My goal: $$\sin x+\sin y=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ 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 after simplifying a bit: $$\frac{e^{ix}-e^{-ix}+e^{iy}-e^{-iy}}{2i}$$ but I don't really know what to do now. I've also tried with the opposite substitution that I used when proving this starting from the right hand side $$\alpha=\frac{x+y}{2}$$ and $$\beta=\frac{x-y}{2}$$ but I'm pretty sstuck. I know I have to factor it to get it closer to our goal, but no clue on how to factor this. Any hints?

• Follow the steps in the answer to the other question, in reverse order.
– mjw
Aug 30 '21 at 1:04
• Alt. hint: $\;\operatorname{cis}(x) = \operatorname{cis}\left(\frac{x+y}{2}+\frac{x-y}{2}\right)=\operatorname{cis}\left(\frac{x+y}{2}\right)\operatorname{cis}\left(\frac{x-y}{2}\right)\,$, then do something similar for $\,\operatorname{cis}(y)\,$ and add the two.
– dxiv
Aug 30 '21 at 1:13

Obviously, you could "cheat" and use the rhs to inform your factorisation. But that's not what you want.

If you write $$a=e^{ix}$$ and $$b=e^{iy}$$, then you have $$\frac{1}{2i}\left(a-\frac 1a + b - \frac1b\right)$$ which is $$\frac1{2abi}(ab^2+a^2b-b-a)$$

One might then recognise the common factor of $$a+b$$, and get

$$\frac1{2abi}(a+b)(ab-1)$$

If I distribute the factor of $$\frac1{ab}$$ evenly between the two brackets, I get

$$\frac1{2i}\frac{a+b}{\sqrt{ab}}\frac{ab-1}{\sqrt{ab}},$$

that is,

$$\frac1{2i}\left(\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{b}}{\sqrt{a}}\right)\left(\sqrt{ab} - \frac{1}{\sqrt{ab}}\right).$$

Remembering that $$a=e^{ix}$$ and $$b=e^{iy}$$, this is

$$\frac1{2i}\left(\sqrt{e^{i(x-y)}}+\sqrt{e^{-i(x-y)}}\right)\left(\sqrt{e^{i(x+y)}} - \sqrt{e^{-i(x+y)}}\right),$$

that is,

$$\frac1{2i}\left({e^{i\frac{x-y}2}}+{e^{-i\frac{x-y}2}}\right)\left({e^{i\frac{x+y}2}} - {e^{-i\frac{x+y}2}}\right)$$

which gives you the right hand side after just a couple more fairly obvious steps.

Edit: yes, I've glossed over the fact that complex numbers have two square roots, and there's no obvious preference to determine what is meant by $$\sqrt{z}$$. Defining $$\sqrt{a}$$ and $$\sqrt{b}$$ carefully would give either the desired identity, or $$\sin x+\sin y=2\sin\left(\frac{x+y}{2}+\pi\right)\cos\left(\frac{x-y}2+\pi\right)$$