Proving Cosine Addition Identity using Complex numbers

Question

I want to see how one can stumble upon the following using complex numbers

$\cos{A} + \cos{B} = 2\cos{\frac{A+B}{2}}\cos\frac{A-B}{2}$

I know getting to angle sum formulas like $\cos{(A+B)}$ and $\sin{(A+B)}$ is easy. Just sub in $(A+B)$ as the angle in Euler's formula.

I know the $\cos{A} + \cos{B}$ formula can be derived by letting $A+B=C$, $A-B=D$, bringing the question into a $\cos{(C+D)} + \cos{(C-D)}$ format and then cancelling out a few terms, you get the formula.

But is there an elegant way to reach it? Using exponential rules and Euler's formula in a very elegant way... I mean something that doesn't require the thought, "Oh let me write it as a sum and difference". Something more straightforward... something that would be a very natural thing to do. For someone who doesn't know the answer in the first place.

EDIT: I'm not talking about a way up the answer to the question. Just a naive travel to the answer itself as if it is unknown. So I'm NOT looking for a way that starts with defining 2cos((A+B)/2)cos((A-B)/2) and works its way up. Rather, the other way around.

Answer 1

Generalizing the usual trick to find the modulus and argument of $e^{ia}+1$, let us divide $e^{iA}+e^{iB}$ by $e^{i\frac{A+B}2}$. The quotient is $e^{i\frac{2A-(A+B)}2}+e^{i\frac{2B-(A+B)}2}$. Then, taking the real and imaginary parts of $$e^{iA}+e^{iB}=e^{i\frac{A+B}2}\left(e^{i\frac{A-B}2}+e^{i\frac{B-A}2}\right)=2e^{i\frac{A+B}2}\cos\frac{A-B}2$$ we get simultaneously $$\begin{cases}\cos A+\cos B&=2\cos\frac{A+B}2\cos\frac{A-B}2,\\ \sin A+\sin B&=2\sin\frac{A+B}2\cos\frac{A-B}2. \end{cases}$$