Using Euler's formula to prove trig identities using "sum to product" technique

[MiddleEast]

Homework Statement

Derive sum to product formulas using Euler formula

Relevant Equations

Euler formula e^ix = cos x + i sin x

Hello,
This is actually not homework.
I was google searching for "proving trig identities from geometric point of view), found one of the result which proves trig identities using Euler formula. I really liked it. Easier, quicker & simple.
But when the author speak about sum to product formulas, he gave unclear hint and did not do the derivation. Please see attached.
Can some1 explain how to do it using Euler? am really struggling with memorizing sum to product identities.
I still can derive them easily from angles sum & difference sin(x+y) ... etc, But I want to know how to do it using Euler formula.

Thanks

 

[pasmith]

What is meant is, for example, $$\begin{split} \cos a \cos b &= \frac{2\cos a \cos b}{2} + \frac{\sin a \sin b - \sin a \sin b}{2} \\ &= \frac{ \cos a \cos b + \sin a \sin b}{2} + \frac{\cos a \cos b - \sin a \sin b}{2} \\ &= \frac{\cos (a - b)}{2} + \frac{\cos (a + b)}{2} \end{split}.$$ But it is easier to start from $$\begin{split} \cos (a + b) &= \cos a \cos b - \sin a \sin b \\ \cos (a - b) &= \cos a \cos b + \sin a \sin b\end{split}$$ and add the two to get an expression for $\cos a \cos b$ or subtract the first from the second to get an expression for $\sin a \sin b$.

 

[MiddleEast]

Thank you for your reply. Appreciated but it is algebraic. I need the one which uses euler formula.

 

[hunt_mat]

Let $$e^{i(x+y)}=e^{ix}e^{iy}=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))$$ Likewise for $e^{i(x-y)}$. Then add and subtract.

 

[WWGD]

As a general comment, the natural log takes you between sums and products, in that ln(ab)=ln(a) +ln(b)