Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ I've seen this identity on examsolutions, but I'm unsure on how to prove it.
$$\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$

Substitute the following: $v := \frac{x+y}{2}$ and $w := \frac{x-y}{2}$
Hence: $$\cos(x) + \cos(y) = \cos(v + w) + \cos(v - w)$$
Using the addition theorem of the cosinus yields:
$$= \cos(v) \cdot \cos(w) - \sin(v) \cdot \sin(w) + \cos(v) \cdot \cos(w) + \sin(v) \cdot \sin(w)$$
$$= 2 \cdot \cos(v) \cdot \cos(w) = 2 \cdot \cos(\frac{x+y}{2}) \cdot \cos(\frac{x-y}{2})$$
