<s>[INST] geometric proof of $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$ I have seen geometric proof of identities
$$\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$$
and
$$\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}$$
By adding two equation, $$2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$$.
But how to prove this by geometry?
Thank you.
 [/INST] 
$$\begin{align}
2 \cos A \cos B &= \cos(A-B)+\cos(A+B) \\[6pt]
2 \sin A \,\sin B &= \cos(A-B)-\cos(A+B)
\end{align}$$
Note. Although not labeled (yet), these identities are also evident:
$$\begin{align}
2 \,\sin A \cos B &= \sin(A+B)+\sin(A-B) \\[6pt]
2 \cos A \,\sin B &= \sin(A+B)-\sin(A-B)
\end{align}$$
</s>