# 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.

• If $\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$ and $\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}$ were proven geometrically, doesn't that mean you have already shown $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$ geometrically? Feb 25 '15 at 16:52

\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}
\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}