Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$ Define $\sin x $ and $\cos x$ via their infinite series:
$$
\sin x = \sum_n (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}, \qquad
\cos x = \sum_n (-1)^n \frac{x^{2n}}{(2n)!}.
$$
Is there a short, clever proof that $\cos(x+y) = \cos x \cos y - \sin x \sin y$ for all real $x,y$?  I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = \cos y$, $f'(0) = - \sin y$. Does anyone know other (preferably slick!) proofs?

Suppose we know that $\sin(x+y) = \sin x \cos y + \cos x \sin y$, then we can do this to get a proof:
$\cos (x + y) \\ =  \sin (x + y + \frac{\pi}{2}) \\ = \sin x \cos (y + \frac{\pi}{2}) + \cos x \sin (y + \frac{\pi}{2}) \\ = \sin x (-\sin y) + \cos x \cos y \\ = \cos x \cos y - \sin x \sin y$
By the way, both have a geometric proof.
