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?

$$\begin{array} {rcl}
\cos(x + y) + i \sin(x + y)& = & e^{i(x + y)} \\
&=& e^{ix}e^{iy} \\
&=& (\cos(x) + i\sin(x))(\cos(y) + i\sin(y)) \\
&=& (\cos(x)\cos(y) - \sin(x)\sin(y)) + i(\sin(x)\cos(y) + \sin(y)\cos(x)) \\
\end{array}$$
Equating real and imaginary parts you get
$$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$
$$\sin(x + y) = \sin(x)\cos(y) + \sin(y)\cos(x)$$
