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?
 A: One way is to use the fact that
$$\cos(\theta) = \dfrac{e^{i \theta}+e^{-i \theta}}2$$
$$\cos(x+y) = \dfrac{e^{i(x+y)}+e^{-i(x+y)}}2 = \left(\dfrac{e^{ix}+e^{-ix}}2 \right) \left(\dfrac{e^{iy}+e^{-iy}}2\right) - \left(\dfrac{e^{ix}-e^{-ix}}{2i} \right) \left(\dfrac{e^{iy}-e^{-iy}}{2i}\right)$$
A: New answer to an old question.  This one, maybe the slickiest of them all, is due to Erhard Schmidt.
Define 
$$f(t)=\cos(x+y-t)\cos(t)-\sin(x+y-t)\sin(t).$$
  Verify $f'(t)=0$, hence $f$ is constant.  Now the desired angle sum identity follows from $f(0)=f(y)$ 
A: This was addressed in the question already. I leave it so that the method is fully explained.
LEMMA Let $f$ be a function with second derivative everywhere such that $f''+f=0$ and $f'(0)=0$; $f(0)=0$. Then $f$ is identically zero everywhere.
P We have that $$f''+f=0$$ Then $$f'f''+ff'=0$$ or $$(f')^2+f^2=C$$
But the initial conditions force $f'^2+f^2=0$ everywhere, which means $f\equiv 0$. $\blacktriangle$.
PROP Let $f$ be a function with second derivative everywhere such that $f''+f=0$, and $f'(0)=a$, $f(0)=b$. Then $$f=a\sin+b\cos $$
P Let $g=f-a\sin+b\cos$. Then $g''+g=0$ and $g'(0)=0$, $g(0)=0$. The lemma implies $g\equiv 0$, so that $f=a\sin+b\cos$. $\blacktriangle$.

Differentiate with respect to one variable and use the uniqueness of the solution of a second degree ODE with initial conditions.
That is, your cosine on the left vetifies $$f''+f=0$$ and $f'(0)=–\sin y$, $f''(0)=\cos y$. Then it must coincide with the unique solution $$f'(0) \sin+f(0)\cos$$ 
A: I always had a hard time to memorize that formula.
But actually, that's not really needed, because there it is an easy way to reconstruct it from the from the laws of exponentiation applied to complex exponentiation:
$$e^{i(x + y)} = e^{ix} \cdot e^{iy}.$$
Using the complex multiplication rule $\operatorname{Re}(ab) = \operatorname{Re}(a)\operatorname{Re}(b) - \operatorname{Im}(a)\operatorname{Im}(b)$, taking the real part gives
$$\operatorname{Re}(e^{i(x + y)}) = \operatorname{Re}(e^{ix})\operatorname{Re}(e^{iy}) - \operatorname{Im}(e^{ix})\operatorname{Im}(e^{iy}).$$
So by $\cos(x) = \operatorname{Re}(e^{ix})$ and $\sin(x) = \operatorname{Im}(e^{ix})$
$$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y).$$
A: The way I learned it as a kid was geometric, and probably looked like the proof seen here  on Wikipedia.
The segment $OP$ has length $1$. We have the $\sin(\alpha + \beta) = PB = PR + RB = \cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)$.
Then, to prove the cosine identity we can use that $\cos(\alpha + \beta) = \sin(\alpha + \beta + \pi/2)$ and use the sine identity.
A: Let $\vec{u},\vec{v}\in\mathbb{R}^2$ unitary vectors such that
$$
\vec{u}=\big(\cos(x),\sin(x)\big)\quad \mbox{ and } \quad \vec{v}=\big(\cos(-y),\sin(-y)\big)
$$ 
Here $x$ and $-y$ are the smallest angle formed between the x-axis and the vectors $\vec{u}$ and $\vec{v}$ respectively. Then 
\begin{align}
\cos\big( x+y\big) = & \cos\big( x-(-y)\big)\\
= & \frac{\vec{u}\bullet\vec{v}}{\|\vec{u}\|\cdot \|\vec{v}\|} \\
= & \vec{u}\bullet\vec{v}\\
= & \cos(x)\cdot\cos(-y)+\sin(x)\cdot\sin(-y)\\
= & \cos(x)\cdot\cos(y)-\sin(x)\cdot\sin(y)\\
\end{align}
A: For variety, here is a different proof. Unfortunately it might be considered circular since it relies on differentiation of trig functions. But maybe you know their derivatives without using this identity, if say $\sin$ and $\cos$ have been defined by their Taylor series.
Apply $\frac{d^2}{dx^2}$ to each side (viewing $y$ as some constant), and you see that each side a solution to  $\frac{d^2}{dx^2}f(x)=-f(x)$
Both sides are in agreement at $x=-y$. Also the first derivatives of each side are in agreement at $x=-y$. Therefore they are the same expression.
A: $$\begin{array} {rcl}
\cos(x + y) + i \sin(x + y)& = & e^{i(x + y)} \\
&=& e^{ix}e^{iy} \\
&=& (\cos(x) + i\sin(x))(\cos(y) + i\sin(y)) \\
&=& (\cos(x)\cos(y) - \sin(x)\sin(y)) + i(\sin(x)\cos(y) + \sin(y)\cos(x)) \\
\end{array}$$
Equating real and imaginary parts you get
$$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$
$$\sin(x + y) = \sin(x)\cos(y) + \sin(y)\cos(x)$$
A: 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.
