Different ways finding the derivative of $\sin$ and $\cos$. I am looking for different ways of differentiating $\sin$ and $\cos$, especially when using the geometric definition, but ways that use other defintions are also welcome. 
Please include the definition you are using. 
I think there are several ways to do this I didn't find, that make clever use of trigonometric identities or derivative facts. I am looking for these for several reasons. Firstly, it is just interesting to see. Secondly, there may be interesting techniques that can also be applied in a clever way to other derivative problems. It is also interesting to see how proofs can come form completely different fields of mathematics or even from physics.
I have included several solutions in the answers. 

Instead of using usual definitions, I tried to start from the trigonometric identities, and 'redefine' them as the special functions which satisfy those identities.

Claim.
Suppose that there exist differentiable functions $f, g:\mathbb{R}\to\mathbb{R}$ such that


*

*$\forall x,y\in \mathbb{R}; \; f(x)g(y)+g(x)f(y)=f(x+y)$

*$\forall x,y\in \mathbb{R}; \; g(x)g(y)-f(x)f(y)=g(x+y)$

*$f(0)=0$, $g(0)=1$, $f'(0)=1$, $g'(0)=0$


Then
$$\forall x\in \mathbb{R}; \; f'(x)=g(x),\; g'(x)=-f(x)$$
Proof) Let $w(x)=g(x)+if(x)$ ($w:\mathbb{R}\to\mathbb{C}$). Then
\begin{aligned}
w(x+y)&=(g(x)g(y)-f(x)f(y))+i(f(x)g(y)+f(y)g(x))\\
&=(g(x)+if(x))(g(y)+if(y))\\
&=w(x)w(y)
\end{aligned}
Since $f$ and $g$ are differentiable, $w$ is also differentiable. So
\begin{aligned}
w'(x)&=\lim_{h\to 0}\frac{w(x+h)-w(x)}{h}\\
&=\lim_{h\to 0}\frac{w(x)w(h)-w(x)}{h}=w(x)\lim_{h\to 0}\frac{w(h)-1}{h}\\
\end{aligned}
$w(0)=g(0)+if(0)=1$, so
$$w'(x)=w(x)\lim_{h\to 0}\frac{w(h)-w(0)}{h-0}=w(x)w'(0)=iw(x)$$
This gives
$$g'(x)+if'(x)=w'(x)=iw(x)=-f(x)+ig(x)$$
$\therefore$ $f'=g$, $g'=-f$. $\quad\square$
