Q:

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. 

A:

Consider the image from this "proof without words",

Asymptotically, the angle between the black radius and the red vertical line is complementary to both angles marked as $\theta$. Thus, asymptotically, those angles are equal, and the two red triangles are similar. Therefore, by similar triangles,
$$
\frac{\mathrm{d}\sin(\theta)}{\mathrm{d}\theta}=\frac{\cos(\theta)}{1}
$$
To get the derivative of $\cos(\theta)$, recall that $\cos(\theta)=\sin\left(\frac\pi2-\theta\right)$ and $\sin(\theta)=\cos\left(\frac\pi2-\theta\right)$. Then the Chain Rule says
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}\theta}\cos(\theta)
&=\frac{\mathrm{d}}{\mathrm{d}\theta}\sin\left(\frac\pi2-\theta\right)\\
&=-\cos\left(\frac\pi2-\theta\right)\\[3pt]
&=-\sin(\theta)
\end{align}
$$

A:

This is a copy of my answer to How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?:

From the above picture, $\arcsin y$ is twice the area of the orange bit. The area of the red bit is ${1 \over 2}y\sqrt{1-y^2}$. The area of the red bit plus the orange bit is $\int_{0}^y \sqrt{1-Y^2} dY$. So $$\arcsin y = 2\int_{0}^y \sqrt{1-Y^2} dY - y\sqrt{1-y^2}.$$ Differentiating with respect to $y$ gives $\frac{d\arcsin y}{dy} = \frac{1}{\sqrt{1-y^2}}$. Using the theorem for the derivative of inverse functions yields $\sin' \theta = \sqrt{1 - \sin^2 \theta} = \cos \theta$.
A similar thing can be done with the arc length definition of $\arcsin$.

A:

This answer uses the following definitions:

$$e^{ix} = \cos(x) + i \sin(x)$$
  $$\sin(-x) = -\sin(x)$$
  $$\cos(-x) = \cos(x)$$

Rearranging gives the complex exponential definition of cosine:
$$ \cos(x) = (e^{ix} + e^{-ix})/2 $$
$$ \Rightarrow \cos'(x) = i(e^{ix} -e^{-ix})/2 $$
The complex exponential definition of sine:
$$ \sin(x) = (e^{ix} -e^{-ix})/2/i $$
Multiply with the complex conjugate
$$ \Rightarrow \sin(x) = (-ie^{ix} + ie^{-ix})/2 $$
It is now trivial to see that $-\sin(x) = \cos'(x)$.
Similar is the derivative of $\sin(x)$.

