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. 

Start with
$$
        \sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\\
        \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)
$$
Differentiate with respect to $y$:
$$
         \sin'(x+y)=\sin(x)\cos'(y)+\cos(x)\sin'(y)\\
         \cos'(x+y)=\cos(x)\cos'(y)-\sin(x)\sin'(y)
$$
Let $y=0$ and note that $\cos'(0)=0$ because $\cos$ has a maximum at $0$:
$$
          \sin'(x)=\cos(x)\sin'(0)\\
          \cos'(x)=-\sin(x)\sin'(0)
$$
So everything comes down to determining the constant $C=\sin'(0)$:
$$
          \sin'(x)=C\cos(x),\;\;\; \cos'(x)=-C\sin(x).
$$
That makes sense because the original identities for $\sin$ and $\cos$ are not affected by scaling $x$ and $y$. The constant $C$ is determined by the system of units you use such as radians vs. degrees. $C=1$ for radians, which requires the standard geometric argument.
