Using the Chain Rule to prove trig derivatives I'm having trouble with this problem, I'm not sure how to tackle it and I was wondering if somebody could set me on the right path. The problem is as follows:
Use the Chain Rule to show that if $\theta$ is measured in degrees, then
$$\frac{d}{d\theta}(\sin\theta^{\circ}) = \frac{\pi}{180^{\circ}}\cos\theta^{\circ}$$
Thanks!
 A: Hint: 
$$\frac{d}{d \theta}\sin(\theta^{\circ}) =  \frac{d}{d\theta} \sin\left(\frac{\pi}{180^{\circ}}  \cdot \theta^{\circ}\right).$$ 
Now, can you apply the chain rule?
A: Assuming that $\sin$ is a function that expects an angle measured in radians for an arguement, I'm going to further assume that the notation $\theta^\circ$ is just a shorthand for $\phi(\theta)$, where $\phi$ is the function that converts degrees to radians. I define it as follows.
$$\phi:R\to R |\phi(\theta) = \frac{\pi}{180}\theta$$
Now its just a simple matter of evaluating
$$\frac{d}{d\theta}\sin(\phi(\theta))=\frac{\sin\phi}{d\phi}\cdot\frac{d\phi(\theta)}{\theta}=\cos\phi\cdot\frac{\pi}{180}=\frac{\pi}{180}\cos\phi(\theta)=\frac{\pi}{180}\cos\theta^\circ$$
A: It is useful to use separate notations for the "classical" trigonometric functions, where the input is in degrees. So let $S(x)$ be the classical sine function, the one that for example gives $S(90)=1$. Let $C(x)$ be the classical cosine function. 
We use $\sin(u)$, $\cos(u)$ for the sine and cosine functions of calculus.
We want to prove that $S'(x)=\frac{\pi}{180}C(x)$.
Note that 
$$S(x)=\sin(\pi x/180),\quad\text{and}\quad C(x)=\cos(\pi x/180).$$
Using the Chain Rule, we find that the derivative of $S(x)$, that is, the derivative of $\sin(\pi x/180)$, is $\frac{\pi}{180}\cos(\pi x/180)$. 
This is easy, just let $u=\pi x/180$. 
But $\cos(\pi x/180)=C(x)$, and we are finished. 
