I'm still struggling to understand why the derivative of sine only works for radians. I had always thought that radians and degrees were both arbitrary units of measurement, and just now I'm discovering that I've been wrong all along! I'm guessing that when you differentiate sine, the step that only works for radians is when you replace $\sin(dx)$ with just $dx$, because as $dx$ approaches $0$ then $\sin(dx)$ equals $dx$ because $\sin(\theta)$ equals $\theta$. But isn't the same true for degrees? As $dx$ approaches $\theta$ degrees then $\sin(dx \,\text{degrees})$ still approaches $0$. But I've come to the understanding that $\sin(dx \,\text{degrees})$ approaches $0$ almost $60$ times slower, so if $\sin(dx \,\text{radians})$ can be replaced with $dx$ then $\sin(dx \,\text{degrees})$ would have to be replaced with $(\pi/180)$ times $dx$ degrees.
But the question remains of why it works perfectly for radians. How do we know that we can replace $\sin(dx)$ with just $dx$ without any kind of conversion applied like we need for degrees? It's not good enough to just say that we can see that $\sin(dx)$ approaches $dx$ as $dx$ gets very small. Mathematically we can see that $\sin(.00001)$ is pretty darn close to $0.00001$ when we're using radians. But let's say we had a unit of measurement "sixths" where there are $6$ of them in a full circle, pretty close to radians. It would also look like $\sin(dx \,\text{sixths})$ approaches $dx$ when it gets very small, but we know we'd have to replace $\sin(dx \,\text{sixths})$ with $(\pi/3) \,dx$ sixths when differentiating. So how do we know that radians work out so magically, and why do they?
I've read the answers to this question and followed the links, and no, they don't answer my question.
Best Answer
Radians, unlike degrees, are not arbitrary in an important sense.
The circumference of a unit circle is $2\pi$; an arc of the unit circle subtended by an angle of $\theta$ radians has arc length of $\theta$.
With these 'natural' units, the trigonometric functions behave in a certain way. Particularly important is $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \quad\quad - (*)$$
Now study the derivative of $\sin$ at $x = a$:
$$\lim_{x \to a} \frac{\sin x - \sin a}{x-a} = \lim_{x \to a}\left( \frac{\sin\left(\frac{x-a}{2}\right)}{(x-a)/2}\cdot \cos\left(\frac{x+a}{2}\right)\right)$$
This limit is equal to $$\cos a$$ precisely because of the limit $(*)$. And $(*)$ is quite different in degrees.