Why does the derivative of sine only work for radians?

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.

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.

It seems to me that the best answer thus far is Simon S's. Others have hinted at the important property:

$$ \lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 $$

Some have simply stated it's important with little reason as to why it's important (specifically in regards to your question about the derivative of $\sin(x)$ equaling the $\cos(x)$). Simon S's answer explained why that limit is important for the derivative. However, what I find lacking is why it is that the limit equals what it equals and what would it equal if we decided to use degrees instead of radians.

At this point, I want to acknowledge that my answer is essentially the same as Simon S's except that I am going to go into gruesome detail.

Before I go into this, there is absolutely nothing wrong with using degrees over radians. It will change what the definition of the derivative of the trigonometric functions are, but it won't change any of our math--it just introduces a tedious factor we always have to carry around.

I am going to use this geometric proof as a way to make sense of the limit above:

There is only one part of the proof that will change if we decide to use degrees as opposed to radians and that is when we find the area of the sector subtended by $\theta$. When we use radians we get: $A_{AB} = \pi 1^2 * \frac{\theta}{2\pi} = \frac{\theta}{2}$--just as they found in the given proof. If however, we use degrees then we will get: $A_{AB} = \theta * \frac{\pi}{360}$. Now this changes their initial inequality which the rest of the proof relies on:

$$ \frac{1}{2}\sin(\theta) \leq \frac{\pi \theta}{360} \leq \frac{1}{2}\tan(\theta) $$

(the others don't change because the sine and tangents equal the same thing regardless of whether or not we use radians or degrees--with a proper, trigonometric definition of each, of course).

We still proceed in the same way (I'm going to be less formal and not worry about the absolute values--although we should technically). We divide everything by $\sin(\theta)$ which since I'm only worrying about the first quadrant won't change the directions of the inequalities:

$$ \frac{1}{2} \leq \frac{\pi \theta}{360 \sin(\theta)} \leq \frac{1}{2\cos(\theta)}\\ \frac{360}{2\pi} \leq \frac{\theta}{\sin(\theta)} \leq \frac{360}{2\pi \cos(\theta)} \\ \frac{\pi}{180} \geq \frac{\sin(\theta)}{\theta} \geq \frac{\pi}{180}\frac{1}{\cos(\theta)} $$

When we plug in $\theta = 0$ (whether radians or degrees) we get $\cos(0) = 1$ and thus we use the squeeze theorem to show that:

$$ \frac{\pi}{180} \leq \lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} \leq \frac{\pi}{180} $$

Therefore, if we use degrees, then:

$$ \lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = \frac{\pi}{180} $$

Going back to Simon S's answer, this gives, as the definition of the derivative for $\sin(x)$:

$$ \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin(x)}{h}\\ \lim_{h \rightarrow 0} \frac{\sin(x)\cos(h) + \sin(h)\cos(x) - \sin(x)}{h} \\ \lim_{h \rightarrow 0} \frac{\sin(h)\cos(x) + \sin(x)(\cos(h) - 1)}{h} $$

This may be a little sloppy, but when $h = 0$ $\cos(h) - 1 = 1 - 1 = 0$, so we can drop the second part and are left with:

*Actually this is extremely sloppy, at this point I would refer back to Simon S's Answer

$$ \lim_{h \rightarrow 0} \frac{\sin(h)}{h}\cos(x) = \cos(x)\lim_{h \rightarrow 0} \frac{\sin(h)}{h} $$

Using our above result we find the following:

$$ \frac{d}{dx}\sin(x) = \frac{\pi}{180}\cos(x) $$

This is what the derivative of $\sin(x)$ is when we use degrees! And yes, this will work fine in a Taylor series where we plug in degrees for the polynomial as opposed to radians (although the Taylor series will look different!).

And hopefully you already realize that this is what the derivative of $\sin(x)$ is when we use degrees, because if we accept that we must use radians, then we must convert our degrees to radians:

$$ \sin(x^\circ) = \sin\left(\frac{\pi}{180}x\right) $$

Now using the chain rule we get:

$$ \frac{d}{dx}\sin(x^\circ) = \frac{\pi}{180}\cos(x^\circ) $$

So the question isn't really why does it only work with radians--it works just as well with degrees except that we get a different definition of the derivative. The reason we prefer radians to degrees is that radians doesn't require this extra factor of $\frac{\pi}{180}$ every single time we differentiate a trigonometric function.