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(θ) equals θ. But isn’t the same true for degrees? As dx approaches θ degrees then sin(dxdegrees) still approaches 0. But I’ve come to the understanding that sin(dxdegrees) approaches 0 almost 60 times slower, so if sin(dxradians) can be replaced with dx then sin(dxdegrees) would have to be replaced with (π/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(dxsixths) approaches dx when it gets very small, but we know we’d have to replace sin(dxsixths) with (π/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.

Source : Link , Question Author : Kyle Delaney , Answer Author : Simon S

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