# Proving the Derivative of cosine and sine functions

In the proof of the derivatives of cosine and sine functions, we used the facts that:

$$\lim\limits_{\Delta x \to 0} \frac{\cos \Delta x - 1}{\Delta x} = 0$$

and

$$\lim\limits_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1.$$

I saw the proof of these two facts but it's said that $x$ here must be in radians, so why it must be measured in radians?

Suppose that $\sin$ is the sine of an angle given in radians, and $\sin_d$ is the same thing, only the input is in degrees. Define $\cos_d$ the same way.
$$\sin_d(x) = \sin\left(\dfrac{\pi}{180} x\right)$$
$$(\sin_d(x))' = \cos\left(\dfrac{\pi}{180}x\right)\cdot \left(\dfrac{\pi}{180}x\right)' = \dfrac{\pi}{180}\cos\left(\dfrac{\pi}{180}x\right) = \dfrac{\pi}{180}\cos_d(x)$$
Radians and degrees are only different by the constant factor $$\frac{180^\circ}{\pi}.$$ This means the proof works as good as in radians, but it is necessary to be consistent!