2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

Then

$$\sin_d(x) = \sin\left(\dfrac{\pi}{180} x\right)$$

and so, taking derivatives, we have to use the chain rule, and

$$ (\sin_d(x))' = \cos\left(\dfrac{\pi}{180}x\right)\cdot \left(\dfrac{\pi}{180}x\right)' = \dfrac{\pi}{180}\cos\left(\dfrac{\pi}{180}x\right) = \dfrac{\pi}{180}\cos_d(x) $$

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.