Why is the derivative of sine the cosine in radians but not in degrees?

Solution 1:

Here's a graph of $y = \sin x$.

The red line is $\sin x$ if $x$ is interpreted as an angle in degrees, and the blue line is $\sin x$ if $x$ is interpreted as an angle in radians.  See the difference?

Here's a zoomed-in version

What is $\sin(1.57\text{ rad}),$ and what is $\sin(1.57˚)?$

Intuitively, you can think of $\sin(x˚)$ as a "stretched out" version of the original graph. As a result, the slope of the tangent flattens, changing the derivative.

Solution 2:

I'm gonna assume that both system of measure exist, and that we all know that $$ \frac{\text{measure in degree}}{360}=\frac{\text{measure in rad}}{2\pi}. $$ I will also take for granted the classic derivative formulas for $\sin(x),\cos(x)\ldots$

It is often useful to think in terms of a new function to see this. There is no particular reason as to why they are not the same except that because that is the way the maths work. Nothing better to understand why something isn't true than to show it is not.

Let's define the degree sine function and call it $S(x)$ and the degree cosine function and call it $C(X)$. These are the functions that take $x$ degrees and return the value of $\sin(x°),\cos(x°)$. $\,S(x)$ corresponds to the red line in Omnomnomnom's answer. Now let us compute the derivative of $S(x)$, based on what we know. $$ \begin{align} \frac{d}{dx}S(x)&=\lim_{h\to 0}\frac{S(x+h)-S(x)}{h}&\\ & = \lim_{h\to 0} \frac{\sin(\frac{2\pi}{360}(x+h))-\sin(\frac{2\pi}{360}x)}{h} &\text{from measure to measure relation}\\ &=\frac{\pi}{180}\cos{\frac{\pi}{180}}=\frac{\pi}{180}C(x).& \text{from classic derivative formula} \end{align} $$ In a similar way, we'd find $C^{'}(x)=-\frac{\pi}{180}S(x)$.

Some not directly related but similar reading about radians vs degrees can be found here and also here.