Derivative of the sine function when the argument is measured in degrees

Question

I'm trying to show that the derivative of $\sin\theta$ is equal to $\pi/180 \cos\theta$ if $\theta$ is measured in degrees. The main idea is that we need to convert $\theta$ to radians to be able to apply the identity $d/dx \sin x = \cos x $. So we need to express $ \sin \theta$ as $$ \sin_{deg} \theta = \sin(\pi \theta /180), $$ where $\sin_{deg}$ is the $\sin$ function that takes degrees as input. Then applying the chain rule yields $$ d/d\theta [ \sin(\pi\theta/180)] = \cos(\pi \theta/180) \pi/180 = \frac{\pi}{180}\cos_{deg}\theta. $$ Is this derivation formally correct?

Answer 1

This annoyed me when I revisited this material years later and had to teach this material to someone else, so I'm posting an answer here. (The other answer is perfectly fine, but I wanted to give a more complete exposition.)

Now here's the thing: you're told to find the derivative of $\sin(\theta)$ when $\theta$ is in degrees. At a first glance, this seems simple: it should just be $\cos(\theta)$. However, this answer is wrong, because you found that $\sin(\theta)$ has derivative $\cos(\theta)$ under the assumption that $\theta$ is measured in radians, and not in degrees.

Here's how you should approach the problem.

Notice that $\sin(\theta)$, when $\theta$ is in degrees or when $\theta$ is in radians, gives two different values. So, in fact, for this problem, writing $\sin(\theta)$ is itself ambiguous, because it isn't clear if $\theta$ is in degrees or radians. (However, for the rest of your studies, you'll likely assume the radian form.)

For the moment, suppose that $\sin_d(\theta)$ denotes $\sin(\theta)$ when $\theta$ is measured in degrees, and $\sin_r(\theta)$ denotes $\sin(\theta)$ when $\theta$ is measured in radians. We apply similar meaning for $\cos_d$ and $\cos_r$.

The motivation for this seems confusing at a first glance: why do we need to do this? It's because when we're calculating $\sin(\theta)$ depending on whether or not $\theta$ is in degrees or radians, we are fundamentally working with two different functions. Assuming we only care about the standard unit circle, when $\theta$ is in degrees, the domain is $[0, 360)$. When $\theta$ is in radians, the domain is $[0, 2\pi)$. Hence why we use $\sin_d$ and $\sin_r$ for these two different cases.

Now, back to the problem: what you are asked to find is $$\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta)\text{.}$$ You know for a fact that $$\dfrac{\text{d}}{\text{d}\theta}\sin_r(\theta) = \cos_r(\theta)\text{.}$$ So now, the problem boils down to this: how can we write $\sin_d$ in terms of $\sin_r$ so that we can apply the chain rule to find $\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta)$?

Recall that $\pi$ radians is $180$ degrees. So, if we have an angle which is $\theta$ degrees, it follows that the equivalent radian angle is $\dfrac{\theta}{180}\pi$. It follows that $$\sin_d(\theta)=\sin_r\left(\dfrac{\theta}{180}\pi \right)\text{.}$$

Hence,

$$\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta) = \dfrac{\text{d}}{\text{d}\theta}\sin_r\left(\dfrac{\theta}{180}\pi \right) = \dfrac{\pi}{180}\cos_r\left(\dfrac{\pi}{180}\theta \right)$$ from an application of the chain rule.

Lastly, note that the angle $\dfrac{\pi}{180}\theta$ is in radians. The equivalent degree measure would be $$\dfrac{\pi}{180}\theta \cdot \dfrac{180}{\pi} = \theta$$ hence, $$\dfrac{\pi}{180}\cos_r\left(\dfrac{\pi}{180}\theta \right) = \dfrac{\pi}{180}\cos_d(\theta)$$ and then we obtain $$\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta) = \dfrac{\pi}{180}\cos_d(\theta)$$ as desired.

Remark: This is an exercise in Stewart's text. I wouldn't expect a typical Calculus I student to be able to do this exercise, given that very little time (in my experience) is spent on functions and given the strange notation (this would look strange to a Calc. I student - remember, most of these people haven't even read proofs) involved. As you might guess, although this problem is interesting, given the way this exercise is usually written and implemented in calculus courses, I'm not a fan. More time should be dedicated to thinking about functions than usually done for a Calc. I course should this problem be assigned as an exercise.

Answer 2

Yes, it is correct, but keep in mind that what you are calculating is:

$$f(x)=\sin(πx/180)$$

$$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\frac{\pi}{180}\cos(\pi x/180),$$ where $x$ is expressed in degrees.

Answer 3

If a chain rule type reasoning is not satisfying, the intuition as to WHY you need radians to get such a derivative ($\frac{d}{dx}\sin(x)=\cos(x)$ when in radians) is because of how "nicely" a radian is defined. Namely 1 rad = the angle (where the vertex is the center of the circle) needed for the arclength (which is swept out by that angle along the circle's perimeter) to have length equal to the radius of the circle.

This definition is also why there are $2\pi$ radians "in a circle" (since $C=2\pi r$). It is a way to go from an angle measurement to a distance measurement. This is why it is used as opposed to degrees (a relatively arbitrary number from history that just has lots of nice factors). Going from angle measurement to a distance measurement is more obviously needed when you consider why we use a unit circle for computing values and identities. Such is often the case too for the geometric proof for the derivative of sine, because then the hypotenuse has length 1 and so the y-coordinate of a point on the circle at angle $\theta$ is simply $\frac{\sin(\theta)}{1}$ because $\sin(\theta)=\frac{\text{opp.}}{\text{hyp.}}=\frac{\text{y-value}}{\text{radius}}$ instead of $y=r\sin\theta$... however keeping the r variable is fine as well, since these factors will end up canceling. This factor-cancelling however is not the case for the angle when $\theta$ is in degrees:

It's not hard to conclude that $\frac{360 \text{ degrees per circle}}{2\pi\text{ radians per circle}}\cdot\frac{1 \text{ radian}}{1\text{ radius swept out}}=\frac{180}{\pi}$ degrees are needed to have one radius length swept out along the circumference. So when you draw the picture to derive the derivative of $\sin(\theta)$, your $\Delta(\theta)$ is only an angle measure, not a length represented in the diagram. The arclength change is instead $\frac{180}{\pi}r\Delta(\theta)$, so that is where the factor comes from when you finish the proof (since $\frac{r\cdot d\sin\theta}{\frac{180}{\pi}r\cdot d\theta}=\lim\frac{r\Delta\sin\theta}{\frac{180}{\pi}r\cdot\Delta\theta}$ and then you get your factor of $\frac{\pi}{180}$ surviving in your formula for $\frac{d\sin\theta}{d\theta}$ but the $r$ factors go away).