Derivative of the $\sin(x)$ when $x$ is measured in degrees

So a classic thing to derive in calculus textbooks is something like a statement as follows

Is $\frac{d}{dx}\sin(u)$ the same as the derivative of $\frac{d}{dx}\sin(x)$ where $u$ is an angle measured in degrees and $x$ is measured in radians? and of course the answer is no because of the chain rule.

Except usually this is ambiguously worded as "Is the derivative of $\sin(u)$, where $u$ is measured in degrees, equal to the derivative of $\sin(x)$ where $x$ is the same angle but measured in in radians?"

Then the texts go on to say something like "No and this why we don't work in degrees and instead chose to work in radians, to avoid all the messy constants that come out of taking derivatives." Am I crazy by thinking this is an odd thing to say that will end up confusing students. If your independent variable was an angle measured in degrees, you are probably more interested in it's derivative with respect to degrees not radians, which would infact be equal at the corresponding degrees and radians of an angle. Is my understanding wrong here. Is what the books say fine? I think at minimum they should at least be clear that we are taking the derivative with respect to radians, no?

Note this is not a duplicate of

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

Even though it is highly related.

• DO not use geometry to define sine :) Oct 14, 2013 at 22:44
• @HagenvonEitzen, why not? What definition do you prefer? Oct 14, 2013 at 22:59
• user1544793, why are you teaching degrees? Oct 14, 2013 at 23:32
• I'm not teaching degrees, this is a standard result derived in most elementary calculus texts. I encourage my students to read the book. Oct 14, 2013 at 23:33
• If the book words the problem incorrectly, you can reword it for them and discourage them from reading that bit... Oct 14, 2013 at 23:58

Whether an angle is measured in degrees or radians is most definitely worth bringing up in a class/lecture so the students can get a solid understanding of derivatives.

UNITS DO MATTER!!

Of course they do. Saying units do not matter is like claiming that $50$ miles per hour is equivalent to 50 feet per hour or 50 miles per second.

For x measured in radians, $d/dx(\sin x) = \cos x$

For x measured in degrees, $d/dx(\sin x) = (\pi/180) \cos x$

A derivative is the slope of a line, the change in the vertical units per change in the horizontal units. When you go from radians to degrees, the change in vertical units remains constant while the change in horizontal units increases tremendously (which, in turn, has the derivative become a lot smaller).

A good way to see this is simply to graph $\sin x$ in degrees. If the derivatives are all the same as they are when measured in radians you should find a tangent line with slope $1$ somewhere (as is true for $\sin x$ at $x=0$ radians). You will not find any tangents with slopes nearly this large when your angles are measured in degrees.

What you guys are claiming is "obvious" is patently false.

The essential problem is writing "$\sin u$, where $u$ is measured in degrees". This doesn't really mean anything. If you want degrees, you need to write them explicitly: $\sin u^\circ$. If you want the sine of an angle, you need to write that: $\sin\angle ABC$ or similar. At some point in geometry or trigonometry class, someone should be teaching that $\vphantom t^\circ=\frac{2\pi}{360}$.

• Very true, and I think this is where the confusion lies with most students. What they mean by $\sin(u)$ is $\sin(u)$ where $u=(\pi/180)x$ where $x$ corresponds to our usual definition of the domain of the sine function. You can still take the derivative with respect to $u$ though and I think its necessary to point out explicitly that this is not what they are doing. The question came up because one student thought this was the same as plugging sin(4) in radian mode into their calculator and plugging sin(4) in degree mode into their calculator and noting they were different. Oct 14, 2013 at 23:49
• although hopefully they teach them degree$=2\pi/360$ ;) Oct 14, 2013 at 23:50
• @user1544793, yes, of course. I briefly forgot that detail of ancient Babylonian geometry. Oct 14, 2013 at 23:52
• @user1544793, what do you mean by that "What they mean" sentence? If any student writes $\sin 30$ to mean $\sin 30^\circ$, they should be dinged a point. It's best to catch bad habits early! Also, students should be strongly discouraged from using their calculators in "degree mode", and if calculators are used for the class they should be shown how to put them in radian mode. Oct 14, 2013 at 23:55
• Oh no thats not what I was saying. We aren't teaching degrees, or using calculators, at all. The student was just making an analogy to the calculator, as to show that he thought the problem was not very insightful. But I pointed out that the analogy didn't quite make sense, although his general conclusion about the problem not being very insightful is probably a reasonable opinion. Oct 15, 2013 at 0:00

Yes, you're describing the same object $x$, but units matter. That is, the units you use to measure that object matter in describing what you mean. Like, 1 meter and 3.2808 feet are practically the same length, but are described through different units of measure.

And in Calculus, radians is a unit of measure that will give you the least amount of headaches (less tracking of constants, etc.) when calculating derivatives, integrals, etc. It's similar to why $\log_e=\ln$, in Calculus, is called the natural logarithm: $\frac{d}{dx}\ln x=\frac{1}{x}$, while $\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$.

In fact, radians (as a measure of angles) is arguably a better measurement to define and use than degrees. Given a circle of radius $r$, if you're interested in the length of the arc of a circle, arc given by an angle $\theta$:

In degrees: the length of the arc of a circle is $r\left(\frac{\theta\pi}{180}\right)$

In radians: the length of the arc of a circle is simply $r\theta$.

Fundamentally, calculating $\frac{d}{dx}\sin x$ boils down to solving the limit $\lim_{x\to 0}\frac{\sin x}{x}$. And what does one mean by $\frac{\sin x}{x}$?

Geometrically, and seeing $x$ in radians, one can say $\frac{\sin x}{x}$ is simply the ratio of the length of a side of a right-triangle ($\sin x$) and the length of an arc of a circle ($x$). Describing $\frac{\sin x}{x}$ this way, we get a ratio of lengths.