I'm having trouble with this problem, I'm not sure how to tackle it and I was wondering if somebody could set me on the right path. The problem is as follows:

Use the Chain Rule to show that if $\theta$ is measured in degrees, then

$$\frac{d}{d\theta}(\sin\theta^{\circ}) = \frac{\pi}{180^{\circ}}\cos\theta^{\circ}$$


  • $\begingroup$ The most important thing you should learn from this is that this is why radian measure is used in calculus. It's the same as the reason why $e$ is the "natural" base for logarithmic and exponential functions. $\endgroup$ – Michael Hardy Jul 14 '14 at 20:02


$$\frac{d}{d \theta}\sin(\theta^{\circ}) = \frac{d}{d\theta} \sin\left(\frac{\pi}{180^{\circ}} \cdot \theta^{\circ}\right).$$

Now, can you apply the chain rule?

  • $\begingroup$ @Hakim: I think the degree symbol should be removed from the theta on the right, since that angle is in radians. You could add an r (as in ${\theta ^r}$) to emphasize the radians [I believe that is a standard]. $\endgroup$ – Rory Daulton Jul 14 '14 at 20:02
  • $\begingroup$ @Rory: in my opinion, correct notation should be $$\frac d{d\theta°}\sin°(\theta°)=\frac d{d\theta°}\sin(\frac\pi{180°}\cdot\theta°).$$ All $\theta$'s are in degrees, and so is the first $\sin$ function. The other one is the ordinary sine function, in radians, such that $\sin'\theta=\cos\theta$. Also possible: $$\frac d{d\theta}\sin°(\theta)=\frac d{d\theta}\sin(\frac\pi{180}\cdot\theta).$$ $\endgroup$ – Yves Daoust Jul 14 '14 at 21:32
  • $\begingroup$ @YvesDaoust Really? Why? I've never seen the $\circ$ for my derivatives when doing them in class/examples. $\endgroup$ – Rivasa Jul 14 '14 at 21:51
  • $\begingroup$ @Link. Neither have I. This is just a convenience to recall that $\theta$ is in degrees. But it is wrong to denote with the same $\sin$ two functions that take arguments in degree and in radians. $\endgroup$ – Yves Daoust Jul 14 '14 at 22:32
  • $\begingroup$ @YvesDaoust I see, that makes sense. $\endgroup$ – Rivasa Jul 14 '14 at 22:35

Assuming that $\sin$ is a function that expects an angle measured in radians for an arguement, I'm going to further assume that the notation $\theta^\circ$ is just a shorthand for $\phi(\theta)$, where $\phi$ is the function that converts degrees to radians. I define it as follows. $$\phi:R\to R |\phi(\theta) = \frac{\pi}{180}\theta$$

Now its just a simple matter of evaluating $$\frac{d}{d\theta}\sin(\phi(\theta))=\frac{\sin\phi}{d\phi}\cdot\frac{d\phi(\theta)}{\theta}=\cos\phi\cdot\frac{\pi}{180}=\frac{\pi}{180}\cos\phi(\theta)=\frac{\pi}{180}\cos\theta^\circ$$


It is useful to use separate notations for the "classical" trigonometric functions, where the input is in degrees. So let $S(x)$ be the classical sine function, the one that for example gives $S(90)=1$. Let $C(x)$ be the classical cosine function. We use $\sin(u)$, $\cos(u)$ for the sine and cosine functions of calculus.

We want to prove that $S'(x)=\frac{\pi}{180}C(x)$.

Note that $$S(x)=\sin(\pi x/180),\quad\text{and}\quad C(x)=\cos(\pi x/180).$$

Using the Chain Rule, we find that the derivative of $S(x)$, that is, the derivative of $\sin(\pi x/180)$, is $\frac{\pi}{180}\cos(\pi x/180)$. This is easy, just let $u=\pi x/180$.

But $\cos(\pi x/180)=C(x)$, and we are finished.

  • $\begingroup$ What about $\cos°(x)$ and $\sin°(x)$ ? $\endgroup$ – Yves Daoust Jul 14 '14 at 21:36
  • $\begingroup$ I thought about but rejected $\text{Sin}$. The notation you suggest is ingenious. I preferred a clean break, and chose to give the degree measure sine a name that would distinguish it from the mathematician's sine. Superscript emphasizes commonality, and here one wants to stress distinctness. $\endgroup$ – André Nicolas Jul 14 '14 at 21:50
  • $\begingroup$ $cosd$ and $sind$ borrowed from FORTRAN ? $\endgroup$ – Yves Daoust Jul 14 '14 at 22:34
  • $\begingroup$ Now this one takes me back! Except that the serious programming I did was in SOAP, Fortran was way too slow for library routines. $\endgroup$ – André Nicolas Jul 14 '14 at 22:43

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