# Geometric intuition for derivatives of basic trig functions

I was inspired by this question to try and come up with geometric proofs for the derivatives of basic trig functions--basically, those that have simple representations on the unit circle ($\sin, \cos, \tan, \sec, \csc, \cot$): I was initially a bit skeptical about how easy it might be, but then I found this very simple proof for $\sin$ and $\cos$; the basic insight can be seen in this picture from an alternative version of the proof I found later: Basically, we use the fact arc $PQ$ and segment $PQ$ are the same as $\Delta\theta\rightarrow 0$, and the former has measure $\Delta\theta$.

Nevertheless, I've had no luck so far getting a proof for $\sec$; I have a feeling the proofs of $sec$ and $tan$ are very closely related, as are the $\csc, \cot$ proofs.

Has anyone seen a proof for the four remaining basic functions anywhere? Perhaps I just haven't drawn the right picture yet.

Another possible avenue is this representation: • Related: This is my three-dimensional "trigonograph" of the derivatives of sine and cosine. As for the other functions ... I have no solution to offer here, but a thought: I've always found it interesting that, just as sine and cosine are derivatives of each other (sign changes notwithstanding), secant and tangent are as well ... except for an "extra" factor of secant: $$\frac{d}{d\theta}\color{red}{\sec\theta} = \color{blue}{\tan\theta}\cdot\sec\theta \qquad \frac{d}{d\theta}\color{blue}{\tan\theta} = \color{red}{\sec\theta}\cdot\sec\theta$$ – Blue May 26 '15 at 16:12
• @Blue the answers below give you the tie you've been looking for--basically, the extra $\sec\theta$ comes from the radius of the circle used in the proof; $\csc\theta$ and $\cot\theta$ show the same switch from the circle of radius $\csc\theta$. And the reason the pairings are like that can be tied back to the Pythagorean trig identities--$\sin^2\theta+\cos^2\theta=1$, $1+\tan^2\theta=\sec^2\theta$, and $1+\cot^2\theta=\csc^2\theta$. – MichaelChirico May 27 '15 at 17:25
• Indeed, I'm familiar with Jim's solution. (Your "summary graphic" is a nice touch.) My "thought" had been intended as a hint in that direction, since I didn't have a postable image readily available. In retrospect, the hinty-ness c/should've been clearer, say, by explicitly suggesting that you seek a "$d\sec$-$d\tan$-$\sec\theta d\theta$" right triangle (which arises pretty naturally from your second "avenue"). I think I got caught-up in colorizing my equations and lost focus. :) – Blue May 27 '15 at 19:30

## 2 Answers

Yes, there are nice geometric explanations of the derivative formulas for all six basic trig functions, which ought to be much more widely known. (I hesitate to use the word "proof" for an argument that uses infinitesimals.) They are all based on the following fact about isosceles triangles:

For an isosceles triangle with small vertex angle $d\theta$, the base length $ds$ satisfies $$ds \;\approx\; r\,d\theta$$ where $r$ is the length of the legs. As you already pointed out, there is a nice geometric explanation of the derivative formulas for $\sin \theta$ and $\cos \theta$ that uses this fact. The following picture shows a right triangle version of the explanation, as opposed to the unit circle version you gave above: The following picture shows a geometric explanation of the derivative formulas for $\sec \theta$ and $\tan \theta$, again using right triangles. Finally, the following picture shows a geometric explanation of the derivative formulas for $\csc \theta$ and $\cot \theta$. This is a direct corollary of @JimBelk's wonderful answer, but I thought it might be nice to have a concise summary graphic which gets all the derivatives at once, which it turns out is pretty simple to do from the second unit circle picture I posted with my question: The blue triangle shows the derivatives of $\sin\theta$ and $\cos\theta$, the yellow triangle shows the derivatives of $\sec\theta$ and $\tan\theta$, and the salmon triangle shows the derivatives of $\csc\theta$ and $\cot\theta$.

It should be straightforward to verify this using similar triangles and a limit argument to show the green segments are equal in length to the arcs of 3 circles of appropriate radii.