Derivatives of Trigonometric Functions

First, let’s present the standard approach. This is from the calculus textbook I teach out of.

der of sinx

This was, as far as I was concerned, the only possible proof. The pedagogical flexibility lay entirely in how to frame the question, how to get students to discover the fact on their own (via graphical techniques), and how to add extra meaning to the result.

The most important question, so I thought for years, was really how one introduces and understands the fact that \lim_{x \to 0} \frac{\sin x}{x}=1. Some textbooks introduce it more or less out of the blue as “an important limit to know” and prove it via the Squeeze Theorem. Others prefer to wait until halfway through the above proof, realizing only then that this limit is important and solving it with a purpose in mind. There is also a difference of opinion as to how much rigor is required to establish the key inequality, that \sin \theta < \theta < \tan \theta. My textbook uses an area argument, but others prove the inequality with a nested sequence of segment inequalities.

My personal preference is for students to encounter \lim_{x \to 0} \frac{\sin x}{x}=1 “naturally” by attempting to graph y=\frac{\sin x}{x} in precalculus, along with other interesting functions like y=x \cdot \sin x, y=x \cdot \cos x, y = x + \sin x, y = e^{-x} \cdot \sin x, and y = \sin(1/x). These are more or less exercises in recognizing the so-called “envelope” of the product or sum of a periodic function and another function and have various scientific applications. The very informal geometric argument for why \lim_{x \to 0} \frac{\sin x}{x}=1 that one encounters in precalc prepares one for the more formal proof in calculus via the Squeeze Theorem.

All of this hard work to prove that \lim_{x \to 0} \frac{\sin x}{x}=1 almost seems to make it the real theorem and leaves \frac{d}{dx} [\sin x] = \cos x as a corollary.

By contrast, consider this:

Proof of derivative of sinx

I’m tempted to make no further comment, since this beautiful and striking diagram so thoroughly and clearly explains why the derivative of sine is cosine. Tiny changes in the sine of an angle are proportional to the cosine of that angle since the red arc length above is effectively a tangent to the circle. I would go so far as to say that until you see a diagram like this, you don’t even really understand the theorem at all. Why don’t we teach the derivative of sine this way? Why is this figure not in all the textbooks? I think I know the answers to these questions. The answers involve a long story about the history of calculus, the banishment of infinitesimals during the quest for rigor, and the abandonment of geometry as a satisfactory basis for analysis. But these diagrams are just too beautiful to give up and it’s cruel of us to keep them hidden from our students.

Here’s another calculus proof:

Proof of derivative of arcsinx

Compare this to the standard treatment you find in textbooks:


Which one of these proofs excites you? Which one makes you really feel like you understand the theorem and why it’s true?

I have created an entire series and I post them here without further comment.

Proof of derivative of tanx

Proof of derivative of arctanx

Proof of derivative of secx

Proof of derivative of arcsecx



6 thoughts on “Derivatives of Trigonometric Functions

  1. Not a great error, but did you mean that the limit of sin(x)/x = 1 as x -> 0 in the paragraph starting with “The most important question.”

    • Viktor, your textbook is brilliant. I will be consuming it in more depth in the coming months and my students will surely benefit. Thank you for sharing so freely.

  2. Excellent post, John! Your figures are terrific! I’m definitely going to try to convey the geometric explanation for the derivatives of trig functions next time I teach Calculus I.

    Also, Grant Sanderson conveys the geometric explanation for the derivative of sin(x) in his new Essence of Calculus, Volume 3:

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