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:

crapderarcsine

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

13 thoughts on “Derivatives of Trigonometric Functions”

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.”

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wrose31 on April 3, 2017 at 7:21 pm said:

I did. Oops. And I kept repeating the error throughout. Fixed now. Thanks.

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10 out of 10 for this, or maybe 11 out of 10.

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Agreed. This is also my perspective in my book:
http://intellectualmathematics.com/calculus/
and my associated rants against the standard approach, such as:
http://intellectualmathematics.com/blog/down-with-the-definition-of-the-derivative/

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Mr. Chase on April 4, 2017 at 10:04 am said:

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.

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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: https://youtu.be/S0_qX4VJhMQ

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John R Chase on August 3, 2017 at 5:23 pm said:

Somehow I missed your reply, Matthew. It’s actually Will Rose, my colleague, who made this guest post. So give him all the credit! But I agree that this is beautiful stuff!

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I think one reason the geometric proofs are avoided is that they’re circular (pun not intended). They depend on the fact that the difference between a chord and the arc subtended by an angle goes to zero as the angle gets small. But that’s the same as saying

$\lim_{\theta \to 0} 2\sin(\frac{\theta}{2}) - \theta = 0$

Which in turn is pretty close to $\lim_{x \to 0} \frac{\sin x}{x}=1$ .

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wrose31 on August 3, 2017 at 12:55 pm said:

Yeah, Simplico, but there’s geometry in all approaches. There has to be, because sine is defined geometrically.

I do think it’s important to learn $\lim_{x \to 0} \frac{\sin x}{x}=1$ as its own theorem. It says something deep about the sine function and codifies what students already know by looking at the sine function near the origin — the slope of the tangent line @ $x=0$ is 1. I used this fact over and over again in my geometric proofs above.

To establish that $\lim_{x \to 0} \frac{\sin x}{x}=1$ , you can just draw a picture of the sine of a very small angle, which will perhaps satisfy 95% of people and bother 5% of people. For those that are bothered, there is the more rigorous proof via the squeeze theorem, but as I mentioned above, that also depends on some geometry, arguably some lower-level and therefore more obvious geometry, but it’s still somewhat involved. Perhaps now, 99% of students will be convinced.

So if you insist on proving $\lim_{x \to 0} \frac{\sin x}{x}=1$ “rigorously” via the squeeze theorem, you’re targeting only those students who find the naive geometric proof unconvincing and the complicated geometric proof convincing.

What I think is unforgivable though, is the abandonment of geometric reasoning in the actual proof of the derivative of sine. As you can see from the above diagram, the proof basically writes itself once you set up the picture. For this one, at least, no knowledge of $\lim_{x \to 0} \frac{\sin x}{x}=1$ is even required. You just use the definition of sine and the definition of the derivative and get to work.

By contrast, the algebraic proof of the derivative of sine is boring, tedious, challenging, and unilluminating. And the proof depends crucially on $\lim_{x \to 0} \frac{\sin x}{x}=1$ , which was proven geometrically, so you haven’t really gotten rid of geometry, you’ve just locked it away in the basement.

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- Simplicio on August 3, 2017 at 5:09 pm said:
  
  Thanks for the reply (and for fixing the mess I made of my latex markup).
  
  I should’ve said in my last comment I use basically your first diagram when teaching Calc I students, so I agree with you that it’s a good way to convince students. The fact that an arc approaches the length of its chord as the sub-tending angle gets small seems pretty intuitive for most people. But the more engaged students usually than ask how we know it’s true, which then requires something equivalent to $lim_{\theta \to 0} \frac{sin(\theta)}{\theta} = 1$ .
  
  It occurs to me a pretty similar proof is to draw a vector tangent to circumference at $\theta = 0$ with magnitude $s$ and call it the derivative of sine at 0. Than it’s easy to see that this magnitude must be the same all around the circle, with just the direction changing. And so the derivative must go as $s*cos(\theta)$ . And it’s easy to convince yourself that s depends on your units for the angle, and so you can find units for which the derivative of sin(x) = cos(x). Though it remains to be shown that the angular units so desired are radians.

In fact, the derivative of the sine reduces the problem to the definition of the radian. Why the radian is defined this way, as arc divided by radius? Because, when you measure the angles in radians you obtain that sin(x)/x -> 1 and then d(sin(x))/dx = cos(x). That’s the real reason. Any other unit (degrees, for instance) produces a constant in front of the cosine (something that the students tend to forget or simply ignore that is a requisite for the angle to be expressed in radians if they want the numerical values of the derivatives to be correct in their numerical applications).

It’s the same as with number e. What is special about it? Why do we use it for powers and not the simpler 2^x or 10^x, or as basis for the logarithms? Because is the exponential whose derivative is the same function.

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Mr. Chase on September 3, 2017 at 3:37 pm said:

Great thoughts, here, Antonio. I agree that these proofs are wrapped up very closely with the definition of radian. The “traditional” proofs from the book are, too, but that gets obscured. The fact that these geometric proofs get us to the definition so quickly is certainly a huge selling point for this approach.

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Can we proof derivative of logarithm functions geometrically?????

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Derivatives of Trigonometric Functions

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