Visualizing Elementary Calculus: Trigonometry

Here we’ll find the derivatives of trigonometric functions. The goal is to reinforce the idea of \textrm{d} as a thing that means “a little bit of” and grant some new insight into why these derivatives are what they are. The first argument is based on the preface of Tristan Needham’s Visual Complex Analysis. I haven’t read the bulk of it, but the preface is good.

This series
I – Introduction
II – Trigonometry

The Sine Function

Let’s find \textrm{d}(\sin\theta) / \textrm{d}\theta. The sine function is the height of a right triangle in the unit circle. We’ll draw it, and add a little change in \theta. This induces a change in \sin\theta. The change in \theta is called \textrm{d}\theta and the change in \sin\theta is called \textrm{d}(\sin\theta).

We show the sine of an angle as the dark blue line. The change in the sine when we change the angle slightly is the light blue line.

The interesting part is \textrm{d}\sin\theta, so we’ll zoom in there in the next picture. Before we do, remember that the arc length along a piece of the unit circle is equal to the angle it subtends. This will tell us the length of the little piece of the circumference near \textrm{d}\sin\theta. Also remember that we’re imagining \textrm{d}\theta to get smaller and smaller, until the two radii in the picture are parallel. We get this:

The interesting region is blown up to large size. The black line d(theta) is part of the edge of the circle. The angles marked are congruent to theta.

The section of the circle is \textrm{d}\theta long. It looks like a straight line because we are zoomed in close, like the horizon at the beach. You can use some geometry to show that the angles marked are congruent to \theta.

Looking at the right triangle formed, we can use the definition of the cosine function to read off

\frac{\textrm{d}(\sin\theta)}{\textrm{d}\theta} = \cos\theta

which is the derivative of the sine function.

Motion on the Unit Circle

Another way to view these derivatives is to imagine a point moving around the outside edge of the unit circle with speed one. Its location as a function of time is (\cos t, \sin t).

Its velocity is tangent to the circle and length one. Let’s draw the velocity vector right at the point, and then also translate it to the origin.

The position of the point is the red vector r. Its velocity is the green tangent v, which has also been copied to the origin.

We want to know the coordinates of \vec{v}. That’s not too hard; \vec{v} is a quarter-circle rotation of \vec{r}. Draw in the components of \vec{r}, and rotate those components to get \vec{v}. The x-component of the position becomes the y-component of the velocity, and the y-component of the position becomes minus one times the x-component of the velocity.

The components of the position get rotated a quarter turn to make the components of the velocity.

The derivative of position is velocity, and so comparing components between the position and velocity vectors, we get

\frac{\textrm{d}(\cos\theta)}{\textrm{d}\theta} = -\sin\theta

\frac{\textrm{d}(\sin\theta)}{\textrm{d}\theta} = \cos\theta


  • Look back at the first derivation we gave that \textrm{d}(\sin\theta)/\textrm{d}\theta = \cos\theta. Rework it to find derivatives of the other five trig functions. You might want to note that one way to interpret \tan\theta and \sec\theta is

The tangent and secant of an angle are side lengths of a right triangle with "adjacent" side length one.


  • Look back at the argument about a dot moving around a circle. Consider a larger circle to find the derivative of 5\sin\theta with respect to \theta. (Answer: 5\cos\theta)
  • Suppose the dot moving around the edge of the circle is going three times as fast. What does this mean for the derivative of \sin(3 t) and \cos(3 t) with respect to t? Remember that the velocity must still be perpendicular to the position, but not necessarily unit length and more. (Answer: the derivative of \sin(3 t) with respect to t is 3\cos(3 t).
  • Suppose the dot is moving at a variable speed v(t) = t, so that it keeps getting faster. Then the y-coordinate of the position is \sin(\frac{1}{2}t^2). Again, the velocity is perpendicular to position, but its length is changing. What is the derivative of \sin(\frac{1}{2}t^2) with respect to t? (Answer: t\cos(\frac{1}{2}t^2)

Tags: , , ,

7 Responses to “Visualizing Elementary Calculus: Trigonometry”

  1. Visualizing Elementary Calculus: Introduction « Arcsecond Says:

    […] « My Brown Big Spiders Visualizing Elementary Calculus: Trigonometry […]

  2. A Non-mathematician’s Non-apology « Arcsecond Says:

    […] Arcsecond « Visualizing Elementary Calculus: Trigonometry […]

  3. Second Linkfest Says:

    […] Eichenlaub: Visualizing Elementary Calculus: Introduction, Visualizing Elementary Calculus: Trigonometry, A Non-mathematician’s Non-apology, My Brown Big […]

  4. Leyan Says:

    What do you use to generate your figures and equations?

  5. Mark Eichenlaub Says:

    Hey Leyan,

    It’s Geogebra, a free app. It’s useful for simple animations, too. I think it strikes a good balance between usability and functionality.

  6. Mark Eichenlaub Says:

    Oh, and the equations are LaTeX, which is built in to blogs. I think you can see the TeX source if you mouse-over.

  7. Visualizing Elementary Calculus: Differentiation Rules 1 « Arcsecond Says:

    […] Series I – Introduction II – Trigonometry III – Differentiation […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: