Here we’ll find the derivatives of trigonometric functions. The goal is to reinforce the idea of 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.
The Sine Function
Let’s find . The sine function is the height of a right triangle in the unit circle. We’ll draw it, and add a little change in . This induces a change in . The change in is called and the change in is called .
The interesting part is , 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 . Also remember that we’re imagining to get smaller and smaller, until the two radii in the picture are parallel. We get this:
The section of the circle is 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 .
Looking at the right triangle formed, we can use the definition of the cosine function to read off
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 .
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.
We want to know the coordinates of . That’s not too hard; is a quarter-circle rotation of . Draw in the components of , and rotate those components to get . 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 derivative of position is velocity, and so comparing components between the position and velocity vectors, we get
- Look back at the first derivation we gave that . Rework it to find derivatives of the other five trig functions. You might want to note that one way to interpret and is
- Look back at the argument about a dot moving around a circle. Consider a larger circle to find the derivative of with respect to . (Answer: )
- Suppose the dot moving around the edge of the circle is going three times as fast. What does this mean for the derivative of and with respect to ? Remember that the velocity must still be perpendicular to the position, but not necessarily unit length and more. (Answer: the derivative of with respect to is .
- Suppose the dot is moving at a variable speed , so that it keeps getting faster. Then the y-coordinate of the position is . Again, the velocity is perpendicular to position, but its length is changing. What is the derivative of with respect to ? (Answer: )