# The derivatives of trig functions

#### Prerequisites

- Trigonometry
- The small angle approximation
- Trig functions for any value of the argument
- What is a derivative, anyway?

To make the most productive use of trig functions in physics, we need to know what their derivatives look like. (You might have guessed this from trig functions for any value of the argument, since there we showed that we were going to want the angles in our trig functions to be a function of time, and in physics we are often differentiating with respect to time to get velocities, accelerations, etc.)

To see what the derivatives of sines and cosines are, look at the power series expansions that were generated in the reading, the small angle approximation. The results for sine and cosine are

$$\sin{\theta} = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - ... $$

$$\cos{\theta} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - ... $$

It's clear how to carry each series on as long as you want. The sine series has alternating odd powers, while the cosine series has alternating even powers.

Since we're doing things that look like a calculus class, let's change our variable to $x$ to make it look more familiar and easier to differentiate. We'll change $x$ back to $\theta$ when we're done.

(This is a "math $x$", not a "physics $x$". It has no dimensions. We'll suppress the idea we are learning that in physics the label we choose tells us about its dimensions. This $x$ is dimensionless, as in math class.)

$$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$

$$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... $$

Since we know that the derivative of $x^n$ with respect to $x$ is just $nx^{n-1}$, we can easily take the derivatives term by term to get

$$\frac{d}{dx}\sin{x} = 1 - 3\frac{x^2}{3!} + 5\frac{x^4}{5!} - ... = \cos(x)$$

$$\frac{d}{dx}\cos{x} = 0 - 2\frac{x}{2!} + 4\frac{x^3}{4!} - ... = -\sin(x) $$

Term by term this matches and from the general pattern you can see that this will always work. (We have used that n! = the product of the integers from 1 up to n, so n/n! = 1/(n-1)!).

This shows why the Taylor series expansion form is so useful! We can do a difficult derivative (that of sin and cos) by using an easy derivative (that of a power)!

Translating back to theta we get the results

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

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

If we take another derivative, we get an interesting result:

$$\frac{d^2}{d\theta^2}\sin{\theta} = -\sin(\theta)$$

$$\frac{d^2}{d\theta^2}\cos{\theta} = -\cos(\theta) $$

The second derivative of a either a sine or cosine is equal to itself with a minus sign. This is reminiscent of the exponential, $f(x) = e^x$, whose first derivative is equal to itself. These properties make sines and cosines (and exponentials) very common in physics applications.

Often, our physical variable will not be $\theta$, but time, with $θ = ω_0t$ where $ω_0$ is a constant with units of radians/sec. We'll then want to get the derivative with respect to time. For this we have to use the chain rule familiar from calculus. (This looks way more obvious when we use the physics way of thinking of the derivative as the ratio of small changes. The $d\theta$ terms then just cancel.)

$$\frac{d}{dt} = \frac{d\theta}{dt}\frac{d}{d\theta}$$

Since $θ = ω_0t$, it follows that $\frac{dθ}{dt} = ω_0$_{ } and we have the useful results

$$\frac{d}{dt}\sin{ω_0t} = \omega_0\cos(ω_0t)$$

$$\frac{d}{dt}\cos{ω_0t} = -\omega_0\sin(ω_0t)$$

(Check the units on these results!)

Joe Redish 3/21/18

Last Modified: May 27, 2019