Sine, cosine, and tangent are of course the workhorse functions of trigonometry. You learn ’em in high school, and if you go on in math and science you never stop using them. Now on many occasions you might have the sine or cosine or tangent of some angle, and you want a way to invert those functions to recover the angle from those values. Let’s take a look at the inverse tangent function:
Now we’ll do one of those things were we skip any motivation and just do some stuff for reasons we’ll get to at the end. Let’s find the derivative of the arctangent function. To start, take the tangent of both sides of the above definition. The tangent undoes the arctangent, so we get:
Differentiate both sides. The derivative of tangent is secant squared, which we could prove if we felt like it, and since f is a function we have to use the chain rule. Doing thus, and recalling that the derivative of x is 1:
Solve for the derivative:
Which isn’t all that helpful except for the fact that there’s a basic trig identity which connects the secant and tangent functions: sec(x)^2 – tan(x)^2 = 1. Substituting:
Ah! But tangent(f) is x, so we can swap that in:
Ok, ok. What’s that good for? Well, if just write that down as our Sunday Function and graph it, we’ll have something to celebrate:
That is the graph of a function, 1/(1+x^2), which has nothing at all to do with circles or geometry in any obvious way. But we know it’s the derivative of the inverse tangent, and conversely the inverse tangent is the integral of that function, so we can easily plug in and see that the area under the curve is pi. Which is nice, because to day is Pi Day. Happy Pi Day!