Inverse Functions: Arcsine (Part 10)

Let’s now switch from functions of the form $y= x^{m/n}$ to the trigonometric functions. We begin with $y = \sin x$ . We will define the function $y = \sin^{-1} x$ using much of the same reasoning that defined $\sqrt{x}$ .

We begin by looking at the graph of $y = \sin x$ .

Of course, we can’t find an inverse for this function as stated. Colloquially, the graph of $f$ fails the horizontal line test. More precisely, there exist two numbers $x_1$ and $x_2$ so that $x_1 \ne x_2$ but $f(x_1) = f(x_2)$ . Indeed, there are infinitely many such pairs.

So how will we find the inverse of $f$ ? Well, we can’t. But we can do something almost as good: we can define a new function $g$ that’s going look an awful lot like $f$ . As before, we will restrict the domain of this new function $g$ so that $g$ satisfies the horizontal line test.

When we did this for the square root function, there were two natural choices for the restricted domain: either $[0,\infty)$ or $(-\infty,0]$ . However, for the sine function, there are plenty of good options from which to choose. Indeed, here are four legitimate options just using the two periods of the sine function shown above. The fourth option is unorthodox, but it nevertheless satisfies the horizontal line test (as long as we’re careful with $\pm 2\pi$ .

So which of these options should we choose? Historically, mathematicians have settled for the interval $[-\pi/2, \pi/2]$ . Why did they settle on this interval? That I can answer with one word: tradition.

So we use the following bijective function (or, using the language that I used when I was a student, a one-to-one and onto function):

$g: [-\pi/2,\pi/2] \to [0,1]$ defined by $g(x) = \sin x$ .

Since this is a bijection, it has an inverse function

$g^{-1}: [0,1] \to [-\pi/2,\pi/2]$

This inverse function is usually denoted as $y = \sin^{-1} x$ or $y = \arcsin x$ . The graph of $y = \sin^{-1} x$ can be produced by reflecting through the line $y=x$ , producing the purple graph below.

One other important note: since $g$ and $g^{-1}$ are inverse functions, it’s always true that $g(g^{-1}(x)) = x$ and $g^{-1}(g(x)) = x$ . However, $g^{-1}$ and $f$ are not inverse functions, and so it’s possible for $g^{-1}(f(x))$ to be something other than $x$ . This parallels the observation that, say, $\sqrt{(-3)^2}$ is not equal to $3$ . We’ll discuss this further in future posts.