Inverse Functions: Arcsecant (Part 26)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of $y = \sec x$ satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of $y = \sec^{-1} x$ uses the interval $[0,\pi]$ — or, more precisely, $[0,\pi/2) \cup (\pi/2, \pi]$ to avoid the vertical asymptote at $x = \pi/2$. This portion of the graph of $y = \sec x$ satisfies the horizontal line test and, conveniently, matches almost perfectly the domain of $y = \cos^{-1} x$. This is perhaps not surprising since, when both are defined, $\cos x$ and $\sec x$ are reciprocals. Since this range of $\sec^{-1} x$ matches that of $\cos^{-1} x$, we have the convenient identity $\sec^{-1} x = \cos^{-1} \left( \displaystyle \frac{1}{x} \right)$

To see why this works, let’s examine the right triangle below. Notice that $\cos \theta = \displaystyle \frac{x}{1} \qquad \Longrightarrow \qquad \theta = \cos^{-1} x$.

Also, $\sec\theta = \displaystyle \frac{1}{x} \qquad \Longrightarrow \qquad \theta = \cos^{-1} \left( \displaystyle \frac{1}{x} \right)$.

This argument provides the justification for $0 < \theta < \pi/2$ — that is, for $x > 1$ — but it still works for $x = 1$ and $x \le -1$.

So this seems like the most natural definition in the world for $\sec^{-1} x$. Unfortunately, there are consequences for this choice in calculus, as we’ll see in tomorrow’s post.