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.

arcsec1

 

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.

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