Mathematics · Ch 2 — Inverse Trigonometric Functions

Graphs of Inverse Trigonometric Functions

4. Graphs of Inverse Trigonometric Functions

The reflection principle. If $g=f^{-1}$ on a branch where $f$ is bijective, then $(a,b)$ lies on the graph of $f$ (restricted to that branch) exactly when $(b,a)$ lies on the graph of $g$ . Swapping the coordinates of every point on a graph is the same geometric operation as reflecting the graph in the line $y=x$ . So each inverse trigonometric function's graph is obtained by reflecting the graph of the corresponding trigonometric function -- restricted to its principal value branch (Section 2) -- in the line $y=x$ .

Graph of $y=\sin^{-1}x$ . Reflecting $y=\sin x$ on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ gives a curve with domain $[-1,1]$ and range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , running from $\left(-1,-\frac{\pi}{2}\right)$ up through the origin $(0,0)$ to $\left(1,\frac{\pi}{2}\right)$ . It is strictly increasing throughout, and since $\sin x$ is an odd function on this branch, so is $\sin^{-1}x$ : $\sin^{-1}(-x)=-\sin^{-1}x$ (proved in Section 5), which shows up as the graph being symmetric about the origin.

Graph of $y=\cos^{-1}x$ . Reflecting $y=\cos x$ on $[0,\pi]$ gives a curve with domain $[-1,1]$ and range $[0,\pi]$ , running from $(-1,\pi)$ down through $\left(0,\frac{\pi}{2}\right)$ to $(1,0)$ . It is strictly decreasing throughout -- the mirror behaviour of $\sin^{-1}x$ -- and, being neither odd nor even, is instead symmetric about the point $\left(0,\frac{\pi}{2}\right)$ , matching the identity $\cos^{-1}(-x)=\pi-\cos^{-1}x$ . …

Figure 1Graph of $y=\sin^{-1}x$

What this figure shows. Shows the graph of $y=\sin^{-1}x$ plotted for $x$ from $-1$ to $1$ on the horizontal axis, with the vertical axis marked at $-\pi/2$ , $0$ and $\pi/2$ . The curve starts at the point $(-1,-\pi/2)$ at the bottom-left, rises smoothly and strictly through the origin $(0,0)$ , and ends at $(1,\pi/2)$ at the top-right -- a strictly increasing curve confined entirely within the horizontal band $x\in[-1,1]$ and the vertical band $y\in[-\pi/2,\pi/2]$ . The curve is steepest near the centre, through the origin, and becomes nearly vertical as $x$ approaches $\pm1$ , reflecting that $\sin x$ itself is flattest (has zero slope) at $x=\pm\pi/2$ . The curve is symmetri …

Figure 2Graph of $y=\cos^{-1}x$

What this figure shows. Shows the graph of $y=\cos^{-1}x$ plotted for $x$ from $-1$ to $1$ on the horizontal axis, with the vertical axis marked at $0$ , $\pi/2$ and $\pi$ . The curve starts at the point $(-1,\pi)$ at the top-left, falls smoothly and strictly through the point $(0,\pi/2)$ on the vertical axis, and ends at $(1,0)$ at the bottom-right -- a strictly decreasing curve, the mirror image in behaviour of $\sin^{-1}x$ 's increasing shape. It is confined to the horizontal band $x\in[-1,1]$ and the vertical band $y\in[0,\pi]$ , and becomes nearly vertical as $x$ approaches $\pm1$ . The curve is symmetric about the point $(0,\pi/2)$ rather than about the origin, match …

Figure 3Graph of $y=\tan^{-1}x$

What this figure shows. Shows the graph of $y=\tan^{-1}x$ plotted for $x$ ranging over a wide interval, roughly $-10$ to $10$ , on the horizontal axis, with the vertical axis marked at $-\pi/2$ , $0$ and $\pi/2$ . Two horizontal dashed lines are drawn at $y=\pi/2$ and $y=-\pi/2$ , marking asymptotes that the curve approaches but never touches as $x\to\infty$ and $x\to-\infty$ respectively. The curve passes through the origin $(0,0)$ , rises steeply through the middle of its domain, and flattens out increasingly as $|x|$ grows large, approaching but never reaching the two horizontal asymptotes -- giving the characteristic S-shaped (sigmoid) curve. The curve is symmetric about the origin -- an odd function -- matching $\tan^{-1}(-x)=-\tan^{-1}x$ , and is defined …

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