Mathematics · Ch 3 — Trigonometric Functions

Inverse Secant Function

3.3.5

Inverse Secant Function

Consider $\sec$ restricted to $[0,\pi]-\left\{\dfrac{\pi}{2}\right\}\to\mathbb{R}-(-1,1)$ . Graphically this restriction is one-one and onto $\mathbb{R}-(-1,1)$ , so its inverse exists, called the inverse secant function, denoted $\sec^{-1}:\mathbb{R}-(-1,1)\to[0,\pi]-\left\{\dfrac{\pi}{2}\right\}$ . For $x\in\mathbb{R}-(-1,1)$ and $\theta\in[0,\pi]-\left\{\dfrac{\pi}{2}\right\}$ , we write $\sec^{-1}x=\theta$ if $\sec\theta=x$ .

Ex. $\sec\pi=-1$ , where $-1\in\mathbb{R}-(-1,1)$ and $\pi\in[0,\pi]-\left\{\dfrac{\pi}{2}\right\}$ , so $\sec^{-1}(-1)=\pi$ : the principal value of $\sec^{-1}(-1)$ is $\pi$ . …

Figure 3.12(a)Fig. 3.12(a) — Graph of y = sec x drawn over a wide domain, with its one-one restricted domain highlighted in bold

$Fig. 3.12(a) — Fig. 3.12(a) — Graph of y = sec x drawn over a wide domain, with its one-one restricted domain highlighted in bold$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Shows the graph of $y=\sec x$ drawn only over $[0,\pi]-\{\pi/2\}$ on the x-axis, split by a vertical asymptote at $x=\pi/2$ into a left branch starting at $y=1$ at $x=0$ and rising to $+\infty$ , and a right branch coming from $-\infty$ and rising to $y=-1$ at $x=\pi$ , together covering $\mathbb{R}-(-1,1)$ exactly on …

Figure 3.12(b)Fig. 3.12(b) — Graph of the inverse function y = sec⁻¹ x (reflection across y = x), with the principal branch in bold

$Fig. 3.12(b) — Fig. 3.12(b) — Graph of the inverse function y = sec⁻¹ x (reflection across y = x), with the principal branch in bold$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Shows the graph of $y=\sec^{-1}x$ over its domain $\mathbb{R}-(-1,1)$ on the x-axis, obtained by reflecting the restricted secant graph in the line $y=x$ ; it forms two branches approaching but never touching the horizontal asymptote $y=\pi/2$ , one lying in $\left[0,\pi/2\right)$ for $x\geq 1$ and one in $\left(\pi/2,\pi\right]$ f …

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