2.8: Limits and continuity of Inverse Trigonometric functions

Inverse functions

Recall that a function $f$ is one-to-one (often written as $1-1$) if it assigns distinct values of $y$ to distinct values of $x$. In other words, if $x_1 \ne x_2$ then $f(x_1 ) \ne f(x_2 )$. Equivalently, $f$ is one-to-one if $f(x_1 ) = f(x_2 )$ implies $x_1 = x_2$. There is a simple horizontal rule for determining whether a function $y=f(x)$ is one-to-one: $f$ is one-to-one if and only if every horizontal line intersects the graph of $y=f(x)$ in the $xy$-coordinate plane at most once (see Figure 5.3.3).

Figure 5.3.3 Horizontal rule for one-to-one functions

If a function $f$ is one-to-one on its domain, then $f$ has an inverse function, denoted by $f^{-1}$, such that $y=f(x)$ if and only if $f^{-1}(y) = x$. The domain of $f^{-1}$ is the range of $f$.

The basic idea is that $f^{-1}$ "undoes'' what $f$ does, and vice versa. In other words,
$$\nonumber \begin{alignat*}{3} f^{-1}(f(x)) ~&=~ x \quad&&\text{for all \(x \) in the domain of \(f \), and}\\ \nonumber f(f^{-1}(y)) ~&=~ y \quad&&\text{for all \(y \) in the range of \(f \).} \end{alignat*}$$

Theorem $\PageIndex{1}$

If $f$ is continuous and one to one, then \(f^{-1}\ is continuous on its domain.

Inverse Trigonometric functions

We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. However, we can restrict those functions to subsets of their domains where they are one-to-one. For example, $y=\sin\;x$ is one-to-one over the interval $\left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$, as we see in the graph below:

For $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ we have $-1 \le \sin\;x \le 1$, so we can define the inverse sine function $y=\sin^{-1} x$ (sometimes called the arc sine and denoted by $y=\arcsin\;(x$) whose domain is the interval $[-1,1]$ and whose range is the interval $\left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$. In other words:

$$\begin{alignat}{3} \sin^{-1} (\sin\;y) ~&=~ y \quad&&\text{for \(-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}\)}\label{eqn:arcsin1}\\ \sin\;(\sin^{-1} x) ~&=~ x \quad&&\text{for \(-1 \le x \le 1\)}\label{eqn:arcsin2} \end{alignat}$$

Summary of Inverse Trigonometric functions

Lets illustrate the summary of Trigonometric functions and Inverse Trigonometric functions in following table:

Trigonometric function	graph of the Trigonometric function	Restricted domain and the range	Inverse Trigonometric function	graph of the Inverse Trigonometric function	Properties
$f(x)=\sin(x)$		$\left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$ and $[-1,1]$	$f^{-1}(x)=\sin^{-1} x$
$f(x)=\cos(x)$		$[0,\pi]$ and $[-1,1]$	$f^{-1}(x)=\cos^{-1} x$		$$\begin{alignat}{3} \cos^{-1} (\cos\;y) ~&=~ y \quad&&\text{for \(0 \le y \le \pi\)}\label{eqn:arccos1}\\ \cos\;(\cos^{-1} x) ~&=~ x \quad&&\text{for \(-1 \le x \le 1\)}\label{eqn:arccos2} \end{alignat}$$
$f(x)=\tan(x)$		$\left( -\frac{\pi}{2},\frac{\pi}{2} \right)$ and $\mathbb{R}$	$f^{-1}(x)=\tan^{-1} x$		$$\begin{alignat}{3} \tan^{-1} (\tan\;y) ~&=~ y \quad&&\text{for \(-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}\)}\label{eqn:arctan1}\\ \tan\;(\tan^{-1} x) ~&=~ x \quad&&\text{for all real \(x\)}\label{eqn:arctan2} \end{alignat}$$
$f(x)=\cot(x)$					$$\begin{alignat}{3} \cot^{-1} (\cot\;y) ~&=~ y \quad&&\text{for \(0 < y < \pi\)}\label{eqn:arccot1}\\ \cot\;(\cot^{-1} x) ~&=~ x \quad&&\text{for all real \(x\)}\label{eqn:arccot2} \end{alignat}$$
$f(x)=\sec(x)$		$[0,\pi]$, with $x\ne \frac{\pi}{2}$ and $\mathbb{R}$			$$\begin{alignat}{3} \csc^{-1} (\csc\;y) ~&=~ y \quad&&\text{for \(-\frac{\pi}{2} \le y \le \frac{\pi}{2} \), \(y \ne 0\)}\label{eqn:arccsc1}\\ \csc\;(\csc^{-1} x) ~&=~ x \quad&&\text{for \(|x| \ge 1\)}\label{eqn:arccsc2} \end{alignat}$$ $$\begin{alignat}{3} \sec^{-1} (\sec\;y) ~&=~ y \quad&&\text{for \(0 \le y \le \pi \), \(y \ne \frac{\pi}{2}\)}\label{eqn:arcsec1}\\ \sec\;(\sec^{-1} x) ~&=~ x \quad&&\text{for \(|x| \ge 1\)}\label{eqn:arcsec2} \end{alignat}$$

Below are examples:

Example $\PageIndex{1}$:

Find $\sin^{-1} \left(\sin\;\frac{\pi}{4}\right)$.

Solution

Since $-\frac{\pi}{2} \le \frac{\pi}{4} \le \frac{\pi}{2}$, we know that $\sin^{-1} \left(\sin\;\frac{\pi}{4}\right) = \boxed{\frac{\pi}{4}}\;$, by Equation $\ref{eqn:arcsin1}$.

Example $\PageIndex{2}$:

Find $\sin^{-1} \left(\sin\;\frac{5\pi}{4}\right)$.

Solution

Since $\frac{5\pi}{4} > \frac{\pi}{2}$, we can not use Equation $\ref{eqn:arcsin1}$. But we know that $\sin\;\frac{5\pi}{4} = -\frac{1}{\sqrt{2}}$. Thus, $\sin^{-1} \left(\sin\;\frac{5\pi}{4}\right) = \sin^{-1} \left( -\frac{1}{\sqrt{2}} \right)$ is, by definition, the angle $y$ such that $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ and $\sin\;y = -\frac{1}{\sqrt{2}}$. That angle is $y=-\frac{\pi}{4}$, since

$$\sin\;\left( -\tfrac{\pi}{4} \right) ~=~ -\sin\;\left( \tfrac{\pi}{4} \right) ~=~ -\tfrac{1}{\sqrt{2}} ~. \nonumber$$

Example $\PageIndex{3}$:

Find $\cos^{-1} \left(\cos\;\frac{\pi}{3}\right)$.

Solution

Since $0 \le \frac{\pi}{3} \le \pi$, we know that $\cos^{-1} \left(\cos\;\frac{\pi}{3}\right) = \boxed{\frac{\pi}{3}}\;$, by Equation $\ref{eqn:arccos1}$.

Example $\PageIndex{4}$:

Find $\cos^{-1} \left(\cos\;\frac{4\pi}{3}\right)$.

Solution

Since $\frac{4\pi}{3} > \pi$, we can not use Equation $\ref{eqn:arccos1}$. But we know that $\cos\;\frac{4\pi}{3} = -\frac{1}{2}$. Thus, $\cos^{-1} \left(\cos\;\frac{4\pi}{3}\right) = \cos^{-1} \left( -\frac{1}{2} \right)$ is, by definition, the angle $y$ such that $0 \le y \le \pi$ and $\cos\;y = -\frac{1}{2}$. That angle is $y=\frac{2\pi}{3}$ (i.e. $120^\circ$). Thus, $\cos^{-1} \left(\cos\;\frac{4\pi}{3}\right) = \boxed{\tfrac{2\pi}{3}}\;$.

Example $\PageIndex{5}$:

Find $\tan^{-1} \left(\tan\;\frac{\pi}{4}\right)$.

Solution

Since $-\tfrac{\pi}{2} \le \tfrac{\pi}{4} \le \tfrac{\pi}{2}$, we know that $\tan^{-1} \left(\tan\;\frac{\pi}{4}\right) = \boxed{\frac{\pi}{4}}\;$, by Equation $\ref{eqn:arctan1}$.

Example $\PageIndex{6}$:

Find $\tan^{-1} \left(\tan\;\pi\right)$.

Solution

Since $\pi > \tfrac{\pi}{2}$, we can not use Equation $\ref{eqn:arctan1}$. But we know that $\tan\;\pi = 0$. Thus, $\tan^{-1} \left(\tan\;\pi\right) = \tan^{-1} 0$ is, by definition, the angle $y$ such that $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$ and $\tan\;y = 0$. That angle is $y=0$. Thus, $\tan^{-1} \left(\tan\;\pi \right) = \boxed{0}\;$.

Example$\PageIndex{7}$:

Find the exact value of $\cos\;\left(\sin^{-1}\;\left(-\frac{1}{4}\right)\right)$.

Solution

Let $\theta = \sin^{-1}\;\left(-\frac{1}{4}\right)$. We know that $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$, so since $\sin\;\theta = -\frac{1}{4} < 0$, $\theta$ must be in QIV. Hence $\cos\;\theta > 0$. Thus,

$$\cos^2 \;\theta ~=~ 1 ~-~ \sin^2 \;\theta ~=~ 1 ~-~ \left( -\frac{1}{4} \right)^2 ~=~\frac{15}{16} \quad\Rightarrow\quad \cos\;\theta ~=~ \frac{\sqrt{15}}{4} ~. \nonumber$$

Note that we took the positive square root above since $\cos\;\theta > 0$. Thus, $\cos\;\left(\sin^{-1}\;\left(-\frac{1}{4}\right)\right) = \boxed{\frac{\sqrt{15}}{4}}\;$.

Example $\PageIndex{8}$:

Show that $\tan\;(\sin^{-1} x) = \dfrac{x}{\sqrt{1 - x^2}}$ for $-1 < x < 1$.

Solution

When $x=0$, the Equation holds trivially, since

$$\nonumber \tan\;(\sin^{-1} 0) ~=~ \tan\;0 ~=~ 0 ~=~ \dfrac{0}{\sqrt{1 - 0^2}} ~.$$

Now suppose that $0 < x < 1$. Let $\theta = \sin^{-1} x$. Then $\theta$ is in QI and $\sin\;\theta = x$. Draw a right triangle with an angle $\theta$ such that the opposite leg has length $x$ and the hypotenuse has length $1$, as in Figure 5.3.10 (note that this is possible since $0 < x < 1$). Then $\sin\;\theta = \frac{x}{1} = x$. By the Pythagorean Theorem, the adjacent leg has length $\sqrt{1 - x^2}$. Thus, $\tan\;\theta = \frac{x}{\sqrt{1 - x^2}}$.

If $-1 < x < 0$ then $\theta = \sin^{-1} x$ is in QIV. So we can draw the same triangle except that it would be "upside down'' and we would again have $\tan\;\theta = \frac{x}{\sqrt{1 - x^2}}$, since the tangent and sine have the same sign (negative) in QIV. Thus, $\tan\;(\sin^{-1} x) = \dfrac{x}{\sqrt{1 - x^2}}$ for $-1 < x < 1$.

Example $\PageIndex{9}$:

Prove the identity $\tan^{-1} x \;+\; \cot^{-1} x ~=~ \frac{\pi}{2}$.

Solution:

Let $\theta = \cot^{-1} x$. Using relations, we have

$$\nonumber \tan\;\left( \tfrac{\pi}{2} - \theta \right) ~=~ -\tan\;\left( \theta - \tfrac{\pi}{2} \right) ~=~ \cot\;\theta ~=~ \cot\;(\cot^{-1} x) ~=~ x ~,$$

by Equation $\ref{eqn:arccot2}$. So since $\tan\;(\tan^{-1} x) = x$ for all $x$, this means that $\tan\;(\tan^{-1} x) = \tan\;\left( \tfrac{\pi}{2} - \theta \right)$. Thus, $\tan\;(\tan^{-1} x) = \tan\;\left( \tfrac{\pi}{2} - \cot^{-1} x \right)$. Now, we know that $0 < \cot^{-1} x < \pi$, so $-\tfrac{\pi}{2} < \tfrac{\pi}{2} - \cot^{-1} x < \tfrac{\pi}{2}$, i.e. $\tfrac{\pi}{2} - \cot^{-1} x$ is in the restricted subset on which the tangent function is one-to-one. Hence, $\tan\;(\tan^{-1} x) = \tan\;\left( \tfrac{\pi}{2} - \cot^{-1} x \right)$ implies that $\tan^{-1} x = \tfrac{\pi}{2} - \cot^{-1} x$, which proves the identity.

Continuity of Inverse Trigonometric functions

Example $\PageIndex{1}$:

Let $f(x)= \frac{3 \sec^{-1} (x)}{4-\tan^{-1}( x)}$. Find the values(if any) for which $f(x)$ is continuous.

Exercise $\PageIndex{1}$

Let $f(x)= \frac{3 \sec^{-1} (x)}{8+2\tan^{-1}( x)}$. Find the values(if any) for which $f(x)$ is continuous.

Answer

Limit of Inverse Trigonometric functions

Theorem $\PageIndex{1}$

$lim_{x \rightarrow \infty} \tan^{-1}( x) = \frac{\pi}{2}$.

$lim_{x \rightarrow -\infty} \tan^{-1}( x) = -\frac{\pi}{2}$.

$lim_{x \rightarrow \infty} \sec^{-1} (x)=\lim _{x \rightarrow \infty} \sec^{-1} (x )= \frac{\pi}{2}$.

Example $\PageIndex{1}$:

Find $\lim_{x \rightarrow \infty} sin\left( 2\tan^{-1}( x)\right)$.

Exercise $\PageIndex{1}$

Find $\lim_{x \rightarrow -\infty} sin\left( 2\tan^{-1}( x)\right)$.

Answer

Contributors and Attributions

• Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2.

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

• Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)