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Mathematics LibreTexts

2.8: Limits and continuity of Inverse Trigonometric functions

This page is a draft and is under active development. 

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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).

alt
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:

alt

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) alt

\left[ -\frac{\pi}{2},\frac{\pi}{2} \right]

and [-1,1]

f^{-1}(x)=\sin^{-1} x alt
f(x)=\cos(x) alt

[0,\pi]

and [-1,1]

f^{-1}(x)=\cos^{-1} x alt \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)

alt

\left( -\frac{\pi}{2},\frac{\pi}{2} \right)

and \mathbb{R}

f^{-1}(x)=\tan^{-1} x

alt \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)

alt \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}

alt

\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 .

alt

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


2.8: Limits and continuity of Inverse Trigonometric functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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