Graph of Inverse Trigonometric Functions, Domain and Range of ITF, Practice Problems and FAQs

Inverse functions of the basic trigonometric functions will give Inverse Trigonometric Functions. They are also called arcus functions or anti-trigonometric functions or cyclometric functions. The inverse functions in trigonometry are used to get the angle from any of the trigonometric ratios. The inverse trigonometric functions have major applications in the field of geometry, engineering and navigation. Let’s look at their graphs in this article.

Table of contents

• Domain and range of trigonometric functions
• Domain and range of inverse trigonometric functions
• Graphs of inverse trigonometric functions
• Practice Problems
• FAQs

Domain and range of Trigonometric Functions:

As you are already familiar with the domain and range of trigonometric functions. So, let’s revisit them before getting to the domain and range of inverse trigonometric functions.

We know that the necessary condition for existence of inverse of a function is that the functions has to be one - one and onto i.e. the function has to be bijective. But in general trigonometric functions in their entire domain are many-to-one functions. So, they are not bijective. So, we have to restrict their domain to make every trigonometric function bijective fort its inverse to exist.

We know that for sine function the domain is the set of all real numbers and range is [–1, 1] which is shown in the below graph as well.

On restricting its domain to $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, it will become one-one

and onto having range [– 1, 1]. Instead, sine function can be restricted to any of the intervals

$\left[-\frac{3\pi }{2},\frac{-\pi }{2}\right]$ or $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$or $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ etc., where each of these will be one-one and onto with range [–1, 1]. Therefore, we can define the inverse of sine function corresponding to each of these intervals.

The inverse of sine function is written as ^-1(arc sine function). Hence, -1 is a function with domain [–1, 1] and any of the intervals $\left[-\frac{3\pi }{2},\frac{-\pi }{2}\right]$ or $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$or $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ as range. Corresponding to each of these intervals, we will get a branch of -1 function.

The branch with range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is called the principal value branch. When we refer

to the function -1, we take it as the function whose domain is [–1, 1] and range is

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

Hence we can write -1 : [–1, 1] → $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. In the same manner we can define all the other inverse trigonometric functions as well.

We have six inverse trigonometric functions: ${sin}^{-1}x, {cos}^{-1}x , {tan}^{-1}x, {cot}^{-1}x,{cosec}^{-1}x,{sec}^{-1}x$ in their respective domains. An inverse trigonometric function gives angle in radians. For example, ${sin}^{-1}x$ is the measure of the angle in radian whose sine is x.

Some important points to keep in mind:

• ${sin}^{-1}x$ is also denoted by arc x.
• $(sinx{)}^{-1}=\frac{1}{sin x}$ (Wherever it exists ) but not equal to ${sin}^{-1}x$

Domain and range of inverse trigonometric functions:

Graphs of inverse trigonometric functions

The graph of an inverse function can be drawn by taking mirror image of the corresponding graph of the original function in restricted domain along the line y=x(i.e., by interchanging its coordinate axes).

Now, let’s draw the graphs of inverse trigonometric functions one by one.

1) Graph of ${sin}^{-1}\mathit{ }\mathit{x}$:

To obtain the graph of the function $y={sin}^{-1} x$, we can simply interchange x and y axes of the graph $f\left(x\right)=sin x$

For the function $f\left(x\right)=\square \sin x,$

Restricted domain: $\left[-\frac{\pi }{2}, \frac{\pi }{2}\right]$

Range: $\left[-1, 1\right]$

Here, we can see that the nature of the graph of f(x) is strictly increasing.

By taking the mirror image of $f\left(x\right)=sin x$ about the line y = x, we get the graph of the inverse of x, i.e., ${sin}^{-1} x$.

Graph of ${sin}^{-1}x$

For the inverse trigonometric function $y={sin}^{-1} x$, we get,

Domain = $\left[-1, 1\right]$

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

Note: For the inverse trigonometric function $y={sin}^{-1} x$, if x is positive, then angle y lies in the first quadrant, and if x is negative, then angle y lies in the fourth quadrant.

2.) Graph of ${cos}^{-1}\mathit{ }\mathit{x}$:

$y={cos}^{-1} x$

For $f\left(x\right)=cos x$,

Restricted domain: $\left[0, \pi \right]$,

Range: $\left[-1, 1\right]$

Here, we can see that the nature of the graph of f(x) is strictly decreasing.

By taking the mirror image of f(x) about the line y = x, we will get the graph of the inverse of f(x), i.e., ${cos}^{-1} x$.

Graph of ${cos}^{-1} x$

So, For $y= {cos}^{-1} x$, we get,

Domain = $\left[-1, 1\right]$

Range = $\left[0, \pi \right]$

3.) Graph of ${tan}^{-1}\mathit{ }\mathit{x}$:

$y={tan}^{-1} x$

For $f\left(x\right)=tan x$,

Restricted domain:$\left(-\frac{\pi }{2}, \frac{\pi }{2}\right)$

Range: ℝ

Here, we can see that the nature of the graph of f(x) is strictly increasing.

By taking the mirror image of f(x) about the line y=x, we get the graph of the inverse

of f(x), i.e., ${tan}^{-1} x$.

Graph of ${tan}^{-1} x$

So, for the inverse trigonometric function y=tan^-1 x, we get,

Domain = ℝ

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

4.) Graph of ${cot}^{-1}\mathit{ }\mathit{x}$:

$y={cot}^{-1} x$

For $f\left(x\right)=cot x$,

Restricted domain = $\left(0, \pi \right)$,

Range = ℝ

Here, we can see that the nature of the graph of f(x) is strictly decreasing.

By taking the mirror image of f(x) about the line y=x, we get the graph of the inverse

of f(x), i.e., ${cot}^{-1} x$.

Graph of ${cot}^{-1} x$

So, for the inverse trigonometric function $y={cot}^{-1} x$ , we get,

Domain = ℝ

Range = $\left(0, \pi \right)$,

5.) Graph of cosec^-1 x:

y=cosec^-1 x

For the function f(x)=cosec x, we get,

Restricted domain = $\left[-\frac{\pi }{2}, \frac{\pi }{2}\right]-\left\{0\right\}$

Range =$\left(-\infty , -1\right]$ U $\left[1, \infty \right)$

By taking the mirror image of f(x) about the line y=x, we get the graph of the inverse

of f(x), i.e., cosec^-1 x.

Graph of cosec^-1 x

So, For y=cosec^-1 x, we get,

Domain = $\left(-\infty , -1\right]$ U $\left[1, \infty \right)$

Range = $\left[-\frac{\pi }{2}, \frac{\pi }{2}\right]-\left\{0\right\}$

6.) Graph of ${sec}^{-1}\mathit{ }\mathit{x}$:

$y={sec}^{-1} x$

For the function $f\left(x\right)=sec x$, we get,

Restricted domain =$\left[0, \pi \right]-\left\{\frac{\pi }{2}\right\}$

Range = $\left(-\infty , -1\right]$ U $\left[1, \infty \right)$

By taking the mirror image of f(x) about the line y=x, we get the graph of the

inverse of f(x), i.e., ${sec}^{-1} x$.

Graph of ${sec}^{-1} x$

So, for $y={sec}^{-1} x$, we get,

Domain = $\left(-\infty , -1\right]$ U $\left[1, \infty \right)$

Range = $\left[0, \pi \right]-\left\{\frac{\pi }{2}\right\}$

Practice Problems

Example 1.

Find domain of ${sin}^{-1}\left(2x+3\right).$.

Solution

We know, for $f\left(x\right)={sin}^{-1} x$, Domain = $\left[-1, 1\right]$ and Range = $\left[-\frac{\pi }{2}, \frac{\pi }{2}\right]$

$\left(2x+3\right)\in \left[-1, 1\right]$

$\Rightarrow -1 \le 2x+3 \le 1$

$\Rightarrow -4 \le 2x \le -2$

$\Rightarrow -2 \le x \le -1$

$\Rightarrow x \in \left[-2, -1\right]$

Example 2.

Find domain of ${cos}^{-1}\left|x\right|$

Solution

We know, for $f\left(x\right)={cos}^{-1}x$, Domain = $\left[-1, 1\right]$

$\left|x\right|\in \left[-1, 1\right]$

$\Rightarrow -1 \le \left|x\right| \le 1$

 $\Rightarrow$ $\left|x\right| \le 1$

$\Rightarrow -1 \le x \le 1$

$\Rightarrow x \in \left[-\mathrm{1,1}\right]$

Example 3.

Find the domain and the range of -1[x], where [.] represents the greatest integer function.

Solution

Given function, $y={cos}^{-1}\left[x\right]$

We know, for the function $y={cos}^{-1}x$, domain is $-1\le x\le 1$

$\Rightarrow -1 \le \left[x\right]\le 1$

$\Rightarrow -1 \le x < 2\Rightarrow x\in \left[-\mathrm{1,2}\right)$

Also,

$-1 \le \left[x\right]\le 1\Rightarrow \left[x\right] = -1, 0, 1$

$\left[x\right] = -1\Rightarrow {cos}^{-1}\left(-1\right)= \pi$

$\left[x\right] = 0\Rightarrow {cos}^{-1}\left(0\right)= \frac{\pi }{2}$

$\left[x\right] = 1\Rightarrow {cos}^{-1}\left(1\right)= 0$

Range of ${cos}^{-1}\left[x\right]$ is $\left\{0, \frac{\pi }{2}, \pi \right\}$ .

Example 4.

Find the domain of ${cos}^{-1}\left(1+x^2\right)$.

Solution

Given, $g\left(x\right)= {cos}^{-1} \left(1+x^2\right)$,

$\Rightarrow 1+x^2\in \left[-1, 1\right]$

$\Rightarrow -1 \le 1+x^2 \le 1$

$-2 \le x^2 \le 0$

$\Rightarrow x \in \left\{0\right\}$

Example 5.

Find the domain and range of $y = {sin}^{-1}\left(e^x\right)$ .

Solution

The domain of ${sin}^{-1}x$ is $-1\le x\le 1$

For ${sin}^{-1}{e}^x$, we have $-1 \le e^x \le 1$

$\Rightarrow 0 < e^x \le 1$ (exponential function cannot be negative)

$x \le 0$

Also,

$0 < e^x \le 1$

$\Rightarrow {sin}^{-1}\left(0\right)< {sin}^{-1}\left(e^x\right)\le {sin}^{-1}1$

$\Rightarrow 0 < {sin}^{-1}\left(e^x\right)\le \frac{\pi }{2}$

Range =$\left(0, \frac{\pi }{2}\right]$

Example 6.

Find the domain of the function $f\left(x\right)=\sqrt{{sin}^{-1}\left(2x\right)+ \frac{\pi }{6}}$

Solution

Step 1:

Given, y = $f\left(x\right)=\sqrt{{sin}^{-1}\left(2x\right)+ \frac{\pi }{6}}$

For the real value of y,

${sin}^{-1}\left(2x\right)+ \frac{\pi }{6} \ge 0$

$\Rightarrow {sin}^{-1}\left(2x\right)\ge -\frac{\pi }{6}$

Step 2:

We know that the maximum value of ${sin}^{-1}x$ is $\frac{\pi }{2}$.

So,

$\Rightarrow \frac{\pi }{2} \ge {sin}^{-1}\left(2x\right)\ge -\frac{\pi }{6}$

$\Rightarrow sin \frac{\pi }{2} \ge 2x \ge sin\left(-\frac{\pi }{6}\right)$

$\Rightarrow 1 \ge 2x \ge -\frac{1}{2}$

$\Rightarrow \frac{1}{2} \ge x \ge -\frac{1}{4}$

$\Rightarrow x ϵ \left[-\frac{1}{4}, \frac{1}{2}\right]$

FAQs

Q1. Are ${sin}^{-1}x$ and ${\left(sin x\right)}^{-1}$ same ?
Answer: No, these two are not same. ${sin}^{-1}x \left(or arcsin x\right)$ is an inverse trigonometric function which gives the value of angle and ${\left(sin x\right)}^{-1}=cosec x$ .

Q2. Is the range of trigonometric functions same as the domain of corresponding inverse trigonometric functions?
Answer: No, domain of inverse trigonometric function is a subset of the corresponding trigonometric function.

Q3. Is the principal value same as the range?
Answer: No, the principal value is a branch or subset of the range.

Q4. Are inverse trigonometric functions periodic?
Answer: No, inverse trigonometric functions are not periodic.

Related video links: https://www.youtube.com/watch?v=9A2tAADJtkA

https://www.youtube.com/watch?v=urKG4K3J4s4