Graph of Inverse Trigonometric Function

The inverse trignometric function is the inverse of trignometric ratios i.e. sin, cos, tan, cot, sec, cosec. By the use of the inverse trignometric function, we can find the angle with respect to trignometric ratios if the value of the function is given. In real life, the inverse trigonometric function is used by architects to determine the angle of bridges and their supports.

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This Story also Contains

1. Inverse Trigonometric Function
2. Graph of inverse trignometric function
3. Graph of Principal Value of function $f^{-1}(f(x))$
4. Conversion of one ITF to another
5. Solved Examples Based on the graph of inverse trigonometric function

In this article, we will cover the concept of Graph of inverse Trigonometric Functions. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of eleven questions have been asked on this topic including two in 2013, two in 2019, one in 2021, and six in 2022.

Inverse Trigonometric Function

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function and vice versa.

Graph of inverse trignometric function

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

${\sin }^{-1}(x)$ is the inverse of the trignometric function sin x.

The function $y=\sin (x)$ is many one so it is not invertible. Now consider the small portion of the function

$\mathrm{y}=\sin \mathrm{x},\mathrm{x}\in [-\frac{\pi }{2},\frac{\pi }{2}]$ and $\mathrm{y}\in [-1,1]$

Which is strictly increasing, Hence, one-one and inverse is $y={\sin }^{-1}(x)$

Domain is $[-1,1]$ and Range is $[\frac{-\pi }{2},\frac{\pi }{2}]$

Graph of $y={\cos }^{-1}(x)$

${\cos }^{-1}(x)$ is the inverse of $\cos x$

Domain is $[-1,1]$ and Range is $[0,\pi ]$

Graph of $y={\tan }^{-1}(x)$

${\tan }^{-1}(x)$ is the inverse of $\tan x$.

Domain is $\mathbb{R}$ and Range is $(\frac{-\pi }{2},\frac{\pi }{2})$

Graph of $y={\cot }^{-1}(x)$

${\cot }^{-1}(x)$ is the inverse of $\cot x$.

Domain is $\mathbb{R}$ and Range is $(0,\pi )$

Graph of $y={\sec }^{-1}(x)$

${\sec }^{-1}(x)$ is the inverse of $\sec x$.

Domain is $\mathbb{R}-(-1,1)$ and Range is $[0,\pi ]-\left\{\frac{\pi }{2}\right\}$

Graph of $y={\mathrm{cosec}}^{-1}(x)$

${\mathrm{cosec}}^{-1}(x)$ is the inverse of $\mathrm{cosec}x$.

Domain is $\mathbb{R}-(-1,1)$ and Range is $[-\frac{\pi }{2},\frac{\pi }{2}]-\{0\}$

Graph of Principal Value of function $f^{-1}(f(x))$

1. Graph of ${\sin }^{-1}(\sin (x))$

$\begin{array}{ll}& y={\sin }^{-1}(\sin (x))\\ \therefore & \sin y=\sin x\\ \Rightarrow & y=n\pi +(-1{)}^{n}x,n\in I\text{ (integer) }\end{array}$
Since, $y\in [-\pi /2,\pi /2]$, we have different expressions for ${\sin }^{-1}(\sin (x))$ for different values of $x$.

From above, we can plot the graph of $y=\sin -1(\sin (x)$ )

$\begin{array}{rl}& \text{ 2. Graph of }{\cos }^{-1}(\cos (x))\\ & y={\cos }^{-1}(\cos (x))\\ & \therefore \cos y=\cos x\\ & \Rightarrow y=2n\pi =x,n\in I\text{ (integer) }\end{array}$
To draw the graph of $y={\cos }^{-1}(\cos (x))$, we draw all the lines of $y=2n\pi =x,n\in I$ for $y\in [0,\pi ]$.
3. Graph of ${\tan }^{-1}(\tan (x))$

The domain of the function is R and range is ( $-\pi /2,\pi /2)$

$\begin{array}{ll}& y={\tan }^{-1}(\tan (x))\\ \therefore & \tan y=\tan x\\ \Rightarrow & y=n\pi +x,n\in I\text{ (integer) }\end{array}$

To draw the graph of $y={\tan }^{-1}(\tan (x))$, we draw all the line of $y=n\pi +x,n\in I$ for $y\in (-\pi /2,\pi /2)$

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

The domain of the function is R and the range is ( $0,\pi$ )

$\begin{array}{ll}& y={\cot }^{-1}(\cot (x))\\ \therefore & \cot y=\cot x\\ \Rightarrow & y=n\pi +x,n\in I\text{ (integer) }\end{array}$
To draw the graph of $y={\cot }^{-1}(\cot (x))$, we draw all the lines of $y=n\pi +x,n\in I$ for $y\in (0,\pi )$.

5. Graph of ${\mathrm{cosec}}^{-1}(\mathrm{cosec}(x))$

Domain of the function is $R-\{n\pi ,n\in l\}$ and range is $[-\pi /2,\pi /2]-\{0\}$
$y={\mathrm{cosec}}^{-1}(\mathrm{cosec}(x))$
$\therefore \mathrm{cosec}y=\mathrm{cosec}x$
or $\sin x=\sin y$
Hence, the graph of ${\mathrm{cosec}}^{-1}(\mathrm{cosec}(x))$ is the same as that of $y={\sin }^{-1}(\sin (x))$, but excluding points $x=n\pi ,n\in I$.
6. Graph of ${\sec }^{-1}(\sec (x))$

Domain of the function is $R-\{(2n+1)\pi /2,n\in l\}$ and range is $[0,\pi ]-\{\pi /2\}$
$y={\sec }^{-1}(\sec (x))$
$\therefore \sec y=\sec x$
or $\cos y=\cos x$
Hence, graph of ${\sec }^{-1}(\sec (x))$ is the same as that of $y={\cos }^{-1}(\cos (x))$, but excluding points $x=(2n+1)\pi /2,n\in l$.

Conversion of one ITF to another

We can write the different inverse trigonometric functions in terms of other inverse trigonometric functions

1. Converting ${\sin }^{-1}x$

If $0<x<1$
Let ${\sin }^{-1}x=\mathrm{\Theta }$. Then $\sin \theta =x$ or $\sin \theta =x/1$

So, we have the right-angled triangle

From the figure
$\begin{array}{rl}& \cos \theta =\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}\\ & \therefore \theta ={\sin }^{-1}\mathrm{x}={\cos }^{-1}\sqrt{1-{\mathrm{x}}^2}\end{array}$
Also, we have $\theta ={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}$

$\begin{array}{rl}& ={\cot }^{-1}\frac{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\\ & ={\sec }^{-1}\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\\ & ={\csc }^{-1}\frac{1}{\mathrm{x}}\end{array}$

2. Converting ${\cos }^{-1}x$

If $0<x<1$
Let ${\cos }^{-1}x=\theta$, implies $\cos \theta =x$
We have the following right-angled triangle

From the figure
$\begin{array}{rl}& \sin \theta =\frac{\sqrt{1-{\mathrm{x}}^2}}{1}=\sqrt{1-{\mathrm{x}}^2}\\ & \therefore \theta ={\cos }^{-1}\mathrm{x}={\sin }^{-1}\sqrt{1-{\mathrm{x}}^2}\end{array}$

Also, we have $\theta ={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$

$\begin{array}{rl}& ={\cot }^{-1}\frac{x}{\sqrt{1-{\mathrm{x}}^2}}\\ & ={\sec }^{-1}\frac{1}{\mathrm{x}}\\ & ={\csc }^{-1}\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\end{array}$

3. Converting ${\tan }^{-1}x$
ff $x>0$
Let ${\tan }^{-1}x=\theta$, implies $\tan \theta =x$

We have the following right-angled triangle

From the figure
$\begin{array}{rl}\sin \theta & =\frac{x}{\sqrt{1+{\mathrm{x}}^2}}\\ \theta & ={\tan }^{-1}\mathrm{x}={\sin }^{-1}\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\end{array}$
Also, we have $\theta ={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}$

$\begin{array}{rl}& ={\cot }^{-1}\frac{1}{\mathrm{x}}\\ & ={\sec }^{-1}\sqrt{1+{\mathrm{x}}^2}\\ & ={\csc }^{-1}\frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{x}}\end{array}$

Relating $f^{-1}(x)$ with $f^{-1}(-x)$
We have the following important results
1. ${\sin }^{-1}(-x)=-{\sin }^{-1}(x)$ for all $x\in [-1,1]$

Proof:

Let $-x\in [-1,1]$, so $x\in [-1,1]$

Also assume ${\sin }^{-1}(-x)=\theta$ (Hence $\mathrm{\Theta }\in [-\pi /2,\pi /2]$ ) ….(i)

Taking sin of both sides

$\begin{array}{rl}& \sin ({\sin }^{-1}(-\mathrm{x}))=\sin (\theta )\\ & -\mathrm{x}=\sin \mathrm{\Theta }\\ \Rightarrow & \mathrm{x}=-\sin \theta \\ \Rightarrow & \mathrm{x}=\sin (-\mathrm{\Theta })\end{array}$

Taking sin^-1 of both sides
$\Rightarrow {\sin }^{-1}(x)={\sin }^{-1}(\sin (-\mathrm{\Theta }))$

$[$ As $\mathrm{\Theta }\in [-\pi /2,\pi /2]$, so $-\mathrm{\Theta }\in [-\pi /2,\pi /2]]$

$\Rightarrow -{\sin }^{-1}(x)=\theta$
From (i) and (ii), we get

$\Rightarrow {\sin }^{-1}(-x)=-{\sin }^{-1}(x)$
Similarly, we can prove
2. ${\tan }^{-1}(-x)=-{\tan }^{-1}(x)$ for all $x\in R$ and
3. ${\mathrm{cosec}}^{-1}(-x)=-{\mathrm{cosec}}^{-1}(x)$ for all $x\in R-(-1,1)$
4. ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}(x)$ for all $x\in [-1,1]$

Proof:

Let $-x\in [-1,1]$

And assume ${\cos }^{-1}(-x)=\mathrm{\Theta }($ Hence $\mathrm{\Theta }\in [0,\pi ])$ ….(i)

Taking cos of both sides

$\begin{array}{ll}& \cos ({\cos }^{-1}(-x))=\cos \theta \\ \therefore & -x=\cos \theta \\ \Rightarrow & x=-\cos \theta \\ \Rightarrow & x=\cos (\pi -\theta )\end{array}$

Taking cos^-1 of both sides

$\begin{array}{rl}& \Rightarrow {\cos }^{-1}(x)={\cos }^{-1}(\cos (\pi -\theta ))\\ & [:\theta \in [0,\pi ],\text{ so }\pi -\theta \in [0,\pi ]]\\ & {\cos }^{-1}(x)=\pi -\theta \\ & \Rightarrow \theta =\pi -{\cos }^{-1}(x)\end{array}$

From (i) and (i), we get

$\Rightarrow {\cos }^{-1}(-x)=\pi -{\cos }^{-1}(x)$

Similarly, we can prove

5. ${\sec }^{-1}(-x)=\pi -{\sec }^{-1}(x)$ for all $x\in R-(-1,1)$
6. ${\cot }^{-1}(-x)=\pi -{\cot }^{-1}(x)$ for all $x\in R$

Relating $f^{-1}(x)$ with $f^{-1}(1/x)$

1. ${\sin }^{-1}\left(\frac{1}{x}\right)={\csc }^{-1}x$ for all $x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$
2. ${\cos }^{-1}\left(\frac{1}{x}\right)={\sec }^{-1}x$ for all $x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$
3. ${\tan }^{-1}\left(\frac{1}{x}\right)=\{\begin{array}{cl}{\cot }^{-1}x& \text{ for }x>0\\ -\pi +{\cot }^{-1}x& \text{ for }x<0\end{array}$

Proof:

1.

$\begin{array}{rl}& \text{ Let }{\csc }^{-1}\mathrm{x}=\theta \\ & \text{ where }\theta \in [-\pi /2,\pi /2]-\{0\}\\ & \text{ and }\mathrm{x}\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })\\ & \therefore \mathrm{x}=\csc \theta \\ & \Rightarrow \frac{1}{\mathrm{x}}=\sin \theta \\ & \Rightarrow {\sin }^{-1}\frac{1}{\mathrm{x}}={\sin }^{-1}(\sin \theta )\\ & \Rightarrow \theta ={\sin }^{-1}\frac{1}{\mathrm{x}}(\text{ as }\theta \in [-\pi /2,\pi /2]-\{0\})\end{array}$

From (i) and (ii) we get

${\sin }^{-1}\left(\frac{1}{x}\right)={\csc }^{-1}x$

2.

$\begin{array}{rl}& \text{ Let }{\sec }^{-1}\mathrm{x}=\theta \\ & \text{ where }\theta \in [0,\pi ]-\{\pi /2\}\\ & \text{ and }\mathrm{x}\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })\\ & \therefore \mathrm{x}=\sec \theta \\ & \Rightarrow \frac{1}{\mathrm{x}}=\cos \theta \\ & \Rightarrow {\cos }^{-1}\frac{1}{\mathrm{x}}={\cos }^{-1}(\cos \theta )\\ & \Rightarrow \theta ={\cos }^{-1}\frac{1}{\mathrm{x}}(\text{ as }\theta \in [0,\pi ]-\{\pi /2\})\end{array}$

From (i) and (ii) we get

${\cos }^{-1}\left(\frac{1}{x}\right)={\sec }^{-1}x$

3.

Let ${\cot }^{-1}\mathrm{x}=\theta$, where $\theta \in (0,\pi )$ and $\mathrm{x}\in \mathrm{R}$

$\begin{array}{ll}\therefore & \mathrm{x}=\cot \theta \\ \Rightarrow & \frac{1}{\mathrm{x}}=\tan \theta \\ \Rightarrow & {\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)={\tan }^{-1}(\tan \theta )\end{array}$

From the graph

$\begin{array}{rl}{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)& =\{\begin{array}{ll}\theta ,& 0<\theta <\pi /2\\ -\pi +\theta ,& \pi /2<\theta <\pi \end{array}\\ & =\{\begin{array}{ll}{\cot }^{-1}x,& 0<{\cot }^{-1}x<\pi /2\\ -\pi +{\cot }^{-1}x,\pi /2<{\cot }^{-1}x<\pi \end{array}\\ & =\{\begin{array}{ll}{\cot }^{-1}x,& x>0\\ -\pi +{\cot }^{-1}x,x<0\end{array}\end{array}$

Recommended Video Based on Graph of Inverse Trigonometric Function:

Solved Examples Based on the graph of inverse trigonometric function

Example 1: $50\tan (3{\tan }^{-1}\left(\frac{1}{2}\right)+2{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right))+4\sqrt{2}\tan (\frac{1}{2}{\tan }^{-1}(2\sqrt{2}))$ is equal to _______________. [JEE MAINS 2022]

Solution: Let ${\tan }^{-1}\left(\frac{1}{2}\right)=\mathrm{A}$ and ${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)=\mathrm{B}$

Using interconversion

$\mathrm{B}={\tan }^{-1}(2)={\cot }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{2}-\mathrm{A}$

$\begin{array}{rl}& \therefore 50\tan (3{\tan }^{-1}\left(\frac{1}{2}\right)+2{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right))\\ & =50\tan (3A+2B\\ & =50\tan (3\mathrm{A}+2(\frac{\pi }{2}-\mathrm{A}))\\ & =50\tan (\pi +\mathrm{A})\\ & =50\tan \mathrm{A}\\ & =50\tan ({\tan }^{-1}\left(\frac{1}{2}\right))\\ & =50\cdot \frac{1}{2}\end{array}$

$=25$ .......(i)

$\begin{array}{rl}& \text{ Now let }{\tan }^{-1}(2\sqrt{2})=c\Rightarrow \tan c=2\sqrt{2}\\ & \Rightarrow \frac{2\tan \left(\frac{c}{2}\right)}{1-{\tan }^2\left(\frac{c}{2}\right)}=2\sqrt{2}\\ & \Rightarrow 2\tan \left(\frac{c}{2}\right)=2\sqrt{2}-2\sqrt{2}{\tan }^2\left(\frac{c}{2}\right)\\ & \Rightarrow \sqrt{2}{\tan }^2\left(\frac{c}{2}\right)+\tan \left(\frac{c}{2}\right)-\sqrt{2}=0\\ & \Rightarrow \tan \left(\frac{c}{2}\right)=\frac{-1+\sqrt{1+8}}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\\ & \Rightarrow \tan (\frac{1}{2}{\tan }^{-1}(2\sqrt{2}))=\frac{1}{\sqrt{2}}\\ & \therefore 4\sqrt{2}\tan (\frac{1}{2}{\tan }^{-1}(2\sqrt{2}))=4\end{array}$

$\text{ Required expression }=25+4=29$

Hence answer is $29$

Example 2: If the inverse trigonometric functions take principal values, then${\cos }^{-1}(\frac{3}{10}\cos ({\tan }^{-1}\left(\frac{4}{3}\right))+\frac{2}{5}\sin ({\tan }^{-1}\left(\frac{4}{3}\right)))$ is equal to. [JEE MAINS 2022]

Solution

$\begin{array}{rl}& {\cos }^{-1}(\frac{3}{10}\cos ({\tan }^{-1}\left(\frac{4}{3}\right))+\frac{2}{5}\sin ({\tan }^{-1}\left(\frac{4}{3}\right)))\\ & {\cos }^{-1}((\frac{3}{10}\times \frac{3}{5})+\frac{2}{5}\times \frac{4}{5})\\ & {\cot }^{-1}\left(\frac{9+16}{50}\right)=\frac{\pi }{3}\end{array}$

Hence, the answer is $\frac{\pi }{3}$

Example 3: The value of ${\tan }^{-1}\left(\frac{\cos \left(\frac{15\pi }{4}\right)-1}{\sin \left(\frac{\pi }{4}\right)}\right)$ is equal to : [JEE MAINS 2022]

Solution

Given,

${\tan }^{-1}\left(\frac{\cos \left(\frac{15\pi }{4}\right)-1}{\sin \frac{\pi }{4}}\right)$

$={\tan }^{-1}\left(\frac{\cos (4\pi -\frac{\pi }{4})-1}{\sin \frac{\pi }{4}}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{\sqrt{2}}-1}{1/\sqrt{2}}\right)$

$={\tan }^{-1}(1-\sqrt{2})=-\tan -1(\sqrt{2}-1)$

$=\frac{-\pi }{8}$

Hence, the answer is $\frac{-\pi }{8}$

Example 4: ${\sin }^{-1}(\sin \frac{2\pi }{3})+{\cos }^{-1}(\cos \frac{7\pi }{6})+{\tan }^{-1}(\tan \frac{3\pi }{4})$ is equal to? [JEE MAINS 2022]

Solution

${\sin }^{-1}(\sin \frac{2\pi }{3})={\sin }^{-1}(\sin \frac{\pi }{3})=\frac{\pi }{3}$

${\cos }^{-1}(\cos \frac{7\pi }{6})={\cos }^{-1}(-\frac{\sqrt{3}}{2})=\frac{5\pi }{6}$

${\tan }^{-1}(\tan \frac{3\pi }{4})={\tan }^{-1}(-1)=\frac{-\pi }{4}$

$\mathrm{Sum}=\frac{\pi }{3}+\frac{5\pi }{6}-\frac{\pi }{4}$

$=\frac{(4+10-3)\pi }{12}=\frac{11\pi }{12}$

Hence, the answer is $\frac{11\pi }{12}$

Example 5: The value of $\tan (2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right))$ is equal to. [JEE MAINS 2021]

Solution: Given,

$\tan (2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right))$
Let ${\tan }^{-1}\left(\frac{3}{5}\right)=A$ and ${\sin }^{-1}\left(\frac{5}{13}\right)=B$

$=\tan (2A+B)$

$=\frac{\tan (2A)+\tan (B)}{1-\tan (2A)\tan B}$

Now $\tan 2.A=\frac{2\tan .A}{1-{\tan }^{2}A}$
As ${\tan }^{-1}\left(\frac{3}{5}\right)=A\Rightarrow \tan A=\frac{3}{5}$
$\Rightarrow \tan 2.4=\frac{2\cdot \frac{3}{5}}{1-{\left(\frac{3}{5}\right)}^2}=\frac{6\cdot 25}{5(25-9)}=\frac{30}{16}=\frac{15}{8}$
and ${\sin }^{-1}\left(\frac{5}{13}\right)=B$
$\Rightarrow \sin B=\frac{5}{13}\Rightarrow \tan B=\frac{5}{12}$
Using (i)

$\tan (2A+B)=\frac{\frac{15}{8}+\frac{5}{12}}{1-\left(\frac{15}{8}\right)\left(\frac{5}{12}\right)}=\frac{220}{21}$

Hence, the answer is $\frac{220}{21}$