Inverse Trigonometric Functions - Formulas, Graph, Domain & Range

Trigonomerty is a branch of mathematics dealing with the angles to the length of the right angled triangle. Inverse Trigonometric functions are fundamental in mathematics, particularly in geometry, calculus, and applied mathematics. They are used to describe relationships involving lengths and angles in right triangles. The inverse trigonometric function graph helps find the domain and its range. Generally, the inverse of sine, cosine, tangent, cosecant, secant, and cotangent are the inverse trigonometric functions.

In this article, we will cover the concepts of the inverse trigonometric function. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 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.

Function

A relation from a set A to a set B is said to be a function from A to B if every element of set A has one and only one image in set B.

Let A and B are two non-empty sets, then a relation from A to B is said to be a function if each element x in A is assigned a unique element f(x) in B, and it is written as

f: A ➝ B and read as f is mapping from A to B.

Trigonometric Function

Trigonometric functions are trigonometric ratios that are adjacentd on ratios of sides in a right triangle: the hypotenuse (the longest side), the adjacent (the side adjacent to a chosen angle), and the opposite (the side opposite the chosen angle). The trigonometric functions are sine($sin$), cosine($cos$), tangent($tan$), cosecant($csc$), secant($sec$) and cotangent($cot$).

The basic formulas to find the trigonometric functions are as follows:

- $\sin \theta=$ opposite/Hypotenuse
- $\cos \theta=$ adjacent/Hypotenuse
- $\tan \theta=$ opposite/adjacent
- $\sec \theta=$ Hypotenuse/adjacent
- $\operatorname{cosec} \theta=$ Hypotenuse/opposite
- $\cot \theta=$ adjacent/opposite

$\begin{array}{|l|l|l|l|l|}
\hline \text { Angle (Degrees) } & \text { Angle (Radians) } & \sin (\theta) & \cos (\theta) & \tan (\theta) \\
\hline 0^{\circ} & 0 & 0 & 1 & 0 \\
\hline 30^{\circ} & \frac{\pi}{6} & \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{3}} \\
\hline 45^{\circ} & \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 1 \\
\hline 60^{\circ} & \frac{\pi}{3} & \frac{\sqrt{3}}{2} & \frac{1}{2} & \sqrt{3} \\
\hline 90^{\circ} & \frac{\pi}{2} & 1 & 0 & \text { Undefined } \\
\hline
\end{array}$

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the trigonometric functions. Trigonometric functions gives the value of the angles while the inverse trigonometric functions yields the angles to the respective trigonometric values. The invese trigonometric are inverse sine($sin^{-1}(x)$ or $arcsin$(x)), inverse cosine($cos^{-1}(x)$ or $arccos$(x)), inverse tangent($tan^{-1}(x)$ or $arctan$(x)), inverse cosecant($csc^{-1}(x)$ or $arccsc$(x)), inverse secant($sec^{-1}(x)$ or $arcsec$(x)) and inverse tangent($tan^{-1}(x)$ or $arctan$(x)).

Domain and range of inverse trigonometric function

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

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

$y=\sin x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and co- domain $=[-1,1]$

For this domain and co-domain this function is one-one and onto, so it is invertible and its inverse is$y=\sin ^{-1}(x)$

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

2. $y=\cos ^{-1}(x)$

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

3. $y=\tan ^{-1}(x)$

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

4. $y=\cot ^{-1}(x)$

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

5. $y=\sec ^{-1}(x)$

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

6. $y=\operatorname{cosec}^{-1}(x)$

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

Summary

Inverse trigonometric functions are essential for solving angles and distance problems. Their properties and applications extend far beyond right triangles, making them indispensable tools in various scientific and engineering disciplines. Its domain and range are important aspects for analyzing the trigonometric function.

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Solved Examples based on the Inverse Trigonometric Functions

Example 1: The domain of the function $f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty,-a] \cup[a, \infty)$then 'a' is equal to:

Solution:
$
f(x)=\sin \left(\frac{|x|+5}{x^2+1}\right)
$
For domain :

$
-1 \leq \frac{|x|+5}{x^2+1} \leq 1
$
Since $|x|+5 \& x^2+1$ is always positive So, $\frac{|x|+5}{x^2+1} \geq 0 \forall x \in \mathbb{R}$

So for domain:

$\begin{aligned} & \frac{|x|+5}{x^2+1} \leq 1 \\ & \Rightarrow|x|+5 \leq x^2+1 \\ & \Rightarrow 0 \leq x^2-|x|-4 \\ & \Rightarrow 0 \leq\left(|x|-\frac{1+\sqrt{17}}{2}\right)\left(|x|-\frac{1-\sqrt{17}}{2}\right) \\ & \Rightarrow|x| \geq \frac{1+\sqrt{17}}{2} \text { or }|x| \leq \frac{1-\sqrt{17}}{2} \text { (Rejected) }\end{aligned}$
$
\Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right] \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right)
$
So, $a=\frac{1+\sqrt{17}}{2}$

Hence, the answer is $\frac{1+\sqrt{ } 17}{2}$.

Example 2: The real-valued function $f(x)=\frac{\operatorname{cosec}^{-1} x}{\sqrt{x-[x]}}$ where $x$ denotes the greatest integer less than or equal to x, is defined for all x belonging to :

1) all non-integers except the interval

2) all real except integers

3) all integers except 0, -1, 1

4) all real except the interval $[-1,1]$

Solution:

$f(x)=\frac{\operatorname{cosec}^{-1} x}{\sqrt{\{x\}}}$

Domain of $\operatorname{cosec}^{-1} x$ is $|x| \geq 1$ and $x-[x]>0 \Rightarrow x \in R-\{I\}$
So $x \in R-I-[-1,1]$

Hence, the answer is the option 1.

Example 3: What is the domain of the function $\mathrm{f}(\mathrm{x})=\sin ^{-1}\left(\log _4 \mathrm{x}^2\right)$ ?

1) $\left[-2,-\frac{1}{2}\right] \cup\left[\frac{1}{2}, 2\right]$

2) $\left[-2,-\frac{1}{2}\right]$

3) $\left[\frac{1}{2}, 2\right]$

4) $[-2,2]$

Solution:

As we have learned in

Inverse Trigonometric Function

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

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

Now,

For f(x) to be defined

$-1 \leq \log _4 x^2 \leq 1$ ........(i) and $x^2>0$ .........(ii)

From (ii); $x \in \mathbb{R}-\{0\}$

From (i), we have

$\begin{aligned} & 4^{-1} \leq x^2 \leq 4^1 \\ & \Rightarrow \frac{1}{4} \leq x^2 \leq 4 \\ & \Rightarrow-2 \leq x \leq \frac{-1}{2} \quad \text { or } \quad \frac{1}{2} \leq x \leq 2\end{aligned}$\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\tan ^{-1}\left(\frac{4}{33}\right)+\ldots$

Hence, the answer is the option 1.

Example 4: The sum of the infinite terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\tan ^{-1}\left(\frac{4}{33}\right)+\ldots$ is equal to

Solution:

$\begin{aligned} \therefore \quad T_r & =\tan ^{-1}\left(\frac{2^{r-1}}{1+2^{2 r-1}}\right) \\ & =\tan ^{-1}\left(\frac{2^r-2^{r-1}}{1+2^r \cdot 2^{r-1}}\right) \\ & =\tan ^{-1}\left(2^r\right)-\tan ^{-1}\left(2^{r-1}\right)\end{aligned}$

$\begin{aligned} \therefore S_n=\sum_{r=1}^n T_r & =\sum_{r=1}^n \tan ^{-1}\left(2^r\right)-\tan ^{-1}\left(2^{r-1}\right) \\ & =\tan ^{-1}\left(2^n\right)-\tan ^{-1}\left(2^0\right) \\ & =\tan ^{-1}\left(2^n\right)-\tan ^{-1}(1) \\ & =\tan ^{-1}\left(2^n\right)-\frac{\pi}{4}\end{aligned}$

Hence,

$\begin{aligned} S_{\infty} & =\tan ^{-1}\left(2^{\infty}\right)-\frac{\pi}{4} \\ & =\tan ^{-1}(\infty)-\frac{\pi}{4} \\ & =\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\end{aligned}$
Hence, the answer is $\frac{\pi}{4}$.

Example 5: For $x \in(-1,1]$, the number of solutions of the equation $\sin ^{-1} x=2 \tan ^{-1}$\begin{aligned} & \sin ^{-1} x=2 \tan ^{-1} x \\ & \sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \\ & \Rightarrow x=\frac{2 x}{1+x^2} \\ & \Rightarrow x\left(\frac{2}{1+x^2}-1\right)=0 \\ & \Rightarrow x=0,1,-1 \text { but }-1 \text { is not included }\end{aligned}$ is equal to ______.

Solution:

$\begin{aligned} & \sin ^{-1} x=2 \tan ^{-1} x \\ & \sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \\ & \Rightarrow x=\frac{2 x}{1+x^2} \\ & \Rightarrow x\left(\frac{2}{1+x^2}-1\right)=0 \\ & \Rightarrow x=0,1,-1 \text { but }-1 \text { is not included }\end{aligned}$

Answer 2 solutions

Hence, the answer is 2.

Summary

Inverse trigonometric functions are essential for solving angles and distance problems. Their properties and applications extend far beyond right triangles, making them indispensable tools in various scientific and engineering disciplines. Its domain and range are important aspects for analyzing the trigonometric function.