Question

What is the domain and range of inverse trigonometric functions?

Answer 1

Hint: The Inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, etc. The inverse trigonometric functions are used to find the angle measure of a right-angled triangle when the measure of two sides of the triangle are known. The conventional symbol used to represent them is ‘arcsin’, ‘arccosine’, ‘arctan’, etc.

Complete step by step answer:
We will now see the domain and range of all the six inverse trigonometric functions in the following order:
(1) ${\sin }^{-1}\left(x\right)$
The domain of ${\sin }^{-1}\left(x\right)$ is equal to the range of $\sin \left(x\right)$. So, it could be written as:
$\Rightarrow D[{\sin }^{-1}(x)]=[-1,1]$
And, the range of ${\sin }^{-1}\left(x\right)$ is equal to the domain of $\sin \left(x\right)$. So, it could be written as:
$\Rightarrow R\left[{\sin }^{-1}\left(x\right)\right]=[-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]$

(2) ${\cos }^{-1}\left(x\right)$
The domain of ${\cos }^{-1}\left(x\right)$ is equal to the range of $\cos \left(x\right)$. So, it could be written as:
$\Rightarrow D[{\cos }^{-1}(x)]=[-1,1]$
And, the range of ${\cos }^{-1}\left(x\right)$ is equal to the domain of $\cos \left(x\right)$. So, it could be written as:
$\Rightarrow R\left[{\cos }^{-1}\left(x\right)\right]=[0,\pi ]$

(3) ${\tan }^{-1}\left(x\right)$
The domain of ${\tan }^{-1}\left(x\right)$ is equal to the range of $\tan \left(x\right)$. So, it could be written as:
$\Rightarrow D[{\tan }^{-1}(x)]=(-\mathrm{\infty },\mathrm{\infty })$
And, the range of ${\tan }^{-1}\left(x\right)$ is equal to the domain of $\tan \left(x\right)$. So, it could be written as:
$\Rightarrow R\left[{\tan }^{-1}\left(x\right)\right]=(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$

(4) ${\cot }^{-1}\left(x\right)$
The domain of ${\cot }^{-1}\left(x\right)$ is equal to the range of $\cot \left(x\right)$. So, it could be written as:
$\Rightarrow D[{\cot }^{-1}(x)]=(-\mathrm{\infty },\mathrm{\infty })$
And, the range of ${\cot }^{-1}\left(x\right)$ is equal to the domain of $\cot \left(x\right)$. So, it could be written as:
$\Rightarrow R\left[{\cot }^{-1}\left(x\right)\right]=(0,\pi )$

(5) $\cos e{c}^{-1}\left(x\right)$
The domain of $\cos e{c}^{-1}\left(x\right)$ is equal to the range of $\cos ec\left(x\right)$. So, it could be written as:
$\Rightarrow D[\cos e{c}^{-1}(x)]=(-\mathrm{\infty },1]\cup [1,\mathrm{\infty })$
And, the range of $\cos e{c}^{-1}\left(x\right)$ is equal to the domain of $\cos ec\left(x\right)$. So, it could be written as:
$\Rightarrow R[\cos e{c}^{-1}\left(x\right)]=[-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]-\left\{0\right\}$

(6) ${\sec }^{-1}\left(x\right)$
The domain of ${\sec }^{-1}\left(x\right)$ is equal to the range of $\sec \left(x\right)$. So, it could be written as:
$\Rightarrow D[{\sec }^{-1}(x)]=(-\mathrm{\infty },1]\cup [1,\mathrm{\infty })$
And, the range of ${\sec }^{-1}\left(x\right)$ is equal to the domain of $\sec \left(x\right)$. So, it could be written as:
$\Rightarrow R\left[{\sec }^{-1}\left(x\right)\right]=[0,\pi ]-\left\{{\displaystyle \frac{\pi }{2}}\right\}$

Note: The inverse functions are basically the mirror image of the fundamental functions. That is, they are identical in shape about the line, $y=x$ . This property is used in problems to plot the graph of these inverse trigonometric functions.