What is the domain and range of inverse trigonometric functions?
Answer
592.8k+ views
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:
$\Rightarrow D\left[ {{\sin }^{-1}}(x) \right]=\left[ -1,1 \right]$
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:
$\Rightarrow R\left[ {{\sin }^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
(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:
$\Rightarrow D\left[ {{\cos }^{-1}}(x) \right]=\left[ -1,1 \right]$
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:
$\Rightarrow R\left[ {{\cos }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]$
(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:
$\Rightarrow D\left[ {{\tan }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
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:
$\Rightarrow R\left[ {{\tan }^{-1}}\left( x \right) \right]=\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
(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:
$\Rightarrow D\left[ {{\cot }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
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:
$\Rightarrow R\left[ {{\cot }^{-1}}\left( x \right) \right]=\left( 0,\pi \right)$
(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:
$\Rightarrow D\left[ \cos e{{c}^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
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:
$\Rightarrow R\left[ \cos e{{c}^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}$
(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:
$\Rightarrow D\left[ {{\sec }^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
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:
$\Rightarrow R\left[ {{\sec }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$
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.
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:
$\Rightarrow D\left[ {{\sin }^{-1}}(x) \right]=\left[ -1,1 \right]$
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:
$\Rightarrow R\left[ {{\sin }^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
(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:
$\Rightarrow D\left[ {{\cos }^{-1}}(x) \right]=\left[ -1,1 \right]$
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:
$\Rightarrow R\left[ {{\cos }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]$
(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:
$\Rightarrow D\left[ {{\tan }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
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:
$\Rightarrow R\left[ {{\tan }^{-1}}\left( x \right) \right]=\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
(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:
$\Rightarrow D\left[ {{\cot }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
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:
$\Rightarrow R\left[ {{\cot }^{-1}}\left( x \right) \right]=\left( 0,\pi \right)$
(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:
$\Rightarrow D\left[ \cos e{{c}^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
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:
$\Rightarrow R\left[ \cos e{{c}^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}$
(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:
$\Rightarrow D\left[ {{\sec }^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
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:
$\Rightarrow R\left[ {{\sec }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$
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.
Recently Updated Pages
Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

