In the previous sections, we have seen all the six inverse trigonometric functions. The following table will help us to memorize the principal branch of each of those functions.
$\begin{array}{cc}{} &{\textbf{Function}} &{\textbf{Domain}} &{\textbf{Range}} &{}\\
{} &{\sin^{-1}} &{[-1,1]} &{\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]} &{}\\
{} &{\cos^{-1}} &{[-1,1]} &{\left[0, \pi \right]} &{}\\
{} &{\csc^{-1}} &{R - (-1,1)} &{\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{0 \}} &{}\\
{} &{\sec^{-1}} &{R - (-1,1)} &{\left[0, \pi \right] - \{\frac{\pi}{2} \}} &{}\\
{} &{\tan^{-1}} &{R} &{\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)} &{}\\
{} &{\cot^{-1}} &{R} &{\left(0, \pi \right)} &{}\\
\end{array}$
Let us write three important points to remember:
1. We denote the inverse trigonometric functions using the superscript '-1'.
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For example, the inverse sine function is denoted as sin-1.
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This should not be confused with (sin x)-1.
(sin x)-1 is $\frac{1}{\sin x}$
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This is applicable to all trigonometric functions.
2. If the branch is not specified, it is understood that, the principal branch is being considered.
3. Consider any one of the six inverse trigonometric functions.
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That function will have only one set as it’s domain.
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But that function will have infinite number of sets as the range.
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If we pick a value from the domain and use it as the input, we will get an output in each of the range sets.
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But the output present in the range corresponding to the principal branch, is considered as the principal value of that function.
Now we will see some solved examples
Solved Example 18.1
Find the principal value of $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
Solution:
1. We are asked to find $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
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Let $x~=~\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
2. So our aim is to find x. It can be done in 4 steps:
(i) $x~=~\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$ is an equation of the form:
$x~=~f(y)~=~\sin^{-1}(y)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{\sqrt2}$.
•
Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~ \sin x$
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In our present case, it is: $y ~=~ \frac{1}{\sqrt2} ~=~ \sin x$
(iii) So we have a trigonometric equation:
$\sin x = \frac{1}{\sqrt2}$
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When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
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In the present case, we do not need to write many steps. We already
know that, $\sin \left(\frac{\pi}{4} \right)~=~\frac{1}{\sqrt2}$
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So we can write: $x~=~\frac{\pi}{4}$
3. Finally, we check whether the value obtained is the principal value. It can be done in 5 steps:
(i) The final answer that we obtained is: $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)~=~\frac{\pi}{4}$
(ii) It is clear that,
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For the given inverse trigonometric function,
♦ The input y is $\frac{1}{\sqrt2}$
♦ The output x is $\frac{\pi}{4}$
(iii) For the $\sin^{-1}$ function:
♦ Domain is [-1,1]
♦ Range corresponding to the principal branch is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$
(iv) The input y falls within [-1,1]. So the input is acceptable.
(v) The output x falls within $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$. So the output obtained is also acceptable.
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We can write:
♦ The output $\frac{\pi}{4}$,
♦ is the principal value of the $\sin^{-1}$ function,
♦ when the input is $\frac{1}{\sqrt2}$.
Solved Example 18.2
Find the principal value of $\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$
Solution:
1. We are asked to find $\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$
•
Let $x~=~\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$
2. So our aim is to find x. It can be done in 4 steps:
(i) $x~=~\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$ is an equation of the form:
$x~=~f(y)~=~\cot^{-1}(y)$
(ii) This is an inverse trigonometric function, where input y = $\frac{-1}{\sqrt3}$.
•
Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~ \cot x$
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In our present case, it is: $y ~=~ \frac{-1}{\sqrt3} ~=~ \cot x$
(iii) So we have a trigonometric equation:
$\cot x = \frac{-1}{\sqrt3}$
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When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11. In the present case, it can be done as shown below:
$\begin{array}{ll}{} &{\cot x} & {~=~} &{\frac{-1}{\sqrt{3}}} &{} \\
{\implies} &{\tan x} & {~=~} &{-\sqrt{3}~~\color{magenta}{\text{- - - (A)}}} &{} \\
{} &{\tan \left(\frac{\pi}{3} \right)} & {~=~} &{\sqrt{3}~~\color{magenta}{\text{- - - (B)}}} &{} \\
{}
&{\left[\tan \left(\pi – \theta \right)\right.} & {~=~} &{\left. - \tan
\theta\right] ~~\color{magenta}{\text{- - - (C)}}} &{} \\
{\implies} &{\tan \left(\pi – \frac{\pi}{3} \right)} & {~=~} &{- \tan \frac{\pi}{3}} &{} \\
{\implies}
&{\tan \left(\frac{2 \pi}{3} \right)} & {~=~} &{-
\sqrt{3}~~\color{magenta}{\text{- - - (D)}}} &{} \\
{\implies} &{x} & {~=~} &{\frac{2 \pi}{3}~~\color{magenta}{\text{- - - (E)}}} &{} \\
\end{array}
$
• Line A:
For simplicity, we convert cot x to tan x. This line gives the modified equation which is to be solved.
• Line B:
We write a basic equation which is closest to the equation to be solved.
• Line C:
We write the trigonometric identity which will help to solve the equation.
This identity is derived from identities 9(c) and 9(d). The list of identities can be seen here.
• Line D:
We get this result from (B).
• Line E:
We get this result by comparing D and A.
3. Finally, we check whether the value obtained is the principal value. It can be done in 5 steps:
(i) The final answer that we obtained is: $\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)~=~\frac{2 \pi}{3}$
(ii) Based on this, we can write:
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For the given inverse trigonometric function,
♦ The input y is $\frac{-1}{\sqrt3}$
♦ The output x is $\frac{2 \pi}{3}$
(iii) For the $\cot^{-1}$ function:
♦ Domain is R
♦ Range corresponding to the principal branch is $\left(0, \pi \right)$
(iv) The input y is a real number. So the input is acceptable.
(v) The output x falls within $\left(0, \pi \right)$. So the output obtained is also acceptable.
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We can write:
♦ The output $\frac{2 \pi}{3}$,
♦ is the principal value of the $\cot^{-1}$ function,
♦ when the input is $\frac{-1}{\sqrt3}$.
The link below gives a few more solved examples:
In the next section, we will see properties of inverse trigonometric functions.
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