21.10 - Derivatives of Inverse Trigonometric Functions

In the previous section, we completed a discussion on derivatives of implicit functions. In this section, we will see derivatives of inverse trigonometric functions.

• We have seen inverse trigonometric functions in chapter 18. Now we want to find the derivatives of those functions.
• We know that, only continuous functions are differentiable. Inverse trigonometric functions are continuous functions. We will see the proof in higher classes.
• At present we will see the method for differentiating those functions.

Let us see an example. It can be written in 8 steps:
1. Given that, f(x) = sin⁻¹x. We want to find f'(x).
2. Let y = sin⁻¹x. Then x = sin y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\sin y)$
⇒ $1~=~ \cos y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}$
4. In the above result, the denominator cos y should not be equal to zero.
• That means, y should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, sin⁻¹x should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, x should not be equal to −1 or 1. 
5. In chapter 18, we saw that, the acceptable domain for the sin⁻¹ function is [−1,1] (details here).
• But from the above step (4), we see that, derivative of the sin⁻¹ function is not defined at −1 and 1.
• So we can write two important points:
(i) For the sin⁻¹ function, the input can be taken from the interval [−1,1].
(ii) Derivative of the sin⁻¹ function is available only in the interval (−1,1).
6. We can eliminate the trigonometric ratios from the result.
• We have:
$\cos^2 y ~=~ 1 - \sin^2 y ~=~ 1 - [\sin(\sin^{-1} x)]^2 ~=~1 - x^2 $
So $\cos y ~=~ \pm \sqrt{1 - x^2}$
7. We have to determine whether cos y is $\sqrt{1 - x^2}$ or $- \sqrt{1 - x^2}$.
• The input for cos y is "y", which is "sin⁻¹x".
• We saw that, sin⁻¹x should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, input for cos y should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• Between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, cosine is +ve.
• So cos y is +ve.
• Thus $\cos y ~=~ \sqrt{1 - x^2}$
8. So we can write:
If y = sin⁻¹ x, then $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}~=~ \frac{1}{\sqrt{1 - x^2}}$

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Now we will see the derivative of inverse cosine function. It can be written in 8 steps:

1. Given that, f(x) = cos⁻¹x. We want to find f'(x).
2. Let y = cos⁻¹x. Then x = cos y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\cos y)$
⇒ $1~=~ -\sin y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}$
4. In the above result, the denominator sin y should not be equal to zero.
• That means, y should not be equal to 0 or $\pi$.
• That means, cos⁻¹x should not be equal to 0 or $\pi$.
• That means, x should not be equal to 1 or −1. 
5. In chapter 18, we saw that, the acceptable domain for the cos⁻¹ function is [−1,1] (details here).
• But from the above step (4), we see that, derivative of the cos⁻¹ function is not defined at −1 and 1.
• So we can write two important points:
(i) For the cos⁻¹ function, the input can be taken from the interval [−1,1].
(ii) Derivative of the cos⁻¹ function is available only in the interval (−1,1).
6. We can eliminate the trigonometric ratios from the result.
• We have:
$\sin^2 y ~=~ 1 - \cos^2 y ~=~ 1 - [\cos(\cos^{-1} x)]^2 ~=~1 - x^2 $
So $\sin y ~=~ \pm \sqrt{1 - x^2}$
7. We have to determine whether sin y is $\sqrt{1 - x^2}$ or $- \sqrt{1 - x^2}$.
• The input for sin y is "y", which is "cos⁻¹x".
• We saw that, cos⁻¹x should not be equal to 0 or π.
• That means, input for sin y should not be equal to 0 or π.
• Between 0 and π, sine is +ve.
• So sin y is +ve.
• Thus $\sin y ~=~ \sqrt{1 - x^2}$
8. So we can write:
If y = cos⁻¹ x, then $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}~=~\frac{-1}{\sqrt{1 - x^2}}$

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Now we will see the derivative of inverse tangent function. It can be written in 4 steps:

1. Given that, f(x) = tan⁻¹x. We want to find f'(x).
2. Let y = tan⁻¹x. Then x = tan y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\tan y)$
⇒ $1~=~ \sec^2 y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{1}{\sec^2 y}~=~\frac{1}{1 + \tan^2 y}~=~\frac{1}{1 + x^2 }$
4. In the above result, the denominator 1 + x² can never become zero.
• That means, x can be any real number.

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• We have seen the derivatives of:
    ♦ inverse sine function
    ♦ inverse cosine function
    ♦ inverse tangent function

• There are three more inverse trigonometric functions:
    ♦ inverse cosecant function
    ♦ inverse secant function
    ♦ inverse cotangent function
        ✰ Derivatives of these three functions can be obtained by writing them in terms of sine, cosine or tangent. For that, we can use the first property of inverse trigonometric functions. Details here.

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Let us see some solved examples.

Solved example 21.37
Find $\frac{dy}{dx}$ if y = 4 cos⁻¹ x − 10 tan⁻¹ x.
Solution:

Solved example 21.38
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, \sqrt{x} \sin^{-1} x$.
Solution:

 

Solved example 21.39
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, \frac{1}{\sin^{-1} x}$
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{d}{dx}\left(\sin^{-1} x \right)^{-1}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{-1 \left(\sin^{-1} x \right)^{-2} \frac{d}{dx}\left(\sin^{-1} x \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{\left(\sin^{-1} x \right)^{2} \sqrt{1 – x^2}}}    \\
\end{array}$

Solved example 21.40
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, x \tan^{-1} \sqrt{x}$
Solution:

 

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Link to a few more solved examples is given below:

Exercise 21.3

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We have completed a discussion on the derivatives of inverse trigonometric functions. In the next section, we will see Exponential functions.

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18.8 - Properties II and III

In the previous section, we saw the first property of inverse trigonometric functions. In this section, we will see the second and third properties.

Property II
• This has 3 parts:
$\begin{array}{cc}{}    &{\text{(i)}}    &{\sin^{-1}\left(-x \right)}    {}={}    &{-\sin^{-1}x},    &{x \in [-1,1]}\\
{}    &{\text{(ii)}}    &{\tan^{-1}\left(-x \right)}    {}={}    &{-\tan^{-1}x},    &{x \in R}\\
{}    &{\text{(iii)}}    &{\csc^{-1}\left(-x \right)}    {}={}    &{-\csc^{-1}x},    &{|x| \ge 1}\\
\end{array}$

• Let us prove part (i):
$\begin{array}{ll}{}    &{\text{Let}~\sin^{-1} \left(-x \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\sin y}    & {~=~}    &{-x}    &{} \\
{\implies}    &{- \sin y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\sin (-y)}    & {~=~}    &{x~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\sin^{-1} x}    & {~=~}    &{-y}    &{} \\
{\implies}    &{-\sin^{-1} x}    & {~=~}    &{y}    &{} \\
{\implies}    &{-\sin^{-1} x}    & {~=~}    &{\sin^{-1} \left(-x \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}$               

◼ Remarks:
• In line (A), we assume that $\sin^{-1}(-x)$ = y
• In line B, we use the identity 1: sin (-𝜃) = - sin 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

• Let us prove part (ii):
$\begin{array}{ll}{}    &{\text{Let}~\tan^{-1} \left(-x \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\tan y}    & {~=~}    &{-x}    &{} \\
{\implies}    &{- \tan y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\tan (-y)}    & {~=~}    &{x~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\tan^{-1} x}    & {~=~}    &{-y}    &{} \\
{\implies}    &{-\tan^{-1} x}    & {~=~}    &{y}    &{} \\
{\implies}    &{-\tan^{-1} x}    & {~=~}    &{\tan^{-1} \left(-x \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}$               

◼ Remarks:
• In line (A), we assume that $\tan^{-1}(-x)$ = y
• In line B, we use the identities 1 and 2 to get: tan (-𝜃) = - tan 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

• Let us prove part (iii):
$\begin{array}{ll}{}    &{\text{Let}~\csc^{-1} \left(-x \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\csc y}    & {~=~}    &{-x}    &{} \\
{\implies}    &{- \csc y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\frac{-1}{\sin y}}    & {~=~}    &{x}    &{} \\
{\implies}    &{-\sin y}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\sin (-y)}    & {~=~}    &{\frac{1}{x}~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\sin^{-1} \left(\frac{1}{x} \right)}    & {~=~}    &{-y}    &{} \\
{\implies}    &{\csc^{-1} x }    & {~=~}    &{-y~\color{magenta}{\text{- - - (C)}}}    &{} \\
{\implies}    &{\csc^{-1} x}    & {~=~}    &{-\csc^{-1} \left(-x \right)~\color{magenta}{\text{- - - (D)}}}    &{} \\
{\implies}    &{-\csc^{-1} x}    & {~=~}    &{\csc^{-1} \left(-x \right)}    &{} \\
\end{array}$

◼ Remarks:
• In line (A), we assume that $\csc^{-1}(-x)$ = y
• In line B, we use the identity 1: sin (-𝜃) = - sin 𝜃
    ♦ List of identities can be seen here.
• In line C, we use part (i) of property I
• In line D, we substitute for y, using the assumption in (A).

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Property III
• This has 3 parts:
$\begin{array}{cc}{}    &{\text{(i)}}    &{\cos^{-1}\left(-x \right)}    {}={}    &{\pi-\cos^{-1}x},    &{x \in [-1,1]}\\
{}    &{\text{(ii)}}    &{\sec^{-1}\left(-x \right)}    {}={}    &{\pi-\sec^{-1}x},    &{|x| \ge 1}\\
{}    &{\text{(iii)}}    &{\cot^{-1}\left(-x \right)}    {}={}    &{\pi-\cot^{-1}x},    &{x \in R}\\
\end{array}$

• Let us prove part (i):

◼ Remarks:
• In line (A), we assume that $\cos^{-1}(-x)$ = y
• In line B, we use the identity 1: cos (π-𝜃) = - cos 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

• Let us prove part (ii):

◼ Remarks:
• In line (A), we assume that $\sec^{-1}(-x)$ = y
• In line B, we use the identity 1: cos (π-𝜃) = - cos 𝜃
    ♦ List of identities can be seen here.
• In line C, we use part (ii) of property I
• In line D, we substitute for y, using the assumption in (A).

• Let us prove part (iii):

◼ Remarks:
• In line (A), we assume that $\cot^{-1}(-x)$ = y
• In line B, we use the identities 9(c) and 9(d) to get: cot (π-𝜃) = - cot 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

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In the next section, we will see fourth and fifth properties.

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18.7 - Properties of Inverse Trigonometric Functions

In the previous section, we saw some solved examples on the six inverse trigonometric functions. In this section, we will see the properties of inverse trigonometric functions.

First we will see how the inputs and outputs are to be denoted. It can be written in 3 steps:  

1. In the discussions so far in this chapter, we gave primary importance to the original trigonometric function.
• For example, the original trigonometric function was:
y = f(x) = sin x
    ♦ Input was denoted as ‘x’.
    ♦ Output was denoted as ‘y’.
2. From the original trigonometric function, we derived the inverse trigonometric function.
• For example:
x = f(y) = sin^-1 y
    ♦ Input was denoted as ‘y’.
    ♦ Output was denoted as ‘x’.
3. We have learned the basic details about inverse trigonometric functions. We can now treat them independently. So from now on wards, the inverse trigonometric function will be our primary function.
    ♦ Input for the inverse trigonometric function will be denoted as ‘x’.
    ♦ Output for the inverse trigonometric function will be denoted as ‘y’.

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Now we will see an interesting case. It can be written in 3 steps:
1. In the previous chapter, we saw composite functions. Let us recall the details:
(i) f: A→B and g: B→C are two functions.
• Then the composite function g(f(x)) connects the domain of f with the codomain of g.
• So we can write:
    ♦ Domain of g(f(x)) is A
    ♦ Codomain of g(f(x)) is C
(ii) Suppose that, g is the inverse of f.
• Then f is mapped from A to B and g is mapped from B to A.
• In such a situation, we can write:
    ♦ Domain of g(f(x)) is A
    ♦ Codomain of g(f(x)) is also A
This is because, g(f(x)) connects the domain of f with the codomain of g.
• Also, g(f(x)) is an identity function. Whatever value we give as input, the output will be that same value.
(Recall that, function of “the inverse of that function” is an identity function)
• Some examples can be seen here.
(iii) Similarly, we can write about f(g(x)):
    ♦ Domain of f(g(x)) is B .
    ♦ Codomain of f(g(x)) is also B.
• f(g(x)) is also an identity function. Whatever value we give as input, the output will be that same value.
2. Now we will apply the above information on sin function and it’s inverse.
(i) Let us write the domain and codomain:
    ♦ sin function is mapped from $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$ to [-1,1]
    ♦ sin^-1 function is mapped from [-1,1] to $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$
(ii) sin(sin^-1 x) will connect the domain of sin^-1 with the codomain of sin.
• That means, sin(sin^-1 x) is mapped from [-1,1] to [-1,1]
• In other words, sin(sin^-1 x) is a function on [-1,1]
• Also, sin(sin^-1 x) is an identity function. Whatever value we give as input, the output will be that same value.
(iii) sin^-1(sin x) will connect the domain of sin with the codomain of sin^-1.
• That means, sin^-1(sin x) is mapped from $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$ to $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$
• In other words, sin^-1(sin x) is a function on $\left[-{\frac{\pi}{2}, \frac{\pi}{2}} \right]$.
• Also, sin^-1(sin x) is an identity function. Whatever value we give as input, the output will be that same value.
3. The above step 2 is applicable to all six trigonometric functions. So we can write 12 identity functions:

$\begin{array}{cc}{}    &{\text{(i)}}    &{\sin \left(\sin^{-1} (x) \right)}    &{\text{(vii)}}    &{\sin^{-1} \left(\sin(x) \right)}    &{}\\
{}    &{\text{(ii)}}    &{\cos \left(\cos^{-1} (x) \right)}    &{\text{(viii)}}    &{\cos^{-1} \left(\cos (x) \right)}    &{}\\
{}    &{\text{(iii)}}    &{\tan \left(\tan^{-1} (x) \right)}    &{\text{(ix)}}    &{\tan^{-1} \left(\tan (x) \right)}    &{}\\
{}    &{\text{(iv)}}    &{\csc \left(\csc^{-1} (x) \right)}    &{\text{(x)}}    &{\csc^{-1} \left(\csc (x) \right)}    &{}\\
{}    &{\text{(v)}}    &{\sec \left(\sec^{-1} (x) \right)}    &{\text{(xi)}}    &{\sec^{-1} \left(\sec (x) \right)}    &{}\\
{}    &{\text{(vi)}}    &{\cot \left(\cot^{-1} (x) \right)}    &{\text{(xii)}}    &{\cot^{-1} \left(\cot (x) \right)}    &{}\\
\end{array}                   
$

In all the above 12 cases, whatever value we give as input, the output will be that same value.

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Now we will see 6 properties of inverse trigonometric functions.

Property I
• This has 3 parts:
\begin{array}{cc}{}    &{\text{(i)}}    &{\sin^{-1}\left(\frac{1}{x} \right)}    {}={}    &{\csc^{-1}x}    &{x \ge 1~\text{or}~x \le -1}\\
{}    &{\text{(ii)}}    &{\cos^{-1}\left(\frac{1}{x} \right)}    {}={}    &{\sec^{-1}x}    &{x \ge 1~\text{or}~x \le -1}\\
{}    &{\text{(iii)}}    &{\tan^{-1}\left(\frac{1}{x} \right)}    {}={}    &{\cot^{-1}x}    &{x > 0}\\
\end{array}

• Let us prove part (i):
$\begin{array}{ll}{}    &{\text{Let}~\sin^{-1} \left(\frac{1}{x} \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\sin y}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\frac{1}{\csc y}}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\csc y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\csc^{-1} x}    & {~=~}    &{y~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\csc^{-1} x}    & {~=~}    &{\sin^{-1} \left(\frac{1}{x} \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• In line (A), we assume that $\sin^{-1}x$ = y
• In line B, we substitute for y, using the assumption in (A).
• When the substitution is done, we get line C. 

• Let us prove part (ii):
$\begin{array}{ll}{}    &{\text{Let}~\cos^{-1} \left(\frac{1}{x} \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\cos y}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\frac{1}{\sec y}}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\sec y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\sec^{-1} x}    & {~=~}    &{y~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\sec^{-1} x}    & {~=~}    &{\cos^{-1} \left(\frac{1}{x} \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• In line (A), we assume that $\cos^{-1}x$ = y
• In line B, we substitute for y, using the assumption in (A).
• When the substitution is done, we get line C.

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◼ In the same way, we can prove part (iii) also.
◼ Note:
For each of the three parts in property I, we see acceptable values of x. For example, for part (i), x ≥ 1 or x ≤ -1. We will see the details about such “acceptable values” in higher classes. At present, all we need to know is that, whenever we see $\csc^{-1} (x)$, we can put $\sin^{-1} \left(\frac{1}{x} \right)$ in it’s place.

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In the next section, we will see the second and third properties of inverse trigonometric functions.

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18.6 - Solved Examples on Inverse Trigonometric Functions

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.

$\begin{array}{cc}{}    &{\textbf{Function}}    &{\textbf{Domain}}    &{\textbf{Range}}    &{}\\
{}    &{\sin^{-1}}    &{[-1,1]}    &{\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]}    &{}\\
{}    &{\cos^{-1}}    &{[-1,1]}    &{\left[0, \pi \right]}    &{}\\
{}    &{\csc^{-1}}    &{R - (-1,1)}    &{\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{0 \}}    &{}\\
{}    &{\sec^{-1}}    &{R - (-1,1)}    &{\left[0, \pi \right] - \{\frac{\pi}{2} \}}    &{}\\
{}    &{\tan^{-1}}    &{R}    &{\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)}    &{}\\
{}    &{\cot^{-1}}    &{R}    &{\left(0, \pi \right)}    &{}\\
\end{array}$

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Let us write three important points to remember:
1. We denote the inverse trigonometric functions using the superscript '-1'.
• For example, the inverse sine function is denoted as sin^-1.
• This should not be confused with (sin x)^-1.
(sin x)^-1 is $\frac{1}{\sin x}$
• 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.
• That function will have only one set as it’s domain.
• But that function will have infinite number of sets as the range.
• 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.
• But the output present in the range corresponding to the principal branch, is considered as the principal value of that function.

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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 $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
• 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:
$y ~=~ f(x) ~=~ \sin x$
• In our present case, it is: $y ~=~ \frac{1}{\sqrt2} ~=~ \sin x$
(iii) So we have a trigonometric equation:
$\sin x = \frac{1}{\sqrt2}$
• 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, we do not need to write many steps. We already know that, $\sin \left(\frac{\pi}{4} \right)~=~\frac{1}{\sqrt2}$
• 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,
• 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.
• 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:
$y ~=~ f(x) ~=~ \cot x$
• In our present case, it is: $y ~=~ \frac{-1}{\sqrt3} ~=~ \cot x$
(iii) So we have a trigonometric equation:
$\cot x = \frac{-1}{\sqrt3}$
• 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}               
$

◼ Remarks:
• 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:
• 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.
• We can write:
   ♦ The output $\frac{2 \pi}{3}$,
   ♦ is the principal value of the $\cot^{-1}$ function,
   ♦ when the input is $\frac{-1}{\sqrt3}$.

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The link below gives a few more solved examples:

Exercise 18.1

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In the next section, we will see properties of inverse trigonometric functions.

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Chapter 18 - Inverse Trigonometric Functions

In the previous section, we completed a discussion on relations and functions. In this chapter, we will see inverse trigonometric functions.

Some basics can be written in 8 steps:
1. Consider the sine function:
f(x) = sin x
• The input x can be any real number. So the domain is R.
• The output lies in the interval [-1,1]. So the codomain is [-1,1]
• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the sine function. It is shown again in fig.18.1 below:

Fig.18.1

2. Let us check whether the sine function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\sin \left(\frac{\pi}{2} \right)~=~1 $
• Let input x = $\frac{5 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{5 \pi}{2} \right)~=~\sin \left(\frac{5 \pi}{2} \right)~=~1 $   
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\sin \left(\frac{-3 \pi}{2} \right)~=~1 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the sine function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.2 below:

Fig.18.2

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain [-1,1].
• There will be always a x value in $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, which will satisfy the equation y = sin x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{sine}:~ \left[\frac{-\pi}{2}, \frac{\pi}{2} \right]~\to~[-1,1]$, defined as f(x) = sin x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ and range is [-1,1]
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = sin x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
(iv) The range of f will be the domain of g. So the domain of g is [-1,1]
(iv) The inverse of sine function is denoted as sin^-1. So we can define the inverse function as:
$\sin^{-1}:~ [-1,1]~\to~\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, defined as x = g(y) = sin^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{2}$
• Then we get an equation: $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\sin^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{2}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\sin x$
• In our present case, it is:
$y~=~\frac{1}{2}~=~\sin x$
(iii) So we have a trigonometric equation:
$\sin x~=~\frac{1}{2}$
• 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, we do not need to write many steps. We already know that, $\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}$
• So we can write:
$x ~=~\frac{\pi}{6}$

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The above 8 steps help us to understand the basics about inverse trigonometric functions. Now we will see a few more details. It can be written in 6 steps:
1. We saw that, the sine function is not a one-one function. But to make it one-one, we restricted the domain to $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$.
2. There are other possible “restricted domains” available.
• $\left[\frac{-3 \pi}{2}, \frac{- \pi}{2} \right]$ is shown in magenta color in fig.18.3 below:

Fig.18.3

• $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ is shown in cyan color in fig,18.3 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the sin^-1 function.
    ♦ The restricted domain $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$ gives the principal branch of the  sin^-1 function.

4. In class 11, we plotted the sine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , \frac{1}{2}  \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , \frac{1}{2}  \right)$ in the set f will become $\left(\frac{1}{2} , \frac{\pi}{6}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.4 below:

Fi.18.4

(v) We see that, the graph is smaller in width but larger in height.
• The graph is smaller in width because:
    ♦ Values to the left of -1 cannot be used as input values.    
    ♦ Also, values to the right of 1 cannot be used as input values.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ principal branch.
• The cyan curve is related to the $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ branch. 
• The magenta curve is related to the $\left[\frac{-3 \pi}{2}, \frac{\pi}{2} \right]$ branch.

5. Consider any function f for which the inverse g exists.
• The graph of g will be the mirror image of the graph of f.
   ♦ The line with equation y=x will be the mirror line.
• We will prove this in the case of sine function. It can be done in 5 steps:
(i) Consider the grid of the graph in fig.18.4 above.
• The x-axis is divided into equal parts, with each part equal to 1 unit.
• The y-axis is also divided into equal parts, with each part equal to π/2 units.
• π/2 = 3.14/2 = 1.57, which is larger than 1. This is the reason why we see larger divisions along the y-axis.
• For the next graph in fig.18.5 below, we will use π/2 for both the axes.

Fig.18.5

(ii) In fig.18.5 above,
    ♦ The sine function is drawn in red color.
    ♦ The sin^-1 function is drawn in green color.
    ♦ The mirror line f(x) = x is drawn in white color.
(iii) Any two convenient points P and Q are marked on the sine function.
• We need to show that, their mirror images P' and Q' lie on the sin^-1 function.
(iv) First we consider point P.
• From P, drop a perpendicular to the mirror line. Let P₁ be the foot of the perpendicular.
• Extend PP₁ so as to intersect with the green curve. Let P' be the point of intersection.
• We can measure and see that, PP₁ = P₁P'.
• Here we see two facts:
    ♦ PP' is perpendicular to the mirror line.
    ♦ PP₁ = P₁P'.
So P' is the mirror image of P.
• We also see that, the coordinates of P' is the interchanged version of P.
(v) Next we consider point Q.
• From Q, drop a perpendicular to the mirror line. Let Q₁ be the foot of the perpendicular.
• Extend QQ₁ so as to intersect with the green curve. Let Q' be the point of intersection.
• We can measure and see that, QQ₁ = Q₁Q'.
• Here we see two facts:
    ♦ QQ' is perpendicular to the mirror line.
    ♦ QQ₁ = Q₁Q'.
So Q' is the mirror image of Q.
• We also see that, the coordinates of Q' is the interchanged version of Q.
• In this way, we can prove that, any point on sin^-1 function is the mirror image of a corresponding point in the sine function.

6. The sin^-1 function is also known as arcsine function.

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We have completed a basic discussion on the sin^-1 function. In the next section, we will see cos^-1 function.

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