If a function $f : X \to Y$ defined as $f(x) = y$ is one-one and onto, then we can define a unique function $g : Y \to X$ such that $g(y) = x$, where $x \in X$ and $y = f(x)$, $y \in Y$. Function $g$ is called the inverse of function $f$. The domain of sine function is $\mathbb{R}$ and function $\sin : \mathbb{R} \to \mathbb{R}$ is neither one-one nor onto. Let sine function be defined from set $A$ to $[-1, 1]$ such that inverse of sine function exists, i.e., $\sin^{-1}x$ is defined from $[-1, 1]$ to $A$. On the basis of the above information, answer the following questions: (i) If $A$ is the interval other than principal value branch, give an example of one such interval. (ii) If $\sin^{-1}(x)$ is defined from $[-1, 1]$ to its principal value branch, find the value of $\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1)$. (iii)(a) Draw the graph of $\sin^{-1}x$ from $[-1, 1]$ to its principal value branch.

![Figure — Part (iii)(a) asks to draw the graph of sin-inverse x on -1,1 to its principal branch; the catalog figure sh](/figures/inverse-trigonometric-functions/graphs-sin-line-fig-2-1.svg) > [!TLDR] > To define an inverse for the sine function, its domain must be restricted to an interval where it is one-one and onto, such as $[-\pi/2, \pi/2]$ (the principal value branch) or other similar intervals like $[\pi/2, 3\pi/2]$. Using the principal value branch, $\sin^{-1}(-1/2) - \sin^{-1}(1)$ evaluates to $-\frac{2\pi}{3}$, and its graph is a reflection of $\sin x$ on $[-\pi/2, \pi/2]$ across $y=x$. The concept of an inverse function is fundamental in mathematics. For a function $f: X \to Y$ to have an inverse, it must be **one-one (injective)** and **onto (surjective)**. * **One-one** means that distinct elements in the domain $X$ map to distinct elements in the codomain $Y$. In simpler terms, if $f(x_1) = f(x_2)$, then $x_1 = x_2$. Graphically, this means any horizontal line intersects the graph at most once. * **Onto** means that every element in the codomain $Y$ is the image of at least one element in the domain $X$. In other words, the range of the function must be equal to its codomain. The standard sine function, $\sin: \mathbb{R} \to \mathbb{R}$, is neither one-one nor onto. * It's **not one-one** because, for example, $\sin(0) = 0$ and $\sin(\pi) = 0$. Many different $x$ values map to the same $y$ value. * It's **not onto** because its range is $[-1, 1]$, which is a proper subset of its codomain $\mathbb{R}$. To define an inverse for the sine function, we must restrict its domain such that it becomes one-one, and we must restrict its codomain to its range, $[-1, 1]$, to make it onto. The problem statement defines $\sin^{-1}x$ as a function from $[-1, 1]$ to a set $A$, where $A$ is a restricted domain of the original sine function. Let's address each part of the question. ### (i) If $A$ is the interval other than principal value branch, give an

If a function f : X → Y defined as f(x) = y is one-one and onto, then we can define a unique function g : Y →

Question

f : X \to Y

(i) If

A

is the interval other than principal value branch, give an example of one such interval.

(ii) If

\sin^{-1}(x)

is defined from

[-1, 1]

to its principal value branch, find the value of

\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1)

. (iii)(a) Draw the graph of

\sin^{-1}x

from

[-1, 1]

to its principal value branch.

CBSECBSE Class XII Board 2024Subjective· 4mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Figure — Part (iii)(a) asks to draw the graph of sin-inverse x on -1,1 to its principal branch; the catalog figure sh

To define an inverse for the sine function, its domain must be restricted to an interval where it is one-one and onto, such as $[-\pi/2, \pi/2]$ (the principal value branch) or other similar intervals like $[\pi/2, 3\pi/2]$ . Using the principal value branch, $\sin^{-1}(-1/2) - \sin^{-1}(1)$ evaluates to $-\frac{2\pi}{3}$ , and its graph is a reflection of $\sin x$ on $[-\pi/2, \pi/2]$ across $y=x$ .

The concept of an inverse function is fundamental in mathematics. For a function $f: X \to Y$ to have an inverse, it must be one-one (injective) and onto (surjective).

One-one means that distinct elements in the domain $X$ map to distinct elements in the codomain $Y$ . In simpler terms, if $f(x_1) = f(x_2)$ , then $x_1 = x_2$ . Graphically, this means any horizontal line intersects the graph at most once.
Onto means that every element in the codomain $Y$ is the image of at least one element in the domain $X$ . In other words, the range of the function must be equal to its codomain.

The standard sine function, $\sin: \mathbb{R} \to \mathbb{R}$ , is neither one-one nor onto.

It's not one-one because, for example, $\sin(0) = 0$ and $\sin(\pi) = 0$ . Many different $x$ values map to the same $y$ value.
It's not onto because its range is $[-1, 1]$ , which is a proper subset of its codomain $\mathbb{R}$ .

To define an inverse for the sine function, we must restrict its domain such that it becomes one-one, and we must restrict its codomain to its range, $[-1, 1]$ , to make it onto. The problem statement defines $\sin^{-1}x$ as a function from $[-1, 1]$ to a set $A$ , where $A$ is a restricted domain of the original sine function.

Let's address each part of the question.

(i) If $A$ is the interval other than principal value branch, give an example of one such interval.

Understanding the requirement for $A$ : For $\sin: A \to [-1, 1]$ to have an inverse, the function must be one-one and onto within this restricted domain $A$ . This means that over the interval $A$ , the sine function must be strictly monotonic (either strictly increasing or strictly decreasing) and its range must cover the entire interval $[-1, 1]$ .
The Principal Value Branch: The most commonly chosen interval for $A$ is $[-\pi/2, \pi/2]$ . In this interval, $\sin x$ is strictly increasing from $-1$ to $1$ , making it one-one, and its range is exactly $[-1, 1]$ , making it onto. This is called the principal value branch.
Finding other intervals: Due to the periodic nature of the sine function, there are infinitely many such intervals. We need an interval of length $\pi$ where the sine function covers its full range $[-1, 1]$ exactly once.
- Consider the interval $[\pi/2, 3\pi/2]$ . In this interval, $\sin x$ decreases from $1$ (at $x=\pi/2$ ) to $-1$ (at $x=3\pi/2$ ). It is strictly decreasing and covers the range $[-1, 1]$ exactly once.
- Another example is $[-3\pi/2, -\pi/2]$ . Here, $\sin x$ decreases from $1$ to $-1$ .
- In general, any interval of the form $[(2n-1)\pi/2, (2n+1)\pi/2]$ for an integer $n$ would work.
Example: An example of an interval for $A$ other than the principal value branch is $[\pi/2, 3\pi/2]$ .

(ii) If $\sin^{-1}(x)$ is defined from $[-1, 1]$ to its principal value branch, find the value of $\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1)$ .

Definition of $\sin^{-1}x$ : When $\sin^{-1}x$ is defined to its principal value branch, it means that for any $x \in [-1, 1]$ , $\sin^{-1}x$ is the unique angle $\theta \in [-\pi/2, \pi/2]$ such that $\sin\theta = x$ .
Evaluating $\sin^{-1}(-1/2)$ :
- We need to find an angle $\theta_1 \in [-\pi/2, \pi/2]$ such that $\sin(\theta_1) = -1/2$ .
- We know that $\sin(\pi/6) = 1/2$ . Since sine is an odd function ( $\sin(-\theta) = -\sin(\theta)$ ), we have $\sin(-\pi/6) = -\sin(\pi/6) = -1/2$ .
- The angle $-\pi/6$ lies within the principal value branch $[-\pi/2, \pi/2]$ .
- Therefore, $\sin^{-1}(-1/2) = -\pi/6$ .
Evaluating $\sin^{-1}(1)$ :
- We need to find an angle $\theta_2 \in [-\pi/2, \pi/2]$ such that $\sin(\theta_2) = 1$ .
- We know that $\sin(\pi/2) = 1$ .
- The angle $\pi/2$ lies within the principal value branch $[-\pi/2, \pi/2]$ .
- Therefore, $\sin^{-1}(1) = \pi/2$ .
Calculating the difference:
- Now, substitute the values we found:

\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1) = -\dfrac{\pi}{6} - \dfrac{\pi}{2}

*   To subtract, find a common denominator:

-\dfrac{\pi}{6} - \dfrac{3\pi}{6} = -\dfrac{\pi + 3\pi}{6} = -\dfrac{4\pi}{6} = -\dfrac{2\pi}{3}

> [!WARNING]
> A common mistake is to forget the principal value branch restriction and give an angle like $7\pi/6$ for $\sin^{-1}(-1/2)$. While $\sin(7\pi/6) = -1/2$, $7\pi/6$ is not in $[-\pi/2, \pi/2]$. Always ensure the output of an inverse trigonometric function is within its defined range.

(iii)(a) Draw the graph of $\sin^{-1}x$ from $[-1, 1]$ to its principal value branch. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q.The value of \tan^{-1}\left(\frac{1}{x}\right) will be:$
(a) \tan x
(b) \cot x
(c) \cot^{-1}\frac{1}{x}
(d) \cot^{-1} x
›Reveal solutionSolution

$\tan^{-1}(1/x)$ and $\cot^{-1}x$ are the same angle for positive $x$ , because tangent and cotangent are reciprocal functions.

Concept: For $x>0$ , the standard identity is

$\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x$ …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{7\pi}{6}$
(c) $\dfrac{1}{\sqrt{3}}$
(d) $\dfrac{5\pi}{6}$
›Reveal solutionSolution

Use periodicity of $\tan$ to bring the angle into the principal range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ of $\tan^{-1}$ .

Since $\tan$ has period $\pi$ : $\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6}-\pi\right) = \tan\dfrac{\pi}{6}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$-1 \le x \le 1$
›Reveal solutionSolution

$\sin$ and $\sin^{-1}$ are inverse functions of each other on $[-1,1]$ , so applying one after the other returns the original input.

For $-1 \le x \le 1$ , $\sin^{-1}x$ is defined and gives an angle $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ such that $\sin\theta = x$ . Applying $\sin$ to this angle simply recovers $x$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) + \tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{12}$
(c) $\dfrac{5\pi}{12}$
(d) $\dfrac{7\pi}{12}$
›Reveal solutionSolution

Bring each angle into the principal-value range using $\sin^{-1}(\sin\theta)=\pi-\theta$ and $\tan^{-1}(\tan\theta)=\theta-\pi$ for $\theta$ in the second quadrant.

Term 1: $\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right)$

The range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ . Since $\dfrac{2\pi}{3}\in\left(\dfrac{\pi}{2},\pi\right)$ lies outside this range, use

$\sin^{-1}(\sin\theta) = \pi-\theta \quad \text{for } \theta\in\left(\tfrac{\pi}{2},\pi\right)$

$\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) = \pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}$

Term 2: $\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$x \in [-1, 1],\ \sin^{-1}(-x) =$
(a) $-\sin^{-1} x$
(b) $\sin^{-1} x$
(c) $-\cos^{-1} x$
(d) $\cos^{-1} x$
›Reveal solutionSolution

$\sin^{-1}$ is odd: $\sin^{-1}(-x)=-\sin^{-1}x$ .

For $x\in[-1,1]$ , the inverse sine is an odd function, so

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$\sin^{-1}\left(\sin \dfrac{2\pi}{3}\right)$
›Reveal solutionSolution

Since $\dfrac{2\pi}{3}$ is outside the principal range of $\sin^{-1}$ , first rewrite $\sin\dfrac{2\pi}{3}$ as the sine of an angle inside $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ .

$\sin\dfrac{2\pi}{3} = \sin\left(\pi - \dfrac{2\pi}{3}\right) = \sin\dfrac{\pi}{3}$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{\pi}{4}$
(c) $\dfrac{\pi}{3}$
(d) $\dfrac{\pi}{2}$
›Reveal solutionSolution

Reduce the angle into the principal branch $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ using the period $\pi$ of $\tan$ .

$\tan^{-1}(\tan\theta)=\theta$ holds only when $\theta$ already lies in the principal value branch $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

Here $\theta=\dfrac{7\pi}{6}$ , which does not lie in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ . Since $\tan$ has period $\pi$ :

$\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6}-\pi\right) = \tan\dfrac{\pi}{6}$ …

Start your 14-day free trial to see the full solution →
CBSE 2023Set E1 markMCQ
$\sin\left(\sin^{-1}\dfrac{1}{2}\right) =$
(a) $1$
(b) $\dfrac{1}{2}$
(c) $\dfrac{\sqrt{3}}{2}$
(d) $0$
›Reveal solutionSolution

$\sin\left(\sin^{-1}\dfrac{1}{2}\right)=\dfrac{1}{2}$ .

Since $\dfrac{1}{2}\in[-1,1]$ , $\sin(\sin^{-1}x)=x$ holds:

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$\sin^{-1}\dfrac13 + \cos^{-1}\dfrac13$
(a) $0$
(b) $1$
(c) $\dfrac{\pi}{2}$
(d) $\pi$
›Reveal solutionSolution

Use the standard identity $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ for any $x\in[-1,1]$ .

Here $x=\dfrac13\in[-1,1]$ , so directly:

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$\sin^{-1}x = \dfrac{\pi}{5}$
(a) $\dfrac{9\pi}{10}$
(b) $\dfrac{7\pi}{10}$
(c) $\dfrac{5\pi}{10}$
(d) $\dfrac{3\pi}{10}$
›Reveal solutionSolution

Use $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ to get $\cos^{-1}x$ directly from $\sin^{-1}x$ .

Given $\sin^{-1}x=\dfrac{\pi}{5}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$\sin^{-1}\dfrac{x}{5}+\sec^{-1}\dfrac{5}{4}=\dfrac{\pi}{2}$
›Reveal solutionSolution

Rewrite $\sec^{-1}$ in terms of $\cos^{-1}$ and use $\sin^{-1}z+\cos^{-1}z=\dfrac{\pi}{2}$ .

$\sec^{-1}\dfrac54=\cos^{-1}\dfrac45$ (since $\sec^{-1}y=\cos^{-1}\frac1y$ ).

Given: $\sin^{-1}\dfrac x5+\cos^{-1}\dfrac45=\dfrac{\pi}{2}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$
(a) $x^{1/3}$
(b) $x^3$
(c) $x$
(d) None of the above
(a) $\dfrac{7\pi}{6}$
(b) $\dfrac{5\pi}{6}$
(c) $\dfrac{\pi}{3}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

$\cos^{-1}(\cos\theta)$ returns the angle in the principal range $[0,\pi]$ having the same cosine; $\tfrac{7\pi}{6}$ lies outside it.

The principal-value range of $\cos^{-1}$ is $[0,\pi]$ , but $\dfrac{7\pi}{6} \notin [0,\pi]$ . Compute the cosine:

$\cos\frac{7\pi}{6} = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt3}{2}.$

We need the angle $\theta \in [0,\pi]$ with $\cos\theta = -\dfrac{\sqrt3}{2}$ . That angle is

$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.$

Hence $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right) = \dfrac{5\pi}{6}$ , option (b).

…

Start your 14-day free trial to see the full solution →

Concept understanding — Inverse Function Relationship

Inverse Function Relationship

The defining equations

When does an inverse exist?

The geometry

Why the restriction bites — the trig case

(i) If $A$ is the interval other than principal value branch, give an example of one such interval.

(ii) If $\sin^{-1}(x)$ is defined from $[-1, 1]$ to its principal value branch, find the value of $\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1)$ .

(iii)(a) Draw the graph of $\sin^{-1}x$ from $[-1, 1]$ to its principal value branch. …

(i) If AAA is the interval other than principal value branch, give an example of one such interval.

(ii) If sin⁡−1(x)\sin^{-1}(x)sin−1(x) is defined from [−1,1][-1, 1][−1,1] to its principal value branch, find the value of sin⁡−1(−12)−sin⁡−1(1)\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1)sin−1(−21​)−sin−1(1).

(iii)(a) Draw the graph of sin⁡−1x\sin^{-1}xsin−1x from [−1,1][-1, 1][−1,1] to its principal value branch. …

🎓Unlock everything free for 14 days

(i) If $A$ is the interval other than principal value branch, give an example of one such interval.

(ii) If $\sin^{-1}(x)$ is defined from $[-1, 1]$ to its principal value branch, find the value of $\sin^{-1}\left(-\dfrac{1}{2}\right) - \sin^{-1}(1)$ .

(iii)(a) Draw the graph of $\sin^{-1}x$ from $[-1, 1]$ to its principal value branch. …

Unlock everything free for 14 days