If $\sin^{-1}x = y$, then (a) 0 \le y \le \pi (b) -\frac{\pi}{2} < y < \frac{\pi}{2} (c) 0 < y < \pi (d) -\frac{\pi}{2} \le y \le \frac{\pi}{2}

If sin⁻¹x = y, then (a) 0 ≤ y ≤ π (b) -(π)/(2) < y – Class 12 Mathematics

Question of 108

\sin^{-1}x = y

(a) 0 \le y \le \pi

(b) -\frac{\pi}{2} < y < \frac{\pi}{2}

(d) -\frac{\pi}{2} \le y \le \frac{\pi}{2}

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2022MCQ· 1mImportance★★★★★

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Concept understanding — Inverse Sine Principal Value

Principal Value of Inverse Sine

The equation $\sin\theta = x$ has infinitely many solutions. If $\sin\theta = \tfrac12$ , then $\theta$ could be $\tfrac{\pi}{6}$ , $\tfrac{5\pi}{6}$ , $\tfrac{13\pi}{6}$ , and so on. To make $\sin^{-1}$ a genuine function, we must agree on one answer. That agreed-upon answer is called the principal value.

Restricting the range

Sine is one-to-one on $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ , and on this interval it climbs through every value from $-1$ to $1$ exactly once. So we define:

\sin^{-1} x = \theta \quad\text{means}\quad \sin\theta = x \ \text{ and }\ \theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right].

Domain: $x \in [-1, 1]$ (sine never exceeds these values).
Principal value range: $\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ .

The principal value is the unique angle in this closed interval whose sine is $x$ .

Reading off values

$\sin^{-1}\left(\tfrac12\right) = \tfrac{\pi}{6}$ , since $\tfrac{\pi}{6} \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and $\sin\tfrac{\pi}{6} = \tfrac12$ .
$\sin^{-1}\left(-\tfrac12\right) = -\tfrac{\pi}{6}$ — the answer can be negative, because the range dips to $-\tfrac{\pi}{2}$ .
$\sin^{-1}(1) = \tfrac{\pi}{2}$ and $\sin^{-1}(0) = 0$ .

Note

$\sin^{-1} x$ is an angle, not a ratio, and it is not $\tfrac{1}{\sin x}$ (that is $\csc x$ ). The $-1$ here means "inverse", not a power.

The classic trap: $\sin^{-1}(\sin x)$

Many students write $\sin^{-1}(\sin x) = x$ automatically. This is true only when $x$ already lies in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ . Otherwise you must return the principal value — the equivalent angle inside the range. …

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The principal-value branch of $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ .

…

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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 51 on this concept.

CBSE 2026Set A1 markMCQ
$\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) =$
(a) $\pi$
(b) $\frac{2\pi}{3}$
(c) $\frac{3\pi}{4}$
(d) $\frac{\pi}{2}$
›Reveal solutionSolution
The sum equals $\frac{3\pi}{4}$ .

Evaluate each principal value:

$\tan^{-1}(1) = \frac{\pi}{4}$ .

$\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ (range $[0,\pi]$ ).

$\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ (range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ ).

Add, using LCD 12: …
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CBSE 2026Set ANNUAL1 markMCQ
$\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$
(a) $\frac{13\pi}{6}$
(b) $\frac{7\pi}{6}$
(c) $\frac{5\pi}{6}$
(d) $\frac{\pi}{6}$
›Reveal solutionSolution

$\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\frac{\pi}{6}$ .

The principal value branch of $\cos^{-1}$ is $[0,\pi]$ . To evaluate $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$ we must first reduce $\frac{13\pi}{6}$ to an angle whose cosine is the same and which lies in $[0,\pi]$ .

$\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$

Since cosine has period $2\pi$ :

$\cos\frac{13\pi}{6}=\cos\left(2\pi+\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$
(a) $\pi/4$
(b) $\pi/6$
(c) $\pi/3$
(d) $\pi/2$
›Reveal solutionSolution

The principal value branch of $\sin^{-1}$ is $[-\pi/2,\pi/2]$ ; the angle in this range whose sine is $1/\sqrt2$ is $\pi/4$ .

We need $\theta \in [-\pi/2,\pi/2]$ such that $\sin\theta = \dfrac{1}{\sqrt2}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. sin⁻¹(sin(2π/3)) is equal to$
(a) 2π/3
(b) π/3
(c) −π/3
(d) None of the above
›Reveal solutionSolution

$2\pi/3$ lies outside the principal range $[-\pi/2,\pi/2]$ of $\sin^{-1}$ , so we must find the angle inside that range with the same sine value.

$\sin(2\pi/3) = \sin(\pi-\pi/3) = \sin(\pi/3) = \dfrac{\sqrt3}{2}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$
(a) $-\frac{\pi}{4}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{4}$
›Reveal solutionSolution

$\sin^{-1}\dfrac{1}{\sqrt2}=\dfrac\pi4$ .

The principal value lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$ . Since $\sin\dfrac\pi4=\dfrac{1}{\sqrt2}$ , th …

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CBSE 2026Set ANNUAL1 mark
$\sin^{-1} x$
›Reveal solutionSolution

The principal-value range of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ .

To make sine invertible it is restricted to $\left[-\dfrac\pi2,\dfrac\pi2\right]$ , on which it is one-one and onto $[-1,1]$ ; hence this is the pr …

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CBSE 2026Set ANNUAL1 markMCQ
$\sin\left[\dfrac{\pi}{2}-\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\right]$
(a) $1$
(b) $\dfrac{1}{3}$
(c) $-1$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

$\sin\!\left[\tfrac{\pi}{2}-\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right]=\cos\!\left(\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right)=\tfrac12$ .

Step 1: $\sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta$ , so the expression equals $\cos\!\left(\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right)$ .

…

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CBSE 2025Set 65/2/11 markMCQ
$y = \sin^{-1}x$
›Reveal solutionSolution
For the inverse sine function, the principal value range is $[-\pi/2, \pi/2]$ . When $x$ is restricted to $[-1, 0]$ , $y$ takes values from $-\pi/2$ up to $0$ , including both endpoints. The correct answer is (B).

The key to this problem is understanding what "principal value" means for inverse trigonometric functions. Unlike regular sine, which is periodic and not one-to-one, $\sin^{-1}x$ (also written as $\arcsin x$ ) is defined as the inverse of the sine function only on a carefully chosen interval where sine is one-to-one. That interval is $[-\pi/2, \pi/2]$ .

So by definition, for any $x$ in $[-1, 1]$ , the value $y = \sin^{-1}x$ is always the unique angle in $[-\pi/2, \pi/2]$ whose sine is $x$ . This is the principal value branch — it's not a choice; it's the definition.

Now the question gives you a further restriction: $x$ is only between $-1$ and $0$ . You're being asked: as $x$ runs through that half of the domain, what part of the principal range does $y$ cover?

Let's work through it.

Recall the principal range of $\sin^{-1}x$ .

The output $y$ always lies in $[-\pi/2, \pi/2]$ . That's the full range for the full domain $[-1, 1]$ .

Identify the endpoints for $x$ in $[-1, 0]$ .

When $x = -1$ , $y = \sin^{-1}(-1)$ . What angle in $[-\pi/2, \pi/2]$ has sine equal to $-1$ ? That's $-\pi/2$ .

When $x = 0$ , $y = \sin^{-1}(0)$ . The angle in $[-\pi/2, \pi/2]$ with sine $0$ is $0$ .

Check monotonicity.

The function $\sin^{-1}x$ is strictly increasing on $[-1, 1]$ . So as $x$ increases from $-1$ to $0$ , $y$ increases from $-\pi/2$ to $0$ . Since the function is continuous and strictly increasing, it hits every value between $-\pi/2$ and $0$ .

Are the endpoints included? …
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CBSE 2025Set 65/4/11 markMCQ
$\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$
›Reveal solutionSolution
To find the principal value of $\sin^{-1}(\sin \theta)$ , we must ensure the angle $\theta$ lies within the principal value range of $\sin^{-1}(x)$ , which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . By adjusting the given angle $-\frac{10\pi}{3}$ to an equivalent angle within this range, we find the principal value is $\boxed{\frac{\pi}{3}}$ .

The problem asks for the principal value of $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$ . This involves understanding the definition of the inverse sine function and its principal value branch.

The inverse sine function, $\sin^{-1}(x)$ (also written as $\arcsin(x)$ ), gives an angle whose sine is $x$ . For $\sin^{-1}(x)$ to be a function, its range must be restricted. By convention, the principal value branch of $\sin^{-1}(x)$ is defined such that its output angle lies in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

This means that for an expression like $\sin^{-1}(\sin \theta)$ , the result is not always simply $\theta$ . It is $\theta$ only if $\theta$ itself is already within the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . If $\theta$ is outside this range, we need to find an equivalent angle $\alpha$ such that $\sin \alpha = \sin \theta$ and $\alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Then, $\sin^{-1}(\sin \theta) = \sin^{-1}(\sin \alpha) = \alpha$ .

Let's apply this concept step-by-step:

Identify the principal value range for $\sin^{-1}(x)$ :

The principal value of $\sin^{-1}(x)$ must lie in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . This is equivalent to angles from $-90^\circ$ to $90^\circ$ .

Analyze the inner angle:

The given angle inside the sine function is $-\frac{10\pi}{3}$ .

We need to evaluate $\sin\left(-\frac{10\pi}{3}\right)$ .

To simplify this, we can add or subtract multiples of $2\pi$ (a full rotation) to find a coterminal angle that is easier to work with.

$-\frac{10\pi}{3} = -\frac{10\pi}{3} + 4\pi$ (since $4\pi = \frac{12\pi}{3}$ )

$= \frac{-10\pi + 12\pi}{3} = \frac{2\pi}{3}$ .

So, $\sin\left(-\frac{10\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right)$ .

Watch out

A common mistake is to directly write $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right) = -\frac{10\pi}{3}$ . This is incorrect because $-\frac{10\pi}{3}$ (which is $-600^\circ$ ) is not in the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (which is $[-90^\circ, 90^\circ]$ ).

Find an equivalent angle within the principal value range:

Now we need to find the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$ . …
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CBSE 2025Set X11 markMCQ
$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$
(a) $\dfrac{\pi}{2}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{4}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Principal value of an inverse-sine — correct option is (c). …

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CBSE 2025Set E1 markMCQ
$\sin\left(\sin^{-1}\frac{2\pi}{3}\right) + \tan^{-1}\left(\tan\frac{3\pi}{4}\right) =$
(a) $\frac{17\pi}{12}$
(b) $\frac{5}{12}\pi$
(c) $\frac{\pi}{12}$
(d) $-\frac{\pi}{12}$
›Reveal solutionSolution

Apply the inverse cancellation and reduce the second term to its principal range; result $\frac{5\pi}{12}$ .

First term: Using $\sin(\sin^{-1}\theta)=\theta$ as intended by the paper, $\sin\left(\sin^{-1}\frac{2\pi}{3}\right)=\frac{2\pi}{3}$ . (Strictly, $\frac{2\pi}{3}>1$ is outside the domain of $\sin^{-1}$ , but the question intends the direct cancellation.)

…

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CBSE 2025Set ANNUAL1 mark
$\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$
›Reveal solutionSolution

The principal value branch of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ .

Since $\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$ and $\dfrac{\pi}{3}\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ , we …

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