(a) Evaluate : 3 sin⁻¹(frac1√(2)) + 2 cos⁻¹(frac√(3)2) + cos⁻¹(0) OR (b) Draw the graph of f(x) = sin⁻¹ x, x ∈

Question

3 \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + 2 \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}(0)

(OR)

(b) Draw the graph of

f(x) = \sin^{-1} x

x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]

. Also, write range of

f(x)

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

Take $\sin x=\tfrac12$ . It has infinitely many solutions: $x=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},-\tfrac{7\pi}{6},\dots$ — every angle whose sine is $\tfrac12$ . So if we want an inverse that returns a single angle for $\sin^{-1}(0.5)$ , we must first agree on one angle to report. A function is allowed only one output per input, and $\sin x$ over all of $\mathbb{R}$ is many-to-one — it fails the horizontal line test and cannot be inverted as it stands.

The idea: restrict to one clean interval

For each trigonometric ratio we restrict the angle to a single standard interval on which the function is one-to-one while still covering its entire range exactly once. On that interval the inverse becomes well-defined and single-valued. That interval — the set of angles the inverse is allowed to return — is the principal value branch (also called the principal value domain).

The interval is chosen to be strictly monotonic, to hit every output once, and to sit as close to $0$ as possible. For sine that is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ , where $\sin$ increases from $-1$ to $1$ .

The principal value branch of an inverse trig function is the interval of angles it returns — the restricted interval on which the original ratio is one-to-one and onto its range.

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

Why the intervals differ …

Part (b)Concept understanding — Inverse Trigonometric Graphs

Inverse Trigonometric Graphs

A trigonometric function such as $\sin x$ takes an angle and returns a ratio. An inverse trig function reverses this: given the ratio, it returns the angle. Their graphs are the trig graphs reflected across the line $y = x$ — but only after a careful restriction.

Why we must restrict first

On its full domain $\sin x$ repeats forever, so $\sin x = 0.5$ has infinitely many solutions and sine fails the horizontal-line test. To invert it we keep only a piece where it is one-to-one. That restricted piece becomes the domain of the inverse; its outputs become the range.

Important

The inverse graph is the mirror image of the restricted original across $y = x$ : every point $(a,b)$ becomes $(b,a)$ .

The three graphs

$\sin^{-1} x$ — restrict $\sin x$ to $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ (strictly increasing).

Domain $[-1,1]$ , range $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ . An S-shaped curve from $(-1,-\tfrac{\pi}{2})$ up through $(0,0)$ to $(1,\tfrac{\pi}{2})$ .

$\cos^{-1} x$ — restrict $\cos x$ to $[0,\pi]$ (strictly decreasing).

Domain $[-1,1]$ , range $[0,\pi]$ . Falls from $(-1,\pi)$ through $(0,\tfrac{\pi}{2})$ to $(1,0)$ .

$\tan^{-1} x$ — restrict $\tan x$ to $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ .

Domain $(-\infty,\infty)$ , range $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ . Passes through $(0,0)$ with horizontal asymptotes $y = \pm\tfrac{\pi}{2}$ .

Function	Domain	Range
$\sin^{-1} x$	$[-1,1]$	$[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$(-\infty,\infty)$	$(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$

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Figure — Graph of y=sin-inverse x for x in -1/sqrt2, 1/sqrt2

The sum of principal values is $\dfrac{19\pi}{12}$ .
The graph of $\sin^{-1}x$ on $\left[-\tfrac{1}{\sqrt2},\tfrac{1}{\sqrt2}\right]$ is the increasing principal branch through the origin, with range $\left[-\tfrac{\pi}{4},\tfrac{\pi}{4}\right]$ .

Part (a)

Every inverse trig function returns the angle from its principal range. Evaluate each term:

$\sin^{-1}\dfrac{1}{\sqrt2}$ : since $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}$ and $\dfrac{\pi}{4}\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ , we get $\dfrac{\pi}{4}$ ; so $3\cdot\dfrac{\pi}{4}=\dfrac{3\pi}{4}$ .
$\cos^{-1}\dfrac{\sqrt3}{2}$ : since $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$ and $\dfrac{\pi}{6}\in[0,\pi]$ , we get $\dfrac{\pi}{6}$ ; so $2\cdot\dfrac{\pi}{6}=\dfrac{\pi}{3}$ .
$\cos^{-1}(0)=\dfrac{\pi}{2}$ .

Add over common denominator $12$ :

$\frac{3\pi}{4}+\frac{\pi}{3}+\frac{\pi}{2}=\frac{9\pi}{12}+\frac{4\pi}{12}+\frac{6\pi}{12}=\frac{19\pi}{12}.$ …

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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 69 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$2\cos^{-1}x = y$
›Reveal solutionSolution
The range of $\cos^{-1}x$ is $[0,\pi]$ , so multiplying by 2 gives $y = 2\cos^{-1}x$ a range of $[0,2\pi]$ . The correct option is (C).

Concept and Intuition

The key to this problem lies entirely in understanding the range of the inverse cosine function. $\cos^{-1}x$ (also written as $\arccos x$ ) is defined as the angle whose cosine is $x$ , and by convention, that angle is always taken from the interval $[0, \pi]$ . This is not arbitrary — it's the standard principal value branch that makes the function one-to-one and therefore invertible.

Once you know that $\cos^{-1}x$ lives between $0$ and $\pi$ (inclusive), finding the range of $y = 2\cos^{-1}x$ is simply a matter of scaling that interval by a factor of 2. No tricky domain restrictions, no sign flips — just multiplication.

Watch out

A common mistake is to confuse the range of $\cos^{-1}x$ with that of $\sin^{-1}x$ (which is $[-\pi/2, \pi/2]$ ). Always recall: $\cos^{-1}x \in [0,\pi]$ , not $[-\pi/2,\pi/2]$ .

Step-by-step solution

Recall the range of $\cos^{-1}x$ The inverse cosine function $\cos^{-1}: [-1,1] \to [0,\pi]$ gives an output angle in radians. This means:

$0 \le \cos^{-1}x \le \pi \quad \text{for all } x \in [-1,1].$

Multiply the inequality by 2 Since 2 is positive, multiplying through preserves the direction of the inequalities:

$2 \cdot 0 \le 2\cos^{-1}x \le 2 \cdot \pi$
which simplifies to:
$0 \le y \le 2\pi.$

Check if every value in $[0,2\pi]$ is actually attained …
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CBSE 2026Set V11 markMCQ
$\tan^{-1} x$
(a) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(b) $(0, \pi)$
(c) $[-1, 1]$
(d) $(-\infty, \infty)$
›Reveal solutionSolution

The tangent function maps $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ onto all of $\mathbb{R}$ , so $\tan^{-1}x$ accepts every real $x$ ; answer (d).

The principal-branch tangent $\tan:\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\to\mathbb{R}$ is a bijection onto $\mathbb{R}$ . Its inverse $\tan^{-1}$ therefore has domain equal to the range of $\tan$ , namely all real …
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CBSE 2026Set CX1 mark
$\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$
›Reveal solutionSolution

$\tan^{-1}\sqrt3=\tfrac{\pi}{3}$ , $\sec^{-1}(-2)=\tfrac{2\pi}{3}$ , giving $-\tfrac{\pi}{3}$ .

Concept: Use the principal-value ranges: $\tan^{-1}\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $\sec^{-1}\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ .

$\tan^{-1}\sqrt3=\dfrac{\pi}{3}\qquad(\tan\tfrac{\pi}{3}=\sqrt3).$ …

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

$\sin^{-1}x$ is defined only where $\sin\theta=x$ has a solution, i.e. for $x\in[-1,1]$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$y = \cos^{-1} x$
(a) $0 \le y \le \pi$
(b) $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
(c) $-\pi \le y \le \pi$
(d) None of these
›Reveal solutionSolution

$\cos^{-1}x$ is defined so that its principal value always lies in $[0,\pi]$ .

The function $\cos x$ is one-one and onto from $[0,\pi]$ to $[-1,1]$ , so its inverse $\cos^{-1}x$ is defined on domain $[-1,1]$ with range (principal value …

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CBSE 2026Set ANNUAL1 markMCQ
$\tan^{-1}(-1)$
(a) $\dfrac{\pi}{4}$
(b) $-\dfrac{\pi}{4}$
(c) $\dfrac{3\pi}{4}$
(d) None of these
›Reveal solutionSolution

The principal value of $\tan^{-1}x$ always lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ .

We need $y$ such that $\tan y = -1$ and $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\cos^{-1} x$
(a) $[0, \pi]$
(b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
(c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
(d) None of these
›Reveal solutionSolution

The principal value branch of $\cos^{-1}x$ is $[0,\pi]$ by definition.

The function $\cos:[0,\pi]\to[-1,1]$ is a bijection, so its inverse $\cos^{-1}:[-1,1]\to[0,\pi]$ is defined w …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Principal value of cos⁻¹(1/2) is:$
(a) π/2
(b) π/3
(c) π/4
(d) π/6
›Reveal solutionSolution

The principal value of $\cos^{-1}x$ lies in $[0,\pi]$ , and $\cos\left(\frac{\pi}{3}\right) = \frac12$ .

We need $\theta \in [0,\pi]$ such that $\cos\theta = \dfrac12$ .

…

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

Find the angle in the principal-value range $[0,\pi]$ of $\cos^{-1}$ whose cosine equals $\dfrac{\sqrt3}{2}$ .

We need $\theta\in[0,\pi]$ (the principal-value branch of $\cos^{-1}$ ) such that

$\cos\theta = \frac{\sqrt3}{2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The principal value of cot⁻¹(-1/\sqrt3) is ................. .$
(a) π/3
(b) π/4
(c) 2π/3
(d) 4π/3
›Reveal solutionSolution

The principal range of $\cot^{-1}$ is $(0,\pi)$ ; find the angle in that range with cotangent $-1/\sqrt3$ .

We know $\cot(\pi/3)=1/\sqrt3$ . Since the given value is negative and the principal range of $\cot^{-1}$ is $(0,\pi)$ , the required angle lies in the second quadrant, where cotangent is negative.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The principal value of tan^{-1}(-\sqrt{3}) is:$
(a)
(i) \pi/3
(b)
(ii) -\pi/3
(c)
(iii) \pi/6
(d)
(iv) -\pi/6
›Reveal solutionSolution
$\tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}$ — option (ii).

Concept. The principal value of $\tan^{-1} x$ is the unique angle $\theta$ lying in the open interval $\left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$ such that $\tan\theta = x$ . This is the standard NCERT/CBSE principal-value branch that the UBSE Class-12 syllabus also follows.

Why this branch. Tangent is one-to-one on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , so exactly one angle there gives each real value.

Steps.

We need $\theta$ with $\tan\theta = -\sqrt{3}$ and $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ . …
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CBSE 20251 markMCQ
$y = \tan x$
›Reveal solutionSolution
The given graph shows a curve that starts at $(-1, \pi)$ , passes through $(0, \frac{\pi}{2})$ , and ends at $(1, 0)$ — this is exactly the principal branch of $y = \cos^{-1} x$ . The correct option is (C).

The key to identifying an inverse trigonometric graph lies in knowing the principal value branches — the restricted domains and ranges that make each inverse function one-to-one. For $\sin^{-1} x$ , the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ; for $\cos^{-1} x$ , it’s $[0, \pi]$ ; for $\tan^{-1} x$ , it’s $(-\frac{\pi}{2}, \frac{\pi}{2})$ . The graph given in the question (not shown here, but described in the options) has a starting point at $x = -1$ , $y = \pi$ , passes through $(0, \frac{\pi}{2})$ , and ends at $(1, 0)$ . That immediately tells you the range is $[0, \pi]$ and the domain is $[-1, 1]$ — the signature of $\cos^{-1} x$ .

Let’s walk through the reasoning step by step.

Eliminate the impossible options first.

Option (A) describes $y = \tan x$ , which is a trigonometric function, not its inverse. The question asks for the graph of the inverse, so (A) is out.

Option (B) describes $y = \sin^{-1} x$ with range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . Its graph starts at $(-1, -\frac{\pi}{2})$ and ends at $(1, \frac{\pi}{2})$ , passing through $(0,0)$ . The given graph passes through $(0, \frac{\pi}{2})$ , not $(0,0)$ , so (B) is incorrect.

Compare the two remaining options: (C) and (D).

Both claim the graph is $y = \cos^{-1} x$ , with the same starting and ending points and the same point $(0, \frac{\pi}{2})$ . They are identical in description. This is a trick — the question likely expects you to notice that (C) and (D) are word-for-word the same. In such multiple-choice questions, if two options are identical, they cannot both be correct; the correct one is the one that matches the graph. Since the description fits $\cos^{-1} x$ perfectly, the answer must be either (C) or (D). But because they are duplicates, the intended correct choice is (C) (often the first occurrence in such lists).

Confirm the properties of $\cos^{-1} x$ . …
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Q.(a) Evaluate : 3sin⁡−1(12)+2cos⁡−1(32)+cos⁡−1(0)3 \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + 2 \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}(0)3sin−1(2​1​)+2cos−1(23​​)+cos−1(0)

Part (a)

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Q.(a) Evaluate : $3 \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + 2 \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}(0)$

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