MathsISC Class 12 · Inverse Trigonometric Functions

Principal Values of Inverse Trigonometric Functions: Domain, Range and Graphs

Updated 15 October 2026 · Fact-checked

The principal value of an inverse trigonometric function is the one angle it returns from its fixed range: sin⁻¹x lies in [−π/2, π/2], cos⁻¹x in [0, π], tan⁻¹x in (−π/2, π/2) and cot⁻¹x in (0, π). To evaluate, find the angle in that branch whose trigonometric value equals the given number.

For −1/2, sin⁻¹ gives −π/6 in the right-half branch, but cos⁻¹ gives 2π/3 because its branch is the upper half, 0 to π.

Illustration drawn for QuestionBuddy.

Understand Principal Values of Inverse Trigonometric Functions

Trigonometric functions are periodic, so they are many-one: $sin \frac{π}{6} = sin \frac{5 π}{6} = \frac{1}{2}$ . A many-one function has no inverse. To define $sin^{- 1}$ we restrict $sin x$ to an interval on which it is one-one and takes every value it can, and invert only that piece. The chosen interval becomes the range of the inverse function, called the principal value branch, and the angle it returns is the principal value.

The standard choices are:

$sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$
$cos^{- 1} : [- 1, 1] \to [0, π]$
$tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2})$
$cot^{- 1} : R \to (0, π)$
$sec^{- 1} : R - (- 1, 1) \to [0, π] - {\frac{π}{2}}$
$cosec^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}] - {0}$

A useful way to remember them: $sin^{- 1}$ , $tan^{- 1}$ and $cosec^{- 1}$ have branches centred on $0$ (so they give negative angles for negative inputs), while $cos^{- 1}$ , $cot^{- 1}$ and $sec^{- 1}$ have branches from $0$ to $π$ (so they give obtuse angles for negative inputs). That is why $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ but $cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ .

Graphs. The graph of $y = sin^{- 1} x$ is the reflection of the piece of $y = sin x$ on $[- \frac{π}{2}, \frac{π}{2}]$ in the line $y = x$ . It runs from $(- 1, - \frac{π}{2})$ to $(1, \frac{π}{2})$ and is increasing. $y = cos^{- 1} x$ runs from $(- 1, π)$ down to $(1, 0)$ and is decreasing. $y = tan^{- 1} x$ is increasing with horizontal asymptotes $y = \pm \frac{π}{2}$ . The ISC 2025 paper asked candidates to identify the graph of $sin^{- 1} x$ and state its domain and range; the examiners noted that many could not write the domain and range.

Inverse of a function, and function of an inverse. Two compositions behave differently.

$sin (sin^{- 1} x) = x$ for every $x \in [- 1, 1]$ . The inner function gives an angle, and sine undoes it.
$sin^{- 1} (sin x) = x$ only for $x \in [- \frac{π}{2}, \frac{π}{2}]$ . Outside, you must find the angle in the branch with the same sine. For example, $sin^{- 1} (sin \frac{3 π}{5}) = π - \frac{3 π}{5} = \frac{2 π}{5}$ , because $sin (π - θ) = sin θ$ and $\frac{2 π}{5}$ is in the branch. Similarly $cos^{- 1} (cos \frac{7 π}{6}) = 2 π - \frac{7 π}{6} = \frac{5 π}{6}$ , and $tan^{- 1} (tan \frac{3 π}{4}) = \frac{3 π}{4} - π = - \frac{π}{4}$ .

Domains of composite arguments. For $sin^{- 1} (g (x))$ or $cos^{- 1} (g (x))$ to be defined, solve $- 1 \leq g (x) \leq 1$ . For $sec^{- 1}$ and $cosec^{- 1}$ , solve $∣ g (x) ∣ \geq 1$ .

Values of mixed expressions. To find $cos (sin^{- 1} \frac{3}{5})$ , let $θ = sin^{- 1} \frac{3}{5}$ , so $sin θ = \frac{3}{5}$ with $θ$ in the first quadrant, and read $cos θ = \frac{4}{5}$ off a 3-4-5 triangle. If the input is negative, the branch tells you the quadrant and hence the sign.

Principal Values of Inverse Trigonometric Functions: formulas to remember

sin⁻¹ and cos⁻¹: $sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}], cos^{- 1} : [- 1, 1] \to [0, π]$; Negative input: sin⁻¹ gives a negative angle, cos⁻¹ an obtuse one.
tan⁻¹ and cot⁻¹: $tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2}), cot^{- 1} : R \to (0, π)$; Open intervals: the end values are never reached.
sec⁻¹ and cosec⁻¹: $sec^{- 1} : ∣ x ∣ \geq 1 \to [0, π] - {\frac{π}{2}}, cosec^{- 1} : ∣ x ∣ \geq 1 \to [- \frac{π}{2}, \frac{π}{2}] - {0}$; The excluded angles are where sec and cosec are undefined.
Function of its inverse: $sin (sin^{- 1} x) = x, cos (cos^{- 1} x) = x, tan (tan^{- 1} x) = x$; Valid for every x in the domain of the inverse function.
Inverse of the function: $sin^{- 1} (sin x) = x only for x \in [- \frac{π}{2}, \frac{π}{2}]$; Similarly cos⁻¹(cos x) = x only on [0, π] and tan⁻¹(tan x) = x only on (−π/2, π/2).
Bringing angles into the branch: $sin (π - θ) = sin θ, cos (2 π - θ) = cos θ, tan (θ - π) = tan θ$; Use these to replace an angle outside the branch by one inside it.

How to solve Principal Values of Inverse Trigonometric Functions questions

Use this routine for every evaluation question on principal values.

1Write the principal value branch of the function you are evaluating.
2If the input is a standard value, recall the acute angle with that trigonometric value (for $\frac{1}{2}$ , $\frac{π}{6}$ for sine, $\frac{π}{3}$ for cosine).
3If the input is negative, place the angle using the branch: subtract for $sin^{- 1}, tan^{- 1}, cosec^{- 1}$ (give $- θ$ ); use $π - θ$ for $cos^{- 1}, cot^{- 1}, sec^{- 1}$ .
4For $sin^{- 1} (sin x)$ and similar, check whether $x$ is in the branch. If not, use a reduction formula to find the angle in the branch with the same value.
5For a value such as $cos (sin^{- 1} a)$ , put $θ = sin^{- 1} a$ , draw a right triangle, and use the branch to decide the sign.
6For domain questions, solve $- 1 \leq g (x) \leq 1$ (or $∣ g (x) ∣ \geq 1$ for sec and cosec) and write the answer as an interval or union of intervals.
7Check that your final angle really lies in the branch before you write it down.

Principal Values of Inverse Trigonometric Functions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Principal value of sin⁻¹

Suppose a phone app measures the slope of a ramp by reading its length along the surface and the height it rises, and reports the angle of slope as $sin^{- 1} \frac{rise}{length}$ .

A ramp 4 m long rises 2 m. A second ramp of the same length rises 0.5 m. What angles does the app report? Why is the principal value of $sin^{- 1}$ the right one here?

Try it first, then show the solutionSolution

First ramp: $sin^{- 1} \frac{2}{4} = sin^{- 1} \frac{1}{2} = \frac{π}{6} = 3 0^{\circ}$ .
Second ramp: $sin^{- 1} \frac{0.5}{4} = sin^{- 1} \frac{1}{8} \approx 0.125$ rad $\approx 7. 2^{\circ}$ .
Both $\frac{π}{6}$ and $\frac{5 π}{6}$ have sine $\frac{1}{2}$ , but a ramp slope must be between $0$ and $\frac{π}{2}$ .
The principal branch $[- \frac{π}{2}, \frac{π}{2}]$ contains exactly the physically sensible angles, so the app gives the right answer.

Answer: About $3 0^{\circ}$ and $7. 2^{\circ}$ ; the principal value is the only one that makes sense for a slope.

Memory hook: A slope angle lives between −90° and 90°: the principal branch of sin⁻¹.

Principal value gives only the first solution

Imagine a Ferris wheel of radius 10 m whose lowest seat is at ground level. After the wheel has turned through $t$ radians from the bottom, a seat is at height $h = 10 - 10 cos t$ metres.

Use $cos^{- 1}$ to find the angle turned when the seat first reaches 15 m. When does it next pass 15 m on the same turn?

Try it first, then show the solutionSolution

$10 - 10 cos t = 15 \Rightarrow cos t = - \frac{1}{2}$ .
Principal value: $t = cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ , on the way up.
Cosine is also $- \frac{1}{2}$ at $2 π - \frac{2 π}{3} = \frac{4 π}{3}$ , on the way down, but this lies outside $[0, π]$ , so $cos^{- 1}$ cannot return it.

Answer: First at $t = \frac{2 π}{3}$ (120°), again at $t = \frac{4 π}{3}$ (240°). The inverse function gives only the first.

Memory hook: cos⁻¹ only sees the first half-turn of the wheel: angles from 0 to π.

Tricks to remember Principal Values of Inverse Trigonometric Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Centred or from zero
sin, tan, cosec: branch centred on 0 (negative input, negative angle). cos, cot, sec: branch from 0 to π (negative input, π minus).
Use it when: Every principal value with a negative input.
Inside-out works, outside-in may not
sin(sin⁻¹x) always returns x. sin⁻¹(sin x) returns x only if x is already in the branch.
Use it when: Spotting trick questions such as sin⁻¹(sin 2).
Reflect in y = x
Each inverse graph is the restricted trigonometric graph reflected in y = x: domain and range swap.
Use it when: Sketching graphs and reading their domain and range.
Triangle for mixed values
For cos(sin⁻¹a), draw a right triangle with opposite side a and hypotenuse 1; the adjacent side is √(1 − a²).
Use it when: Evaluating expressions like tan(sin⁻¹(3/5)).

Principal Values of Inverse Trigonometric Functions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Principal value with a negative input · 1 mark · MCQBoard level

The principal value of $cos^{- 1} (- \frac{3}{2})$ is:
(a) $- \frac{π}{6}$
(b) $\frac{π}{6}$
(c) $\frac{5 π}{6}$
(d) $\frac{7 π}{6}$

Try it first, then show the solutionSolution

$cos \frac{π}{6} = \frac{3}{2}$ , and $cos^{- 1} (- x) = π - cos^{- 1} x$ , so the value is $π - \frac{π}{6} = \frac{5 π}{6}$ , which lies in $[0, π]$ .

Answer: (c) $\frac{5 π}{6}$ .

Example 2: A sum of three principal values · 2 marks · Short answerBoard level

Find the value of $tan^{- 1} (1) + cos^{- 1} (- \frac{1}{2}) + sin^{- 1} (- \frac{1}{2})$ .

Try it first, then show the solutionSolution

$tan^{- 1} 1 = \frac{π}{4}$ ; $cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ ; $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ . (1½ marks)
Sum $= \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6} = \frac{3 π + 8 π - 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4}$ . (½ mark)

Answer: $\frac{3 π}{4}$ .

Example 3: sin⁻¹(sin x) outside the branch · 2 marks · Short answerHarder board level

Find the value of $sin^{- 1} (sin \frac{3 π}{5})$ .

Try it first, then show the solutionSolution

$\frac{3 π}{5}$ is greater than $\frac{π}{2}$ , so it is not in $[- \frac{π}{2}, \frac{π}{2}]$ and the answer is not $\frac{3 π}{5}$ . (1 mark)
$sin \frac{3 π}{5} = sin (π - \frac{3 π}{5}) = sin \frac{2 π}{5}$ , and $\frac{2 π}{5}$ is in the branch. So the value is $\frac{2 π}{5}$ . (1 mark)

Answer: $\frac{2 π}{5}$ .

Example 4: cos⁻¹(cos x) and tan⁻¹(tan x) · 3 marks · Short answerHarder board level

Evaluate $cos^{- 1} (cos \frac{7 π}{6}) + tan^{- 1} (tan \frac{3 π}{4})$ .

Try it first, then show the solutionSolution

$\frac{7 π}{6} \in / [0, π]$ . Since $cos \frac{7 π}{6} = cos (2 π - \frac{7 π}{6}) = cos \frac{5 π}{6}$ and $\frac{5 π}{6} \in [0, π]$ , the first term is $\frac{5 π}{6}$ . (1 mark)
$\frac{3 π}{4} \in / (- \frac{π}{2}, \frac{π}{2})$ . Since $tan \frac{3 π}{4} = tan (\frac{3 π}{4} - π) = tan (- \frac{π}{4})$ , the second term is $- \frac{π}{4}$ . (1 mark)
Sum $= \frac{5 π}{6} - \frac{π}{4} = \frac{10 π - 3 π}{12} = \frac{7 π}{12}$ . (1 mark)

Answer: $\frac{7 π}{12}$ .

Example 5: Domain of a composite inverse function · 3 marks · Short answerHarder board level

Find the domain of (i) $f (x) = sin^{- 1} (2 x - 1)$ and (ii) $g (x) = cos^{- 1} (x^{2} - 4)$ .

Try it first, then show the solutionSolution

(i) $- 1 \leq 2 x - 1 \leq 1 \Rightarrow 0 \leq 2 x \leq 2 \Rightarrow 0 \leq x \leq 1$ . Domain $[0, 1]$ . (1 mark)
(ii) $- 1 \leq x^{2} - 4 \leq 1 \Rightarrow 3 \leq x^{2} \leq 5$ . (1 mark)
$x^{2} \geq 3$ gives $∣ x ∣ \geq 3$ , and $x^{2} \leq 5$ gives $∣ x ∣ \leq 5$ . Domain $[- 5, - 3] \cup [3, 5]$ . (1 mark)

Answer: (i) $[0, 1]$ ; (ii) $[- 5, - 3] \cup [3, 5]$ .

Example 6: Graph, branch and values together · 5 marks · Long answerHOTS / competency

(i) State the domain and range of $y = sin^{- 1} x$ and describe how its graph is obtained from $y = sin x$ .
(ii) Find $tan (2 sin^{- 1} \frac{3}{5})$ .
(iii) Find $sin^{- 1} (sin 2)$ , where 2 is in radians.

Try it first, then show the solutionSolution

(i) Domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ . Restrict $y = sin x$ to $[- \frac{π}{2}, \frac{π}{2}]$ , where it is one-one and onto $[- 1, 1]$ , and reflect that piece in the line $y = x$ . The graph rises from $(- 1, - \frac{π}{2})$ through the origin to $(1, \frac{π}{2})$ . (2 marks)
(ii) Let $θ = sin^{- 1} \frac{3}{5}$ , so $θ$ is acute with $tan θ = \frac{3}{4}$ . Then $tan 2 θ = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{3/2}{7/16} = \frac{24}{7}$ . (2 marks)
(iii) $2 > \frac{π}{2} \approx 1.571$ , so 2 is outside the branch. $sin 2 = sin (π - 2)$ and $π - 2 \approx 1.14$ is inside it. So $sin^{- 1} (sin 2) = π - 2$ . (1 mark)

Answer: (i) Domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ , reflection in $y = x$ ; (ii) $\frac{24}{7}$ ; (iii) $π - 2$ .

Where students lose marks in Principal Values of Inverse Trigonometric Functions

The slips examiners see most often, and what to write instead.

Mistake: Giving a value outside the branch
An angle with the right sine or cosine is not necessarily the principal value.
Wrong working
$cos^{- 1} (- \frac{1}{2}) = - \frac{π}{3}$ .
Right working
$cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3} \in [0, π]$ .
Do this: Write the branch first and check the answer against it.
Usually costs: 1 mark in Section A.
Mistake: Cancelling sin⁻¹(sin x) to x
The cancellation is valid only when x lies in the branch.
Wrong working
$sin^{- 1} (sin \frac{3 π}{5}) = \frac{3 π}{5}$ .
Right working
$sin^{- 1} (sin \frac{3 π}{5}) = sin^{- 1} (sin \frac{2 π}{5}) = \frac{2 π}{5}$ .
Do this: Ask "is x in the branch?" before cancelling.
Usually costs: The whole part.
Mistake: Not knowing the domain and range of sin⁻¹
The ISC 2025 examiners reported that many candidates could not write the domain and range of inverse sine in a graph question.
Do this: Learn the six branches as a table and practise sketching each graph from its end points.
Usually costs: 1 to 2 marks in a case-based question.
Mistake: Wrong sign in mixed values
For a negative input the angle is in a different quadrant, which changes the sign of other ratios.
Wrong working
$sin (cos^{- 1} (- \frac{3}{5})) = - \frac{4}{5}$ .
Right working
$cos^{- 1} (- \frac{3}{5})$ is in $(\frac{π}{2}, π)$ , where sine is positive, so the value is $\frac{4}{5}$ .
Do this: Use the branch to fix the quadrant before reading other ratios from the triangle.
Usually costs: 1 mark.

Exam tips

Learn the branch table as two groups of three: sin, tan, cosec centred on zero; cos, cot, sec from 0 to π.
In a graph question, label the end points of the inverse graph, such as $(- 1, - \frac{π}{2})$ and $(1, \frac{π}{2})$ , and mark the asymptotes for $tan^{- 1}$ and $cot^{- 1}$ .
Write answers in radians as multiples of π; examiners do not expect degrees unless asked.
For a composite domain, show the double inequality on one line; it is usually the method mark.
In a 1-mark MCQ, if two options have the right trigonometric value, the one inside the branch is the answer.

Practice questions on Principal Values of Inverse Trigonometric Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The range of $cot^{- 1} x$ is:
(a) $[0, π]$
(b) $(0, π)$
(c) $(- \frac{π}{2}, \frac{π}{2})$
(d) $[- \frac{π}{2}, \frac{π}{2}]$
Hint
cot is undefined at 0 and π.
Answer
(b) $(0, π)$ .
Q2 · 1 mark · Very short answerBoard level
Write the principal value of $cosec^{- 1} (- 2)$ .
Hint
cosec⁻¹ behaves like sin⁻¹ for negative inputs.
Answer
$- \frac{π}{6}$ .
Q3 · 2 marks · Short answerHarder board level
Find the value of $sec^{- 1} (- 2) + cot^{- 1} (- 1)$ .
Hint
Both give π minus an acute angle.
Answer
$sec^{- 1} (- 2) = π - \frac{π}{3} = \frac{2 π}{3}$ ; $cot^{- 1} (- 1) = π - \frac{π}{4} = \frac{3 π}{4}$ . Sum $= \frac{17 π}{12}$ .
Q4 · 1 mark · Very short answerHarder board level
Find $sin (\frac{π}{3} - sin^{- 1} (- \frac{1}{2}))$ .
Hint
$sin^{- 1} (- \frac{1}{2})$ is negative.
Answer
$sin (\frac{π}{3} + \frac{π}{6}) = sin \frac{π}{2} = 1$ .
Q5 · 2 marks · Short answerHarder board level
Evaluate $cos (sin^{- 1} \frac{3}{5} + cos^{- 1} \frac{12}{13})$ .
Hint
Use cos(A + B) with two right triangles.
Answer
$cos A = \frac{4}{5}$ , $sin A = \frac{3}{5}$ , $cos B = \frac{12}{13}$ , $sin B = \frac{5}{13}$ . Value $= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$ .
Q6 · 2 marks · Short answerHOTS / competency
For which values of $x$ in $[0, 2 π]$ is $sin^{- 1} (sin x) = π - x$ ?
Hint
$π - x$ must lie in $[- \frac{π}{2}, \frac{π}{2}]$ .
Answer
Since $sin (π - x) = sin x$ , the equation holds exactly when $π - x \in [- \frac{π}{2}, \frac{π}{2}]$ , that is, $x \in [\frac{π}{2}, \frac{3 π}{2}]$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a robot arm has a sensor that reports only $cos θ$ , where $θ$ is the angle the arm has turned from its rest position, and the control software converts the reading with $cos^{- 1}$ .
(i) The sensor reads $- 0.5$ . What angle does the software report? (1 mark)
(ii) The arm has actually turned through $\frac{4 π}{3}$ . What does the software report? (1 mark)
(iii) State the set of angles the software can ever report. (1 mark)
(iv) What extra reading would let the software tell $\frac{2 π}{3}$ from $\frac{4 π}{3}$ ? (1 mark)
Hint
cos⁻¹ returns values only in [0, π].
Answer
(i) $cos^{- 1} (- 0.5) = \frac{2 π}{3}$ . (ii) $cos \frac{4 π}{3} = - 0.5$ , so it again reports $\frac{2 π}{3}$ . (iii) $[0, π]$ . (iv) The sign of $sin θ$ : it is positive at $\frac{2 π}{3}$ and negative at $\frac{4 π}{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the principal value branch of $tan^{- 1} x$ .Show answer
$(- \frac{π}{2}, \frac{π}{2})$ .
2.Find $sin^{- 1} (- \frac{3}{2})$ .Show answer
$- \frac{π}{3}$ .
3.Find $cot^{- 1} (- 3)$ .Show answer
$π - \frac{π}{6} = \frac{5 π}{6}$ .
4.Is $sin (sin^{- 1} 0.7) = 0.7$ ?Show answer
Yes; $sin (sin^{- 1} x) = x$ for all $x \in [- 1, 1]$ .
5.Find $tan^{- 1} (tan \frac{5 π}{4})$ .Show answer
$\frac{5 π}{4} - π = \frac{π}{4}$ .
6.What is the domain of $sec^{- 1} x$ ?Show answer
$R - (- 1, 1)$ , that is $x \leq - 1$ or $x \geq 1$ .

Principal Values of Inverse Trigonometric Functions: frequently asked questions

What is a principal value branch?

It is the interval of angles to which a trigonometric function is restricted so that it becomes one-one and onto, making the inverse function possible. The inverse function always returns an angle from this interval, called the principal value.

Why is the range of cos⁻¹x [0, π] and not [−π/2, π/2]?

On [−π/2, π/2] cosine is not one-one (cos(−θ) = cos θ) and never takes negative values. On [0, π] it decreases steadily from 1 to −1, so it is one-one and takes every value in [−1, 1] exactly once.

Is sin⁻¹x the same as 1/sin x?

No. sin⁻¹x is the inverse sine, an angle whose sine is x. The reciprocal 1/sin x is cosec x. Writing (sin x)⁻¹ makes the reciprocal clear.

How do I evaluate tan⁻¹(tan x) when x is not in the branch?

Tangent has period π, so add or subtract multiples of π until the angle lies in (−π/2, π/2). That angle is the answer. For example, tan⁻¹(tan 3π/4) = 3π/4 − π = −π/4.

Are graphs of inverse trigonometric functions in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists the definition, domain, range, principal value branch and graphs of inverse trigonometric functions, and a 2025 question asked candidates to identify a graph and state its domain and range.