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Chapter 2: Inverse Trigonometric Functions

Class 12 • Mathematics • Chapter 2

Inverse Trigonometric Functions

Going backwards from a ratio to the angle, with one agreed answer every time.

CBSE / NCERT Aligned

School Revise

Chapter Roadmap

Why Restrict the Domain • sin and cos Inverses • tan and cot Inverses • sec and cosec Inverses • Finding Principal Values • Elementary Properties • Key Results • Applications

Why Inverse Trigonometric Functions Matter

Ordinary trigonometry takes an angle and gives a ratio: sin 30° = 1/2. Very often we need to go the other way, starting from the ratio 1/2 and asking which angle produced it. That reverse journey is what inverse trigonometric functions do, and they appear whenever an angle has to be recovered from a measurement, such as the angle of a ramp from its height and length.

There is a catch. Because sin, cos and tan repeat, infinitely many angles share the same ratio, so the reverse is not automatically a function. To fix this we agree on a single stretch of angles, the principal value branch, for each inverse. This chapter sets out those branches, shows how to read off the one correct answer (the principal value), and develops a small set of properties that simplify expressions quickly.

Key idea

A trigonometric function is not one-one over all angles, so we restrict its domain to make it invertible. The agreed output range of each inverse is its principal value branch, and the single answer it gives is the principal value.

Key Terms You Must Know

Term

Meaning

Example

Inverse trig function

Gives the angle whose trig ratio is a given value.

sin⁻¹(1/2) = π/6

Principal value branch

The agreed output range that makes the inverse a function.

sin⁻¹ has range [−π/2, π/2]

Principal value

The single value the inverse returns, inside its branch.

cos⁻¹(0) = π/2

Domain

The set of inputs the inverse will accept.

sin⁻¹ accepts [−1, 1]

Range

The set of outputs (angles) the inverse can give.

tan⁻¹ gives (−π/2, π/2)

Odd function

A function with f(−x) = −f(x).

sin⁻¹(−x) = −sin⁻¹x

Complementary pair

Two inverses adding to π/2 for valid x.

sin⁻¹x + cos⁻¹x = π/2

Core Concepts, Step by Step

1. Why We Restrict the Domain

A function can be reversed only when it is one-one, that is when each output comes from just one input. But sin x takes the value 1/2 at 30°, at 150°, and at countless other angles, so over all angles it is far from one-one. To build an inverse we therefore restrict sin to a stretch of angles on which it is one-one and still reaches every value from −1 to 1. The agreed stretch becomes the range of the inverse, called its principal value branch.

2. The Inverses of sin and cos

For sin⁻¹ we restrict sin to angles from −π/2 to π/2. So sin⁻¹ has domain [−1, 1] and range [−π/2, π/2]. For cos⁻¹ we restrict cos to angles from 0 to π, giving domain [−1, 1] and range [0, π]. Reading them is straightforward: sin⁻¹(1/2) asks ‘which angle in [−π/2, π/2] has sine 1/2?’, and the answer is π/6.

Graphs of sin⁻¹x and cos⁻¹x on the domain [−1, 1]

Graphs of inverse sine and inverse cosine

3. The Inverses of tan and cot

tan⁻¹ accepts every real number (since tan reaches all values) and returns an angle in the open interval (−π/2, π/2). cot⁻¹ also accepts every real number and returns an angle in (0, π). These two are very common in calculus, because tan⁻¹ turns up whenever we integrate expressions of the form 1/(1 + x²), which you will meet later in the year.

Graph of tan⁻¹x with horizontal asymptotes y = ±π/2

Graph of inverse tangent

4. The Inverses of sec and cosec

sec⁻¹ accepts values with |x| ≥ 1 and returns an angle in [0, π] other than π/2. cosec⁻¹ accepts |x| ≥ 1 and returns an angle in [−π/2, π/2] other than 0. They are used less often than the first four, but they complete the family. The table below collects the domain and principal value branch of all six, which is the single most important thing to know in this chapter.

Domains and principal value branches of the six inverses

Function

Domain (input)

Principal value branch (output)

sin⁻¹ x

[−1, 1]

[−π/2, π/2]

cos⁻¹ x

[−1, 1]

[0, π]

tan⁻¹ x

all real x

(−π/2, π/2)

cot⁻¹ x

all real x

(0, π)

sec⁻¹ x

|x| ≥ 1

[0, π] except π/2

cosec⁻¹ x

|x| ≥ 1

[−π/2, π/2] except 0

5. Finding Principal Values

To find a principal value, ask which angle inside the correct branch gives the required ratio. For example, to find cos⁻¹(−1/2), look for an angle in [0, π] whose cosine is −1/2; that angle is 2π/3. A common slip is to give an angle outside the branch, so always check that your answer lies within the range from the table. Negative inputs often push the answer into the second quadrant for cos⁻¹, or below zero for sin⁻¹ and tan⁻¹.

6. Elementary Properties

A handful of properties simplify almost every expression. The inverses of sin, tan and cosec are odd, so sin⁻¹(−x) = −sin⁻¹x and tan⁻¹(−x) = −tan⁻¹x. Each function pairs with its co-function to give π/2: sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2, and sec⁻¹x + cosec⁻¹x = π/2. Reciprocal inputs swap function and co-function, for example sin⁻¹(1/x) = cosec⁻¹x for |x| ≥ 1.

Key Results & Proofs

Three short results do most of the simplifying in this chapter, and each follows from the definition of the inverse.

Result 1: sin⁻¹x + cos⁻¹x = π/2

Statement. For every x in [−1, 1], sin⁻¹x + cos⁻¹x = π/2.

Proof

Set sin⁻¹x equal to an angle, then use the complement relation.

Let sin⁻¹x = θ, so sin θ = x with θ in [−π/2, π/2]. – name the first angle

x

=

sin θ = cos(π/2 − θ)

since sine equals cosine of the complement

Now π/2 − θ lies in [0, π], the branch of cos⁻¹. – check the branch

cos⁻¹x

=

π/2 − θ = π/2 − sin⁻¹x

take cos⁻¹ of both sides

sin⁻¹x + cos⁻¹x

=

π/2

rearrange

The same idea gives tan⁻¹x + cot⁻¹x = π/2 and sec⁻¹x + cosec⁻¹x = π/2.

Result 2: sin⁻¹ is an Odd Function

Statement. For every x in [−1, 1], sin⁻¹(−x) = −sin⁻¹x.

Proof

Use the fact that sine itself is an odd function.

Let sin⁻¹x = θ, so sin θ = x with θ in [−π/2, π/2]. – name the angle

sin(−θ)

=

−sin θ = −x

sine is an odd function

Also −θ lies in [−π/2, π/2], the branch of sin⁻¹. – check the branch

sin⁻¹(−x)

=

−θ = −sin⁻¹x

take sin⁻¹ of both sides

tan⁻¹ and cosec⁻¹ are odd in the same way; cos⁻¹ and sec⁻¹ are not.

Result 3: tan⁻¹x + cot⁻¹x = π/2

Statement. For every real x, tan⁻¹x + cot⁻¹x = π/2.

Proof

The same complement trick used for sine and cosine works here.

Let tan⁻¹x = θ, so tan θ = x with θ in (−π/2, π/2). – name the angle

x

=

tan θ = cot(π/2 − θ)

tangent equals cotangent of the complement

And π/2 − θ lies in (0, π), the branch of cot⁻¹. – check the branch

cot⁻¹x

=

π/2 − θ = π/2 − tan⁻¹x

take cot⁻¹ of both sides

tan⁻¹x + cot⁻¹x

=

π/2

rearrange

So tan⁻¹(1) + cot⁻¹(1) = π/4 + π/4 = π/2, as expected.

Worked Examples

Example 1

Question: Find the principal value of sin⁻¹(1/2).

▶ Show full working

Ask which angle in [−π/2, π/2] has sine 1/2.

sin(π/6) = 1/2, and π/6 lies in [−π/2, π/2]. – find the angle in the branch

Answer: sin⁻¹(1/2) = π/6.

Example 2

Question: Find the principal value of cos⁻¹(−1/2).

▶ Show full working

Ask which angle in [0, π] has cosine −1/2.

cos(2π/3) = −1/2, and 2π/3 lies in [0, π]. – find the angle in the branch

Answer: cos⁻¹(−1/2) = 2π/3.

Example 3

Question: Find the principal value of tan⁻¹(1).

▶ Show full working

Ask which angle in (−π/2, π/2) has tangent 1.

tan(π/4) = 1, and π/4 lies in (−π/2, π/2). – find the angle in the branch

Answer: tan⁻¹(1) = π/4.

Example 4

Question: Find the principal value of sin⁻¹(−1/√2).

▶ Show full working

Use the branch [−π/2, π/2]; a negative input gives a negative angle.

sin(−π/4) = −1/√2, and −π/4 lies in the branch. – find the angle

Answer: sin⁻¹(−1/√2) = −π/4.

Example 5

Question: Find the principal value of cos⁻¹(1).

▶ Show full working

Ask which angle in [0, π] has cosine 1.

cos(0) = 1, and 0 lies in [0, π]. – find the angle

Answer: cos⁻¹(1) = 0.

Example 6

Question: Find the principal value of tan⁻¹(−√3).

▶ Show full working

Use the branch (−π/2, π/2).

tan(π/3) = √3, so tan(−π/3) = −√3. – use the odd property

−π/3 lies in the branch. – check the branch

Answer: tan⁻¹(−√3) = −π/3.

Example 7

Question: Evaluate sin⁻¹(1/2) + cos⁻¹(1/2).

▶ Show full working

Use the complementary property, or add the two principal values.

sin⁻¹(1/2) = π/6 and cos⁻¹(1/2) = π/3. – find each value

sum

=

π/6 + π/3 = π/2

add the angles

Answer: The value is π/2, agreeing with sin⁻¹x + cos⁻¹x = π/2.

Example 8

Question: Evaluate sin⁻¹(sin(2π/3)).

▶ Show full working

Since 2π/3 is outside the branch [−π/2, π/2], simplify the inside first.

sin(2π/3) = √3/2. – evaluate the inside

sin⁻¹(√3/2) = π/3, which is in the branch. – take the inverse

Answer: The value is π/3, not 2π/3.

Example 9

Question: Evaluate cos⁻¹(cos(7π/6)).

▶ Show full working

Since 7π/6 is outside [0, π], simplify the inside first.

cos(7π/6) = −√3/2. – evaluate the inside

cos⁻¹(−√3/2) = 5π/6, which is in [0, π]. – take the inverse

Answer: The value is 5π/6, not 7π/6.

Example 10

Question: Find the principal value of sec⁻¹(2).

▶ Show full working

Ask which angle in [0, π] (not π/2) has secant 2.

sec θ = 2 means cos θ = 1/2. – rewrite using cosine

cos(π/3) = 1/2, and π/3 is in the branch. – find the angle

Answer: sec⁻¹(2) = π/3.

Example 11

Question: Find the principal value of cosec⁻¹(2).

▶ Show full working

Ask which angle in [−π/2, π/2] (not 0) has cosecant 2.

cosec θ = 2 means sin θ = 1/2. – rewrite using sine

sin(π/6) = 1/2, and π/6 is in the branch. – find the angle

Answer: cosec⁻¹(2) = π/6.

Example 12

Question: Show that tan⁻¹(1) + cot⁻¹(1) = π/2.

▶ Show full working

Find each principal value and add.

tan⁻¹(1) = π/4. – first value

cot⁻¹(1) = π/4, since cot(π/4) = 1 and π/4 is in (0, π). – second value

sum

=

π/4 + π/4 = π/2

add

Answer: The sum is π/2, agreeing with tan⁻¹x + cot⁻¹x = π/2.

Where You Meet This in Real Life

Angles from measurements

To find the angle of a ramp, roof or staircase from its height and base, you take an inverse tangent of the ratio.

Navigation and surveying

Bearings and elevation angles are recovered from horizontal and vertical distances using inverse trigonometric functions.

Physics

The launch angle of a projectile, or the angle of refraction of light, is often found by inverting a trigonometric ratio.

Computer graphics

Rotating and positioning objects on screen needs angles computed from coordinates, which is exactly what these inverses provide.

Calculus ahead

The inverse tangent appears when integrating 1/(1 + x²), so this chapter feeds directly into the integration you study later.

Practice Sets A to D

Practice Set A – Basics

A1. Find the principal value of sin⁻¹(0).

▶ Reveal full working

Which angle in the branch has sine 0?

sin(0) = 0, and 0 is in [−π/2, π/2].

Answer: 0.

A2. Find the principal value of cos⁻¹(0).

▶ Reveal full working

Which angle in [0, π] has cosine 0?

cos(π/2) = 0, and π/2 is in [0, π].

Answer: π/2.

A3. Find the principal value of tan⁻¹(0).

▶ Reveal full working

Which angle in the branch has tangent 0?

tan(0) = 0, and 0 is in (−π/2, π/2).

Answer: 0.

A4. Find the principal value of sin⁻¹(1).

▶ Reveal full working

Which angle in the branch has sine 1?

sin(π/2) = 1, and π/2 is in the branch.

Answer: π/2.

Practice Set B – Conceptual

B1. Why must we restrict the domain of sine before defining sin⁻¹?

▶ Reveal full working

Think about one-one functions.

Over all angles sine repeats, so it is not one-one.

Restricting it makes it one-one, so an inverse can exist.

Answer: Because sine is not one-one over all angles; restricting makes it invertible.

B2. State the principal value branch (range) of cos⁻¹.

▶ Reveal full working

Recall the table.

cos⁻¹ returns angles from 0 to π.

Answer: [0, π].

B3. State the principal value branch of tan⁻¹.

▶ Reveal full working

Recall the table.

tan⁻¹ returns angles in the open interval (−π/2, π/2).

Answer: (−π/2, π/2).

B4. Is sin⁻¹(sin x) always equal to x?

▶ Reveal full working

Think about the branch.

It equals x only when x lies in [−π/2, π/2].

Otherwise you must reduce the angle into the branch first.

Answer: No, only when x is in [−π/2, π/2].

Practice Set C – Application / Numerical

C1. Find the principal value of cos⁻¹(−1/√2).

▶ Reveal full working

Which angle in [0, π] has cosine −1/√2?

cos(3π/4) = −1/√2, and 3π/4 is in [0, π].

Answer: 3π/4.

C2. Find the principal value of tan⁻¹(−1).

▶ Reveal full working

Use the branch (−π/2, π/2).

tan(−π/4) = −1, and −π/4 is in the branch.

Answer: −π/4.

C3. Find the principal value of sin⁻¹(−1/2).

▶ Reveal full working

Use the odd property of sin⁻¹.

sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6.

Answer: −π/6.

C4. Find the principal value of sec⁻¹(√2).

▶ Reveal full working

Rewrite with cosine.

sec θ = √2 means cos θ = 1/√2.

cos(π/4) = 1/√2, and π/4 is in the branch.

Answer: π/4.

Practice Set D – HOTS / Multi-step

D1. Evaluate sin⁻¹(sin(3π/4)).

▶ Reveal full working

Simplify the inside first, since 3π/4 is outside the branch.

sin(3π/4) = √2/2. – evaluate the inside

sin⁻¹(√2/2) = π/4, which is in the branch. – take the inverse

Answer: π/4.

D2. Evaluate cos⁻¹(cos(4π/3)).

▶ Reveal full working

Simplify the inside, since 4π/3 is outside [0, π].

cos(4π/3) = −1/2. – evaluate the inside

cos⁻¹(−1/2) = 2π/3, which is in [0, π]. – take the inverse

Answer: 2π/3.

D3. Prove that tan⁻¹x + cot⁻¹x = π/2.

▶ Reveal full working

Let tan⁻¹x = θ and use the complement relation.

tan θ = x = cot(π/2 − θ). – complement relation

π/2 − θ lies in (0, π), the branch of cot⁻¹. – check the branch

cot⁻¹x

=

π/2 − tan⁻¹x

take cot⁻¹

tan⁻¹x + cot⁻¹x

=

π/2

rearrange

Answer: Proved.

D4. Evaluate sin⁻¹(1/2) + cos⁻¹(1/2) + tan⁻¹(1).

▶ Reveal full working

Find each principal value and add.

sin⁻¹(1/2) = π/6, cos⁻¹(1/2) = π/3, tan⁻¹(1) = π/4. – three values

sum

=

π/6 + π/3 + π/4 = π/2 + π/4

add the first two, then the third

sum

=

3π/4

combine

Answer: 3π/4.

Chapter Summary

Everything in One Glance

Why Inverses Need Branches

Trig functions repeat, so we restrict the domain to make each one-one and invertible.

sin and cos Inverses

sin⁻¹: domain [−1,1], range [−π/2,π/2]; cos⁻¹: domain [−1,1], range [0,π].

tan and cot Inverses

tan⁻¹: all x, range (−π/2,π/2); cot⁻¹: all x, range (0,π).

sec and cosec Inverses

Both need |x| ≥ 1; sec⁻¹ in [0,π] except π/2, cosec⁻¹ in [−π/2,π/2] except 0.

Principal Value

Always pick the angle inside the correct branch; check it against the table.

Properties

Odd: sin⁻¹(−x) = −sin⁻¹x. Complementary: sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2.

Are You Exam-Ready?

8-Point Exam Quick-Check

Find the principal value of cos⁻¹(1/2).

Find the principal value of sin⁻¹(√3/2).

Find the principal value of tan⁻¹(√3).

State the domain and range of cos⁻¹.

Evaluate sin⁻¹(1) + cos⁻¹(1).

Find the principal value of cosec⁻¹(2).

Evaluate cos⁻¹(cos(5π/3)).

State the value of tan⁻¹x + cot⁻¹x.

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Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions, Complete Notes and Practice

These free Class 12 Maths Chapter 2 Inverse Trigonometric Functions notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.