Inverse Trigonometric Functions – Class 12 Maths Chapter 2 Notes & NCERT PDF

Class 12MathematicsChapter 2NCERT book: Mathematics Part I - Class 12

Notes and a simple summary of Chapter 2 of the NCERT Class 12 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 2: Inverse Trigonometric Functions

Introduction

Chapter 2 of NCERT Class 12 Mathematics (Part I) builds on Chapter 1: only one-one and onto functions have inverses. Trigonometric functions are periodic, so they are not one-one on their natural domains. By restricting the domain we make each of them bijective, and then define its inverse. These inverses are used widely in calculus (integration) and in science and engineering. The rationalised chapter has 2.1 Introduction, 2.2 Basic Concepts and 2.3 Properties of Inverse Trigonometric Functions.

Key Concepts

2.2 Basic Concepts — Principal Value Branches

Each inverse function is defined on the restricted domain where the original function is bijective. The principal value branch gives the standard range:

sin⁻¹ x: domain [−1, 1], range [−π/2, π/2]

cos⁻¹ x: domain [−1, 1], range [0, π]

cosec⁻¹ x: domain R − (−1, 1), range [−π/2, π/2] − {0}

sec⁻¹ x: domain R − (−1, 1), range [0, π] − {π/2}

tan⁻¹ x: domain R, range (−π/2, π/2)

cot⁻¹ x: domain R, range (0, π)

Principal value: the value of an inverse trigonometric function that lies in its principal value branch.

Notation warning: sin⁻¹x is not (sin x)⁻¹. In fact (sin x)⁻¹ = 1/sin x = cosec x.

Graphs: the graph of y = sin⁻¹x is the reflection of the restricted graph of y = sin x in the line y = x, since domain and range interchange. The book draws each principal branch in this way.

When no branch is mentioned, the principal value branch is meant.

2.3 Properties

For suitable values of the domain:

y = sin⁻¹x ⇒ x = sin y, and x = sin y ⇒ y = sin⁻¹x.

sin(sin⁻¹x) = x for x ∈ [−1, 1].

sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2]. This is the source of many exam traps.

Similar results hold for the other five functions within their branches.

The book's worked results use substitution:

sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹x for −1/√2 ≤ x ≤ 1/√2. Put x = sin θ, so the expression becomes sin⁻¹(sin 2θ) = 2θ.

The same expression equals 2 cos⁻¹x for 1/√2 ≤ x ≤ 1.

Expressions like tan⁻¹(cos x/(1 − sin x)) are simplified by writing cos x and 1 − sin x in half-angles. This one reduces to π/4 + x/2 on the given interval.

Worked Examples

Principal value of sin⁻¹(1/2). sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2], so the answer is π/6.

Principal value of cos⁻¹(−1/2). cos(2π/3) = −1/2 and 2π/3 ∈ [0, π], so the answer is 2π/3. A common error is −π/3, which lies outside [0, π].

tan⁻¹(−√3) = −π/3, and tan⁻¹(−1) = −π/4, since the range is (−π/2, π/2).

cos⁻¹(1/2) + 2 sin⁻¹(1/2) = π/3 + 2(π/6) = 2π/3.

sin⁻¹(sin 3π/5). 3π/5 is not in [−π/2, π/2]. Since sin(3π/5) = sin(π − 3π/5) = sin(2π/5), and 2π/5 is in range, the answer is 2π/5, not 3π/5.

cos⁻¹(cos 7π/6). 7π/6 is not in [0, π]. Since cos(7π/6) = cos(5π/6), the answer is 5π/6.

tan⁻¹(1) + cot⁻¹(−1) = π/4 + 3π/4 = π, because cot⁻¹ has range (0, π).

Formula / Key-Facts Sheet

Ranges: sin⁻¹ [−π/2, π/2]; cos⁻¹ [0, π]; tan⁻¹ (−π/2, π/2); cot⁻¹ (0, π); sec⁻¹ [0, π] − {π/2}; cosec⁻¹ [−π/2, π/2] − {0}.

sin(sin⁻¹x) = x for x ∈ [−1, 1]; sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2].

sec⁻¹2 = π/3, cosec⁻¹2 = π/6, cot⁻¹(√3) = π/6.

Substitution trick: put x = sin θ, cos θ or tan θ to simplify.

Board Exam Focus

1-mark MCQs: principal values (Exercise 2.1), for example the value of tan⁻¹√3 − sec⁻¹(−2).

2–3 marks: write in simplest form, such as tan⁻¹(√(1 + x²) − 1)/x. Use x = tan θ, then half-angle formulas.

"Find the value of" questions of the form sin⁻¹(sin x) or cos⁻¹(cos x) outside the branch. Always bring the angle into the principal range first.

Domain questions, such as the domain of sin⁻¹(2x): solve −1 ≤ 2x ≤ 1, giving [−1/2, 1/2].

Key Terms

Inverse trigonometric function (प्रतिलोम त्रिकोणमितीय फलन)

Principal value (मुख्य मान)

Principal value branch (मुख्य मान शाखा)

Domain (प्रांत)

Range (परिसर)

Restriction of domain (प्रांत का प्रतिबंधन)

Periodic function (आवर्ती फलन)

Branch (शाखा)

Reflection in y = x (रेखा y = x में प्रतिबिंब)

Common Mistakes

Treating sin⁻¹x as 1/sin x.

Giving cos⁻¹(−1/2) as −π/3. The cos⁻¹ range is [0, π], so the answer is 2π/3.

Writing sin⁻¹(sin 3π/5) = 3π/5 without checking the branch.

Using range (−π/2, π/2) for cot⁻¹. Its range is (0, π).

Forgetting the restrictions under which simplification formulas hold.

💡 Key Learning Points

✓Explain why trigonometric functions need restricted domains to have inverses.
✓Recall the domain and principal value branch of all six inverse trigonometric functions.
✓Find principal values correctly, especially for negative arguments.
✓Use sin(sin⁻¹x) = x and sin⁻¹(sin x) = x with their exact validity conditions.
✓Simplify expressions using substitutions such as x = sin θ or x = tan θ.

👨‍🏫 Teaching Tips

→Draw y = sin x, highlight [−π/2, π/2], then reflect it in y = x to build sin⁻¹ live.
→Make a 'range wheel' poster: the four quadrant ranges students must memorise.
→Run rapid-fire principal-value drills with negative arguments.
→Show the sin⁻¹(sin 3π/5) trap and have students find the equivalent in-range angle.
→Link to Chapter 1: invertible ⇔ bijective, so restriction is needed.
→Practise substitution proofs (x = sin θ) as a fixed three-line template.
→Use graphing software to compare sin⁻¹(sin x) with y = x over a wide interval.

📋 Assessment Questions

State the principal value branch of cos⁻¹x and cot⁻¹x.
Find the principal value of cos⁻¹(−1/2).
Evaluate sin⁻¹(sin 3π/5).
Find the value of tan⁻¹(1) + cot⁻¹(−1).
Show that sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹x for −1/√2 ≤ x ≤ 1/√2.

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