Why is the range of cos⁻¹ chosen as [0, π] instead of [−π/2, π/2] like sin⁻¹?

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CBSE Class 12 Mathematics★ 5-8 marks · CBSE Class 12 Mathematics Chapter 2

Inverse Trigonometric Functions

If sin(x) = y, what is x in terms of y? The answer is sin⁻¹(y) — but with a TWIST: since trig functions are NOT one-one, we must RESTRICT their domains to define inverses. CBSE Class 12 Mathematics Chapter 2.

⏱ 18 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1State the domain and principal value branch (range) for all six inverse trigonometric functions
2Evaluate inverse trigonometric expressions by converting to standard forms
3Apply identities: sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; sec⁻¹x + cosec⁻¹x = π/2
4Apply identities for tan⁻¹: tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) when xy < 1
5Simplify composite expressions like sin(cos⁻¹x), tan(sin⁻¹x) by using substitution

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Why this chapter matters

Inverse trigonometric functions appear throughout calculus — differentiation, integration, differential equations. Knowing the domains/ranges of sin⁻¹, cos⁻¹, tan⁻¹ and the key identities (sin⁻¹x + cos⁻¹x = π/2) is essential for avoiding errors in later chapters. The identities are high-yield exam targets.

Inverse Trigonometric Functions

"Inverse trigonometric functions take a RATIO and return an ANGLE."

1. Chapter Overview

Trigonometric functions (sin, cos, tan) are PERIODIC — they repeat values. Therefore: they are NOT ONE-ONE on their full domains. To define INVERSES, we RESTRICT their domains to intervals where they ARE one-one. This chapter covers: sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, cosec⁻¹ — their domains, ranges, principal values, graphs, and properties.

2. Principal Value Branches

Function	Domain (input)	Range (Principal Value — output)
sin⁻¹ x	[-1, 1]	[-π/2, π/2]
cos⁻¹ x	[-1, 1]	[0, π]
tan⁻¹ x	R (all real)	(-π/2, π/2)
cot⁻¹ x	R	(0, π)
sec⁻¹ x	(-∞, -1] ∪ [1, ∞)	[0, π] — {π/2}
cosec⁻¹ x	(-∞, -1] ∪ [1, ∞)	[-π/2, π/2] — {0}

3. Key Properties

Basic Identities

sin(sin⁻¹ x) = x for x ∈ [-1, 1]. sin⁻¹(sin x) = x for x ∈ [-π/2, π/2].
cos⁻¹(-x) = π — cos⁻¹ x
sin⁻¹ x + cos⁻¹ x = π/2 (for x ∈ [-1, 1])
tan⁻¹ x + cot⁻¹ x = π/2 (for all x ∈ R)

Conversion Formulas

sin⁻¹ x = cos⁻¹ √(1 — x²) = tan⁻¹ [x/√(1 — x²)]
tan⁻¹ x + tan⁻¹ y = tan⁻¹ [(x + y)/(1 — xy)]. Watch the quadrant!

4. Exam Focus

Principal values — domains and ranges of ALL 6 inverse trig functions
sin⁻¹ x + cos⁻¹ x = π/2. tan⁻¹ x + cot⁻¹ x = π/2.
tan⁻¹ x + tan⁻¹ y identity.
Simplifying expressions involving inverse trig functions.

5. Conclusion

Inverse trig functions are ESSENTIAL for integration, differential equations, and countless applications:

They take a RATIO (sine of an angle) and return the ANGLE
The key: KNOW YOUR PRINCIPAL VALUE RANGES. 'If you don't know the range of sin⁻¹, you can't solve the problem. Period.'

'Inverse trig functions are the bridge from geometry to algebra — turning angles into numbers and numbers back into angles.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Domains and Principal Value Branches

sin⁻¹x: domain [−1, 1], range [−π/2, π/2]. cos⁻¹x: domain [−1, 1], range [0, π]. tan⁻¹x: domain ℝ, range (−π/2, π/2). cot⁻¹x: domain ℝ, range (0, π). sec⁻¹x: domain ℝ−(−1,1), range [0,π]−{π/2}. cosec⁻¹x: domain ℝ−(−1,1), range [−π/2,π/2]−{0}.

The RANGE (principal value branch) is what makes inverse trig functions single-valued. Memory: sin⁻¹ range is [−π/2, π/2] (centred at 0, ±π/2). cos⁻¹ range is [0, π] (non-negative). tan⁻¹ range is OPEN: (−π/2, π/2) (endpoints excluded because tan is undefined there).

Complementary Angle Identities

sin⁻¹x + cos⁻¹x = π/2 for x ∈ [−1,1]. tan⁻¹x + cot⁻¹x = π/2 for x ∈ ℝ. sec⁻¹x + cosec⁻¹x = π/2 for |x| ≥ 1. These follow from the complementary angle property: if sin θ = x, then cos(π/2−θ) = x, so cos⁻¹x = π/2 − θ = π/2 − sin⁻¹x.

Highest-frequency exam identity: sin⁻¹x + cos⁻¹x = π/2. MCQ: 'If sin⁻¹(3/5) = α, find cos⁻¹(3/5).' Answer: π/2 − α. No calculation needed — just the identity.

tan⁻¹ Addition/Subtraction Formulas

tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) if xy < 1. tan⁻¹x + tan⁻¹y = π + tan⁻¹((x+y)/(1−xy)) if xy > 1 and x > 0. tan⁻¹x − tan⁻¹y = tan⁻¹((x−y)/(1+xy)) if xy > −1. DOUBLE ANGLE: 2tan⁻¹x = tan⁻¹(2x/(1−x²)) if |x| < 1.

The condition xy < 1 (for the basic formula) is the most commonly ignored. When xy > 1 and x,y > 0, add π to the formula. Exam problems test this edge case — forgetting the π gives a wrong answer.

sin⁻¹ and cos⁻¹ Identities

sin⁻¹(−x) = −sin⁻¹x. cos⁻¹(−x) = π − cos⁻¹x. tan⁻¹(−x) = −tan⁻¹x. sin(sin⁻¹x) = x for x ∈ [−1,1]. cos(cos⁻¹x) = x for x ∈ [−1,1]. sin⁻¹(sin θ) = θ for θ ∈ [−π/2, π/2]. cos⁻¹(cos θ) = θ for θ ∈ [0, π].

KEY TRAP: sin⁻¹(sin θ) = θ ONLY if θ is in [−π/2, π/2]. If θ is outside this range, reduce it first. Example: sin⁻¹(sin(3π/4)) ≠ 3π/4. Since 3π/4 ∉ [−π/2, π/2], we use sin(3π/4) = sin(π − 3π/4) = sin(π/4). So sin⁻¹(sin(3π/4)) = π/4.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Writing sin⁻¹(1/2) = 5π/6

✓ sin⁻¹(1/2) = π/6, NOT 5π/6. While sin(5π/6) = 1/2 is true, the PRINCIPAL VALUE BRANCH of sin⁻¹ is [−π/2, π/2], and 5π/6 is NOT in this range. Only π/6 is in [−π/2, π/2] with sin(π/6) = 1/2. Principal value is always chosen from the specified range.

WATCH OUT

✗ Applying tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) without checking xy < 1

✓ The formula tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) is valid ONLY when xy < 1. If xy > 1 and x,y > 0, the correct formula is: tan⁻¹x + tan⁻¹y = π + tan⁻¹((x+y)/(1−xy)). Exam problems deliberately set xy > 1 to catch students who forget this condition.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 2.1 and 2.2

Exercise 2.1 and 2.2

Finding principal values; proving identities using sin⁻¹+cos⁻¹=π/2; tan⁻¹ addition formulas; simplifying composite expressions; miscellaneous problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· principal-value

Find the principal values of: (i) sin⁻¹(−√3/2) (ii) cos⁻¹(cos(7π/6)) (iii) tan⁻¹(−√3).

▶ Show solution

(i) sin⁻¹(−√3/2): Since sin(−π/3) = −√3/2 and −π/3 ∈ [−π/2, π/2], the principal value is −π/3. (ii) cos⁻¹(cos(7π/6)): 7π/6 ∉ [0, π] (the principal range). cos(7π/6) = cos(2π − 7π/6) = cos(5π/6). Now 5π/6 ∈ [0, π], so cos⁻¹(cos(7π/6)) = cos⁻¹(cos(5π/6)) = 5π/6. (iii) tan⁻¹(−√3): Since tan(−π/3) = −√3 and −π/3 ∈ (−π/2, π/2), the principal value is −π/3.

Q2MEDIUM· identity-application

Prove that tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).

▶ Show solution

LHS = tan⁻¹(1/2) + tan⁻¹(2/11). Check: xy = (1/2)(2/11) = 1/11 < 1. So we can use the addition formula: tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)). tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹((1/2 + 2/11)/(1 − 1/11)) = tan⁻¹((11/22 + 4/22)/(10/11)) = tan⁻¹((15/22)/(10/11)) = tan⁻¹(15/22 × 11/10) = tan⁻¹(165/220) = tan⁻¹(3/4) = RHS. ✓

Q3HARD· long-answer

Prove that: sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = π.

▶ Show solution

Let α = sin⁻¹(12/13), β = cos⁻¹(4/5). From α = sin⁻¹(12/13): sin α = 12/13, so cos α = 5/13, tan α = 12/5. From β = cos⁻¹(4/5): cos β = 4/5, so sin β = 3/5, tan β = 3/4. Now find α + β using tan: tan(α+β) = (tan α + tan β)/(1 − tan α tan β) = (12/5 + 3/4)/(1 − (12/5)(3/4)) = (48/20 + 15/20)/(1 − 36/20) = (63/20)/((20−36)/20) = (63/20)/(−16/20) = −63/16. Since tan α = 12/5 > 0 and α ∈ [0, π/2], and tan β = 3/4 > 0 and β ∈ [0, π/2], we have α+β ∈ (0, π). Since tan(α+β) = −63/16 < 0, and α+β ∈ (0, π), we have α+β ∈ (π/2, π). Therefore α+β = π − tan⁻¹(63/16) (since tan(π−θ) = −tan θ). So: α + β + tan⁻¹(63/16) = π. That is: sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = π. ✓

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•sin⁻¹: domain [−1,1], range [−π/2, π/2]. cos⁻¹: domain [−1,1], range [0, π].
•tan⁻¹: domain ℝ, range (−π/2, π/2). Open endpoints.
•sin⁻¹x + cos⁻¹x = π/2. tan⁻¹x + cot⁻¹x = π/2.
•tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) when xy < 1
•If xy > 1 and x,y > 0: formula gives tan⁻¹(...) + π
•sin⁻¹(−x) = −sin⁻¹x. cos⁻¹(−x) = π − cos⁻¹x.
•sin⁻¹(sin θ) = θ ONLY if θ ∈ [−π/2, π/2]. Must reduce otherwise.
•Composite expressions: sin(cos⁻¹x) = √(1−x²). Use right triangle or identity.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-8 marks

Question type	Marks each	Typical count	What it tests
MCQ / Short (principal values)	1-2	1-2	Find principal value; evaluate composite expressions; apply sin⁻¹x + cos⁻¹x = π/2
Proof / Identity	3-6	1	Prove an identity using tan⁻¹ addition/subtraction formulas; prove composite equals a constant

Prep strategy

Memorise all 6 domains and ranges: sin⁻¹ → [−π/2, π/2]; cos⁻¹ → [0, π]; tan⁻¹ → (−π/2, π/2). These determine all principal value answers.
For tan⁻¹ addition: ALWAYS check xy < 1 before applying the formula. Determine the quadrant if xy > 1 to add π.
For sin⁻¹(sin θ) or cos⁻¹(cos θ): first check if θ is in the principal branch. If not, reduce to an equivalent angle IN the principal branch before evaluating.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Navigation and GPS

GPS devices use inverse trigonometric functions constantly. When your phone calculates the bearing (direction angle) from your location to a destination, it uses atan2(Δy, Δx) — the two-argument arctangent (a generalisation of tan⁻¹). Every time you navigate, inverse trig functions determine your direction. The range restrictions (principal value branch) prevent ambiguity — ensuring your phone gives you one direction, not multiple.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

For principal value questions: state the range first, then find the value in that range. 'sin⁻¹(−√3/2) — range is [−π/2, π/2]. sin(−π/3) = −√3/2 and −π/3 ∈ [−π/2, π/2]. Therefore answer is −π/3.'
For identity proofs: always convert tan⁻¹ sums using the formula, then simplify the resulting fraction. Show xy check. Arrive at the target expression.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Explore the Gregory-Leibniz series: π/4 = 1 − 1/3 + 1/5 − 1/7 + ... = tan⁻¹(1). This infinite series comes directly from the Maclaurin series for tan⁻¹x at x=1, connecting inverse trig to infinite series — one of the most elegant results in mathematics
Study the Machin formula: π/4 = 4·tan⁻¹(1/5) − tan⁻¹(1/239), used historically for computing π to many decimal places — a direct application of the tan⁻¹ addition formula

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Inverse Trigonometry)High

CUET (Mathematics)Medium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The range must be chosen so that cos is ONE-ONE on that interval (otherwise the inverse is not well-defined). cos is one-one (strictly decreasing) on [0, π]. It is NOT one-one on [−π/2, π/2] because cos(π/4) = cos(−π/4) = √2/2, for example. The range [0, π] is chosen to make cos one-one. Note this means cos⁻¹x is always non-negative (its range is [0, π]).

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