Chapter 2 of 16

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ICSE Class 11 Mathematics★ ISC Class 11 Mathematics · Internal assessment + foundation for ISC Class 12 · Board: ISC

Sets, Relations, Functions, Trigonometry & Complex Numbers

The foundations of higher mathematics: SET THEORY, RELATIONS and FUNCTIONS, TRIGONOMETRIC functions (identities, equations, graphs), and COMPLEX NUMBERS (algebra, Argand plane, polar form, De Moivre's Theorem). ISC Class 11 Mathematics.

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

1Perform set operations (union, intersection, complement, De Morgan's laws) and solve 2-set and 3-set problems using Venn diagrams and the inclusion-exclusion formula
2Classify functions as injective, surjective, or bijective; find composite functions f∘g and verify the conditions for a function to be invertible
3State the domain and range of all six trigonometric functions; apply compound angle, double angle, and allied angle formulas to simplify expressions and prove identities
4Solve general trigonometric equations and express solutions in the form nπ + (−1)ⁿα or 2nπ ± α
5Perform complex number algebra (addition, multiplication, division, conjugate); represent complex numbers in polar form and apply De Moivre's theorem to find powers and roots; use cube roots of unity

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Why this chapter matters

ISC Class 11 Mathematics builds the abstract foundations that all Class 12 calculus and algebra depend on. Sets and functions introduce the rigorous language of mathematics — domain, range, bijection, invertibility — that reappears in Class 12 integration and matrices. Trigonometric functions at Class 11 are far deeper than Class 10: graphs, domain/range, compound angle formulas, and general solutions of equations are tested independently and as tools for solving calculus problems at Class 12. Complex numbers (Argand plane, polar form, De Moivre's Theorem, cube roots of unity) appear regularly in ISC and JEE exams and require algebraic fluency.

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Sets, Relations, Trigonometric Functions & Complex Numbers

1. Sets

Types and Operations

Union (A ∪ B): Elements in A OR B. Intersection (A ∩ B): In BOTH. Difference (A — B): In A but NOT B.
Complement (A′): Elements NOT in A. De Morgan's Laws: (A ∪ B)′ = A′ ∩ B′. (A ∩ B)′ = A′ ∪ B′.
Cardinal Number: n(A ∪ B) = n(A) + n(B) — n(A ∩ B). Power Set: 2ⁿ subsets.

Venn Diagrams — Visual representation. Essential for solving 2-set and 3-set problems.

2. Relations and Functions

Relations

A relation R from A to B is a SUBSET of A × B. Domain (first elements). Range (second elements).

Functions

A function f: A → B assigns EXACTLY ONE element of B to EACH element of A.

Types of Functions

Type	Definition
One-One (Injective)	Different inputs → DIFFERENT outputs. f(x₁)=f(x₂) ⇒ x₁=x₂.
Onto (Surjective)	Range = Codomain. Every element of B is 'hit.'
Bijective	BOTH one-one AND onto. INVERTIBLE.

Composition: (f ∘ g)(x) = f(g(x)). Inverse: f⁻¹ exists iff f is BIJECTIVE.

3. Trigonometric Functions

Angle Measurement

Degrees (360° in a circle). Radians (2π in a circle). π radians = 180°.

Domain and Range

Function	Domain	Range
sin x, cos x	R	[-1, 1]
tan x	R — {(2n+1)π/2}	R
cot x	R — {nπ}	R
sec x	R — {(2n+1)π/2}	(−∞,−1] ∪ [1,∞)
cosec x	R — {nπ}	(−∞,−1] ∪ [1,∞)

Fundamental Identities

sin²θ + cos²θ = 1. 1 + tan²θ = sec²θ. 1 + cot²θ = cosec²θ.

Compound Angle Formulas

sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB ∓ sinA sinB
tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB)

Multiple Angle

sin2A = 2sinA cosA. cos2A = cos²A — sin²A = 2cos²A — 1 = 1 — 2sin²A.

Trigonometric Equations

General solutions: sinθ=0 → θ=nπ. cosθ=0 → θ=(2n+1)π/2. tanθ=0 → θ=nπ. sinθ=sinα → θ=nπ+(−1)ⁿα.

4. Complex Numbers

Definition: z = a + ib, where a,b ∈ R, i = √(−1), i² = −1.

Algebra

Addition: (a+ib) + (c+id) = (a+c) + i(b+d)
Multiplication: (a+ib)(c+id) = (ac−bd) + i(ad+bc)
Division: Multiply numerator and denominator by CONJUGATE (a−ib).

Conjugate: z̄ = a — ib. |z| = √(a² + b²). zz̄ = |z|².

Argand Plane — x-axis = Real axis. y-axis = Imaginary axis.

Polar Form: z = r(cos θ + i sin θ), r = |z|, θ = arg(z) = tan⁻¹(b/a) (adjusted for quadrant).

De Moivre's Theorem

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). Used to find powers and roots of complex numbers.

Cube Roots of Unity (ω)

1, ω = (−1+i√3)/2, ω² = (−1−i√3)/2. 1 + ω + ω² = 0. ω³ = 1.

Key formulas & results

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

Sets — Operations and Cardinal Number

OPERATIONS: A ∪ B (union — in A OR B). A ∩ B (intersection — in BOTH). A − B (difference — in A but NOT B). A′ (complement — NOT in A). DE MORGAN'S LAWS: (A ∪ B)′ = A′ ∩ B′. (A ∩ B)′ = A′ ∪ B′. CARDINAL NUMBER: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). For three sets: n(A ∪ B ∪ C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C). POWER SET: If |A| = n, then power set has 2ⁿ subsets.

Venn diagram problems: always start by filling in the intersection of all three sets first, then work outward. In ISC set problems, the most common error is double-counting elements in the intersection.

Functions — Classification and Composition

FUNCTION: f: A → B assigns exactly one element of B to each element of A. ONE-ONE (INJECTIVE): f(x₁) = f(x₂) ⇒ x₁ = x₂. Graphically: passes horizontal line test. ONTO (SURJECTIVE): Range = Codomain — every element of B has a preimage. BIJECTIVE: Both one-one AND onto — therefore INVERTIBLE. COMPOSITION: (f ∘ g)(x) = f(g(x)). Note: f ∘ g ≠ g ∘ f in general (not commutative). But (f ∘ g) ∘ h = f ∘ (g ∘ h) (associative). INVERSE: f⁻¹ exists iff f is BIJECTIVE. (f ∘ f⁻¹)(x) = x = (f⁻¹ ∘ f)(x).

To prove a function is bijective: prove one-one SEPARATELY (assume f(x₁) = f(x₂) and show x₁ = x₂) and onto SEPARATELY (for any y in codomain, find x in domain such that f(x) = y). Only then conclude invertibility.

Trigonometry — Identities and General Solutions

FUNDAMENTAL: sin²θ + cos²θ = 1. 1 + tan²θ = sec²θ. 1 + cot²θ = cosec²θ. COMPOUND ANGLES: sin(A±B) = sinA cosB ± cosA sinB. cos(A±B) = cosA cosB ∓ sinA sinB. tan(A±B) = (tanA ± tanB)/(1 ∓ tanA tanB). DOUBLE ANGLE: sin2A = 2sinA cosA. cos2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A. tan2A = 2tanA/(1 − tan²A). PRODUCT TO SUM: 2sinA cosB = sin(A+B) + sin(A−B). GENERAL SOLUTIONS: sinθ = sinα → θ = nπ + (−1)ⁿα. cosθ = cosα → θ = 2nπ ± α. tanθ = tanα → θ = nπ + α.

HALF-ANGLE FORMS (for integration use): sin²θ = (1 − cos2θ)/2. cos²θ = (1 + cos2θ)/2. These are used constantly in Class 12 integration. DOMAIN AND RANGE: sin, cos have domain ℝ, range [−1,1]. tan has domain ℝ − {odd multiples of π/2}, range ℝ.

Complex Numbers — Algebra, Polar Form and De Moivre

z = a + ib. i = √(−1). i² = −1. i³ = −i. i⁴ = 1 (cycles of 4). CONJUGATE: z̄ = a − ib. MODULUS: |z| = √(a² + b²). zz̄ = |z|². DIVISION: multiply numerator and denominator by conjugate of denominator. ARGAND PLANE: Real axis (x), Imaginary axis (y). Point (a, b) represents z = a + ib. POLAR FORM: z = r(cosθ + i sinθ). r = |z|. θ = arg(z) = tan⁻¹(b/a) (adjust for quadrant). DE MOIVRE'S THEOREM: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ). CUBE ROOTS OF UNITY: 1, ω = (−1+i√3)/2, ω² = (−1−i√3)/2. Properties: 1 + ω + ω² = 0. ω³ = 1.

Key complex number results: |z₁z₂| = |z₁||z₂|. arg(z₁z₂) = arg(z₁) + arg(z₂). For z⁻¹: use z̄/|z|². The triangle inequality: |z₁ + z₂| ≤ |z₁| + |z₂|. Cube roots of unity are used to simplify expressions like (1+ω)³ = (−ω²)³ = −ω⁶ = −1.

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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

✗ Confusing f∘g with g∘f (thinking composition is commutative)

✓ Function composition is NOT commutative in general. (f∘g)(x) = f(g(x)) — apply g FIRST, then f. (g∘f)(x) = g(f(x)) — apply f FIRST, then g. For example, if f(x) = x² and g(x) = x+1: (f∘g)(x) = (x+1)² but (g∘f)(x) = x²+1. These are different functions. To remember: in f∘g, the RIGHTMOST function acts first (g acts on x, then f acts on the result).

WATCH OUT

✗ Writing general solution of sinθ = sinα as θ = 2nπ ± α (mixing it up with cosine)

✓ General solutions: SINE: sinθ = sinα → θ = nπ + (−1)ⁿα. This handles both θ = α and θ = π − α. COSINE: cosθ = cosα → θ = 2nπ ± α (both +α and −α are solutions since cosine is even). TANGENT: tanθ = tanα → θ = nπ + α (period π). Memory aid: Sine has period 2π but the formula uses nπ due to the (−1)ⁿ factor; cosine formula uses 2nπ because you add ±α.

WATCH OUT

✗ Calculating arg(z) as tan⁻¹(b/a) without checking the quadrant

✓ tan⁻¹(b/a) only gives the correct argument when z is in the FIRST quadrant (a > 0, b > 0). For other quadrants: Q2 (a < 0, b > 0): arg(z) = π − tan⁻¹(|b/a|). Q3 (a < 0, b < 0): arg(z) = −π + tan⁻¹(|b/a|) (or equivalently, π + tan⁻¹(b/a) taking the negative form). Q4 (a > 0, b < 0): arg(z) = −tan⁻¹(|b/a|). Always DRAW the Argand plane point to identify the quadrant before computing arg(z).

Practice problems

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

Q1EASY· sets-venn

In a survey of 100 people, 60 read newspaper A, 40 read newspaper B, and 20 read both. Find (a) those who read only A, (b) those who read only B, (c) those who read neither.

▶ Show solution

(a) Only A = n(A) − n(A∩B) = 60 − 20 = 40. (b) Only B = n(B) − n(A∩B) = 40 − 20 = 20. (c) n(A∪B) = n(A) + n(B) − n(A∩B) = 60 + 40 − 20 = 80. Neither = total − n(A∪B) = 100 − 80 = 20. Summary: 40 read only A, 20 read only B, 20 read both, 20 read neither. Check: 40 + 20 + 20 + 20 = 100 ✓

Q2MEDIUM· trigonometry-identity

Prove that: (sin 3A − sin A)/(cos A − cos 3A) = cot 2A.

▶ Show solution

Use sum-to-product formulas. Numerator: sin 3A − sin A = 2 cos[(3A+A)/2] · sin[(3A−A)/2] = 2 cos 2A · sin A. Denominator: cos A − cos 3A = −2 sin[(A+3A)/2] · sin[(A−3A)/2] = −2 sin 2A · sin(−A) = −2 sin 2A · (−sin A) = 2 sin 2A · sin A. Therefore: (sin 3A − sin A)/(cos A − cos 3A) = (2 cos 2A · sin A)/(2 sin 2A · sin A) = cos 2A/sin 2A = cot 2A. ∎

Q3HARD· complex-numbers

If 1, ω, ω² are the cube roots of unity, prove that (1 − ω + ω²)(1 + ω − ω²) = 4.

▶ Show solution

We know: 1 + ω + ω² = 0, so ω + ω² = −1, and ω³ = 1. Compute (1 − ω + ω²): use ω² = −1 − ω (from 1 + ω + ω² = 0). So 1 − ω + ω² = 1 − ω + (−1 − ω) = −2ω. Compute (1 + ω − ω²): 1 + ω − ω² = 1 + ω − (−1−ω) = 1 + ω + 1 + ω = 2 + 2ω = 2(1 + ω) = 2(−ω²) = −2ω². Therefore: (1 − ω + ω²)(1 + ω − ω²) = (−2ω)(−2ω²) = 4ω³ = 4 × 1 = 4. ∎

5-minute revision

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

•De Morgan's: (A∪B)′ = A′∩B′. (A∩B)′ = A′∪B′. Power set of n elements = 2ⁿ subsets.
•n(A∪B) = n(A)+n(B)−n(A∩B). For three sets add all three, subtract pairwise, add triple intersection.
•Bijective = one-one + onto = invertible. Composition: (f∘g)(x) = f(g(x)) — g acts first.
•sin²θ+cos²θ=1. 1+tan²θ=sec²θ. 1+cot²θ=cosec²θ.
•sin(A+B) = sinA cosB + cosA sinB. cos(A+B) = cosA cosB − sinA sinB.
•General solution: sinθ=sinα → θ=nπ+(−1)ⁿα. cosθ=cosα → θ=2nπ±α. tanθ=tanα → θ=nπ+α.
•i²=−1. i³=−i. i⁴=1. Cycles of 4. |z|=√(a²+b²). zz̄=|z|².
•De Moivre: (cosθ+i sinθ)ⁿ = cos(nθ)+i sin(nθ).
•Cube roots of unity: 1, ω, ω². 1+ω+ω²=0. ω³=1. Use to simplify expressions.
•arg(z): use tan⁻¹(b/a) then ADJUST for quadrant — Q2: π−angle; Q3: −π+angle; Q4: −angle.

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Where this shows up in the real world

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

Going beyond the textbook

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

Research Russell's Paradox and Axiomatic Set Theory (ZFC) — Bertrand Russell's 1901 paradox (the set of all sets that do not contain themselves) destroyed naive set theory and forced the development of axiomatic foundations. Investigate the Zermelo-Fraenkel axioms with Choice (ZFC) and how they avoid paradoxes.
Investigate Euler's Identity e^(iπ) + 1 = 0 — often called the most beautiful equation in mathematics. It connects five fundamental constants (0, 1, e, i, π) and three operations (addition, multiplication, exponentiation). Derive it from Euler's formula e^(ix) = cos x + i sin x using De Moivre's theorem.
Explore the Fundamental Theorem of Algebra (Gauss, 1799) — every polynomial of degree n with complex coefficients has exactly n complex roots (counting multiplicity). This guarantees that complex numbers are 'algebraically complete' — no further number system is needed to solve polynomial equations.
Research Fourier Series — periodic functions can be expressed as infinite sums of sines and cosines: f(x) = a₀ + Σ(aₙ cos nx + bₙ sin nx). The trig identities you learn in Class 11 are the foundation. Fourier analysis powers digital signal processing, MP3 compression, and quantum mechanics.

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Questions students ask

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

A function f: A → B is ONTO (surjective) if every element of the CODOMAIN B has at least one preimage in A — meaning Range = Codomain. METHOD 1 (Algebraic): For a given y in B, set f(x) = y and solve for x. If you can always find a valid x in A for every y in B, the function is onto. METHOD 2 (Range check): Find the range of f(x) and check if it equals the entire codomain. Example: f(x) = x² from ℝ → ℝ. Range = [0, ∞) ≠ ℝ (negative y values have no preimage). So NOT onto. BUT f(x) = x² from ℝ → [0,∞): range = [0,∞) = codomain. NOW it is onto. Key insight: whether a function is onto depends on BOTH the formula and the CODOMAIN specified. Changing the codomain changes the answer.

PRINCIPAL VALUE: The unique solution in a specified restricted range — typically [−π/2, π/2] for sine/tan and [0, π] for cosine. E.g., sin x = 1/2 has principal value x = π/6 (within [−π/2, π/2]). GENERAL SOLUTION: ALL solutions across the entire real line, expressed using the integer n. Since trig functions are PERIODIC, infinitely many solutions exist. For sin x = 1/2: general solution is x = nπ + (−1)ⁿ(π/6), which generates π/6, 5π/6, π/6 + 2π, 5π/6 + 2π, etc. ISC Class 11 exams usually ask for GENERAL SOLUTION. If the question specifies a range (0 ≤ x ≤ 2π), find all solutions within that range from the general formula.

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