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.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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

TypeDefinition
One-One (Injective)Different inputs → DIFFERENT outputs. f(x₁)=f(x₂) ⇒ x₁=x₂.
Onto (Surjective)Range = Codomain. Every element of B is 'hit.'
BijectiveBOTH 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

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

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Sets, Relations, Functions, Trigonometry & Complex Numbers?

3 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

3 questions~2 min

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.

ICSE marks blueprint

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

Where this shows up in the real world

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

Set theory is the mathematical foundation of databases

Set theory is the mathematical foundation of databases — SQL queries use union (OR), intersection (AND), and complement (NOT)

Function composition is the formal basis of all programmi…

Function composition is the formal basis of all programming: function pipelines, chained method calls, and functional programming are all f∘g in practice

Compound angle trigonometric formulas are used in signal …

Compound angle trigonometric formulas are used in signal processing, Fourier analysis (breaking complex signals into sine/cosine components), and AC circuit analysis

Complex numbers are essential in electrical engineering

Complex numbers are essential in electrical engineering — impedance, phasors, and AC circuit analysis are all done using complex arithmetic

De Moivre's Theorem in polar form is used in physics for …

De Moivre's Theorem in polar form is used in physics for quantum mechanical wave functions and in engineering for rotating machinery analysis

Exam strategy

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

1
For set problems in ISC: always draw a Venn diagram. For function proofs: state your method ('To prove f is one-one, assume f(x₁) = f(x₂)...'), then carry through algebraically, then conclude. For trigonometric identity proofs: start from the more complex side and work toward the simpler — do NOT work on both sides simultaneously. For general solutions: check your formula by substituting specific values of n (n=0, 1, −1) and verifying the solutions make sense. For complex number problems: sketch the Argand plane point before computing modulus/argument.

Going beyond the textbook

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

STRETCH
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.
STRETCH
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.
STRETCH
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.
STRETCH
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.

Where else this chapter is tested

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

ISC Class 11 Mathematics School Exam
ISC Class 12 Mathematics Board
JEE Main Mathematics
JEE Advanced Mathematics
BITSAT Mathematics

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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