CHAPTER ONE

Elementary Set Theory

A set is a well-defined collection of distinct objects considered as a single entity. The objects belonging to the set are called its elements or members.

The following are sets:

1. The vowels in the English alphabet: {a, e, i, o, u}.
2. The positive integers less than 6: {1, 2, 3, 4, 5}.
3. The solutions of x² - 5x + 6 = 0: {2, 3}.
4. The months having exactly 30 days: {April, June, September, November}.

A set records membership, not frequency. An element either belongs to a set or it does not. Writing an element several times does not alter the set.

Do not confuse ∈ with ⊆. The symbol ∈ relates an element to a set, whereas ⊆ relates one set to another set.

For example, if A = {1,2,3}, then 1 ∈ A, while {1} ⊆ A.

Let A be the set of positive odd integers less than 10. Then

The set of positive even integers less than 12 may be written as

In roster form, this is

An empty set is a set containing no element. It is denoted by ∅ or { }.

∅ and {∅} are not the same set. The set ∅ has no element, whereas {∅} has one element, namely the empty set itself.

A universal set is the set containing all objects under consideration in a particular discussion. It is commonly denoted by U.

Two sets A and B are equal if and only if they contain exactly the same elements.

Let

Then A = B. The different order of listing does not affect equality.

Consider

Each set contains three elements. They are therefore equivalent in cardinality, but they are not equal because their elements are different.

A set A is a subset of a set B if every element of A is also an element of B. This is written as A ⊆ B.

If

then A ⊂ B because all members of A occur in B and A ≠ B.

Textbooks are not completely uniform in their use of ⊂. Some use it for any subset, while others reserve it for a proper subset. In this material, ⊆ denotes “subset” and ⊂ denotes “proper subset.”

The power set of a set A is the set of all subsets of A. It is denoted by P(A) or sometimes 𝒫(A).

Question: Find the power set of A = {a,b}.

Given: A = {a,b}.

Required: P(A).

Solution:

The subsets of A are:

• the empty set ∅;
• the singleton set {a};
• the singleton set {b};
• the set {a,b} itself.

Answer: P(A) = {∅, {a}, {b}, {a,b}}.

Question: How many subsets does A = {2,4,6,8,10} have?

Given: n = 5.

Required: Number of subsets.

Formula:

Where: n is the number of elements in the original set.

Solution:

Answer: The set has 32 subsets.

The cardinality of a finite set is the number of distinct elements in the set. The cardinality of A is commonly denoted by n(A) or |A|.

Question: Determine the cardinality of B = {1,2,2,3,3,3,4}.

Given: B = {1,2,2,3,3,3,4}.

Required: n(B).

Solution: Repeated entries do not count as additional elements. Hence the set is simply

Therefore,

Answer: The cardinality of B is 4.

1. A set is a well-defined collection of distinct objects called elements or members.
2. Membership is represented by ∈, while non-membership is represented by ∉.
3. Sets may be represented by roster form, set-builder form or an unambiguous verbal description.
4. Important types of sets include empty, singleton, finite, infinite, universal and disjoint sets.
5. Two sets are equal when they contain exactly the same elements.
6. Equivalent finite sets contain the same number of elements, although their elements may differ.
7. A ⊆ B means every element of A is also an element of B.
8. A proper subset of a set is a subset that is not equal to the original set.
9. The power set P(A) consists of all subsets of A.
10. A finite set containing n elements has 2ⁿ subsets.
11. The cardinality of a finite set is the number of distinct elements it contains.

Review Exercises

1. Define a set and explain why the phrase “well-defined” is important.
2. Write the set of positive multiples of 3 less than 20 in roster form and set-builder form.
3. State whether each of the following is finite, infinite, singleton or empty where applicable:
   1. {7}
   2. {x ∈ ℕ : x < 0}
   3. {1,2,3,...,100}
   4. {2,4,6,8,...}
4. If A = {1,2,3}, list all the subsets of A.
5. Find the power set of {a,b}.
6. A set contains six elements. How many subsets does it have? How many of these are proper subsets?
7. Distinguish clearly between equal sets and equivalent finite sets, giving one example of each.
8. Let A = {2,4,6} and B = {1,2,3,4,5,6}. Determine whether A ⊆ B and justify your answer.
9. Explain the difference between ∅ and {∅}. State the cardinality of each.
10. If P(A) contains 128 elements, determine the number of elements in A.

CHAPTER TWO

Operations on Sets and Venn Diagrams

The union of two sets A and B is the set of all elements that belong to A, to B, or to both sets. It is denoted by A ∪ B.

Let

Then

Notice that 3 and 4 are written only once. Repetition does not affect the membership of a set.

The intersection of two sets A and B is the set containing all elements that belong simultaneously to both A and B. It is denoted by A ∩ B.

If

then

Only 4 and 8 occur in both sets.

If

then

The difference of A and B, written A \ B or A - B, is the set of elements that belong to A but do not belong to B.

Let

Then

while

Let A be a subset of a universal set U. The complement of A is the set of all elements of U that do not belong to A. It is denoted by A^c or A′.

Let

and

Then

A complement cannot be determined correctly unless the universal set is known or clear from the context. For example, if A = {1,2}, the elements of A^c depend entirely on the chosen universal set.

Question: Verify the first De Morgan law for

Required: Verify that (A ∪ B)^c = A^c ∩ B^c.

Solution:

Therefore, relative to U,

Also,

and

Hence,

Thus,

Answer: The identity is verified for the given sets.

A Venn diagram is a diagrammatic representation of sets and their relationships, usually using a rectangle for the universal set and closed curves, commonly circles, for subsets.

Two overlapping sets A and B within a universal set. The overlapping region represents A ∩ B.

Question: In a class, 32 students study Mathematics, 25 study Physics, and 12 study both subjects. How many students study at least one of the two subjects?

Given:

n(M) = 32

n(P) = 25

n(M ∩ P) = 12

Required: n(M ∪ P).

Formula:

Where: M ∪ P represents students studying Mathematics or Physics or both.

Solution:

Answer: 45 students study at least one of the two subjects.

Question: A survey of 80 students showed that 45 students use an electronic textbook, 38 use printed textbooks, and 20 use both. Determine:

1. the number who use electronic textbooks only;
2. the number who use printed textbooks only;
3. the number who use at least one of the two forms;
4. the number who use neither form.

Given:

Required: The four specified categories.

Solution:

(a) Electronic textbooks only

(b) Printed textbooks only

(c) At least one form

(d) Neither form

Answer: Electronic only = 25; printed only = 18; at least one = 63; neither = 17.

A useful check is to add the mutually exclusive regions:

Since this equals the total number surveyed, the allocation is consistent.

Question: In a group of 70 students, 42 offer Economics and 36 offer Mathematics. If every student offers at least one of these subjects, how many offer both?

Given:

Required: n(E ∩ M).

Formula:

Solution:

Answer: 8 students offer both Economics and Mathematics.

Unless a problem explicitly says otherwise, a statement such as “20 students study Mathematics and Physics” normally includes students who may also study a third subject. Thus, in a three-set problem, n(M ∩ P) includes the triple intersection M ∩ P ∩ C.

Question: In a survey of university students, 50 study Mathematics, 45 study Physics and 40 study Chemistry. Twenty study both Mathematics and Physics, 18 study both Mathematics and Chemistry, 15 study both Physics and Chemistry, while 8 study all three subjects. How many students study at least one of the three subjects?

Given:

Required: n(M ∪ P ∪ C).

Formula:

Solution:

Answer: 90 students study at least one of the three subjects.

Question: Among 100 students, 55 take Mathematics, 50 take Physics and 45 take Chemistry. Twenty-five take Mathematics and Physics, 20 take Mathematics and Chemistry, 18 take Physics and Chemistry, and 10 take all three subjects. Determine:

1. the number taking Mathematics and Physics only;
2. the number taking only Mathematics;
3. the number taking at least one subject;
4. the number taking none of the three subjects.

Given:

Solution:

(a) Mathematics and Physics only

The stated intersection M ∩ P includes the 10 students who take all three subjects. Therefore,

(b) Mathematics only

First calculate the pairwise-only regions involving Mathematics:

Therefore,

(c) At least one subject

(d) None of the subjects

Answer: Mathematics and Physics only = 15; Mathematics only = 20; at least one subject = 97; none = 3.

If 30 students belong to A and 25 belong to B, it is not generally correct to conclude that 55 students belong to A ∪ B. Students belonging to both sets would have been counted twice.

A ∪ B includes elements in either set or both; A ∩ B contains only elements common to both.

In a three-set problem, A ∩ B normally includes members that may also belong to C. If the question asks for those in A and B only, the triple intersection must be excluded.

A complement is always relative to a universal set. Without identifying the universe, the complement may be ambiguous.

“Neither A nor B” means elements outside both sets, namely (A ∪ B)^c. It does not mean A ∩ B.

1. The union A ∪ B contains elements belonging to A, B, or both.
2. The intersection A ∩ B contains elements common to both sets.
3. Two sets are disjoint when their intersection is empty.
4. The difference A \ B contains elements in A that are not in B.
5. The complement A^c consists of elements of the universal set that are not in A.
6. Union and intersection satisfy commutative, associative, distributive and other standard laws.
7. De Morgan's laws are (A ∪ B)^c = A^c ∩ B^c and (A ∩ B)^c = A^c ∪ B^c.
8. Venn diagrams represent sets and their relationships visually.
9. For two finite sets, n(A ∪ B)=n(A)+n(B)-n(A ∩ B).
10. For three finite sets, pairwise intersections are subtracted and the triple intersection is added back once.
11. In practical Venn diagram problems, mutually exclusive regions should be identified carefully before totals are calculated.

Review Exercises

1. Let A={1,2,3,4,5} and B={4,5,6,7,8}. Find:
   1. A ∪ B
   2. A ∩ B
   3. A \ B
   4. B \ A
2. Given U={1,2,3,4,5,6,7,8,9,10} and A={1,3,5,7,9}, find A^c.
3. If n(A)=28, n(B)=35 and n(A∩B)=12, find n(A∪B).
4. In a class of 60 students, 35 study French, 28 study Arabic and 12 study both languages. Find the number who:
   1. study French only;
   2. study Arabic only;
   3. study at least one language;
   4. study neither language.
5. In a survey of 120 people, 75 like football, 60 like basketball and 35 like both. Determine the number who like:
   1. football only;
   2. basketball only;
   3. at least one of the sports;
   4. neither sport.
6. State De Morgan's two laws for sets and verify one of them using suitable finite sets.
7. In a group of students, 48 take Mathematics, 40 take Statistics and 15 take both subjects. If 20 students take neither subject, determine the total number of students in the group.
8. In a survey, 60 people read newspaper A, 50 read newspaper B, and 30 read both. If 90 people were surveyed, determine how many read neither newspaper.
9. In a class, 40 students take Mathematics, 35 take Physics and 30 take Chemistry. Fifteen take Mathematics and Physics, 12 take Mathematics and Chemistry, 10 take Physics and Chemistry, and 5 take all three. Find the number of students who take at least one of the three subjects.
10. A survey of 150 students showed that 80 use platform A, 70 use platform B and 60 use platform C. Thirty use both A and B, 25 use both A and C, 20 use both B and C, while 10 use all three platforms.
    1. How many use at least one platform?
    2. How many use none of the three platforms?
    3. How many use A and B only?
    4. How many use only A?

CHAPTER THREE

The Real Number System

The real number system consists of all numbers that can be represented as points on the real number line. It includes both rational and irrational numbers.

Since 5 + 8 = 13 and 4 × 7 = 28, the results remain in ℕ.

The integers are the positive whole numbers, zero and their negative counterparts.

Zero is an integer and a whole number. It is neither positive nor negative.

A rational number is any number that can be expressed in the form p/q, where p and q are integers and q ≠ 0.

1/4 = 0.25 is terminating.

2/3 = 0.666... is recurring.

5/11 = 0.454545... is recurring.

Question: Express 0.777... as a rational number.

Given: x = 0.777...

Required: Express x as a fraction.

Solution:

Multiply both sides by 10:

Subtract the first equation from the second:

Therefore,

Answer: 0.777... = 7/9.

Question: Express 0.272727... as a fraction.

Given: x = 0.272727...

Required: Rational form of x.

Solution:

Since two digits repeat, multiply by 100:

Subtract:

Answer: 0.272727... = 3/11.

An irrational number is a real number that cannot be expressed as a ratio p/q of two integers with q ≠ 0.

√16 = 4, which is rational.

√49 = 7, which is rational.

√7 is irrational because 7 is not a perfect square.

The set of real numbers, denoted by ℝ, is the set containing all rational and all irrational numbers.

Question: Classify each of the following numbers as specifically as possible: 8, -5, 3/4, √25, √10, 0.125, π.

Solution:

• 8 is natural, whole, integer, rational and real.
• -5 is integer, rational and real.
• 3/4 is rational and real.
• √25 = 5, so it is natural, whole, integer, rational and real.
• √10 is irrational and real.
• 0.125 = 1/8, so it is rational and real.
• π is irrational and real.

Zero has no multiplicative inverse because there is no real number x for which 0 × x = 1. Division by zero is therefore undefined.

Question: Express -2 ≤ x < 5 in interval notation.

Given: -2 ≤ x < 5.

Required: Interval notation.

Solution:

Since -2 is included, use a square bracket at the left endpoint. Since 5 is excluded, use a round bracket at the right endpoint.

Answer: [-2,5).

The absolute value of a real number x, denoted by |x|, is its distance from zero on the real number line.

|7| = 7.

|-7| = 7.

|0| = 0.

Question: Solve |x| = 6.

Required: Values of x.

Principle: If |x| = a and a > 0, then x = a or x = -a.

Solution:

Answer: x = ±6.

Question: Solve |2x-1| = 7.

Given: |2x-1|=7.

Required: Values of x.

Solution:

There are two cases:

or

From the first equation,

From the second equation,

Answer: x=4 or x=-3.

A surd is an irrational root written in exact radical form, such as √2, √5 or ∛7.

Question: Simplify √72.

Given: √72.

Required: Simplest surd form.

Solution:

Answer: 6√2.

Question: Rationalise 3/√5.

Given: 3/√5.

Required: Equivalent expression with rational denominator.

Solution:

Answer: 3√5/5.

Terminating decimals and recurring decimals are rational. For example, 0.125 = 1/8 and 0.333... = 1/3.

√36 = 6 is rational. A square root becomes irrational only when it cannot be reduced to a rational number.

Expressions such as 5/0 are undefined. Zero cannot be used as a divisor.

On the real number line, -8 < -3, not the other way round.

1. The natural numbers are the positive counting numbers.
2. Whole numbers consist of zero and the positive counting numbers.
3. Integers include negative whole numbers, zero and positive whole numbers.
4. A rational number can be expressed as p/q, where p and q are integers and q ≠ 0.
5. Rational numbers have terminating or recurring decimal expansions.
6. Irrational numbers cannot be written as ratios of integers and have non-terminating, non-repeating decimal expansions.
7. Rational and irrational numbers together form the real numbers.
8. The principal inclusion relationship is ℕ ⊂ W ⊂ ℤ ⊂ ℚ ⊂ ℝ.
9. Real numbers satisfy closure, commutative, associative and distributive properties.
10. The real number line gives a geometric representation of ordering among real numbers.
11. Intervals may be open, closed, half-open or infinite.
12. The absolute value of a number is its distance from zero.
13. Surds are exact irrational root expressions and may often be simplified or rationalised.

Review Exercises

1. Define each of the following: natural number, integer, rational number, irrational number and real number.
2. Classify each number as specifically as possible:
   1. 12
   2. -9
   3. 0
   4. 7/11
   5. √49
   6. √13
   7. π
3. Express 0.666... as a fraction.
4. Express 0.181818... as a fraction.
5. Explain why every integer is rational.
6. State the commutative, associative and distributive laws for real numbers.
7. Express -4 < x ≤ 6 in interval notation.
8. Write the interval [2,9) as an inequality.
9. Solve |x|=11.
10. Solve |3x+2|=8.
11. Simplify √48.
12. Simplify 2√3 + 5√3 - √3.
13. Rationalise 4/√7.
14. State whether each statement is true or false and justify your answer:
    1. Every rational number is an integer.
    2. Every integer is rational.
    3. Every irrational number is real.
    4. Every square root is irrational.

CHAPTER FOUR

Mathematical Induction

Mathematical induction is a method of proof used to establish that a proposition P(n) is true for all integers n from a specified starting value onward.

The assumption that P(k) is true is called the inductive hypothesis. It is not an assumption that the entire theorem is already proved. It is a temporary assumption about one arbitrary case, used to establish the next case.

Checking only the base case does not prove the result. Likewise, proving only the inductive step without establishing the first case is incomplete.

Question: Prove by mathematical induction that

for every positive integer n.

Required: Proof by mathematical induction.

Solution:

Base case: n = 1

Left-hand side:

Right-hand side:

Therefore the result is true for n=1.

Inductive hypothesis

Assume the formula is true for n=k. Then

Inductive step

For n=k+1, the left-hand side becomes

Using the inductive hypothesis:

Factor out k+1:

This is exactly the required form with n=k+1, because

Conclusion: Therefore, by mathematical induction,

for all positive integers n.

Question: Prove that

for all positive integers n.

Solution:

Base case: n = 1

The statement is true for n=1.

Inductive hypothesis

Assume

Inductive step

The next odd number after 2k-1 is

Therefore, for n=k+1,

Using the inductive hypothesis:

Thus the statement is true for k+1.

Conclusion: By mathematical induction,

for every positive integer n.

Question: Prove that

Solution:

Base case: n = 1

Hence the result is true for n=1.

Inductive hypothesis

Assume

Inductive step

For n=k+1,

Using the inductive hypothesis:

Factor out k+1:

Place the terms over a common denominator:

Simplify:

Factor the quadratic expression:

Hence,

Since 2k+3 = 2(k+1)+1, this is the required expression

Conclusion: Therefore the formula holds for all positive integers n.

An integer a is divisible by a non-zero integer m if there exists an integer q such that a = mq.

Question: Prove that 7ⁿ-1 is divisible by 6 for every positive integer n.

Required: Show that 6 | (7ⁿ-1).

Solution:

Base case: n = 1

which is divisible by 6.

Inductive hypothesis

Assume that 7^k-1 is divisible by 6. Then for some integer m,

Therefore,

Inductive step

Consider

Then

Using 7^k=6m+1,

Since 7m+1 is an integer, 7^k+1-1 is divisible by 6.

Conclusion: By mathematical induction, 7ⁿ-1 is divisible by 6 for all positive integers n.

Question: Prove that 5ⁿ-1 is divisible by 4 for all positive integers n.

Solution:

Base case: n = 1

which is divisible by 4.

Inductive hypothesis

Assume

for some integer m. Then

Inductive step

Hence the expression is divisible by 4.

Conclusion: 5ⁿ-1 is divisible by 4 for every positive integer n.

Question: Prove that 2ⁿ ≥ n+1 for all positive integers n.

Solution:

Base case: n = 1

and

Hence 2 ≥ 2, so the statement is true for n=1.

Inductive hypothesis

Assume

Inductive step

Then

Using the inductive hypothesis,

Since k ≥ 1,

Therefore,

This is precisely the required statement for n=k+1, because

Conclusion: Hence 2ⁿ ≥ n+1 for all positive integers n.

Question: Prove that n² ≥ 2n-1 for all positive integers n.

Solution:

Base case: n = 1

and

Therefore the statement is true for n=1.

Inductive hypothesis

Assume

Inductive step

We need to prove

Expand the left-hand side:

By the inductive hypothesis,

For k ≥ 1,

But

Hence

Conclusion: The result holds for all positive integers n.

Question: Prove that 2ⁿ > n² for all integers n ≥ 5.

Solution:

Base case: n = 5

while

Therefore, 32 > 25, so the result is true for n=5.

Inductive hypothesis

Assume for some integer k ≥ 5 that

Inductive step

Using the inductive hypothesis,

It remains to compare 2k² with (k+1)².

We require

Expanding:

This is equivalent to

For k ≥ 5, this inequality is certainly true; for example,

Therefore,

Conclusion: Hence 2ⁿ > n² for every integer n ≥ 5.

The base case must match the first integer for which the theorem is claimed. If a result is stated for n ≥ 4, checking only n=1 does not establish the required starting point.

Verifying P(1), P(2), P(3) and P(4) may suggest a pattern, but it does not prove that the statement holds for every positive integer.

The inductive implication P(k) ⇒ P(k+1) does not by itself establish any particular case. A starting value must first be proved.

In the inductive step, one may assume P(k), but one must not assume P(k+1). The purpose of the step is to prove P(k+1).

A valid induction proof normally requires the k-case assumption to establish the k+1 case. If the hypothesis is never used, the argument may not be an induction proof at all.

In summation problems, it is often best to begin the inductive step with the left-hand side for k+1, split off the final term, and then use the inductive hypothesis.

Question: Prove by mathematical induction that

for every positive integer n.

Solution:

Base case: n = 1

True for n=1.

Inductive hypothesis

Assume

Inductive step

For n=k+1,

Using the inductive hypothesis,

Factor:

This is exactly

Conclusion: Hence 2+4+...+2n=n(n+1) for all positive integers n.

Question: Prove that

Solution:

Base case: n = 1

The statement is true for n=1.

Inductive hypothesis

Assume

Inductive step

Using the inductive hypothesis:

Factor:

This is the required expression for n=k+1.

Conclusion: The result holds for all positive integers n.

Question: Prove that 8ⁿ-1 is divisible by 7 for every positive integer n.

Solution:

Base case: n = 1

which is divisible by 7.

Inductive hypothesis

Assume

for some integer m. Thus,

Inductive step

Therefore 8^k+1-1 is divisible by 7.

Conclusion: Hence 8ⁿ-1 is divisible by 7 for every positive integer n.

1. Mathematical induction is a rigorous method for proving statements about consecutive integers.
2. The method requires a base case and an inductive step.
3. The inductive hypothesis assumes that the proposition is true for an arbitrary integer k.
4. The inductive step proves that truth at k implies truth at k+1.
5. A correct induction proof must include a clear conclusion.
6. Induction is commonly used to prove summation formulae.
7. It can also establish divisibility properties and inequalities.
8. If a proposition begins at an integer other than 1, the base case must use that specified starting integer.
9. Checking several numerical cases does not replace a proof.
10. The inductive hypothesis must not be confused with the statement to be proved for k+1.

Review Exercises

1. State the principle of mathematical induction clearly.
2. Explain the meaning of each of the following:
   1. base case;
   2. inductive hypothesis;
   3. inductive step.
3. Prove by mathematical induction that
   1+2+3+...+n = n(n+1)/2.
4. Prove by mathematical induction that
   2+4+6+...+2n = n(n+1).
5. Prove that
   1+3+5+...+(2n-1)=n²
   for every positive integer n.
6. Prove that
   1+4+7+...+(3n-2)=n(3n-1)/2.
7. Prove that 6ⁿ-1 is divisible by 5 for every positive integer n.
8. Prove that 9ⁿ-1 is divisible by 8 for every positive integer n.
9. Prove that 3ⁿ ≥ 2n+1 for every positive integer n.
10. Prove that n²+n is even for every positive integer n.
11. Prove that
    1+2+2²+...+2ⁿ = 2ⁿ⁺¹-1
    for every non-negative integer n.
12. A student verifies a proposition for n=1,2,3,4,5 and concludes that it is true for every positive integer. Explain why this is not a valid proof.
13. A theorem is claimed to hold for all integers n ≥ 4. Which value should be used in the base step, and why?
14. Identify and correct the logical error in the following statement: “Assume the theorem is true for k+1. Therefore it is true for k+1.”

CHAPTER FIVE

Real Sequences

A real sequence is an ordered list of real numbers arranged according to a definite rule or pattern. Each number in the sequence is called a term.

A sequence is ordered. Thus, 1, 2, 3 and 3, 2, 1 are different sequences even though they contain the same numbers.

Consider the sequence

The general term is

Hence the 20th term is

Question: Given the sequence defined by a_n = 3n - 2, find the first five terms.

Given: a_n = 3n - 2.

Required: The first five terms.

Formula:

Where: n = 1,2,3,4,5.

Solution:

Answer: The first five terms are 1, 4, 7, 10, 13.

A finite sequence is a sequence having a definite last term.

An infinite sequence is a sequence that continues indefinitely and has no final term.

An arithmetic sequence, also called an arithmetic progression, is a sequence in which the difference between any term and the preceding term is constant.

Consider

The differences are

Therefore, the common difference is

Question: Find the 25th term of the arithmetic sequence 4, 7, 10, 13, ....

Given:

Required: a₂₅.

Formula:

Solution:

Answer: The 25th term is 76.

Question: Which term of the arithmetic sequence 8,13,18,23,... is 103?

Given:

Required: n.

Formula:

Solution:

Answer: 103 is the 20th term.

Question: The 5th term of an arithmetic sequence is 18 and the 11th term is 42. Find the first term and the common difference.

Given:

Required: a and d.

Formula:

Solution:

For the 5th term:

For the 11th term:

Subtract the first equation from the second:

Substitute into a+4d=18:

Answer: First term a=2; common difference d=4.

An arithmetic mean between two numbers is a number inserted between them so that the resulting terms form an arithmetic sequence.

Question: Insert one arithmetic mean between 12 and 28.

Given: a=12, b=28.

Required: Arithmetic mean A.

Formula:

Solution:

Answer: The arithmetic mean is 20.

Question: Insert four arithmetic means between 5 and 30.

Given: a=5, b=30, m=4.

Required: Four arithmetic means.

Formula:

Solution:

The sequence is therefore

Answer: The arithmetic means are 10, 15, 20 and 25.

A geometric sequence, or geometric progression, is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed non-zero constant.

Consider

Then

Hence the common ratio is

Question: Find the 8th term of the geometric sequence 3,6,12,24,....

Given:

Required: a₈.

Formula:

Solution:

Answer: The 8th term is 384.

Question: Which term of 5,10,20,40,... is 640?

Given:

Required: n.

Formula:

Solution:

Since 128=2⁷,

Answer: 640 is the 8th term.

Question: The 3rd term of a geometric sequence is 12 and the 6th term is 96. Find the common ratio and first term.

Given:

Required: a and r.

Formula:

Solution:

For the 3rd term:

For the 6th term:

Divide the second equation by the first:

Substitute into ar²=12:

Answer: First term a=3; common ratio r=2.

A geometric mean between two positive numbers is a number inserted between them so that the three numbers form a geometric sequence.

Question: Find the positive geometric mean between 4 and 25.

Given: a=4, b=25.

Required: G.

Formula:

Solution:

Answer: The positive geometric mean is 10.

Question: A student saves ₦5,000 in the first month and increases the amount saved each subsequent month by ₦1,500. How much does the student save in the 12th month?

Given:

Required: Amount saved in the 12th month.

Formula:

Solution:

Answer: The student saves ₦21,500 in the 12th month.

Question: A culture initially contains 200 bacteria and doubles during each fixed growth period. How many bacteria are present at the beginning of the 7th period?

Given:

Required: 7th term.

Formula:

Solution:

Answer: There are 12,800 bacteria at the beginning of the 7th period.

An arithmetic sequence is identified by repeated addition or subtraction. A geometric sequence is identified by repeated multiplication or division.

The nth term formulae are a+(n-1)d and ar^n-1. The exponent or number of common differences is n-1, not n.

A sequence such as 1,4,9,16,25,... is neither arithmetic nor geometric because it has neither a constant difference nor a constant ratio.

• A sequence is an ordered list of terms.
• The nth term gives the value of a sequence at position n.
• A finite sequence has a final term, while an infinite sequence continues indefinitely.
• An arithmetic sequence has a constant common difference.
• The nth term of an arithmetic sequence is a_n=a+(n-1)d.
• An arithmetic mean between a and b is (a+b)/2.
• A geometric sequence has a constant common ratio.
• The nth term of a geometric sequence is a_n=ar^n-1.
• The positive geometric mean between positive numbers a and b is √(ab).
• Sequences may be described explicitly or recursively.
• Arithmetic sequences model constant additive change, while geometric sequences model constant multiplicative change.

Review Exercises

1. Define a sequence and explain how it differs from a set.
2. Write the first six terms of the sequence a_n=4n+1.
3. Find the 30th term of 7,11,15,19,....
4. Which term of 3,8,13,18,... is 98?
5. The 4th term of an arithmetic sequence is 17 and the 10th term is 41. Find the first term and common difference.
6. Insert five arithmetic means between 3 and 27.
7. Find the 9th term of 2,6,18,54,....
8. Which term of 4,12,36,108,... is 972?
9. The 2nd term of a geometric sequence is 6 and the 5th term is 162. Find the common ratio and first term.
10. Find the positive geometric mean between 9 and 25.
11. Determine whether each sequence is arithmetic, geometric or neither:
    1. 4,9,14,19,...
    2. 3,12,48,192,...
    3. 1,4,9,16,...
12. A worker earns ₦80,000 in the first month of a temporary contract and receives an increase of ₦5,000 each month. Find the monthly earning in the 10th month.
13. A quantity begins at 500 and decreases by a factor of 1/2 during each stage. Find the 8th term of the resulting geometric sequence.

CHAPTER SIX

Real Series

A series is the indicated sum of the terms of a sequence.

3,6,9,12,... is an arithmetic sequence, whereas 3+6+9+12+... is an arithmetic series.

The expression

means

Question: Evaluate ∑_k=1⁴(2k+1).

Given: General term 2k+1, with k=1 to 4.

Required: The value of the sum.

Solution:

Answer: 24.

A finite series contains a finite number of terms.

An infinite series is a sum containing infinitely many terms.

An arithmetic series is the sum of the terms of an arithmetic sequence.

Question: Find the sum of the first 30 terms of the arithmetic sequence 4,7,10,13,....

Given:

Required: S₃₀.

Formula:

Solution:

Answer: The sum is 1425.

Question: Find the sum of the integers from 1 to 100.

Given:

Required: S₁₀₀.

Formula:

Solution:

Answer: The sum is 5050.

Question: Find the sum of the arithmetic series 7+12+17+...+102.

Given:

Required: Sum of the series.

Solution:

First find the number of terms:

Now use

Answer: The sum is 1090.

Question: A student saves ₦10,000 in the first month and increases the monthly saving by ₦2,000 each month. Find the total amount saved in 12 months.

Given:

Required: Total savings after 12 months.

Formula:

Solution:

Answer: The total amount saved is ₦252,000.

A geometric series is the sum of the terms of a geometric sequence.

Question: Find the sum of the first 8 terms of 3+6+12+24+....

Given:

Required: S₈.

Formula:

Solution:

Answer: The sum is 765.

Question: Find the sum of the first 6 terms of 64+32+16+8+....

Given:

Required: S₆.

Formula:

Solution:

Answer: The sum of the first six terms is 126.

Question: Find the sum to infinity of

Given:

Required: S_∞.

Formula:

Condition: |r|<1. Here, |1/2|<1, so the series converges.

Solution:

Answer: The sum to infinity is 16.

Question: Find the sum to infinity of 12-6+3-3/2+....

Given:

Required: S_∞.

Formula:

Condition:

Solution:

Answer: The sum to infinity is 8.

An infinite series is said to converge if its sequence of partial sums approaches a finite real number.

An infinite series diverges if its partial sums do not approach a finite limit.

1+1/2+1/4+1/8+... converges because |r|=1/2<1.

1+2+4+8+... diverges because r=2.

1-1+1-1+... does not converge because r=-1, so |r|=1.

The formula S_∞=a/(1-r) must not be used unless |r|<1.

Question: Express 0.777... as a fraction using an infinite geometric series.

Given:

Required: Fractional form.

Identify the geometric series:

Formula:

Solution:

Answer: 0.777...=7/9.

Question: A ball is dropped from a height of 10 m. After each bounce it rises to 3/5 of the height from which it fell. Find the total vertical distance travelled if the bouncing continues indefinitely.

Given:

Required: Total vertical distance.

Reasoning:

The initial fall of 10 m occurs only once. Thereafter, each rebound height is travelled twice: once upward and once downward.

The rebound heights form

which is an infinite geometric series with

Formula:

Solution:

This 15 m represents the total of all upward rebound heights. These heights are also travelled downward, so

Answer: The total vertical distance travelled is 40 m.

Question: A machine worth ₦800,000 loses 10% of its value each year. What is its value after 4 years under this model?

Given:

Required: Value after four successive annual reductions.

Formula:

Solution:

Answer: The machine is worth ₦524,880 after 4 years under the stated depreciation model.

a_n usually denotes the nth term, whereas S_n denotes the sum of the first n terms.

Always identify whether the pattern is based on a constant difference or a constant ratio before choosing a formula.

When only the first and last terms are given, it may first be necessary to determine the number of terms before applying a sum formula.

The infinite geometric series formula is valid only when |r|<1.

In bouncing-ball problems, the original fall normally occurs only once. Rebound heights are usually traversed both upward and downward.

• A series is the sum of the terms of a sequence.
• A finite series has finitely many terms, while an infinite series continues indefinitely.
• Sigma notation provides a compact method of writing sums.
• An arithmetic series is formed from an arithmetic sequence.
• The sum of the first n terms of an arithmetic series is S_n=n(a+l)/2.
• An equivalent arithmetic-series formula is S_n=n[2a+(n-1)d]/2.
• A geometric series is formed from a geometric sequence.
• For r≠1, the finite geometric sum is S_n=a(1-rⁿ)/(1-r).
• An infinite geometric series converges only when |r|<1.
• For a convergent infinite geometric series, S_∞=a/(1-r).
• Recurring decimals and repeated percentage-change models provide useful applications of geometric ideas.
• In applied problems, formulas must be interpreted according to the actual physical or financial quantities being accumulated.

Review Exercises

1. Distinguish between a sequence and a series, giving one example of each.
2. Expand and evaluate:
   ∑_k=1⁶(3k-1).
3. Find the sum of the first 25 terms of 6,10,14,18,....
4. Find the sum of
   9+14+19+...+104.
5. Find the sum of the integers from 20 to 80 inclusive.
6. A worker saves ₦15,000 in the first month and increases the amount by ₦2,500 monthly. Find the total saved in the first 10 months.
7. Find the sum of the first 7 terms of
   2+6+18+54+...
8. Find the sum of the first 8 terms of
   128+64+32+16+...
9. Find the sum to infinity of
   9+3+1+1/3+...
10. Determine whether each series converges or diverges:
    1. 1+1/3+1/9+...
    2. 2+4+8+16+...
    3. 5-2.5+1.25-...
    4. 1-1+1-1+...
11. Express 0.444... as a fraction by interpreting it as an infinite geometric series.
12. Express 0.272727... as a fraction.
13. A ball is dropped from a height of 20 m and rebounds to one-half of its previous height after every impact. Find the total distance travelled if the motion continues indefinitely.
14. A piece of equipment is worth ₦1,200,000 and depreciates by 8% each year. Calculate its value after 5 years under the geometric depreciation model.
15. Explain why the expression S_∞=a/(1-r) cannot be used for the series 3+6+12+24+....

CHAPTER SEVEN

Theory of Quadratic Equations

A quadratic equation in one variable is an equation that can be written in the form

where a, b and c are real constants and a ≠ 0.

The equation

becomes

A root or solution of a quadratic equation is a value of the variable that makes the equation true.

Consider

If x = 2,

If x = 3,

Therefore, 2 and 3 are roots.

If

then

Question: Solve x² - 7x + 12 = 0 by factorisation.

Given:

Required: Values of x.

Solution:

We need two numbers whose product is 12 and whose sum is -7. These numbers are -3 and -4.

Therefore,

Hence,

Answer: x = 3 or x = 4.

Question: Solve 2x² + 7x + 3 = 0.

Given:

Required: Roots of the equation.

Solution:

Multiply the coefficient of x² by the constant:

We need two numbers whose product is 6 and whose sum is 7. They are 6 and 1.

Group the terms:

Thus,

Answer: x=-1/2 or x=-3.

Question: Solve x² + 6x - 7 = 0 by completing the square.

Given:

Required: Values of x.

Solution:

Move the constant term to the right:

Half the coefficient of x is 3, so add 3² = 9 to both sides:

Take square roots:

Hence,

Answer: x=1 or x=-7.

Question: Solve 2x² - 8x - 6 = 0 by completing the square.

Solution:

Divide through by 2:

Move the constant term:

Half of -4 is -2, and its square is 4:

Hence,

Answer: x=2+√7 or x=2-√7.

Question: Solve 3x² - 5x - 2 = 0 using the quadratic formula.

Given:

Required: Roots of the equation.

Formula:

Solution:

Therefore,

or

Answer: x=2 or x=-1/3.

Question: Solve 2x² + 4x + 7 = 0.

Given:

Formula:

Solution:

Since the quantity under the square root is negative, the equation has no real roots. In the complex number system,

Hence,

Answer: The equation has no real roots; its complex roots are -1±(i√10)/2.

The expression

is called the discriminant of the quadratic equation ax²+bx+c=0.

Question: Determine the nature of the roots of 2x² - 3x + 5 = 0.

Given:

Required: Nature of roots.

Formula:

Solution:

Since Δ<0, the equation has no real roots.

Answer: The equation has two complex conjugate roots.

For x²-5x+6=0,

so the roots are distinct and rational.

For x²-2x-1=0,

so the roots are distinct and irrational.

Question: Without solving the equation, find the sum and product of the roots of 3x²-11x+6=0.

Given:

Required: α+β and αβ.

Formula:

Solution:

Answer: Sum of roots = 11/3; product of roots = 2.

Question: If α and β are roots of 2x²-5x+1=0, find α²+β².

Given:

Required: α²+β².

Formula:

Solution:

Answer: α²+β²=21/4.

Question: Form the quadratic equation whose roots are 3 and -5.

Given:

Required: Quadratic equation.

Solution:

Therefore,

Answer: x²+2x-15=0.

Question: Form the quadratic equation whose roots are 2+√3 and 2-√3.

Solution:

Sum:

Product:

Therefore,

Answer: x²-4x+1=0.

Question: Find the value of k for which x²+kx+9=0 has equal roots.

Given:

Condition:

Solution:

Answer: k=6 or k=-6.

Question: Solve

Solution:

Multiply through by 2:

Factorise:

Hence,

Answer: x=4 or x=-1.

Question: The length of a rectangle is 3 m greater than its width. If its area is 40 m², find its dimensions.

Given:

Let the width be x m. Then the length is x+3 m.

Required: Length and width.

Formula:

Solution:

Thus,

Since a physical length cannot be negative, x=5.

Therefore,

Answer: The dimensions are 5 m by 8 m.

Question: The product of two consecutive positive integers is 72. Find the integers.

Given:

Let the smaller integer be x. The next integer is x+1.

Solution:

Thus,

The positive solution is x=8.

Answer: The integers are 8 and 9.

• A quadratic equation has the general form ax²+bx+c=0, where a≠0.
• Roots are values of the variable that satisfy the equation.
• Quadratic equations may be solved by factorisation, completing the square or the quadratic formula.
• The quadratic formula is x=[-b±√(b²-4ac)]/(2a).
• The discriminant is Δ=b²-4ac.
• If Δ>0, the equation has two distinct real roots; if Δ=0, it has equal real roots; if Δ<0, it has no real roots.
• If the roots are α and β, then α+β=-b/a and αβ=c/a.
• A quadratic equation with roots α and β can be written as x²-(α+β)x+αβ=0.
• Conditions on the roots can be analysed through the discriminant, sum and product of roots.
• Quadratic equations provide useful mathematical models for many geometric and numerical problems.

CHAPTER EIGHT

Binomial Theorem

A binomial is an algebraic expression containing exactly two terms joined by addition or subtraction.

The numerical multipliers occurring in the expansion of (a+b)ⁿ are called binomial coefficients.

For a positive integer n, the factorial n! is defined by

Question: Evaluate C(7,3).

Given:

Required: C(7,3).

Formula:

Solution:

Answer: C(7,3)=35.

For any non-negative integer n,

Question: Expand (x+2)⁴.

Given:

Required: Full expansion.

Formula:

Solution:

Answer:

Question: Expand (2x+y)³.

Given:

Formula:

Solution:

Answer:

The (r+1)th term of the expansion of (a+b)ⁿ is

Question: Find the general term in the expansion of (2x+3)⁷.

Given:

Formula:

Solution:

Answer:

Question: Find the 5th term in the expansion of (x+2)⁸.

Given:

Required: Fifth term.

Since the general term is T_r+1, the fifth term corresponds to

Formula:

Solution:

Answer: The fifth term is 1120x⁴.

Question: Find the 4th term in the expansion of (2x+3y)⁶.

Given:

For the fourth term,

Solution:

Answer: T₄=4320x³y³.

Question: Find the middle term in the expansion of (x+2)⁸.

Given:

The expansion contains

terms. Therefore, the middle term is the 5th term.

Thus r=4.

Answer: The middle term is 1120x⁴.

Question: Identify the two middle terms of (a+b)⁷.

Since the expansion contains 8 terms, the middle terms are the 4th and 5th terms.

For the fourth term, r=3:

For the fifth term, r=4:

Answer: The two middle terms are 35a⁴b³ and 35a³b⁴.

Question: Find the coefficient of x³ in the expansion of (2x+1)⁷.

General Term:

The power of x is 7-r. For the term containing x³,

Therefore,

Answer: The coefficient of x³ is 280.

Question: Find the term independent of x in (x²+2/x)⁶.

General Term:

Simplify the power of x:

For the term independent of x,

Therefore,

Answer: The term independent of x is 240.

Question: Expand (x-2)⁵.

Given:

Solution:

Answer:

Question: Use the binomial theorem to evaluate 101⁴.

Given:

Required: 101⁴.

Formula:

Solution:

Answer: 101⁴=104,060,401.

Question: Use a binomial approximation to estimate (1.01)⁵ to first order.

Given:

Formula:

Solution:

Answer: (1.01)⁵≈1.05 to first order.

• A binomial is an algebraic expression consisting of two terms.
• The coefficients in powers of a binomial are called binomial coefficients.
• Factorial notation is defined by n!=n(n-1)...2·1, with 0!=1.
• The binomial coefficient is C(n,r)=n!/[r!(n-r)!].
• Pascal's triangle provides binomial coefficients for successive non-negative integer powers.
• The binomial theorem is (a+b)ⁿ=∑C(n,r)a^n-rb^r.
• The general term is T_r+1=C(n,r)a^n-rb^r.
• An expansion of (a+b)ⁿ contains n+1 terms.
• If n is even, there is one middle term; if n is odd, there are two middle terms.
• The general term can be used to locate a specified power, coefficient or constant term without performing the full expansion.
• For expressions involving subtraction, the negative second term must retain its sign throughout the expansion.
• The theorem can be applied to numerical evaluation, approximation and identities involving binomial coefficients.

CHAPTER NINE

Complex Numbers and the Argand Diagram

A complex number is a number of the form a+bi, where a and b are real numbers and i is the imaginary unit satisfying i²=-1.

so that

Question: Simplify i²⁷.

Given: Exponent = 27.

Required: Simplified value.

Principle: Powers of i repeat every four powers.

Solution:

Hence,

Answer: i²⁷=-i.

If

then

7=7+0i is purely real.

-4i=0-4i is purely imaginary.

If

then

Question: If 2x+(y-1)i=8+5i, find x and y.

Required: Values of x and y.

Solution:

Equate real parts:

Equate imaginary parts:

Answer: x=4, y=6.

Question: Simplify (4+7i)+(3-5i).

Solution:

Answer: 7+2i.

Question: Simplify (6-2i)-(1+4i).

Solution:

Answer: 5-6i.

Question: Simplify (3+2i)(4-i).

Solution:

Since i²=-1,

Therefore,

Answer: 14+5i.

If z=a+bi, the complex conjugate of z is

If z=4+3i, then z̄=4-3i.

If w=-2-5i, then w̄=-2+5i.

Question: Find z z̄ for z=5-2i.

Given:

Solution:

Answer: z z̄=29.

Question: Simplify

Required: Express the answer in the form a+bi.

Solution:

The conjugate of 2-i is 2+i. Therefore,

The numerator is

The denominator is

Hence,

Answer: 2/5+(11/5)i.

The modulus of the complex number z=a+bi is

Question: Find the modulus of z=3+4i.

Given: a=3, b=4.

Formula:

Solution:

Answer: |z|=5.

An Argand diagram is a coordinate plane used to represent complex numbers geometrically. The horizontal axis represents the real part and the vertical axis represents the imaginary part.

The argument of a non-zero complex number z is the angle θ measured from the positive real axis to the line segment joining the origin to the point representing z.

Question: Find the principal argument of z=1+√3i.

Given:

Formula:

Solution:

The point lies in the first quadrant, and tan(π/3)=√3.

Answer: Arg(z)=π/3.

Question: Find the principal argument of z=-1+√3i.

Solution:

The reference angle satisfies

Since the point lies in the second quadrant,

Answer: Arg(z)=2π/3.

Question: Express z=1+i in polar form.

Given:

Required: Polar form.

Modulus:

Argument:

Since the number lies in the first quadrant,

Therefore,

Answer: √2[cos(π/4)+i sin(π/4)].

Question: Express 4[cos(π/6)+i sin(π/6)] in the form a+bi.

Solution:

Hence,

Answer: 2√3+2i.

Question: Find the distance between z₁=2+3i and z₂=-1+i.

Formula:

Solution:

Answer: The distance is √13.

• A complex number has the form a+bi, where i²=-1.
• The real and imaginary parts of z=a+bi are a and b, respectively.
• Two complex numbers are equal when their corresponding real and imaginary parts are equal.
• Complex numbers can be added, subtracted and multiplied using ordinary algebra together with i²=-1.
• The conjugate of a+bi is a-bi.
• Division is performed by multiplying numerator and denominator by the conjugate of the denominator.
• The modulus of a+bi is √(a²+b²).
• The Argand diagram represents a complex number as a point whose coordinates are its real and imaginary parts.
• The argument of a non-zero complex number is the directed angle from the positive real axis to its representing vector.
• The polar form of a complex number is r(cos θ+i sin θ).
• The distance between two complex numbers is the modulus of their difference.

CHAPTER TEN

De Moivre's Theorem and Roots of Unity

Question: Multiply

and

Solution:

Multiply the moduli:

Add the arguments:

Hence,

Answer: 6[cos(5π/12)+i sin(5π/12)].

If n is an integer, then

Question: Find (1+i)⁸.

Given:

Required: z⁸.

Convert to polar form:

Thus,

Formula:

Solution:

Therefore,

Answer: (1+i)⁸=16.

Question: Evaluate (√3+i)⁶.

Convert to polar form:

Since the number lies in the first quadrant,

Hence,

Applying De Moivre's theorem:

Answer: -64.

Question: Simplify [cos(π/3)+i sin(π/3)]^-2.

Solution:

Answer: -1/2-(√3/2)i.

If

then its n distinct nth roots are

where

Question: Find the cube roots of 8.

Given:

Required: Three cube roots.

Formula:

for k=0,1,2.

For k = 0:

For k = 1:

For k = 2:

Answer:

The nth roots of unity are the complex numbers satisfying

Question: Simplify 1+ω+ω²+ω³+ω⁴.

Given:

Solution:

Since

and

the expression becomes

Using 1+ω+ω²=0,

Answer: 1+ω.

Question: Find all fourth roots of unity.

Formula:

For k=0,1,2,3:

Answer: 1, i, -1, -i.

An nth root of unity is called primitive if its powers generate all the nth roots of unity before returning to 1.

For the fourth roots of unity, i and -i are primitive fourth roots because their successive powers generate all four roots. The root -1 is not primitive because (-1)²=1.

Question: Use De Moivre's theorem to express cos 4θ in powers of cos θ.

Solution:

From

the real part is

Substitute

After simplification,

Answer:

Question: Solve z⁴=-16.

Given:

Required: Four fourth roots.

Formula:

where k=0,1,2,3.

Arguments:

Hence the roots are

In Cartesian form,

Answer: The four roots are ±√2±i√2 with the corresponding sign combinations shown above.

• A non-zero complex number may be written in polar form as r(cos θ+i sin θ).
• When complex numbers are multiplied in polar form, their moduli are multiplied and their arguments are added.
• When they are divided, their moduli are divided and their arguments are subtracted.
• De Moivre's theorem states that (cos θ+i sin θ)ⁿ=cos(nθ)+i sin(nθ) for integer n.
• For z=r(cos θ+i sin θ), zⁿ=rⁿ[cos(nθ)+i sin(nθ)].
• The nth roots of r(cos θ+i sin θ) have modulus r^1/n and arguments (θ+2kπ)/n.
• An nth-degree root problem produces n distinct complex roots when the original number is non-zero.
• The nth roots of unity satisfy zⁿ=1 and lie on the unit circle.
• Successive nth roots of unity are separated by an angle of 2π/n.
• The cube roots of unity are 1,ω,ω², where 1+ω+ω²=0.
• Roots of unity form regular polygons in the Argand plane.
• De Moivre's theorem can also be used to derive multiple-angle trigonometric identities and solve equations involving complex powers.

CHAPTER ELEVEN

Circular Measure and Trigonometric Functions

Circular measure is the measurement of angles in radians. It is based on the relationship between the length of an arc of a circle and the radius of that circle.

One radian is the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

To convert degrees to radians:

To convert radians to degrees:

Question: Convert 150° to radians.

Given: θ = 150°.

Required: Angle in radians.

Formula:

Solution:

Answer: 150° = 5π/6 radians.

Question: Convert 7π/12 radians to degrees.

Formula:

Solution:

Answer: 105°.

where:

• s = length of the arc;
• r = radius of the circle;
• θ = angle subtended at the centre, measured in radians.

The formula s=rθ requires θ to be in radians. If the angle is supplied in degrees, it must first be converted to radians.

Question: A circle has radius 12 cm. Find the length of the arc subtended by an angle of π/3 radians.

Given:

Required: Arc length s.

Formula:

Solution:

Answer: 4π cm, approximately 12.57 cm.

Question: An arc of a circle of radius 8 cm has length 20 cm. Find the angle subtended at the centre in radians.

Given:

Formula:

Solution:

Answer: 2.5 radians.

where θ is measured in radians.

Question: Find the area of a sector of radius 9 cm that subtends an angle of 2π/3 radians.

Given:

Required: Area of the sector.

Formula:

Solution:

Answer: 27π cm², approximately 84.82 cm².

Question: The terminal side of an angle passes through the point P(-3,4). Find sin θ, cos θ and tan θ.

Given:

First find r:

Solution:

Answer:

Question: Determine the signs of sin 210°, cos 210° and tan 210°.

Solution:

Since 180°<210°<270°, the angle lies in Quadrant III.

In Quadrant III, sine and cosine are negative, while tangent is positive.

Answer:

A reference angle is the acute positive angle between the terminal side of an angle and the x-axis.

Question: Evaluate sin 150°.

Solution:

The angle 150° lies in Quadrant II. Its reference angle is

Sine is positive in Quadrant II. Therefore,

Answer: 1/2.

Question: Evaluate cos 225°.

Solution:

The angle lies in Quadrant III and has reference angle

Cosine is negative in Quadrant III. Hence,

Answer: -√2/2.

A function f is periodic if there exists a positive number T such that

for every value of θ in the domain. The smallest positive such value of T is called the fundamental period.

Question: Evaluate sin 750°.

Solution:

Subtract two complete revolutions:

Therefore,

Answer: 1/2.

Question: Evaluate cos(-60°).

Solution:

Cosine is an even function. Hence,

Answer: 1/2.

Question: A wheel of radius 0.35 m rotates through an angle of 8 radians without slipping. Find the distance travelled along the ground.

Given:

Required: Distance travelled.

Formula:

Solution:

Answer: The wheel travels 2.8 m.

Question: Find the length of a chord in a circle of radius 10 cm if the chord subtends an angle of 60° at the centre.

Given:

Formula:

Solution:

Answer: 10 cm.

• Circular measure expresses angles in radians.
• One radian is the angle subtended at the centre of a circle by an arc whose length equals the radius.
• 180°=π radians and 360°=2π radians.
• Degrees are converted to radians by multiplying by π/180.
• Radians are converted to degrees by multiplying by 180/π.
• For an angle measured in radians, arc length is s=rθ.
• The area of a sector is A=1/2 r²θ.
• The basic trigonometric functions are sine, cosine and tangent; cosecant, secant and cotangent are their reciprocals.
• Coordinate definitions extend trigonometric functions to angles of any magnitude.
• The signs of trigonometric functions depend on the quadrant containing the terminal side of the angle.
• A reference angle enables the exact value of a trigonometric function to be determined from a corresponding acute angle.
• Sine and cosine have fundamental period 2π, while tangent has fundamental period π.
• Sine and tangent are odd functions, while cosine is an even function.

CHAPTER TWELVE

Trigonometric Identities, Addition and Factor Formulae

A trigonometric identity is an equation involving trigonometric functions that is true for every value of the variable for which both sides of the equation are defined.

Question: Verify that

Solution:

Begin with the left-hand side:

Using

we obtain

This is the right-hand side. Therefore, the identity is verified wherever the original expression is defined.

Question: Simplify

Solution:

Using

we have

Answer: tan θ, for values of θ for which the original expression is defined.

Question: Verify that

Solution:

Combining the fractions on the left:

Since

we obtain

Hence the identity is verified.

Question: Find the exact value of sin 75°.

Given:

Formula:

Solution:

Answer:

Question: Find the exact value of cos 15°.

Given:

Formula:

Solution:

Answer: (√6+√2)/4.

Question: Find the exact value of tan 15°.

Formula:

Solution:

Multiplying numerator and denominator by √3 gives

Rationalising,

Answer: 2-√3.

Question: If sin θ=3/5 and θ is acute, find sin2θ and cos2θ.

Given:

Since θ is acute, cos θ>0. Therefore,

For sin2θ:

For cos2θ:

Answer:

Question: Use a half-angle formula to obtain the exact value of sin22.5°.

Given:

Formula:

because 22.5° lies in the first quadrant.

Solution:

Answer:

Question: If cos θ=1/2, evaluate cos3θ using the triple-angle formula.

Formula:

Solution:

Answer: cos3θ=-1.

Question: Express 2sin5x cos3x as a sum.

Formula:

Solution:

Answer: sin8x+sin2x.

Question: Express sin70°+sin30° as a product and hence evaluate it exactly.

Formula:

Solution:

Thus the required product form is

Since sin70°=cos20° and sin30°=1/2, an alternative exact representation is

Answer: 2sin50°cos20°.

Question: Express 3sinθ+4cosθ in the form Rsin(θ+α).

Given:

Required: R and α.

Solution:

Also,

Hence,

Therefore,

where α=tan^-1(4/3).

Question: Find the maximum and minimum values of 5sinθ+12cosθ.

Formula:

Solution:

Answer: Maximum value = 13; minimum value = −13.

Question: Solve sinθ=1/2 for 0°≤θ≤360°.

Solution:

The reference angle is 30°.

Sine is positive in Quadrants I and II. Therefore,

Answer: 30° and 150°.

Question: Solve 2cosθ+1=0 for 0°≤θ<360°.

Solution:

The reference angle is 60°.

Cosine is negative in Quadrants II and III. Therefore,

Answer: 120° and 240°.

Question: Solve tanθ=-1 for 0°≤θ<360°.

Solution:

The reference angle is 45°. Tangent is negative in Quadrants II and IV.

and

Answer: 135° and 315°.

Question: Solve

for 0°≤θ≤360°.

Solution:

Factorise:

Therefore,

or

For sinθ=1/2:

For sinθ=1:

Answer:

Question: Solve cos2θ=0 for 0≤θ<2π.

Solution:

Since cosine is zero when

we obtain

Within 0≤θ<2π, the values are

Answer:

Question: Given that tanθ=3/4 and θ is acute, find sin2θ without first finding θ.

Given:

Required: sin2θ.

The double-angle formula can be written entirely in terms of tangent:

Solution:

Answer: 24/25.

Question: Solve sin(θ+30°)=1/2 for 0°≤θ≤360°.

Solution:

Let

The equation becomes

Over the corresponding interval 30°≤x≤390°, the solutions are

Therefore,

Answer: θ=120° or 360°.

Question: Evaluate exactly

Formula:

Solution:

Therefore,

Answer: 1/4.

• A trigonometric identity is true for every value for which both sides are defined, whereas a trigonometric equation is true only for particular values.
• The fundamental Pythagorean identity is sin²θ+cos²θ=1.
• Two related identities are 1+tan²θ=sec²θ and 1+cot²θ=cosec²θ.
• Compound-angle formulae express the sine, cosine and tangent of A±B in terms of the individual angles.
• The sine double-angle formula is sin2θ=2sinθcosθ.
• The cosine double-angle formula can be written as cos2θ=cos²θ-sin²θ, 2cos²θ-1, or 1-2sin²θ.
• The tangent double-angle formula is tan2θ=2tanθ/(1-tan²θ).
• Half-angle formulae follow from rearranging the cosine double-angle identities.
• Triple-angle formulae include sin3θ=3sinθ-4sin³θ and cos3θ=4cos³θ-3cosθ.
• Product-to-sum formulae convert products of trigonometric functions into sums or differences.
• Sum-to-product formulae convert sums and differences into products and are also known as factor formulae.
• An expression a sinθ+b cosθ can be represented by a single trigonometric function with amplitude √(a²+b²).
• Trigonometric equations must be solved with careful attention to quadrants, periodicity and the specified interval.