Franklin Kome Amoo (Lead)
Elliot Attipoe (Co-Author)
writetofranky@yahoo.com
Mathematics For Computing I – INF109D
Course Outline
Course Overview
Course Objective
1.1.0. Unit 1.
Unit Outline
Introduction to Set Theory
Session 1 – Definition of Sets and Standard Mathematical Notations
Session 2 – Representation of Sets
Session 3 – Classification of Sets
Session 4 – Subsets, Power Sets and Universal Sets
Session 5 – Compliment Intersection, and Union of Sets
Session 6 – Cartesian Product of Two Sets
2
Unit 1.
1.0
Course Overview:
Hello learners, we are happy to welcome you to this course, Mathematics for Computing I. This
is the beginning of a journey that will help us learn to use the main tools for solving problems in
computing and information technology. This course will provide you with the knowledge and
understanding of the tools for designing solutions to many problems in the real world. Most of
the problems we would like to solve by using computers would also require skilful use of these
tools. The tools are based on some of the various mathematical theories and concepts we will
learn in this course. We will learn to use our ideas and knowledge of the mathematical theories to
formulate relations in the form of equations, formulas, functions, models, assumptions, to
perform appropriate experiments on available data. The experiments can be used to test and
validate the models or the equations as solutions that can lead to useful conclusions or enable us
to make appropriate recommendations based on the proposed solutions to the problem. A good
understanding and application of the tools will depend on our good understanding of
mathematical theories and concepts.
3
1.1.0. Unit Objective
We will present some basic definitions of terminologies and some standard mathematical
notations to prepare the learners for the course. This will be done by explaining in general terms,
the concept of mathematics, and how it can be used to design and solve problems, to conduct
experiments, and to draw conclusions from experiments.
In this unit we will learn to understand some of the common, but important terminologies and
notations in mathematics. We will learn the number system and use it to identify the different
types of numbers. Therefore, we take this opportunity to welcome you to the first unit in this
course. This unit discusses the basic concepts of set theory. We hope you will find the material
useful and beneficial in your career path and in your professional life.
Mathematics is important because it has many applications in our life, irrespective of our
profession. Mathematics and its various concepts and applications are used by many
professionals in many and diverse fields, including scientists, statisticians, mathematicians, data
scientist, engineers, teachers, researchers, economists, students, computer programmers and
application developers. We can be sure that some study of mathematics will open several
opportunities for you to have a fulfilling career.
Let us get into the details.
The first unit has six distinct sessions, but they are related to one another, in that one session
leads to the other in a smooth transition. The first session serves as the introductory part. It
focuses on set theory for identifying items that share some common properties for common uses.
The second session is the application of the set theory for developing relations that can be
viewed as machines that accept input to produce outputs. These machines are mathematically
known as functions. We will build on functions data presentation with some of the popular
methods and styles of data display in the third session. The fourth session begins with
measurements and determination of some of the popular and common data values of central
nature. The fifth session treats the type of data that vary from the centre values. We conclude the
unit in the sixth session by considering data of relative positions to other data.
We can expect to learn some exciting concepts in this session.
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Unit 1: Set Theory
Session 1 – Definition of Sets and Standard Mathematical Notations
1.1.0
Introduction
Session 2 – Representation of Sets
1.2.1. Two Methods for Representing Sets
1.2.2. The Roster Method
1.2.3. The Set-builder Form
Session 3 - Classification of Sets
1.3.1. Finite and Infinite Sets
1.3.1.1. Finite Set:
1.3.1.2. Infinite Set:
1.3.2. Empty (Null) Set
1.3.3. Singleton Set
1.3.4. Equal and Equivalent Sets
1.3.4.1.
Equal Sets
1.3.4.2.
Equivalent Sets
1.3.5. Disjoint Sets
Session 4 – Subsets, Power Sets and Universal Sets
1.4.1. Subsets
1.4.2. Power Sets
1.4.3. The Universal Sets
1.4.4. Venn Diagrams
1.4.5. Difference of Sets
Session 5 – Compliment Intersection, and Union of Sets
1.5.1. Compliment of a Set.
1.5.2. Intersection of Sets.
1.5.3
The Union of Sets
5
Session 6 – Cartesian Product of Two Sets
1.6.1. Cartesian Product of Sets
6
At the end of this unit, the student should be able to:
1.0. Define sets and present them in their different forms
2.0. Define different types of sets such as finite and infinite sets, empty sets, singleton set,
equivalent set, equal set, subsets, and their various examples, display sets in different forms
3.0. How to set theory to define relations
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Session 1 – Definition of Sets and Standard Mathematical Notations
Dear student. We welcome you to the first session of Unit 1 of this course. The title of this unit is
Set Theory. In this first session, we will learn about different types of objects, including symbols
for the different types of numbers. We can group or make a collection of these objects and
numbers according to their common characteristics and features. The concept of groups and
collections of the objects and the types of numbers will form the basis for defining different
types of sets. A good understanding of the concept is needed for understanding the topics in the
remaining session of this unit. We hope you will find the lessons useful and beneficial. Please
enjoy the lessons.
Prerequisite knowledge.
Number systems, concept of ordered pairs.
Learning Objectives.
By the end of this session, students should be able to
1.
2.
3.
4.
Identify a collection of items as a set of some sort.
Identify the different types of symbols for the number system.
Identify the popular symbols for sets.
Define a set and its contents
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1.1.0
Introduction
We begin by listing some examples of items
(i)
(ii)
Collection of tall students in your
college
Collection of honest persons in
your community
(iii) Collection of interesting books in
your college library
(iv) Collection of intelligent students in
your college
(i)
Collection of those students of your college
whose height is more than 180cm
(ii) Collection of those people in your
community who have never been found
involved in any theft case
(iii) Collection of Mathematics books in your
college library
(iv) Collection of those student in your college
who have secured more than 80% marks in
annual examination
In all the collections in the left column of the table, the terms tallness, interesting, honesty, and
intelligence are not well defined. These notions vary from individual to individual. Therefore,
those collections cannot be considered as sets.
The collections in the right column “height”, “more than 180cm.”, “Mathematics books”, “never
been found involved in theft case”, and “marks more than 80%” are well defined properties.
Therefore, those collections can be considered as sets.
A set is a collection of objects, and these objects are called elements of the set. Another idea of a
set is, any collection of distinct objects, that are called elements, is a set. If a collection is a set,
then each object of this collection is said to be an element of the set. For example, the deck of
cards is a set whose elements are the cards. The participants in a party form a set, whose
elements are the people such as Kojo, Akua, Daniel, Robert, Juliet, and Linda; (we can denote
the set of party participants by the letter 𝑃).
A set is usually denoted by a capital letter of the English alphabet and its elements are denoted
by small letters.
For example, the set, 𝐴 = {Toy elephant, packet of sweets, magazines}
In mathematics, various sets of numbers are especially important: the set of real numbers,
denoted by ℝ; the set of rational numbers, denoted by ℚ; the set of integers, denoted by ℤ or 𝕀;
the set of non-negative integers, denoted by ℤ+ 𝑜𝑟 ℤ+.
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If 𝑆 is a set, the notation 𝑎 ∈ 𝑆 means that 𝑎 is an element of 𝑆, and 𝑏 ∉ 𝑆 means that 𝑏 is not an
element of 𝑆. For example, if ℤ represents the set of integers, then −3 ∈ ℤ but 𝜋 ∉ ℤ.
The following are some standard notations to represent different sets of numbers:
1.
2.
3.
4.
5.
6.
ℕ:
𝕎:
ℤ or 𝕀:
ℤ+ :
ℤ−
ℚ:
7.
8.
9.
ℝ:
ℂ:
𝜙
The set of natural numbers, examples are {1,2, 3, 4, … }
The set of whole numbers, examples are {… , −2, −1, 0, 1, 2, … }
The set of integers, examples are {−123, −51, 1, 77}
The set of positive integers, examples are {3, 89, 123, 9876}
The set of negative integers, examples are {−55, −234, −7654}
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3 1 4
The set of rational numbers, examples are {… , − 4 , − 5 , 2 , 5 , … }
The set of real numbers, examples are {… , −2.345, −0.45, 0.987, 3.123, . . }
The set of complex numbers, examples are { 3 + 4𝑖, 2 − 7𝑖 }
The empty set
Other frequently used and common symbols are:
1
∈:
“belongs to” or “is a member of” or “is in”
2.
∉:
“does not belong to” or “is not a member of” or “is not in”
3.
∃:
“there exists”
4.
“there does not exist”
5.
∀
“for all”
For example, ℕ is the set of natural numbers. We know that 3 is a natural number but −3 is not a
natural number. We can write them in symbolic form as 3 ∈ ℕ and −3 ∉ ℕ.
The number of elements of a set 𝐴 (also called the cardinality of 𝐴) is denoted by |𝐴|. Thus if
Summary
In this session, we were introduced to the concept of sets. We learned about the different types of
numbers and their symbols. We learned that a collection of items or group of objects must be
well defined according to a common characteristic or a common feature to define a set.
We hope you find the lesson useful. Please test yourself with the following exercises. It will help
you to assess how you are progressing on the course.
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Self-Assessment.
1. Define a set.
(A) Collection of objects that share the same meaning
(B) Collection of objects that have similar characteristics
(C) Collection of symbols with the same meanings
(D) Collection of objects that have different characteristics
2. What is the name given to the content of a set?
(A) Set factors / Set items
(B) Set features / Set characteristics
(C) Set members / Set elements
(D) Set contents / Set lists
3. In set notation, names of sets are usually represented by
(A) Small letters of the English alphabets
(B) Small symbols of sets
(C) Set notations
(D) Capital letters of the English alphabets
4. What is the meaning of the symbol −23 ∈ ℤ ?
(A) −23 ∈ ℤ means −23 is a member of the set of rational numbers
(B) −23 ∈ ℤ means −23 is a member of the set of negative numbers
(C) −23 ∈ ℤ means −23 is a member of the set of negative integers
(D) −23 ∈ ℤ means −23 is a member of the set of integers
5. What is the meaning of the symbol 𝜙?
(A) The phi set of items
(B) The null or empty set
(C) The symbolic phi set
(D) Properly defined set of items
Answers to Self-Assessment
1.
2.
3.
4.
5.
B
C
D
D
B
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1.2.0. Session 2 – Representation of Sets
1.2.1. Two Methods for Representing Sets
There are two methods for representing a set. They are the (i) The Roster method (Tabular Form)
and the set builder form. Some sets can be described by listing their elements within a pair of
curly braces (curly {} brackets), with each element separated from the other element by a
comma.
1.2.2. The Roster Method
In this method a set is represented by listing all its elements, separating them by commas and
enclosing them in a pair of curly brackets.
Example 1.2.2.1
Let the set 𝑉 be the set of vowels of the English alphabets. It can be written in a Roster form as
𝑉 = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢}
Example 1.2.2.2
Let the set 𝐴 be the set of natural numbers less than 10, then we can represent the set 𝐴 in roster
form as 𝐴 = {1, 2, 3, 4, 5, 6, 7, 8, 9} in Roster form.
Note:
To write a set in Roster form, the elements are not to be repeated to reflect the number of times
an element occurs in the list. All the elements are taken as distinct. For example, if the set 𝑀
represents set of letters in the word “mathematics”, then the set 𝑀 in Roster form is written as
𝑀 = {𝑚, 𝑎, 𝑡, ℎ, 𝑒, 𝑖, 𝑐, 𝑠}
You would have noticed that the letters 𝑚, 𝑎, and 𝑡 appear more than once in the word, they are
listed only once in the roster form
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1.2.3. The Set-builder Form
In set-builder form, we do not list the elements of the of the set. The set is represented by
choosing a common property that fittingly describes the set.
Example 1.2.3.1.
Let the set 𝑉 be the set of vowels of the English alphabets. We can write the set 𝑉 in set builder
form as 𝑉 = {𝑥 ∶ 𝑥 𝑖𝑠 𝑎 𝑣𝑜𝑤𝑒𝑙 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑎𝑙𝑝ℎ𝑎𝑏𝑒𝑡}. Another way of presenting a set in a
set-builder form is 𝑉 = {𝑥 | 𝑥 𝑖𝑠 𝑎 𝑣𝑜𝑤𝑒𝑙 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑎𝑙𝑝ℎ𝑎𝑏𝑒𝑡}, which is read as “V is the
set of all 𝑥 such that 𝑥 is a vowel of the English alphabets”
Example 1.2.3.1
Let the set 𝐴 be the set of natural numbers less than 10. We can write the set 𝐴 as follows:
𝐴 = {𝑥 ∶ 𝑥 ∈ ℕ 𝑎𝑛𝑑 1 ≤ 𝑥 < 10}
OR
𝐴 = {𝑥 | 𝑥 ∈ ℕ 𝑎𝑛𝑑 1 ≤ 𝑥 < 10}
Exercises
1. Write the following sets in set-builder form.
a. 𝐴 = {−3, −2, −1, 0, 1, 2, 3}
𝐴 = {𝑥: 𝑥 ∈ ℤ, 𝑎𝑛𝑑 − 3 ≤ 𝑥 ≤ 3}
b. 𝐵 = { 3, 6, 9, 12}
𝐵 = {𝑥: 𝑥 ∈ 3𝑛, 𝑎𝑛𝑑 𝑛 ∈ ℕ, 1 ≤ 𝑛 ≤ 4}
2. Write the following in Roster form.
a. 𝐶 = {𝑥 ∶ 𝑥 ∈ ℕ 𝑎𝑛𝑑 50 ≤ 𝑥 ≤ 60}
b. 𝐷 = { 𝑥: 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑥 2 − 5𝑥 + 6 = 0}
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Self-Assessment
1. Interpret the following set: 𝐴 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(A) The set 𝐴 is the set of natural numbers less than 10
(B) The set 𝐴 is the set of integers less than 10
(C) The set 𝐴 is the set of natural numbers less than or equal to 10
(D) The set 𝐴 is the set of integers less than 11
2. Write the following in Roster Form: 𝑀 = {𝑥: 𝑥 = 2𝑛 + 1 𝑎𝑛𝑑 𝑛 ∈ ℕ, 𝑛 ≤ 4 }
(A) 𝑀 = {3, 5, 9, 11}
(B) 𝑀 = {3, 5, 7, 9}
(C) 𝑀 = {3, 5, 7, 10}
(D) 𝑀 = {5, 7, 9, 11}
3. Write the following in set-builder form:
𝐵 = {−2, −1, 0, 1, 2, 3, 4, 5}
(A) 𝐵 = {𝑥: 𝑥 ∈ 𝕀 𝑎𝑛𝑑 − 2 < 𝑥 ≤ 5}
(B) 𝐵 = {𝑥: 𝑥 ∈ 𝕀 𝑎𝑛𝑑 − 3 < 𝑥 < 5}
(C) 𝐵 = {𝑥: 𝑥 ∈ 𝕀 𝑎𝑛𝑑 − 2 ≤ 𝑥 < 5}
(D) 𝐵 = {𝑥: 𝑥 ∈ 𝕀 𝑎𝑛𝑑 − 2 ≤ 𝑥 ≤ 5}
4. Write the following in Roster form: 𝐷 = {𝑥: 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑥 2 + 5𝑥 − 24 = 0}
(A) 𝐷 = { 3, −8}
(B) 𝐷 = {3, 8}
(C) 𝐷 = {−3, −8}
(D) 𝐷 = {−3, 8}
5. Write in Roster form the set: 𝑀 = {𝑥: 𝑥 𝑖𝑠 𝑠𝑒𝑡 𝑜𝑓 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝑀𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑠}
(A) 𝑀 = {𝑀, 𝑎, 𝑡, ℎ, 𝑒, 𝑚, 𝑎, 𝑡, 𝑖, 𝑐, 𝑠}
(B) 𝑀 = {𝑚, 𝑎, 𝑡, ℎ, 𝑒, 𝑖, 𝑐, 𝑠}
(C) 𝑀 = {𝑚, 𝑎, 𝑡, ℎ, 𝑒, 𝑚, 𝑎, 𝑡, 𝑖, 𝑐, 𝑠}
(D) 𝑀 = {𝑚, 𝑎, 𝑡, ℎ, 𝑖, 𝑐, 𝑠}
Answers to Self-Assessment
1.
2.
3.
4.
5.
C
B
D
A
B
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1.3.0. Session 3 - Classification of Sets
In this session we will learn to classify sets into different types of set. Among them are, finite
and infinite sets, equal and equivalent sets, empty or null sets, disjoint-sets and singleton sets.
We will learn about their various properties and how to identify each one of them.
1.3.1. Finite and Infinite Sets
Let the sets 𝐴 and 𝐵 be defined as
𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟} and
𝐵 = { 𝑥 ∶ 𝑥 𝑖𝑠 𝑎 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑜𝑓 𝑦𝑜𝑢𝑟 𝑐𝑜𝑙𝑙𝑒𝑔𝑒}
As we can see in the two sets above, elements of the set 𝐴 does not have an end. If we make any
attempt to count all the elements in set 𝐴, it may take us a long time to count to the largest
possible natural number, ∞, but the elements in the set 𝐵 has an end. We can count all the
students in our college. The set 𝐴 said to be an infinite set, and the set 𝐵 is said to be a finite set.
Definitions:
1.3.1.1. Finite Set:
A set is said to be finite if its elements can be counted to its last element.
1.3.1.2. Infinite Set:
A set is said to be infinite if it is not possible to count to its last element.
1.3.2. Empty (Null) Set
The empty set, denoted by 𝜙, is the set that contains no element.
Consider the following sets.
𝐴 = { 𝑥: 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑥 2 + 1 = 0}
𝐵 = { 𝑥 ∶ 𝑥 𝑖𝑠 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 7 𝑎𝑛𝑑 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 5 }
15
The set 𝐴 defines its elements 𝑥 as real numbers. There is no real number whose square is −1 as
we rearrange the expression 𝑥 2 + 1 = 0 as 𝑥 2 = −1
Similarly, there is no number such that the number is greater than 7 and the same number is less
than 5.
Such a set is said to be a null (an empty) set. It is denoted by the symbol void, 𝜙 or {} or ∅
A set which has no element is said to be a null/void set, and it is denoted by 𝜙.
1.3.3. Singleton Set
Consider the following set.
𝐴 = { 𝑥 ∶ 𝑥 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟}
There is only one prime number that is an even number, the number 2. Therefore, the set 𝐴 has
only one element, the element 2 in the set. Such a set is said to be singleton. The set is 𝐴 = {2}.
A set which has only one element is known as singleton set.
1.3.4. Equal and Equivalent Sets
Consider the following examples:
(i)
(ii)
𝐴 = {1, 2, 3} ,
𝐷 = {1, 2, 3},
𝐵 = {2, 1, 3}
𝐸 = {𝑎, 𝑏, 𝑐}
In example (i), Sets 𝐴 and 𝐵 have the same elements. Such sets are said to be equal sets and it is
written as 𝐴 = 𝐵
In example (ii), set 𝐷 and 𝐸 have the same number of elements but the elements are different.
Such sets are said to be equivalent sets, and they are written as 𝐷 ≡ 𝐸.
Definition:
1.3.4.1.
Equal Sets: - Two sets 𝐴 and 𝐵 are said to be equal if they have the same number
of elements and the elements are also the same in each of the set
1.3.4.2.
Equivalent Sets: - Two sets 𝐴 and 𝐵 are said to be equivalent sets if they have the
same number of elements in each set.
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1.3.5. Disjoint Sets
Two sets are said to be disjoint if they do not have any common element. For example, the set
𝐴 = {1, 3, 5} and the set 𝐵 = {2, 4, 6} are disjoint sets.
Example 1.3.5.1.
Given that set 𝐴 = {2, 4} and set 𝐵 = {𝑥 ∶ 𝑥 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥 2 + 6𝑥 + 8 = 0 }. Are 𝐴 and 𝐵
disjoint sets?
Solution 1.3.5.1:
We have been given the set 𝐴 = {2, 4}, the set 𝐵 = {𝑥 ∶ 𝑥 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥 2 + 6𝑥 + 8 = 0 }
For us to establish whether the two sets are disjoint or not, we need to solve the quadratic
equation 𝑥 2 + 6𝑥 + 8 = 0 in the set 𝐵.
We solve by expressing the quadratic in the form 𝑥 2 + 2𝑥 + 4𝑥 + 8 = 0
𝑥 2 + 2𝑥 + 4𝑥 + 8 = 0
𝑥(𝑥 + 2) + 4(𝑥 + 2) = 0
We find the common factor, 𝑥 + 2 is a common factor, we factorize it out. We arrive at
(𝑥 + 2)(𝑥 + 4) = 0
We can say
(𝑥 + 2) = 0
OR
(𝑥 + 4) = 0
The solution says
𝑥 = −2,
𝑥 = −4
The set 𝐴 = {2, 4} and the set 𝐵 = {−2, −4}.
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Clearly, the sets 𝐴 and 𝐵 are disjoint since they do not have any common element.
Example 1.3.5.2.
If the set 𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑣𝑜𝑤𝑒𝑙 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑎𝑙𝑝ℎ𝑎𝑏𝑒𝑡} and set 𝐵 = {𝑦: 𝑦 ∈ ℕ 𝑎𝑛𝑑 𝑦 ≤ 5} .
Is
(i)
(ii)
𝐴=𝐵
𝐴≡𝐵
Solution 1.3.5.2.
The set 𝐴 = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢}, and the set 𝐵 = {1, 2, 3, 4, 5}
Although each set has five elements, the elements are not the same, they are different. However,
they are equivalent.
𝐴 ≠ 𝐵 but 𝐴 ≡ 𝐵
Example 1.3.5.3.
Which of the following sets
𝐴 = {𝑥 ∶ 𝑥 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑎 𝑙𝑖𝑛𝑒} and 𝐵 = {𝑦 ∶ 𝑦 ∈ ℕ 𝑎𝑛𝑑 𝑦 ≤ 50 }
are finite or infinite?
Solution 1.3.5.3.
The number of points on a line is uncountable (cannot be counted), the set 𝐴 is an infinite set, but
we can count the number of natural numbers up to fifty, so the set 𝑩 is a finite set.
Example 1.3.5.4.
Which of the following sets
𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑥 2 − 1 = 0 }
𝐵 = {𝑥: 𝑥 ∈ ℤ 𝑎𝑛𝑑 − 2 ≤ 𝑥 ≤ 2 }
are empty?
18
Solution 1.3.5.4.
The set 𝐴 consists of those irrational numbers that satisfy 𝑥 2 − 1 = 0. If we solve for 𝑥 in the
expression 𝑥 2 − 1 = 0, we get 𝑥 = ±1, but ±1 are not rational numbers. Therefore, the set 𝐴 is
an empty set.
The set 𝐵 = {−2, 1, 0, 1, 2}. The set 𝐵 is not an empty set. It has five elements.
Example 1.3.5.5.
Which of the following sets are singleton?
𝐴 = {𝑥 ∶ 𝑥 ∈ ℤ 𝑎𝑛𝑑 𝑥 − 2 = 0 } and 𝐵 = {𝑦: 𝑦 ∈ ℝ 𝑎𝑛𝑑 𝑦 2 − 2 = 0}
Solution 1.3.5.5.
The set 𝐴 contains those integers which are the solutions of 𝑥 − 2 = 0 or 𝑥 = 2. ∴ 𝐴 = {2}. It
implies that the set 𝐴 is a singleton set
The set 𝐵 is a set of those real numbers which are solutions of 𝑦 2 − 2 = 0 or
𝑦 = ±√2 ∴ 𝐵 = {−√2, √2}. The set 𝐵 has more than one element. In fact, it has two elements.
Thus, the set 𝐵 is not a singleton set.
Exercises for Practice
Ex 1.
Write each of the following sets in roster form.
(i)
(ii)
𝐴 = {𝑥: 𝑥 ∈ ℤ 𝑎𝑛𝑑 − 5 ≤ 𝑥 ≤ 0} = {−5, −4, −3, −2, −1, 0}
𝐵 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑙𝑒𝑡𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝑏𝑎𝑛𝑎𝑛𝑎 } = {𝑏, 𝑎, 𝑛} = {𝑎, 𝑏, 𝑛} = {𝑛, 𝑎, 𝑏}
Ex 2.
Write each of the following sets in the set builder form.
(i)
(ii)
𝐴 = {3, 6, 9, … , ∞}
𝐵 = {2, 3, 5, 7}
Ex 3. In the following sets, check whether 𝐴 = 𝐵 or 𝐴 ≡ 𝐵
(i)
𝐴 = {𝑎},
𝐵 = {𝑥: 𝑥 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟}.
19
𝐴 = {1,2,3,4},
(ii)
𝐵 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑙𝑒𝑡𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝑔𝑢𝑎𝑣𝑎}
Self-Assessment
1. Describe the relationship between the set 𝐴 = {1, 3, 5} and the set 𝐵 = {2, 4, 6}.
(A) The set 𝐴 is a disconnected set from of the set 𝐵
(B) The set 𝐴 and the set 𝐵 are equal sets
(C) The set 𝐵 is an empty from the set 𝐴
(D) The set 𝐴 and the set 𝐵 are disjoint sets
2. What are the members of the set 𝐵 = {𝑦: 𝑦 ∈ ℝ 𝑎𝑛𝑑 𝑦 2 − 2 = 0}?
(A) 𝐵 = {−2, 2}
(B) 𝐵 = {−2, −√2, +√2, 2}
(C) 𝐵 = {−√2, √2}
(D) 𝐵 = {−2, −√2, 0, +√2, 2}
3. Describe the set 𝐴 = {1, 2, 3, 4, … }
(A) An infinite set of natural numbers
(B) An infinite set of integers
(C) A finite set of natural integers
(D) A finite set of whole numbers
4. Write the set 𝐵 = {𝑥: 𝑥 ∈ ℤ 𝑎𝑛𝑑 − 2 ≤ 𝑥 ≤ 2 } in roster form
(A) 𝐵 = {−2, −1, 1, 2}
(B) 𝐵 = {−2, −1, 0, 1, 2}
(C) 𝐵 = {−1, 0, 1, 2}
(D) 𝐵 = {−2, −1, 0, 1}
5. Write the set 𝐴 = {𝑥: 𝑥 = 2𝑛 + 3, 0 ≤ 𝑛 < 4} in a roster form
(A) 𝐴 = {0, 3, 5, 7, 9, 11 }
(B) 𝐴 = {0, 3, 5, 7, 9}
(C) 𝐴 = {3, 5, 7, 9, 11}
(D) 𝐴 = {3, 5, 7, 9}
Answers to Self-Assessments
1.
2.
3.
4.
5.
D
C
A
B
D
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1.4.0. Session 4 – Subsets, Power Sets and Universal Sets
1.4.1. Subsets
Let the set 𝐴 be the set containing all students at your college, UCC, and the set 𝐵, the set of all
students taking INF109D of UCC. In this example, each element of the set 𝐵 is also an element
of the set 𝐴. Note that the set 𝐵, has elements that are also the elements of the set 𝐴, but the set
𝐴, has more other elements that are not in set 𝐵. Such a set, 𝐵 in our example, is said to be the
subset of the set 𝐴.
Symbolically, we write the set 𝐵 is a subset of the set 𝐴 as 𝐵 ⊆ 𝐴
Consider
𝐷 = {1, 2, 3, 4, … … } and 𝐸 = {… . . , −3, −2, −1, 0, 1, 2, 3, … . . }
Here we can see that each element of the set 𝐷 is also an element of the set 𝐸. Therefore, we
conclude that 𝐷 ⊆ 𝐸
If sets 𝐴 and 𝐵 are any two sets such that each element of the set 𝐴 is an element of the set 𝐵
also, then the set 𝐴 is said to be a subset of the set 𝐵
Special Notes
i.
ii.
iii.
iv.
Each set is a subset of itself i.e., 𝐴 ⊆ 𝐴
Null set has no element, so the condition of becoming a subset is automatically
satisfied. Therefore, a null set is a subset of every set.
If 𝐴 ⊆ 𝐵 and 𝐵 ⊆ 𝐴, then 𝐴 = 𝐵.
If 𝐴 ⊆ 𝐵 and 𝐴 ≠ 𝐵 then 𝐴 is said to be a proper subset of 𝐵 and 𝐵 is said to be a
super set of 𝐴. That is, 𝐴 ⊂ 𝐵 or 𝐵 ⊃ 𝐴.
Example 1.4.1.1
If 𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 5} and 𝐵 = {𝑦: 𝑦 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟} then
is 𝐵 a proper subset of 𝐴?
21
Solution 1.4.1.1.
The given sets are 𝐴 = {2, 3} and 𝐵 = {2}
We can see that 𝐵 ⊆ 𝐴 and 𝐵 ≠ 𝐴. We write that 𝐵 ⊂ 𝐴 and say that 𝐵 is a proper subset of 𝐴.
Example 1.4.1.2
If 𝐴 = {1, 2, 3, 4} , and 𝐵 = {2, 3, 4, 5}. Is 𝐴 ⊆ 𝐵 or 𝐵 ⊆ 𝐴 ?
Solution 1.4.1.2
In this example, 1 ∈ 𝐴 but 1 ∉ 𝐵, ⇒ 𝐵  𝐴. Neither 𝐴 is a subset of 𝐵 nor 𝐵 is a subset of 𝐴 .
They are not equal.
Example 1.4.1.3
If 𝐴 = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢} and 𝐵 = {𝑒, 𝑖, 𝑜, 𝑢, 𝑎}. Is 𝐴 ⊆ 𝐵 or 𝐵 ⊆ 𝐴 or both?
Solution 1.4.1.3.
In the given sets, each element of set 𝐴 is also an element of set 𝐵.
∴ 𝐴 ⊆ 𝐵 ……(i)
and each element of set 𝐵 is an element of set 𝐴 also.
∴ 𝐵 ⊆ 𝐴 ….. (ii)
From (i) and (ii) ⇒ 𝐴 = 𝐵
1.4.2. Power Sets
Let 𝐴 = {𝑎, 𝑏}
The subset of 𝐴 are the sets 𝜙, {}, {𝑎}, {𝑏}, and {𝑎, 𝑏}.
If we consider these subsets as elements of a new set 𝐵, (say) then 𝐵 = {𝜙, {𝑎}, {𝑏}, {𝑎, 𝑏}}. Then
we say the new set 𝐵 is said to be the power set of 𝐴.
22
Notation:
Power set of a set 𝐴 is denoted by 𝑃(𝐴). The power set of a set 𝐴 is the set of all subsets of the
given set.
Example 1.4.2.1
Write the power set of each of the following sets:
(i)
(ii)
𝐴 = {𝑥: 𝑥 ∈ ℝ, 𝑎𝑛𝑑 𝑥 2 + 7 = 0}
𝐵 = {𝑦: 𝑦 ∈ ℕ 𝑎𝑛𝑑 1 ≤ 𝑦𝑦 > 1 = 2, 3, 4, 5, … ..
𝑦 ≥ 1 = 1, 2, 3, 4, 5 ….
Solution 1.4.2.1
(i)
(ii)
From the given equation, 𝑥 2 + 7 = 0 , when we rearrange the equation, we will get
𝑥 2 = −7. The condition says 𝑥 is a real number. There is no real number whose
square is less than zero or a negative number. The subset is the empty set or the null
set. 𝐴 = 𝜙 (Null Set) is the only subset for 𝐴. ∴ 𝑃(𝐴) = {𝜙}
The elements of the set 𝐵 can be written as 𝐵 = {1, 2, 3}. The subsets of 𝐵 are
𝜙, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}. The power set of 𝐵 can be written as
𝑃(𝐵) = {𝜙, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
1.4.3. The Universal Set
Consider the following sets.
𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑜𝑓 𝑦𝑜𝑢𝑟 𝑐𝑜𝑙𝑙𝑒𝑔𝑒}
𝐵 = {𝑦: 𝑦 𝑖𝑠 𝑎 𝑚𝑎𝑙𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑜𝑓 𝑦𝑜𝑢𝑟 𝑐𝑜𝑙𝑙𝑒𝑔𝑒}
𝐶 = {𝑧: 𝑧 𝑖𝑠 𝑎 𝑓𝑒𝑚𝑎𝑙𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑜𝑓 𝑦𝑜𝑢𝑟 𝑐𝑜𝑙𝑙𝑒𝑔𝑒}
𝐷 = {𝑎: 𝑎 𝑖𝑠 𝑎 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑜𝑓 𝐼𝑁𝐹109𝐷 𝑜𝑓 𝑦𝑜𝑢𝑟 𝑐𝑜𝑙𝑙𝑒𝑔𝑒}
We can see that the sets 𝐵, 𝐶 and 𝐷 are subsets of the set 𝐴.
The set 𝐴 can be considered as the universal set for this example. Universal sets are generally
denoted as 𝑈. In any problem, a set 𝑈is said to be a universal set if all the sets in the problem are
identified as subsets of 𝑈
23
Special Note:
(i)
(ii)
Universal set does not mean a set containing all objects in the universe.
A set which is a universal set for one problem may not be a universal for another
problem.
Example 1.4.3.1
Which of the following sets can be considered as a universal set?
𝑋 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟}
𝑌 = {𝑦: 𝑦 𝑖𝑠 𝑎 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟}
𝑍 = {𝑧: 𝑧 𝑖𝑠 𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟}
Solution 1.4.3.1
From the given sets, negative integers and natural numbers are also real numbers. We can say
that sets 𝑌 and 𝑍 are subsets of the set 𝑋. Therefore, the set 𝑋 is the universal set for this problem
1.4.4. Venn Diagrams
It is sometimes convenient to represents sets in a diagram, where the universal set is usualy
represented as a rectangle, and the other sets which are the subsets are represented as circles (or
ovals) inside the rectangle.
For example, if 𝑈 = {1, 2, 3, 4, 5}, 𝐴 = {2, 4} and 𝐵 = {1, 3}, then we can represent these sets as
the following Venn Diagram
U
B
A
1
2
3
4
5
24
1.4.5. Difference of Sets
Consider the sets,
𝐴 = {1, 2, 3, 4, 5} and 𝐵 = {2, 4, 6}
A new set that has the elements that are in the set 𝐴 but not in the set 𝐵 is said to be the
difference of the sets 𝐴 and 𝐵 and it is denoted by 𝐴 − 𝐵.
Therefore, from the given sets, 𝐴 − 𝐵 = {1, 2, 3, 4, 5} − {2, 4, 6} = { 1, 3, 5}
Similarly, a set of those elements that are in the set 𝐵 but not in the set 𝐴 is said to be the
difference of 𝐵 and 𝐴, and it is denoted by 𝐵 − 𝐴 .
Therefore, 𝐵 − 𝐴 = {6}.
In general, if 𝐴 and 𝐵 are any two sets then,
𝐴 − 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵}
𝐵 − 𝐴 = {𝑥: 𝑥 ∈ 𝐵 𝑎𝑛𝑑 𝑥 ∉ 𝐴}
Difference of two sets can be represented by using Venn Diagrams as:
A
A
B
B-A
A-B
B
B-A
A-B
When A and B are not disjoint sets
When A and B are disjoint sets
Figure 1.4.5a
Figure 1.4.5b
25
Self-Assessments
1. Let the set 𝐴 = {𝑎, 𝑏, 𝑐}. What is the power set of 𝐴?
a. 𝑃(𝐴) = {𝜙, {𝑎}, {𝑏}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑏, 𝑐}, {𝑎, 𝑏, 𝑐}}
b. 𝑃(𝐴) = {𝜙, {𝑎}, {𝑏}, {𝑐}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑎, 𝑏, 𝑐}}
c. 𝑃(𝐴) = {𝜙, {𝑎}, {𝑏}, {𝑐}, {𝑎, 𝑏}, {𝑏, 𝑐}, {𝑎, 𝑏, 𝑐}}
d. 𝑃(𝐴) = {𝜙, {𝑎}, {𝑏}, {𝑐}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑏, 𝑐}, {𝑎, 𝑏, 𝑐}}
2. Let the set 𝑆 = {1, 2, 3, 4, 5, 6} and the set 𝑇 = {3, 4 ,5, 6}. What is the set 𝑆 − 𝑇?
a. 𝑆 − 𝑇 = {1, 3}
b. 𝑆 − 𝑇 = {3, 1}
c. 𝑆 − 𝑇 = {2, 3}
d. 𝑆 − 𝑇 = {1, 2}
3. Let the set 𝑈 = {1, 2, 3, 4, 5} and the set 𝐴 = {1, 3, 5}. Describe the relationship between
𝑈 and 𝐴
a. 𝐴 ⊂ 𝑈
b. 𝐴 = 𝑈
c. 𝐴 ⊆ 𝑈
d. 𝐴 ∈ 𝑈
4. Let the set 𝐴 = {𝑎, 𝑏, 𝑐}, and the set 𝐵 = {𝑏, 𝑐, 𝑎}. Describe the relationship between the
two sets 𝐴 and 𝐵.
a. 𝐴 ⊂ 𝐵
b. 𝐴 = 𝐵
c. 𝐴 ≡ 𝐵
d. 𝐴 ∈ 𝐵
5.
Answers to Self-Assessments
1.
2.
3.
4.
5.
D
D
A
B
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1.5.0. Session 5 – Compliments, Intersection, and Union of Sets
1.5.1. Compliment of a Set.
Let 𝑋 denote the universal set and 𝑌, 𝑍 be its subsets where:
𝑋 = {𝑥: 𝑥 𝑖𝑠 𝑎𝑛𝑦 𝑚𝑒𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑚𝑖𝑙𝑦}
𝑌 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑚𝑎𝑙𝑒 𝑚𝑒𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑚𝑖𝑙𝑦}
𝑍 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑓𝑒𝑚𝑎𝑙𝑒 𝑚𝑒𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑚𝑖𝑙𝑦}
𝑋 − 𝑌 is a set having female members of the family
𝑋 − 𝑍 is a set having male members of the family
𝑋 − 𝑌 is said to be the compliment of 𝑌 and is usually denoted by 𝑌′ or 𝑌 𝑐
𝑋 − 𝑍 is said to be compliment of 𝑍 and denoted by 𝑍′ or 𝑍 𝑐
If 𝑈 is the universal set and 𝐴 is its subset, then the compliment of 𝐴 is a set of those elements
which are in 𝑈 which are not in 𝐴. It is denoted by 𝐴′ or 𝐴𝑐 .
The compliment of a set can be represented using a Venn diagram as
U
Ac
or
A’
Figure 1.5.0
Special Note:
(i)
Difference of two sets can be found even if none is a subset of the other but
compliment of a set can be found only when the set is a subset of some universal set.
27
(ii)
𝜙 𝑐 = 𝑈.
(iii)
𝑈𝑐 = 𝜙
Example 1.5.1.1
Given that,
𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 10}
𝐵 = {𝑥: 𝑥 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 10}
Find
(i)
𝐴−𝐵
(ii) 𝐵 − 𝐶
(iii) is 𝐴 − 𝐵 = 𝐵 − 𝐴?
Solution 1.5.1.1
The sets given are 𝐴 = {2, 4, 6, 8, 10} and 𝐵 = {1, 3, 5, 7, 9}
Therefore,
(i)
(ii)
(iii)
𝐴 − 𝐵 = {2, 4, 6, 8, 10}
𝐵 − 𝐴 = {1, 3, 5, 7, 9}
From the solutions (i) and (ii) 𝐴 − 𝐵 ≠ 𝐵 − 𝐴
Example 1.5.1.2
Let 𝑈 be the universal set and 𝐴 its subset where:
𝑈 = {𝑥 ∶ 𝑥 ∈ ℕ 𝑎𝑛𝑑 𝑥 ≤ 10}
𝐴 = {𝑦: 𝑦 𝑖𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 10}
Find (i) 𝐴𝑐
(ii) Represent 𝐴𝑐 in a Venn diagram.
Solution 1.5.1.2
We have been given the sets 𝑈 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and 𝐴 = { 2, 3, 5, 7}
(i)
(ii)
𝐴𝑐 = 𝑈 − 𝐴 = {1, 4, 6, 8, 9, 10}
The Venn diagram is below.
28
U
1
4
9
10
6
8
2 3
5
7
Ac
1.5.2
Intersection of Sets
Consider the sets 𝐴 = {1, 2, 3, 4} and 𝐵 = {2, 4, 6}
We can see from the two sets that they have some elements in common. Both sets have the
elements 2 and 4 in them. Sets that have some common elements are said to intersect. Those
elements they have in common can be placed in a different set, say, 𝐶. The new set, 𝐶, that
contains the common elements is said to be the intersection of the sets 𝐴 and 𝐵. The intersection
of the set 𝐴 and the set 𝐵 is denoted by 𝐴 ∩ 𝐵.
In the above example, 𝐴 ∩ 𝐵 = {2, 4}
If 𝐴 and 𝐵 are two sets, then the set of those elements which belong to both sets is said to be the
intersection of 𝐴 and 𝐵. It is denoted by 𝐴 ∩ 𝐵. The intersection of 𝐴 and 𝐵 is the set 𝐴 ∩ 𝐵
consisting of all elements that are in both 𝐴 and 𝐵. In other words, 𝐴 ∩ 𝐵 is the common part of
𝐴 and 𝐵
𝐴 ∩ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵}
A
B
𝐴∩𝐵
B-A
A-B
When A and B are not disjoint sets
Figure 1.5.2
29
Special Note:
If 𝐴 ∩ 𝐵 = 𝜙 then 𝐴 and 𝐵 are said to be disjoint sets. We can refer to 𝐹𝑖𝑔𝑢𝑟𝑒 1.4.5𝑏 in session
the previous session, 𝑆𝑒𝑠𝑠𝑖𝑜𝑛 1.4.5 for a Venn diagram for a disjoint set.
Example 1.5.2.1
Given that,
𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝐾𝑖𝑛𝑔 𝑜𝑢𝑡 𝑜𝑓 52 𝑝𝑙𝑎𝑦𝑖𝑛𝑔 𝑐𝑎𝑟𝑑𝑠} and
𝐵 = {𝑦: 𝑦 𝑖𝑠 𝑎 𝑆𝑝𝑎𝑑𝑒 𝑜𝑢𝑡 𝑜𝑓 52 𝑝𝑙𝑎𝑦𝑖𝑛𝑔 𝑐𝑎𝑟𝑑𝑠}
Find
(i)
𝐴∩𝐵
(ii)
Represent 𝐴 ∩ 𝐵 by a Venn diagram
Solution
(i)
There are only four kings out of 52 playing cards. Therefore, the set 𝐴 has only four
elements,
𝐴 = {𝐾𝑖𝑛𝑔 𝑜𝑓 𝑆𝑝𝑎𝑑𝑒𝑠, 𝐾𝑖𝑛𝑔 𝑜𝑓𝐻𝑒𝑎𝑟𝑡𝑠, 𝐾𝑖𝑛𝑔 𝑜𝑓 𝐶𝑙𝑢𝑏𝑠, 𝐾𝑖𝑛𝑔 𝑜𝑓 𝐷𝑖𝑎𝑚𝑜𝑛𝑑𝑠}
The set 𝐵 has 13 elements since there are 13 cards of spades. Also, there is only one
king out of the 13 cards of spade. The king of spades is the only common element for
the sets 𝐴 and 𝐵.
∴ 𝐴 ∩ 𝐵 = {𝐾𝑖𝑛𝑔 𝑜𝑓 𝑆𝑝𝑎𝑑𝑒𝑠}
(ii)
For convenience, we describe the Venn diagram based on the knowledge we have
acquired in this session. We draw a rectangle for the universal sets of 52 playing
cards. We will draw two ovals (2 circles) inside the rectangle to intersect as shown in
the 𝐹𝑖𝑔𝑢𝑟𝑒 1.5.2. The set of intersection that represents the “King of Spades” is the
shaded portion where the two sets (the ovals) meet.
30
1.5.3
The Union of Sets
Let us consider the following examples.
Let 𝐴 be a set having all players of the national men football team and 𝐵 be a set
having all players of the national women football team. We can see clearly that the
two sets 𝐴 and 𝐵, are disjoint sets. The union of the two sets is a set having all players
of both teams and it is denoted by 𝐴 ∪ 𝐵.
Let the set 𝐷 be the set having all players of the football team and the set 𝐸 be the set
having all players of the Hockey team of your college. Suppose three players are
common to both teams then the union of 𝐷 and 𝐸 is a set of all players of both teams,
but the three common players are to be written once only.
(i)
(ii)
If 𝐴 and 𝐵 are only two sets then the union of 𝐴 and 𝐵 is the set of those element which belong
to 𝐴 and 𝐵. If 𝐴 and 𝐵 are sets, their union 𝐴 ∪ 𝐵 is the set that consists of all elements that are
in 𝐴 or 𝐵 (or in both)
In set builder form:
𝐴 ∪ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}
OR
𝐴 ∪ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 − 𝐵 𝑜𝑟 𝑥 ∈ 𝐵 − 𝐴 𝑜𝑟 𝑥 ∈ 𝐴 ∩ 𝐵}
U
U
A
A
B
B
𝐴∪𝐵
𝐴∪𝐵
Figure 1.5.3b
Figure 1.5.3a
31
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴 − 𝐵) + 𝑛(𝐵 − 𝐴) + 𝑛(𝐴 ∩ 𝐵)
OR
𝑛(𝐴 ∪ 𝐵 ) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)
Where 𝑛(𝐴 ∪ 𝐵) stands for the number of elements in 𝐴 ∪ 𝐵 and so on.
Example 1.5.3.1
Let
𝐴 = {𝑥: 𝑥 ∈ ℤ+ 𝑎𝑛𝑑 𝑥 ≤ 5 }
𝐵 = {𝑦: 𝑦 𝑖𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 10}
Find (i) 𝐴 ∪ 𝐵
(ii) represent 𝐴 ∪ 𝐵 using Venn Diagram
Solution 1.5.3.1
(i)
We have been given,
𝐴 = {1, 2, 3, 4, 5} ,
𝐵 = {2, 3, 5, 7}
∴ 𝐴 ∪ 𝐵 = {1, 2, 3, 4, 5, 7}
(ii)
We describe here the Venn diagram for convenience and allow the learners to
practice the drawing.
We proceed to draw a rectangle to represent the universal set. We also draw two
intersecting ovals (circles) inside the rectangle for the subsets 𝐴 = {1, 2, 3, 4, 5} and
𝐵 = {2, 3, 5, 7}. We note that there are 3 elements that are common to the two sets.
The intersection of the sets 𝐴 ∩ 𝐵 = {2, 3, 5}. We draw the two ovals such that, the
intersection contains the three elements {2, 3, 5}. The remaining elements will be
placed in the sets such that the elements {1, 4} are in 𝐴, but outside 𝐵, and the
element 7 in set 𝐵 but outside 𝐴.
32
We hope the learner will be able to follow the description and draw the Venn diagram
representation of the sets.
Example 1.5.3.2.
If 𝑆 = {1, 2, 3, 4, 5}, 𝑇 = {4, 5, 6, 7} and 𝑉 = {6, 7, 8}, find the sets 𝑆 ∪ 𝑇, 𝑆 ∩ 𝑇, and 𝑆 ∩ 𝑉
Solution 1.5.3.2.
The given sets are 𝑆 = {1, 2, 3, 4, 5}, 𝑇 = {4, 5, 6, 7} and 𝑉 = {6, 7, 8}
𝑆 ∪ 𝑇 = {1, 2, 3, 4, 5, 6, 7}
all elements in 𝑆 or 𝑇
𝑆 ∩ 𝑇 = {4, 5}
elements common to both 𝑆 and 𝑇
𝑆 ∩ 𝑉 = 𝜙 𝑜𝑟 {}
𝑆 and 𝑉 have no elements in common
Self-Assessment
Answers to Self-Assessments
1.
2.
3.
4.
5.
D
C
A
B
D
33
1.6.0. Session 6 – Cartesian Product of Two Sets
1.6.1. Cartesian Product of Sets
Consider the two sets 𝐴 and 𝐵 where 𝐴 = {1, 2} and 𝐵 = {3, 4, 5}.
The set of all ordered pairs of elements of 𝐴 and 𝐵 is the set,
{(1,3), (1,4), (1, 5), (2, 3), (2, 4), (2, 5)}.
This set is denoted by 𝐴 × 𝐵, and it is called the cartesian product of sets 𝐴 and 𝐵 i.e
𝐴 × 𝐵 = {(1, 2), (1, 4), (1, 5), (2,3), (2, 4), (2, 5)}
The cartesian product of the sets 𝐵 and 𝐴 is denoted by,
𝐵 × 𝐴 = {(3,1), (3,2), (4,1), (4, 2), (5, 1), (5,2)}
It can be seen that 𝐴 × 𝐵 ≠ 𝐵 × 𝐴
In a set builder form:
𝐴 × 𝐵 = {(𝑎, 𝑏): 𝑎 ∈ 𝐴 𝑎𝑛𝑑 𝑏 ∈ 𝐵}
𝐵 × 𝐴 = {(𝑏, 𝑎): 𝑏 ∈ 𝐵 𝑎𝑛𝑑 𝑎 ∈ 𝐴}
Note:
If 𝐴 = 𝜙 or 𝐵 = 𝜙 or 𝐴, 𝐵 = 𝜙
Then 𝐴 × 𝐵 = 𝐵 × 𝐴 = 𝜙
Example 1.6.0.1
Let 𝐴 = {𝑎, 𝑏, 𝑐}, 𝐵 = {𝑑, 𝑒} , 𝐶 = {𝑎, 𝑑}
Find
(i) 𝐴 × 𝐵
(ii) 𝐵 × 𝐴
(v) (𝐴 ∩ 𝐵) × 𝐶
(iii) 𝐴 × (𝐵 ∪ 𝐶)
(vi) 𝐴 × (𝐵 − 𝐶)
Solution 1.6.0.1
34
(iv) (𝐴 ∩ 𝐶) × 𝐵
(i)
(ii)
(iii)
(iv)
(v)
(vi)
𝐴 × 𝐵 = {(𝑎, 𝑑), (𝑎, 𝑒), (𝑏, 𝑑), (𝑏, 𝑒), (𝑐, 𝑑), (𝑐, 𝑒) }
𝐵 × 𝐴 = {(𝑑, 𝑎), (𝑑, 𝑏), (𝑑, 𝑐), (𝑒, 𝑎), (𝑒, 𝑏), (𝑒, 𝑐)}
𝐴 = {𝑎, 𝑏, 𝑐}, (𝐵 ∪ 𝐶) = {𝑎, 𝑑, 𝑒} ⇒ 𝐴 × (𝐵 ∪ 𝐶) =
{(𝑎, 𝑎), (𝑎, 𝑑), (𝑎, 𝑒), (𝑏, 𝑎), (𝑏, 𝑑), (𝑏, 𝑒), (𝑐, 𝑎), (𝑐, 𝑑), (𝑐, 𝑒)}
(𝐴 ∩ 𝐵) × 𝐵 = 𝐴 ∩ 𝐶 = {𝑎} × 𝐵 = {𝑎} × {𝑑, 𝑒} = {(𝑎, 𝑑), (𝑎, 𝑒)}
𝐴 ∩ 𝐵 = 𝜙, 𝐶 = {𝑎, 𝑑}, ⇒ (𝐴 ∩ 𝐵) × 𝐶 = 𝜙 × (𝑎, 𝑑) = 𝜙
𝐴 × (𝐵 − 𝐶) = {𝑎, 𝑏, 𝑐} × {𝑒} = {(𝑎, 𝑒), (𝑏, 𝑒), (𝑐, 𝑒)}
35
Unit 2.0.
Session 1.
Definition of Relations, Domain, and Range of Relations
Session 2.
Definitions of Functions, Domain of Functions, and Range of Function
Session 3.
Types of Functions
Session 4.
Composition of Functions
Session 5.
Inverse of Functions
Session 6.
Graphs (Curves) of Functions
36
Overview of the Unit.
1. Understand the concept of functions and identify different types of functions, and their
characteristics
2. Understand Exponents, Roots and Radicals
37
The Concepts of Functions.
By the end of this unit, we hope to learn to understand the terms, function, domain, range, oneto-one, many-to-one functions and how to evaluate functions. We will learn to find simplified
expressions or functions.
38
2.1.0
Session 1 - Definition of Relations, Domain, and Range of Relations
2.1.1
Definition of Relations
Consider the following sample sets:
𝐴 = {𝐾𝑤𝑎𝑑𝑤𝑜, 𝐾𝑜𝑓𝑖, 𝐾𝑜𝑟𝑘𝑢, 𝐴𝑙ℎ𝑎𝑠𝑠𝑎𝑛}
𝐵 = {𝐸𝑠𝑖, 𝐷𝑧𝑖𝑓𝑎, 𝐹𝑢𝑠𝑒𝑖𝑛𝑎}
Suppose Esi has two brothers Kwadwo and Kofi, Dzifa has one brother Korku, and Fuseina has
one brother, Alhassan. If we define a relation 𝑅 “is a brother of” between the elements of the sets
𝐴 and 𝐵 above, then we can say,
Kwadwo R Esi,
Kofi R Esi,
Korku R Dzifa,
Alhassan R Fuseina
If we omit the R between the two names, we can write them in the form of ordered pairs as
follows:
(Kwadwo, Esi), (Kofi, Esi), (Korku, Dzifa), (Alhassan, Fuseina)
The above ordered pairs can be rewritten in the form of a set 𝑅 of ordered pairs as follows:
𝑅 = {(Kwadwo, Esi), (Kofi, Esi), (Korku, Dzifa), (Alhassan, Fuseina)}
We can see from the relation 𝑅 that 𝑅 ⊆ 𝐴 × 𝐵, i.e, 𝑅 = {(𝑎, 𝑏): 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵 𝑎𝑛𝑑 𝑎𝑅𝑏}
If 𝐴 and 𝐵 are any two sets, then a relation 𝑅 from 𝐴 to 𝐵 is a subset of 𝐴 × 𝐵.
If
(i)
(ii)
(iii)
(iv)
𝑅 = 𝜙,
𝑅 is called a void relation.
𝑅 = 𝐴×𝐵
𝑅 is called a universal relation.
If 𝑅 is a relation defined from 𝐴 to 𝐴, it is called a relation defined on 𝐴
𝑅 = {(𝑎, 𝑎) ∀ 𝑎 ∈ 𝐴} is called the identity relation.
39
2.1.2.
Domain and Range of a Relation
If 𝑅 is a relation between any two sets, then the set of its first elements (components) of all the
ordered pairs of 𝑅 is called the Domain and the second set of all the ordered pairs of 𝑅 is called
the Range of the given relation.
From our previous sample sets of brothers and sisters, we have,
Domain = {𝐾𝑤𝑎𝑑𝑤𝑜, 𝐾𝑜𝑓𝑖, 𝐾𝑜𝑟𝑘𝑢, 𝐴𝑙ℎ𝑎𝑠𝑠𝑎𝑛}
Range = {𝐸𝑠𝑖, 𝐷𝑧𝑖𝑓𝑎, 𝐹𝑢𝑠𝑒𝑖𝑛𝑎}
Example 2.1.2.1.
Given that 𝐴 = {2, 4, 5, 6, 7}, 𝐵 = {2, 3}. 𝑅 is a relation from 𝐴 to 𝐵 by,
𝑅 = {(𝑎, 𝑏): 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵 and 𝑎 is divisible by 𝑏}
Find
(i)
(ii)
(iii)
(iv)
𝑅 in the roster form
Domain of 𝑅
Range of 𝑅
Represent R diagrammatically.
Solution 2.1.2.1.
(i)
(ii)
(iii)
(iv)
We begin by identifying the elements in 𝐴 that are divisible by the elements in 𝐵, to
form the pairs that we need. Here we have the elements in 𝐴 = {2, 4, 6} that are
divisible by the elements in 𝐵 = {2, 3}. Our desired set of pairs is
𝑅 = {(2, 2), (4, 2), (6, 2), (6, 3)}
The Domain of 𝑅 are the elements in the first elements that are divisible by the
elements in the second set. Domain of 𝑅 = {2, 4, 6}
The Range of the Domain are those elements in the second set that can divide (is
divisible by) the elements in the first set, the domain. 𝑅 = {2, 3}
We represent the relation by drawing two ovals (circles) that stand next to each other.
We make a list of all the elements 𝐴 = { 2, 4, 5, 6, 7}, in one of the ovals to the left,
and we label it the domain, and another oval to the right we will label as the range
with the elements 𝐵 = {2, 3}. We draw line from each element from the domain to its
matching image in the range as shown in Figure 2.1.1 below.
40
2
2
4
R
5
6
3
7
Figure 2.1.1
Example 2.1.2.2.
If 𝑅 is a relation ‘is greater than’ from 𝐴 to 𝐵, where 𝐴 = {1, 2, 3, 4, 5} and 𝐵 = {1, 2, 6}. Find
𝑅 in the roster form
(i)
(ii)
Domain of 𝑅
(iii) Range of 𝑅
Solution 2.1.2.2.
Here we want to identify all elements in set 𝐴 that are greater than the elements in set
𝐵, and pair them in a roster form:
𝑅 = {(2,1), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}
The Domain of 𝑅 = {2, 3, 4, 5}
The Range of 𝑅 = {1, 2}
(i)
(ii)
(iii)
Answers to Self-Assessments
1.
2.
3.
4.
5.
D
C
A
B
D
41
2.2.0. Session 2 - Definition of Functions, Domain of Functions, and Range of Function
Overview
What does the word function mean in everyday life? If a person attempts to switch on a
television, if we start the engine of a car, if we perform any action on any object, if we carry out
any action with a purpose, we say any of those activities are functions in everyday life. We can
think of a function as an assigned purpose or activity.
However, in this unit of the course, we are more interested in mathematical examples of
functions.
2.2.1. Function as a Rule
A rule such as, ADD 5 denoted by + 5, SUBTRACT 2 denoted by - 2, DIVIDE by 10 denoted /
10, MULTIPLY by 4 dented by x 4, are types of mathematical functions.
A function is a rule. Every function must be given a name. We will use letters such as 𝑓, 𝑔, ℎ, …
to represent names of functions. We can use the letter 𝑓 to represent a rule as follows:
"𝑓"
is the rule
“do something, say square a number”.
When we write 𝑓(2), we mean apply the rule 𝑓 to the number in the parenthesis, or the brackets.
Applying the rule, we obtain 𝑓(2) = 22 = 4, and so on to any number.
For example, if we state the rule, 𝑓(𝑥) → 𝑥 + 5, we mean the function 𝑓 is to add 5 to any given
value of 𝑥.
The function notation 𝑓(𝑥) → 𝑥 + 5 for adding 5 to any given number for the variable 𝑥 is
normally denoted by 𝑓 as follows:
𝑓(𝑥) = 𝑥 + 5
What do we mean when we write the notation 𝑓(𝑥) = 𝑥 + 5?
We mean that the function, 𝑓, takes 𝑥 to 𝑥 + 5. The aim is to add 5 to any given value of 𝑥. The
function 𝑓(𝑥) is pronounced as “𝑓 of 𝑥”
42
2.2.2. Definition of a Function:
A function 𝑓 is a rule that assigns to each element 𝑥 in a set 𝐴 exactly one element, called 𝑓(𝑥),
in a set 𝐵.
We generally consider functions for which sets 𝐴 and 𝐵 are sets of real numbers. The symbol
𝑓(𝑥) is pronounced as "𝑓 𝑜𝑓 𝑥", or "𝑓 𝑎𝑡 𝑥" and is called the value of 𝑓 at 𝑥, or the image of 𝑥
under 𝑓.
The set 𝐴, where we start from is called the domain of the function. The set where we arrive at is
called the range of the function, 𝑓, which is the set of all possible values of 𝑓(𝑥) as 𝑥 varies
throughout the domain.
The term set means a collection of objects.
The domain and the range do not just consist of a few selected points or items, but they could be
all the real numbers that exist.
The variable 𝑥 in the notation 𝑓(𝑥) = 𝑥 + 5 is called the independent variable, because 𝑓(𝑥)
depends on (𝑥).
What does the term function mean in mathematics?
A function is a rule that gives every 𝑥 in the domain only one value in the range, 𝑓(𝑥).
Example 2.2.2.1.
Let 𝑓(𝑥) = 𝑥 2 + 1, and evaluate the following:
(a)
(b)
(c)
(d)
(e)
(f)
𝑓(−3)
𝑓(3)
𝑓(−2)
𝑓(2)
𝑓(−1)
𝑓(1)
43
Solution 2.2.2.1.
The function says, for every value of 𝑥, square it and add 1, therefore
(a)
(b)
(c)
(d)
(e)
(f)
𝑓(−3) = (−3)2 + 1 = 9 + 1 = 10
𝑓(3) = (3)2 + 1 = 9 + 1 = 10
𝑓(−2) = (−2)2 + 1 = 4 + 1 = 5
𝑓(2) = (2)2 + 1 = 4 + 1 = 5
(−1) = (−1)2 + 1 = 1 + 1 = 2
(1)
= (1)2 + 1 = 1 + 1 = 2
Example 2.2.2.2.
Solution 2.2.2.2.
44
Consider the relation 𝑓: {(𝑎, 1), (𝑏, 2), (𝑐, 3), (𝑑, 5)}.
A
B
f
a
1
b
2
c
3
d
4
5
Figure 2.2.2
In the relation, we see that each element of 𝐴 has a unique image in 𝐵. This relation 𝑓 from a set
𝐴 to another set 𝐵 where every element of 𝐴 has a unique image in 𝐵 is defined as a function
from 𝐴 to 𝐵. We observe that in a function no two ordered pairs have the same first element.
We also see that ∃ (there exists) an element ∈ 𝐵, i.e., 4 which does not have its preimage in 𝐴.
Thus here:
(i)
(ii)
the set 𝐵 will be termed as co-domain and
the set {1, 2, 3, 5} is called the range.
From the above we can conclude that the range is a subset of the co-domain
Symbolically, this function can be written as 𝑓: 𝐴 → 𝐵
It is helpful to think of a function as a machine that accepts an input and produces an output.
Diagram of function as a machine
x as input
f
Figure 2.2.3
45
f(x) as output
If 𝑥 is in the domain of the function 𝑓, then when 𝑥 enters the machine, it is accepted as an input
and the machine processes the input 𝑥 to produce an output 𝑓(𝑥) according to the rule of the
function.
Therefore, we can think of the domain as the set of all possible inputs and the range as the set of
all possible outputs.
Example 2.2.2.3.
Which of the following relations are functions from 𝐴 to 𝐵? Write their domain and range. If it is
not a function, give reasons why they are not functions.
(a) {(1, −2), (3, 7), (4, −6), (8, 1)}, 𝐴 = {1, 3, 4, 8}, 𝐵 = {−2, 7, −6, 1, 2}
(b) {(1, 0), (1, −1), (2, 3), (4, 10)}, 𝐴 = {1, 2, 4}, 𝐵 = {0, −1, 3, 10}
(c) {(𝑎, 𝑏), (𝑏, 𝑐), (𝑐, 𝑏), (𝑑, 𝑐)}, 𝐴 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}, 𝐵 = {𝑏, 𝑐}
1
1
1
1 1
1
(d) {(2, 2) , (3, 3) , … , (10, 10)} , 𝐴 = {1, 2, 3, 4}, 𝐵 = {2 , 3 , … , 11}
Solution 2.2.2.3.
(a) The relation {(1, −2), (3, 7), (4, −6), (8, 1)}, 𝐴 = {1, 3, 4, 8}, 𝐵 = {−2, 7, −6, 1, 2} is a
function. The Domain = {1, 3, 4, 8} and the Range = {−2, 7, −6, 1}
(b) The relation {(1, 0), (1, −1), (2, 3), (4, 10)}, 𝐴 = {1, 2, 4}, 𝐵 = {0, −1, 3, 10} is not a
function, because the first two ordered pairs have the same first elements, i.e. the element
1, in the first set has more than one image in the second set
(c) The relation {(𝑎, 𝑏), (𝑏, 𝑐), (𝑐, 𝑏), (𝑑, 𝑐)}, 𝐴 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}, 𝐵 = {𝑏, 𝑐} is not a relation
because the Domain = {𝑎, 𝑏, 𝑐, 𝑑} ≠ 𝐴, the Range = {𝑏, 𝑐}
1
1
1
1 1
1
(d) The relation {(2, 2) , (3, 3) , … , (10, 10)} , 𝐴 = {1, 2, 3, 4}, 𝐵 = {2 , 3 , … , 10} is not a
function because the Domain = {2, 3, 4, 5, 6, 7, 8, 9, 10} ≠ 𝐴 = {1, 2, 3, 4}, the Range =
1 1 1 1 1 1 1 1 1
{2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10}
2.2.3. More Examples on Domain and Range
Let us consider some more examples of functions that are only defined for certain subset of the
set of real numbers.
46
Example 2.2.3.1.
Find the domain of each of the following functions:
1
1
(a) 𝑦 = 𝑥
1
(b) 𝑦 = 𝑥−2
(c) 𝑦 = (𝑥+2)(𝑥−3)
Solution 2.2.3.1.
A rational expression is not defined when the denominator is 0
1
(a) The function 𝑦 = 𝑥 can be described by the following set of ordered pairs
1
1
{… . , (−2, − 2) , (−1, −1), (1,1), (2, 2) , … . }. We can see that 𝑥 can take all real number
values except 0, because the corresponding image, i.e.,
1
0
is not defined for real numbers.
Therefore, the Domain = ℝ − {0} [Set of all real numbers except 0] and Range = ℝ −
{0}
1
(b) For the function, 𝑦 =
, 𝑥 can take all real numbers except 2, because the
𝑥−2
1
corresponding image, i.e., = 2−2 does not exist for real numbers. Therefore, the Domain=
ℝ − {2}
1
(c) For the function, 𝑦 = (𝑥+2)(𝑥−3) , 𝑦 is defined for all real numbers except for 𝑥 = −2 and
for 𝑥 = 3. Therefore, the Domain = ℝ − {−2, 3}
47
2.3.0. Session 3 – Types of Functions
2.3.1. Classification of Functions
Let 𝑓 be a function from 𝐴 to 𝐵. If every element of the set 𝐵 is the image of at least one element
of the set 𝐴, i.e., if there is no unpaired element in the set 𝐵 then we say that the function 𝑓 maps
the set 𝐴 onto the set 𝐵. Otherwise, we say that the function maps the set 𝐴 into the set 𝐵.
Functions for which each element of a set 𝐴 is mapped to a different element of the set 𝐵 are said
to be one-to-one.
One-to one function
1
A
2
B
3
C
4
Figure 2.2.4a
The Domain is = { 𝐴, 𝐵, 𝐶}
The Co-Domain is = {1, 2, 3, 4}
The Range is = {1, 2, 3}
48
A function can map more than one element of the set 𝐴 to the same element of the set 𝐵. Such a
type of function is said to be many-to-one.
Many-to one function
1
A
2
B
3
C
4
Figure 2.2.4b
The Domain is = { 𝐴, 𝐵, 𝐶}
The Co-Domain is = {1, 2, 3, 4}
The Range is = {1, 3}
Relations that are one-to-many can occur, but they are not functions.
2.3.2. Some Special Functions
2.3.2.1. Even Functions
A function is said to be an even function, if for each 𝑥 of the domain,
𝑓(−𝑥) = 𝑓(𝑥)
Examples of even functions are:
(i)
(ii)
(iii)
If 𝑓(𝑥) = 𝑥 2 then 𝑓(−𝑥) = (−𝑥)2 = 𝑥 2 = 𝑓(𝑥)
If 𝑓(𝑥) = cos 𝑥 then 𝑓(−𝑥) = cos(−𝑥) = cos 𝑥 = 𝑓(𝑥)
If 𝑓(𝑥) = |𝑥| then 𝑓(−𝑥) = |−𝑥| = |𝑥| = 𝑓(𝑥)
49
2.3.2.2 Odd Functions
A function is said to be odd if for each 𝑥 𝑓(−𝑥) = −𝑓(𝑥)
Examples of odd functions include:
(i)
(ii)
If 𝑓(𝑥) = 𝑥 3 then 𝑓(−𝑥) = (−𝑥)3 = −𝑥 3 = −𝑓(𝑥)
If 𝑓(𝑥) = sin 𝑥 then 𝑓(−𝑥) = sin(−𝑥) = − sin 𝑥 = −𝑓(𝑥)
2.3.2.3 Polynomial Functions
Any function defined with degree 𝑛, of the form, 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 is a
polynomial function, where 𝑛 is a nonnegative integer and 𝑎𝑛 ≠ 0. We shall revisit polynomials
in Unit 3. A simple definition of polynomials is good for now.
Examples of polynomial functions are:
(i)
(ii)
𝑓(𝑥) = 3𝑥 2 − 4𝑥 − 2, a quadratic function
𝑓(𝑥) = 𝑥 3 − 5𝑥 2 − 𝑥 + 5, a cubic function
2.3.2.4 Reciprocal Functions
1
Functions of the type 𝑦 = 𝑥 , 𝑥 ≠ 0 is called a reciprocal function.
2.3.2.5. Exponential Functions
A function defined in the form 𝑦 = 𝑓(𝑥) = 𝑒 𝑥 , where 𝑥 is any real number is called an
exponential function.
2.3.2.6. Logarithmic Functions
Functions that involve logarithms are logarithmic functions. The exponential function 𝑓(𝑥) =
𝑦 = 𝑒 𝑥 can be rewritten in logarithmic form as 𝑥 = log 𝑒 𝑦. The function 𝑦 = log 𝑒 𝑥 is the
inverse of 𝑦 = 𝑒 𝑥 . We shall learn how to determine the inverses of functions in a moment. The
base of a function is usually not written if the base is 𝑒.
We will usually write log 𝑒 𝑥 as log 𝑥. It is also common to write log 𝑒 𝑥 as ln 𝑥 which
pronounced as natural logarithm.
50
2.3.2.7. Rational Functions
𝑔(𝑥)
Functions of the type 𝑓(𝑥) = ℎ(𝑥) , where ℎ(𝑥) ≠ 0 and 𝑔(𝑥) and ℎ(𝑥) are linear or polynomial
functions and are known as rational functions
For example, 𝑓(𝑥) =
𝑥 2− 4
𝑥+1
, 𝑥 ≠ 1 is a rational function.
51
2.4.0. Session 4 – Composition of Functions
Consider the two functions below:
𝑦 = 2𝑥 + 1,
𝑧 = 𝑦 + 1,
𝑥 ∈ {1, 2, 3}
𝑦 ∈ {3, 5, 7}
The function 𝑧 is defined in terms of 𝑦, and the function 𝑦 is also defined in terms of 𝑥.
Functions of such nature are known as composite functions. The function 𝑧 operates on the
function 𝑦. The function 𝑦, in turn, operates on 𝑥. Symbolically, the composition of the functions
𝑧 and 𝑦 is written as 𝑧(𝑦(𝑥)) or 𝑧𝑜𝑦, which implies that 𝑦 operates on 𝑥, and 𝑧 operates on 𝑦.
Example 2.4.0.1.
Let 𝑓(𝑥) = 3𝑥 + 1 and 𝑔(𝑥) = 𝑥 2 + 2
Obtain the function 𝑓𝑜𝑔(𝑥) = 𝑓(𝑔(𝑥)).
Solution 2.4.0.1.
The above composite function, says, the function 𝑓 accepts the function 𝑔 as input. The rule of
the function 𝑔 says that, for every value of 𝑥, obtain the square of 𝑥 and add 2 to the square of 𝑥,
i.e., 𝑔(𝑥) = 𝑥 2 + 2
The rule for function 𝑓 says that, for every value of 𝑥, multiply 𝑥 by 3, i.e., 3𝑥 and add 1 to 3𝑥,
i.e., 𝑓(𝑥) = 3𝑥 + 1
The overall solution for the composite function follows:
Step 1:
The function 𝑔(𝑥) = 𝑥 2 + 2 becomes the 𝑥 for the function 𝑓(𝑥) , as follows:
i.e., 𝑓(𝑔(𝑥)) = 𝑓(𝑥 2 + 2)
Step 2:
We apply the rule for the function 𝑓, which says, for every 𝑥 , multiply 𝑥 by 3 and add 1. It
follows here:
We multiply 𝑥 2 + 1 by 3 and add 1 as follows:
52
𝑓(𝑔(𝑥)) = 3(𝑥 2 + 2) + 1
which after expansion will produce
𝑓(𝑔(𝑥)) = 3𝑥 2 + 6 + 1 = 3𝑥 2 + 7
(i)
The above rules and steps are applicable to the solution of all forms of composite functions.
We can solve for 𝑔(𝑓(𝑥)) in a similar way.
Example 2.4.0.2.
Let 𝑓(𝑥) = 3𝑥 + 1 and 𝑔(𝑥) = 𝑥 2 + 2
Obtain the function 𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥))
Solution 2.4.0.2.
We begin by making 𝑓(𝑥) = 3𝑥 + 1 the 𝑥 for the function 𝑔(𝑥) = 𝑥 2 + 2 as follows:
𝑔(𝑓(𝑥)) = (3𝑥 + 1)2 + 2
Using the binomial expansion for (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2
We follow with the expansion and simplification as follows:
𝑔(𝑓(𝑥)) = (9𝑥 2 + 6𝑥 + 1) + 2
Simplifying, we obtain,
𝑔(𝑓(𝑥)) = 9𝑥 2 + 6𝑥 + 3
(ii)
Comparing Example 2.4.0.1 and Example 2.4.0.2 we can conclude that 𝑓𝑜𝑔(𝑥) ≠ 𝑔𝑜𝑓(𝑥)
It is important to note that, given any function 𝑓(𝑥) or 𝑔(𝑥), we can operate the function 𝑓(𝑥) on
itself, i.e., 𝑓(𝑓(𝑥)) or 𝑓𝑜𝑓 and 𝑔(𝑔(𝑥)) or 𝑔𝑜𝑔 to obtain another or a different function(s).
53
2.4.1. The Algebra of Functions
𝑓
Let 𝑓 and 𝑔 be functions with domains 𝐴 and 𝐵. Then the functions 𝑓 + 𝑔, 𝑓 − 𝑔, 𝑓𝑔, and 𝑔 are
defined as follows:
(i)
(ii)
(iii)
(𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥)
(𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥)
(𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥)
(iv)
(𝑔) (𝑥) = 𝑔(𝑥)
𝑓
𝑓(𝑥)
Domain 𝐴 ∩ 𝐵
Domain 𝐴 ∩ 𝐵
Domain 𝐴 ∩ 𝐵
Domain {𝑥 ∈ 𝐴 ∩ 𝐵 | 𝑔(𝑥) ≠ 0}
2.4.2. Composition of Three Functions
It is possible to take the composition of three or more functions. For example, the composite
function 𝑓 𝑜 𝑔 𝑜 ℎ is formed by first applying ℎ, then 𝑔 and then 𝑓 as follows:
(𝑓 𝑜 𝑔 𝑜 ℎ)(𝑥) = 𝑓 (𝑔(ℎ(𝑥)))
Exercises:
Find 𝑓𝑜𝑔, 𝑔𝑜𝑓, 𝑓𝑜𝑓and 𝑔𝑜𝑔 for the following functions:
(i)
(ii)
(iii)
1
𝑓(𝑥) = 𝑥 2 + 2 and 𝑔(𝑥) = 1 − 1−𝑥 , 𝑥 ≠ 1
𝑓(𝑥) = 𝑥 2 − 4 and 𝑔(𝑥) = 2𝑥 + 5
𝑓(𝑥) = 𝑥 2 + 3 and 𝑔(𝑥) = 𝑥 − 2
Solution (i) 𝑓𝑜𝑔(𝑥)
(i)
1
Given 𝑓(𝑥) = 𝑥 2 + 2 and 𝑔(𝑥) = 1 − 1−𝑥 , 𝑥 ≠ 1
2
1
𝑓𝑜𝑔(𝑥) = 𝑓 (1 − (
)) + 2
1−𝑥
𝑓𝑜𝑔(𝑥) = (12 − 2(1) (
54
1
1 2
)+(
) )+2
1−𝑥
1−𝑥
𝑓𝑜𝑔(𝑥) = (1 −
𝑓𝑜𝑔(𝑥) =
2
1
+
)+2
1 − 𝑥 (1 − 𝑥)2
(1 − 𝑥)2 − 2(1 − 𝑥) + 1 + 2(1 − 𝑥)2
(1 − 𝑥)2
3(1 − 𝑥)2 − 2(1 − 𝑥) + 1
𝑓𝑜𝑔(𝑥) =
(1 − 𝑥)2
𝑓𝑜𝑔(𝑥) = 3 −
55
2
1
+
1 − 𝑥 (1 − 𝑥)2
2.5.0. Session 5 – Inverse of Functions
The inverse of a function is a rule that acts on the output of the function and produces the
corresponding input. This means that the inverse “undoes” or reverses what the function does.
Not all functions have inverses; those that have inverses are called one-to-one. The inverse of a
function exists if the function is a one-to-one and an onto function.
2.5.1. Definition of a One-to-One Function
A function with domain 𝐴 is called a one-to-one function if no two elements of 𝐴 have the same
image, that is 𝑓(𝑥1 ) ≠ 𝑓(𝑥2 ), where 𝑥1 ≠ 𝑥2
Another way or an equivalent way of writing the condition for a one-to-one function is
If 𝑓(𝑥1) = 𝑓(𝑥2), then 𝑥1 = 𝑥2
2.5.2. The Inverse of a Function.
One-to-one functions are important because they are the functions that possess inverse functions
according to the following definition.
2.5.3. Definition of the Inverse of a Function.
Let 𝑓 be a one-to-one function with domain 𝐴 and range 𝐵. Then its inverse function 𝑓 −1 has
domain 𝐵 and range 𝐴 and is defined by 𝑓 −1 (𝑦) = 𝑥 ⟺ 𝑓(𝑥) = 𝑦 for any 𝑦 ∈ 𝐵
This definition says that if 𝑓 takes 𝑥 to 𝑦, then 𝑓 −1 takes 𝑦 back to 𝑥. If 𝑓 were not one-to-one,
then 𝑓 −1 would not be defined uniquely. From the definition, we have
Domain of 𝑓 −1 = range of 𝑓
Range of 𝑓 −1 = domain of 𝑓
56
Examples 2.5.3.1.
Find the inverses for the following values, if:
(i)
𝑓(1) = 5
(ii)
𝑓(3) = 7 and
(iii)
𝑓(8) = −10
Solution 2.5.3.1.
From the definition of function inverse, if a function 𝑓 takes 𝑥 to 𝑦, then its inverse 𝑓 −1 takes 𝑦
back to 𝑥. Therefore,
(i)
(ii)
(iii)
𝑓 −1 (5) = 1 because 𝑓(1) = 5
𝑓 −1 (7) = 3 because 𝑓(3) = 7
𝑓 −1 (−10) = 8 because 𝑓(8) = −10
2.5.4. Inverse Function Property
Let 𝑓 be a one-to-one function with domain 𝐴 and range 𝐵. The inverse 𝑓 −1 satisfies the
following cancellation properties:
𝑓 −1 (𝑓(𝑥)) = 𝑥 for every 𝑥 in 𝐴
𝑓(𝑓 −1 (𝑥)) = 𝑥 for every 𝑥 in 𝐵
Conversely, any function 𝑓 −1 satisfying these equations is the inverse of 𝑓.
2.5.5. How To Find the Inverse of a One-to-One Function
Step 1: Write the function in the form 𝑦 = 𝑓(𝑥)
Step 2: Solve the equation, 𝑦 = 𝑓(𝑥) for 𝑥 in terms of 𝑦 (if possible)
Step 3: Interchange 𝑥 and 𝑦, to make 𝑦 the subject of the equation. The resulting equation is 𝑦 =
𝑓 −1 (𝑥).
Note that Steps 2 and 3 can be reversed. In other words, we can interchange 𝑥 and 𝑦 first, and
then solve for 𝑦 in terms of 𝑥
57
Example 2.5.5.1
Find the inverse of the function 𝑓(𝑥) = 3𝑥 − 2
Solution 2.5.5.1
Step 1. We write the 𝑦 = 𝑓(𝑥) as follows:
𝑦 = 3𝑥 − 2
Step 2. We solve for 𝑥 in terms of 𝑦, by adding 2 to both sides, i.e.,
𝑦 + 2 = 3𝑥 − 2 + 2
and we get
3𝑥 = 𝑦 + 2
We divide both sides by 3 , and we get,
𝑥=
𝑦+2
3
𝑦=
𝑥+2
3
Step 3: We interchange 𝑥 and 𝑦 to get,
Therefore, the inverse of the function 𝑓(𝑥) = 3𝑥 − 2 is
𝑦=
𝑥+2
3
Example 2.5.5.2.
Find the inverse of the function 𝑓(𝑥) =
𝑥 5 −3
2
58
Solution 2.5.5.2.
As usual we write 𝑦 = 𝑓(𝑥) =
𝑦=
𝑥 5 −3
2
and solve for 𝑥
𝑥 5 −3
2
Multiply both sides or multiply through by 2 to obtain,
2𝑦 = 𝑥 5 − 3, solving for 𝑥, we add 3 to both sides (and switch sides) as follows:
𝑥 5 = 2𝑦 + 3
Taking fifth root of both sides, we have,
𝑥 = (2𝑦 + 3)1/5
We interchange 𝑥 and 𝑦 to obtain,
𝑦 = (2𝑥 + 3)1/5
Therefore, the inverse function is
5
𝑓 −1 = (2𝑥 + 3)1/5 = √2𝑥 + 3
Example 2.5.5.3.
Finding the inverse of a rational function, 𝑓(𝑥) =
59
2𝑥+3
𝑥−1
Solution 2.5.5.3.
We proceed by writing 𝑓(𝑥) =
2𝑥+3
𝑥−1
as 𝑦 =
2𝑥+3
𝑥−1
𝑦=
and solve for 𝑥.
2𝑥 + 3
𝑥−1
Multiply through by 𝑥 − 1 to get,
𝑦 (𝑥 − 1) = 2𝑥 + 3
Expand the equation to obtain,
𝑦𝑥 − 𝑦 = 2𝑥 + 3
Bring the terms in 𝑥 together on the LHS,
𝑦𝑥 − 2𝑥 = 𝑦 + 3
Factor 𝑥 out to get,
𝑥(𝑦 − 2) = 𝑦 + 3
Divide by 𝑦 − 2 to get,
𝑥=
𝑦+3
𝑦−2
Interchange 𝑥 and 𝑦 to get,
60
𝑦=
𝑥+3
𝑥−2
Therefore, the inverse function is,
𝑓 −1 =
𝑥+3
𝑥−2
Exercises
Find the inverse function of 𝑓 if,
(i)
(iv)
𝑓(𝑥) =
𝑥−2
𝑥+2
(ii)
𝑓(𝑥) = √2𝑥 − 1 (v)
𝑓(𝑥) =
1+3𝑥
5−2𝑥
(iii)
3
𝑓(𝑥) = 1 − 𝑥 (vi)
61
𝑓(𝑥) =
4𝑥−2
3𝑥+1
𝑓(𝑥) = √2 + 5𝑥
2.6.0. Session 6 - Graphs (Curves) of Functions
When a mathematical equation is known, coordinates may be calculated for a limited range of
values, and the equation may be represented pictorially as a graph within the range of the
calculated values. It is sometimes useful to show all the characteristics or the features of an
equation. In this case a sketch depicting the equation can be drawn, in which all the important
features are shown, but accurate drawing or plotting of all the points is less important. This
technique is called ‘curve sketching’, and can involve the use of calculus, with, for example,
calculations involving turning points.
If, say, 𝑦 depends on, say, 𝑥 then 𝑦 is said to be a function of 𝑥, and the relationship is expressed
as 𝑦 = 𝑓(𝑥); just as we have seen earlier in this unit.
In the relation 𝑦 = 𝑓(𝑥), 𝑥 is called the independent variable and 𝑦 is the dependent variable.
In science and engineering, corresponding values are obtained as a result of tests or experiments.
We now present a brief description of some standard curves, some of which we have met earlier
in this unit.
2.6.1. The Straight Line
The general equation of a straight line is 𝑦 = 𝑚𝑥 + 𝑐, where 𝑚 is the gradient and 𝑐 is the 𝑦-axis
intercept.
2.6.2. Quadratic Graphs
The general equation of a quadratic graph is 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, and its shape is that of a
parabola.
The simplest example of a quadratic graph is, 𝑦 = 𝑥 2, which is a parabola that opens upwards
2.6.3. Graph of the Cubic Equations
The general equation of the cubic graph is 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑.
The simplest example of the cubic graph is 𝑦 = 𝑥 3
2.6.4. Trigonometric Functions
They are the graphs of the type, 𝑦 = sin 𝜃, 𝑦 = cos 𝜃, 𝑦 = tan 𝜃 etc
62
2.6.5. Circles
The simplest equation of a circle is 𝑥 2 + 𝑦 2 = 𝑟 2 , with center at the origin and radius 𝑟.
More generally, the equation of a circle with centre (𝑎, 𝑏), and radius 𝑟, is given by:
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
2.6.6. Ellipse
The equation of an ellipse is:
𝑥2 𝑦2
+
=1
𝑎2 𝑏2
In the drawing the ellipse, the shape can be in two forms.
It can be drawn such that the longer side lies horizontally on the 𝑥-axis, whereas the shorter side
lies vertically on the 𝑦-axis.
It can be drawn such that the longer side stands vertically on the 𝑦-axis whereas the shorter side
lies horizontally on the 𝑥-axis.
If the length 𝐴𝐵 spans entire width of the ellipse along the 𝑥-axis, then 𝐴𝐵 is called the major
axis.
Also, if the length 𝐶𝐷 spans entire height of the ellipse along the 𝑦-axis, then 𝐶𝐷 is called the
minor axis.
The length 𝑎 is half the length 𝐴𝐵 and lies along the 𝑥-axis as the semi-major axis, whereas the
length 𝑏 is the semi-minor axis.
2.6.7. The Hyperbola
The general equation of a hyperbola is
𝑥2 𝑦2
−
=1
𝑎2 𝑏2
The curve is usually seen to be symmetrical about the both the 𝑥- and 𝑦-axes. The distance
between the curves is given by 2𝑎
63
2.6.8. Rectangular Hyperbola
𝑐
The equation of a rectangular hyperbola is 𝑥𝑦 = 𝑐 or 𝑦 = 𝑎
2.6.9. Logarithmic Functions
The general equation of the logarithmic functions are of the form 𝑦 = ln 𝑥 and 𝑦 = lg 𝑥
2.6.10. Exponential Functions
The general equation of the exponential functions are of the form 𝑦 = 𝑒 𝑥
2.6.11. Polar Curves
The equation of a polar curve is of the form 𝑟 = 𝑓(𝜃).
Examples of polar curves are 𝑟 = 𝑎 sin 𝜃 and 𝑟 = 𝑏 cos 𝜃
64
Unit 3.0.
Quadratics, Polynomial and Rational Functions
Session 1.
Quadratic Functions and Quadratic Equations
Session 2.
The Standard Form of Quadratic Function.
Session 3.
Polynomials
Session 4.
Rational Expressions
Session 5.
Exponents and Radicals
Session 6.
Radicals and Properties of Roots
65
3.1.0. Session 1 – Quadratic Functions and Quadratic Equations
Objectives.
In this session, we will learn about quadratic functions and their various properties. We will learn
how to complete the square on the quadratic expression 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 . We will learn to express
quadratic functions in their standard forms. We will use the standard forms to locate the
maximum and minimum points of the function, and the points where the graph cuts the 𝑥 and 𝑦
axes. We will learn how to use the minimum and maximum points of the function to sketch the
graphs of 𝑦 = 𝐴𝑥 2 + 𝐵𝑥 + 𝑐. We will learn to use standard formulas to compute the minimum
and maximum values of quadratic functions
Let us get into quadratic functions.
3.1.1. General Form Quadratic Function
A quadratic function is a polynomial of degree 2. Quadratic functions are also known as
quadratic equations.
The general form of a quadratic function is of the form:
𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐,
𝑎≠0
If we take 𝑎 = 1 and 𝑏 = 𝑐 = 0 in the quadratic function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, we get the
quadratic function 𝑓(𝑥) = 𝑥 2 , whose graph is a parabola. In fact, the graph of any quadratic
function is a parabola.
3.1.2. Solving Quadratic Equations
Quadratic equations are second-degree equations of the form 𝑥 2 + 2𝑥 − 3 = 0 or 2𝑥 2 + 3 = 5𝑥.
3.1.3. Quadratic Equations
A quadratic equation is an equation of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎, 𝑏 and 𝑐 are real
numbers with 𝑎 ≠ 0.
Some quadratic equations can be solved by factoring and using the following basic property of
real numbers.
66
3.1.4. The Zero-Product Property
𝐴𝐵 = 0
if and only if
𝐴=0
𝐵=0
or
This means if we can factor the left-hand side of a quadratic (or other) equation, then we can
solve it by setting each factor equal to 0 in turn. This method works only when the right-hand
side of the equation is 0.
Example 3.1.4.1. Solving a Quadratic Equation by Factoring.
Solve the equation 𝑥 2 + 5𝑥 = 24.
Solution 3.1.4.1.
We begin by rewriting the equation so that the right-hand side is 0.
𝑥 2 + 5𝑥 = 24
We subtract 24 from both sides
𝑥 2 + 5𝑥 − 24 = 24 − 24
𝑥 2 + 5𝑥 − 24 = 0
The equation is now of the form, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎 = 1, 𝑏 = 5, and 𝑐 = −24
We factor the resulting equation, by looking for two factors of -24 whose sum or those factors of
−24 that add up to 5 in the quadratic equation.
We found −3 and 8, i.e., −3 + 8 = 5 and −3 × 8 = −24
We substitute the two factors in the place of 5𝑥 as in the following expression
𝑥 2 − 3𝑥 + 8𝑥 − 24 = 0
67
We proceed by factoring out the common terms, common terms of 𝑥 2 and 3𝑥 is 𝑥; the common
term of 8𝑥 and 24 is 8. We get the following:
𝑥(𝑥 − 3) + 8(𝑥 − 3) = 0
We do a further factorisation by factoring out the common term, (𝑥 − 3)
(𝑥 − 3)(𝑥 + 8) = 0
We apply the zero-product property at this point
𝑥−3 =0
𝑥+8 = 0
OR
It implies 𝑥 = 3
𝑥 = −8
OR
The solutions are 𝑥 = 3 and 𝑥 = −8
Note why one side of the equation must be 0.
If we had proceeded by factorising the equation as 𝑥(𝑥 + 5) = 24, we would not get anywhere
finding the solutions, because 24 can be factorised in infinitely many ways such as
1
2
(6 × −4), (−6 × 4), (2 × 48) , (− 5 × 60), etc
A quadratic equation of the form 𝑥 2 − 𝑐 = 0, where 𝑐 is a positive constant, will factorise as
(𝑥 − √𝑐)(𝑥 + √𝑐) = 0, so the solutions are 𝑥 = √𝑐 and 𝑥 = −√𝑐 . It is common to abbreviate
the solutions 𝑥 = √𝑐 and 𝑥 = −√𝑐 as 𝑥 = ±√𝑐.
68
3.1.5. Other Methods for Solving Quadratic Equations
Example 3.1.5.1.
Expand (𝑥 + 1)2 + 5.
Solution 3.1.5.1.
How do we expand the expression? We begin by expanding the expression inside the brackets,
(𝑥 + 1)2.
We will use the standard expansion for (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2 for expanding the square of
the sum of any two values.
Let (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2
(3.1.7.1)
⇒ (𝑥 + 1)2 + 5 = (𝑥 2 + 2𝑥 + 1) + 5
⇒ (𝑥 + 1)2 + 5 = 𝑥 2 + 2𝑥 + 6
The above example shows that 𝑥 2 + 2𝑥 + 6 can be written as (𝑥 + 1)2 + 5.
In this section, we are given the expanded version 𝑥 2 + 2𝑥 + 6 and we try to rewrite the
expanded version as (𝑥 + 1)2 + 5. In other words, we go from the solution to the problem.
Expressions of this nature
𝑥 2 + 2𝑥 + 6 = (𝑥 + 1)2 + 5
is an example of an identity.
The term identity means that the two mathematical expressions are equal for all values of the
variables. The following are some other examples of identities:
69
𝑥 2 + 4𝑥 + 1 = (𝑥 + 2)2 − 3
5 2 17
𝑥 − 5𝑥 + 2 = (𝑥 − ) −
2
4
2
7 2 3
𝑥 + 7𝑥 + 13 = (𝑥 + ) +
2
4
2
Can you notice any relationship between the Left-Hand Side (LHS) and the Right-Hand Side
(RHS) of the above expressions?
If we look at the coefficients of 𝑥 of the expressions on the LHS, and if we look at the number
inside the brackets on the RHS, you will notice that the number inside the brackets is half the
coefficient of the variable 𝑥 on the LHS. Look at them again.
In the first example, half of 4 is 2, in the second example, half of -5 is
7
−5
2
, and in the third
example, half of 7 is 2 .
Let us consider the example 𝑥 2 + 4𝑥 + 1 = (𝑥 + 2)2 − 3
We know that the number 2 inside the bracket is half of the coefficient of 𝑥, which is 4, but
where did the −3 on the RHS come from?
Let us apply equation 2.6.8.1., (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2 with 𝑎 = 𝑥, and 𝑏 = 2, we will get
(𝑥 + 2)2 = 𝑥 2 + 2(𝑥)(2) + 22 = 𝑥 2 + 4𝑥 + 4
We have,
𝑥 2 + 4𝑥 + 1 = 𝑥 2 + 4𝑥 + 4
We know that 𝑥 2 + 4𝑥 + 4 = (𝑥 + 2)2
∴ 𝑥 2 + 4𝑥 + 1 = (𝑥 + 2 )2
70
We subtract the square of half the coefficient of 𝑥, (the coefficient of 𝑥 is 4), and half of 4 is 2.
Subtracting the square of 2, which is 4 from the RHS of the equation.
𝑥 2 + 4𝑥 + 1 = (𝑥 + 2 )2 − 4 + 1
𝑥 2 + 4𝑥 + 1 = (𝑥 + 2)2 − 3
Similarly,
5 2
5 2
𝑥 − 5𝑥 + 2 = (𝑥 − ) − ( ) + 2
2
2
2
5 2 25
𝑥 − 5𝑥 + 2 = (𝑥 − ) −
+2
2
4
2
5 2 25 + 8
𝑥 − 5𝑥 + 2 = (𝑥 − ) −
2
4
2
5 2 17
𝑥 2 − 5𝑥 + 2 = (𝑥 − ) −
2
4
Also,
7 2
7 2
𝑥 + 7𝑥 + 13 = (𝑥 + ) − ( ) + 13
2
2
2
7 2 49
𝑥 2 + 7𝑥 + 13 = (𝑥 + ) −
+ 13
2
4
7 2 49 + 52
𝑥 + 7𝑥 + 13 = (𝑥 + ) −
2
4
2
7 2 3
𝑥 2 + 7𝑥 + 13 = (𝑥 + ) +
2
4
71
The above method is called “completing the square” and it results in an algebraic identity. The
following expansions will be helpful. The learners will do well to keep the following expansions
in mind:
(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2
(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏2
(𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏2
Example 3.1.5.2.
Complete the square on the following quadratic expressions.
(a) 𝑥 2 − 8𝑥 + 3
(b)
2𝑥 2 − 5𝑥 + 3
10 − 6𝑥 − 3𝑥 2
(c)
Solution 3.1.5.2.
(a) 𝑥 2 − 8𝑥 + 3 𝑏 = −8, half of 𝑏 = −4
𝑥 2 − 8𝑥 + 3 = (𝑥 − 4)2 − 16 + 3
(𝑥 − 4)2 − 13
(b) 2𝑥 2 − 5𝑥 + 3. We need to make sure the coefficient of 𝑥 2 is 1 by factoring 2 out
2𝑥 2 − 5𝑥 + 3 = 2 [𝑥 2 −
5𝑥 3
+ ]
2 2
5 2
5 2 3
2 [(𝑥 − ) − ( ) + ]
4
4
2
5 2 1
2 [(𝑥 − ) − ]
4
16
5 2
1
2 (𝑥 − 4) − 8 after expanding and cancelling out the 2.
(c) 10 − 6𝑥 − 3𝑥 2 . Here again, we need to make the coefficient of 𝑥 2 become 1 by
factorising the −3 out,
10 − 6𝑥 − 3𝑥 2 = 10 − 3[2𝑥 + 𝑥 2 ]
72
= 10 − 3[(𝑥 + 1)2 − 1] [by completing the square]
= 10 − 3(𝑥 + 1)2 + 3 [by expanding]
= 13 − 3(𝑥 + 1)2
3.1.6. Solving a Simple Quadratic Equations
The solutions of the equation 𝑥 2 = 𝑐 are 𝑥 = √𝑐 and 𝑥 = −√𝑐
Example 3.1.6.1.
Solve the equations
(a) 𝑥 2 = 5
(b) (𝑥 − 4)2 = 5
Solution 3.1.6.1.
(a) From the principle in 3.1.5., we get 𝑥 = ±√5 from 𝑥 2 = 5
(b) From (𝑥 − 4)2 = 5
We can take the square root of each side of the equation.
(𝑥 − 4)2 = 5
Taking square root of each side, we have
𝑥 − 4 = ±√5
Add 4 to each side
𝑥 − 4 + 4 = ±√5 + 4
𝑥 = 4 ± √5
The solutions are 𝑥 = 4 + √5 and 𝑥 = 4 − √5
As we saw in the preceding example, 3.1.5.1., if a quadratic equation is of the form
73
(𝑥 ± 𝑎)2 = 𝑐, then we can solve it by taking the square root of each side. In an equation of this
form, the left-hand side is a perfect square: the square of a liner expression in 𝑥. If a quadratic
does not factor readily, then we can solve it using the technique of “completing the square.”
This means that we add a constant to an expression to make it a perfect square.
For example, to make the expression 𝑥 2 − 6𝑥 a perfect square, we must add 9 since
𝑥 2 − 6𝑥 + 9 = (𝑥 − 3)2.
3.1.7. The Method of Completing the Square
An expression such as 𝑥 2 or (𝑥 + 2)2 or (𝑥 − 3)2 is called a perfect square.
If 𝑥 2 = 3 then 𝑥 = ±√3.
If (𝑥 + 2)2 = 5 then 𝑥 + 2 = ±√5 and 𝑥 = −2 ± √5
If (𝑥 − 3)2 = 8 then 𝑥 − 3 = ±√8 and 𝑥 = 3 ± √8
Hence if a quadratic equation can be rearranged so that one side of the equation is a perfect
square and the other side of the equation is a number, then the solution of the equation is readily
obtained by taking the square roots of each side as we have seen in the above examples.
The process of rearranging one side of the quadratic equation to obtain a perfect square before
we solve it is known as “completing the square”.
(𝑥 + 𝑎)2 = 𝑥 2 + 2𝑎𝑥 + 𝑎2
Therefore, to make the quadratic expression 𝑥 2 + 2𝑎𝑥 into a perfect square, it is necessary to add
2𝑎 2
(half the coefficient of 𝑥)2 i.e., ( 2 ) or 𝑎2 .
74
3 2
For example, 𝑥 2 + 3𝑥 becomes a perfect square by adding (2) , i.e.,
3 2
3 2
𝑥 2 + 3𝑥 + ( ) = (𝑥 + )
2
2
𝑏 2
In general, to make any expression of the form 𝑥 2 + 𝑏𝑥 a perfect square, add (2) , the square of
half the coefficient of 𝑥. This gives the perfect square
𝑏 2
𝑏 2
𝑥 + 𝑏𝑥 + ( ) = (𝑥 + )
2
2
2
The method is demonstrated in the following worked examples.
3.1.8. Solving Quadratic Equations by Completing the Square
Example 3.1.8.1.
Solve each of the equation below by “completing the square”
(a)
(b)
(c)
(d)
2𝑥 2 + 5𝑥 = 3
2𝑥 2 + 9𝑥 + 8 = 0
𝑥 2 − 8𝑥 + 13 = 0
3𝑥 2 − 12𝑥 + 6 = 0
Solution 3.1.8.1.
The procedure is as follows:
1. Rearrange the equation so that all terms are on the same side of the equals sign (and the
coefficient of 𝑥 2 term is positive)
Hence 2𝑥 2 + 5𝑥 − 3 = 0
2. Make the coefficient of the 𝑥 2 term unity (equal to 1). In this case this is achieved by
dividing throughout by 2.
Hence
75
2𝑥 2 5𝑥 3
+
− =0
2
2 2
5
3
𝑥2 + 𝑥 − = 0
2
2
3. Rearrange the equations so that the 𝑥 2 and 𝑥 terms are on one side of the equals sign and
the constant is on the other side. Hence,
5
3
𝑥2 + 𝑥 =
2
2
4. We add to both sides of the equation (half the coefficient of 𝑥 )2. In this case the
5
coefficient of 𝑥 is 2.
5 2
Half the coefficient squared is therefore (4) .
Therefore,
5
5 2 3
5 2
𝑥 + 𝑥+( ) = +( )
2
4
2
4
2
The LHS is now a perfect square, i.e.,
5 2 3
5 2
(𝑥 + ) = + ( )
4
2
4
5. We can evaluate the RHS. Therefore,
5 2 3 25 24 + 25 49
(𝑥 + ) = +
=
=
4
2 16
16
16
6. At this point, we take the square root of both sides of the equation (remembering that the
square root of a number gives a ± answer). Therefore,
5 2
49
√(𝑥 + ) = √
4
16
𝑥+
5
7
=±
4
4
7. We solve the resulting simple equation. Therefore,
76
5 7
𝑥=− ±
4 4
i.e.,
5 7 2 1
𝑥=− + = =
4 4 4 2
and
5 7
12
𝑥=− − =−
= −3
4 4
4
1
Hence 𝑥 = or 𝑥 = −3 are the roots of the equation 2𝑥 2 + 5𝑥 = 3
2
(b) 2𝑥 2 + 9𝑥 + 8 = 0
2𝑥 2 + 9𝑥 + 8 = 0
We begin by making the coefficient of 𝑥 2 unity to give us:
9
𝑥2 + 𝑥 + 4 = 0
2
And rearranging gives us
9
𝑥 2 + 𝑥 = −4
2
We add (half the coefficient of 𝑥)2 to both sides to give:
9
9 2
9 2
𝑥 2 + 𝑥 + ( ) = −4 + ( )
2
4
4
9
9 2
9 2
𝑥 + 𝑥+( ) = ( ) −4
2
4
4
2
77
The LHS is now a perfect square, therefore:
9 2 81
17
(𝑥 + ) =
−4 =
4
16
16
We take square roots of both sides, and we get
9 2
17
√(𝑥 + ) = √
4
16
𝑥+
9
17
= ±√ = ±√1.0625 = ±1.0308
4
16
Hence
9
𝑥 = − ± 1.0308
4
i.e.,
9
𝑥 = − + 1.0308 = −2.25 + 1.0308 = −1.2192 ≈ 1.22
4
9
𝑥 = − − 1.0308 = −3.2808 ≈ −3.28
4
(c)
𝑥 2 − 8𝑥 + 13 = 0
We proceed by subtracting 13 from each side
78
𝑥 2 − 8𝑥 + 13 − 13 = 0 − 13
𝑥 2 − 8𝑥 = −13
We now complete the square.
When completing the square, make sure that the coefficient of 𝑥 2 is 1. If it is not 1, it must be
factored from both terms that contain 𝑥 as in the following
𝑏
𝑎𝑥 2 + 𝑏𝑥 = 𝑎 (𝑥 2 + 𝑥)
𝑎
Then complete the square inside the parentheses. Remember that the term added inside the
parentheses is multiplied by 𝑎
𝑥 2 − 8𝑥 = −13
We now complete the square by adding the square of half of coefficient of 𝑥, which is
8 2
(2) = 42 = 16 to each side
𝑥 2 − 8𝑥 + 16 = −13 + 16
This gives the perfect square
(𝑥 − 4)2 = 3
We take the square root of both sides
𝑥 − 4 = ±√3
Add 4 to each side
79
𝑥 = 4 ± √3
The solutions are
𝑥 = 4 + √3 and 𝑥 = 4 − √3
(d)
3𝑥 2 − 12𝑥 + 6 = 0
After subtracting 6 from each side of the equation, we must factor the coefficient of 𝑥 2 , (the 3)
from the left side to put the equation in the correct form for completing the square.
We are given the equation 3𝑥 2 − 12𝑥 + 6 = 0
We subtract 6 from each side
3𝑥 2 − 12𝑥 + 6 − 6 = 0 − 6
3𝑥 2 − 12𝑥 = −6
We factor out 3 from the left-hand side
3(𝑥 2 − 4𝑥) = −6
We complete he square from this point on, by adding (−2)2 = 4 inside the parentheses. Note
that everything inside the parentheses is multiplied by 3, this means that we are adding 3 × 4 =
12 to the left side of the equation. Therefore, we must add 12 to the right-hand side as well
Adding 4 by completing the square
3(𝑥 2 − 4 + 4) = −6 + (3 × 4)
We get the perfect square
80
3(𝑥 − 2)2 = 6
3(𝑥 2 − 2. 𝑥. 2 + 2.2)2 = 6
We divide each side by 3 to get
(𝑥 − 2)2 = 2
Taking square root of each side
𝑥 − 2 = ±√2
𝑥 = 2 ± √2
The solutions are 𝑥 = 2 + √2 and 𝑥 = 2 − √2
81
3.2.0. Session 2 - The Standard Form of Quadratic Function.
A quadratic function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 can be expressed in the standard form,
𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 by completing the square
The graph of a quadratic 𝑓 is a parabola with vertex (ℎ, 𝑘); the parabola opens upward if 𝑎 > 0
or downward if 𝑎 < 0.
Example 3.2.0.1
Express 𝑓(𝑥) = 2𝑥 2 − 12𝑥 + 23 in standard form.
Solution 3.2.0.1.
We express in standard form by completing the square. We need to convert the coefficient of 𝑥 2
to 1. We do it by factorising it from the terms involving 𝑥 before we complete the square.
𝑓(𝑥) = 2𝑥 2 − 12𝑥 + 23
Factor out 2 from the 𝑥-terms:
𝑓(𝑥) = 2(𝑥 2 − 6𝑥) + 23
We start completing the square by adding the square of half the coefficient of 𝑥, i.e.,
6 2
(2) = 32 = 9, in the parenthesis and subtract (2 × 9) outside. The 2 is the value factored out
to make the coefficient of 𝑥 2 become 1
𝑓(𝑥) = 2(𝑥 2 − 6𝑥 + 9) + 23 − (2)(9)
After factorising and simplifying, we get:
𝑓(𝑥) = 2(𝑥 − 3)2 + 5
The standard form is 𝑓(𝑥 ) = 2(𝑥 − 3)2 + 5 with vertex (3, 5)
82
3.2.1. Minimum and Maximum Values of a Quadratic Function
If a quadratic function has vertex (ℎ, 𝑘), then the function has a minimum value at the vertex if
its graph opens upward; it has a maximum value at the vertex (ℎ, 𝑘), if the graph opens
downwards.
3.2.2. Maximum or Minimum Values of Quadratic Functions.
Let 𝑓 be a quadratic function with standard form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘. The maximum or
minimum value of 𝑓 occurs at 𝑥 = ℎ.
If 𝑎 > 0, then the minimum value of 𝑓 is 𝑓(ℎ) = 𝑘
If 𝑎 < 0, then the maximum value of 𝑓 is 𝑓(ℎ) = 𝑘
Example 3.2.2.1.
Let 𝑓(𝑥) = 5𝑥 2 − 30𝑥 + 49
(i)
(ii)
(iii)
Express 𝑓 in standard form
Sketch the graph of 𝑓
Find the minimum value of 𝑓
Solution 3.2.2.1.
(i)
We proceed by completing the square. First, we want to convert the coefficient of 𝑥 2
to 1
𝑓(𝑥) = 5𝑥 2 − 30𝑥 + 49
We begin by factorising 5 from the 𝑥-terms.
𝑓(𝑥) = 5(𝑥 2 − 6𝑥) + 49
Add the square of half of 6, i.e., 32 = 9 inside of the parentheses, and subtract (5 × 9 ) outside
the parentheses
𝑓(𝑥) = 5(𝑥 2 − 6𝑥 + 9) + 49 − (5)(9)
83
We finish by factorising and simplifying to obtain,
𝑓(𝑥) = 5(𝑥 − 3)2 + 4
The graph is a parabola that has its vertex at (3, 4) and opens upward. With the given
information we sketch the graph on any 𝑥 − 𝑦 axes.
The coefficient of 𝑥 2 is greater than 0, positive, therefore 𝑓 has a minimum value.
The minimum value is at 𝑓(ℎ) = 𝑓(3) = 4
(ii)
(iii)
Example 3.2.2.2.
Consider the quadratic function 𝑓(𝑥) = −𝑥 2 + 𝑥 + 2
(a) Express 𝑓 in standard form
(b) Sketch the graph of 𝑓
(c) Find the maximum value of 𝑓
Solution 3.2.2.2.
(a) As usual, we need to complete the square to express the function in standard form
We are given 𝑓(𝑥) = −𝑥 2 + 𝑥 + 2
We make the coefficient of 𝑥 2 term positive by factorising −1 out from the 𝑥-terms to
get
− (𝑥 2 − 𝑥 ) + 2
1 2
1
We complete the square by adding the square of half of −1 which is (− 2) = 4 inside
1
the brackets (parentheses), and subtract (−1) × 4 outside the brackets (the parentheses) to
get
1
1
− (𝑥 2 − 𝑥 + 4) + 2 − (−1) × 4
We factorise and simplify the expression to get
1 2
1
− (𝑥 − ) + 2 +
2
4
1 2 8+1
− (𝑥 − ) +
2
4
84
1 2 9
− (𝑥 − ) +
2
4
This is the standard form
(b) From the standard form, it shows that the graph is a parabola that opens downward and
1 9
has the vertex ( , ). To help us sketch the graph, we can find the intercepts on the 𝑥 − 𝑦
2 4
axes.
To determine the intercept on the 𝑦-axis, we solve the function for 𝑓(0) to obtain
𝑓(0) = −02 + 0 + 2 = 2
The intercept on the 𝑦-axis is 𝑦 = 𝑓(0) = 2
To find the intercepts on the 𝑥-axis, we set the function 𝑓(𝑥) to zero as follows:
𝑦 = 𝑓(𝑥) = −𝑥 2 + 𝑥 + 2 = 0
Multiply 𝑓(𝑥) by −1 to obtain
𝑥2 − 𝑥 − 2 = 0
and we factorise to obtain
(𝑥 − 2)(𝑥 + 1) = 0
Therefore, the 𝑥-intercepts are 𝑥 = 2 and 𝑥 = −1.
We sketch the graph with the points (2, 0), (−1, 0) and (0, 2)
(c) Since the coefficient of 𝑥 2 is negative or less than zero, the function 𝑓 has a maximum
1
9
2
4
value, which occurs at the 𝑓 ( ) =
85
Expressing a quadratic function in its standard form helps us to sketch its graph as well as to find
its maximum and minimum values. If we are interested in finding the maximum or minimum
value, then a formula is available for doing so. The formula is obtained by completing the square
for the general quadratic function as follows:
𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
We make the coefficient of 𝑥 2 = 1 by factorising 𝑎 out from the 𝑥-terms to obtain
𝑏
= 𝑎 (𝑥 2 + 𝑥) + 𝑐
𝑎
𝑏
Complete the square, by adding the square of half of coefficient of 𝑥, (the coefficient of 𝑥) is (𝑎)
𝑏
𝑏
2
𝑏2
and its half is (2𝑎), whose square is (2𝑎) = 4𝑎2 inside the brackets, and subtract the same,
𝑏2
𝑎 (4𝑎2) from outside the brackets
The expression becomes
𝑏
𝑏2
𝑏2
= 𝑎 (𝑥 2 + 𝑎 𝑥 + 4𝑎2) + 𝑐 − 𝑎 (4𝑎2)
Factorising further, it becomes
= 𝑎 (𝑥 +
𝑏 2
𝑏2
) +𝑐−
2𝑎
4𝑎
𝑏
𝑏2
The equation is now in standard form with ℎ = − (2𝑎) and 𝑘 = 𝑐 − (4𝑎)
Since the maximum or minimum value occurs at 𝑥 = ℎ, we have the following results.
86
3.2.3. Maximum or Minimum Value of a Quadratic Function
The maximum or minimum value of a quadratic function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 occurs at
𝑥= −
𝑏
2𝑎
𝑏
If 𝑎 > 0, then the minimum value is 𝑓 (− 2𝑎).
𝑏
If 𝑎 < 0, then the maximum value is 𝑓 (− 2𝑎).
Example 3.2.3.1.
Finding Maximum and Minimum Values of Quadratic Functions
(a) 𝑓(𝑥) = 𝑥 2 + 4
(b) 𝑔(𝑥) = −2𝑥 2 + 4𝑥 − 5
Solution 3.2.3.1.
(a) This function, 𝑓(𝑥) = 𝑥 2 + 4 is a quadratic function with 𝑎 = 1, and 𝑏 = 4. The
maximum of minimum value occurs at
𝑏
4
𝑥=−
=−
= −2
2𝑎
2×1
Since 𝑎 = 1 > 0, the function has a minimum value.
The minimum value is 𝑓(−2) = (−2)2 + 4(−2) = 4 − 8 = −4
(b) This function 𝑔(𝑥) = −2𝑥 2 + 4𝑥 − 5 is a quadratic function with 𝑎 = −2, and 𝑏 = 4.
Therefore, the maximum or minimum occurs at
𝑏
4
𝑥 = − 2𝑎 = − 2×(−2) = 1
Since 𝑎 < 0, the function has the maximum value
𝑓(1) = −2(1)2 + 4(1) − 5
𝑓(1) = −2 + 4 − 5 = −3
87
3.3.0. Session 3 - Polynomials
3.3.1. Polynomial Functions
Any function defined with degree 𝑛, of the form, 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 is a
polynomial function, where 𝑛 is a nonnegative integer and 𝑎𝑛 ≠ 0. The numbers 𝑎0 , 𝑎1 , 𝑎2 , . . , 𝑎𝑛
are called coefficients of the polynomial. The number 𝑎0 is the constant coefficient or constant
term. The number 𝑎𝑛 , the coefficient of the highest power, is the leading coefficient, and the
term 𝑎𝑛 𝑥 𝑛 is the leading term.
Polynomial functions are usually referred to as polynomials.
3.3.2. Zeros of Polynomials
If 𝑃 is a polynomial function, then 𝑐 is called a zero of 𝑃 if 𝑃(𝑐) = 0. In other words, the zeros
of 𝑃 are the solutions of the polynomial equation 𝑃(𝑥) = 0. Note that if 𝑃(𝑐) = 0, then the
graph of 𝑃 has an 𝑥-intercept at 𝑥 = 𝑐, so the 𝑥-intercepts of the graph are the zeros of the
function.
3.3.3. Real Zeros of Polynomials
If 𝑃 is a polynomial and 𝑐 is a real number, then the following are equivalent:
1.
2.
3.
4.
𝑐 is a zero of 𝑃
𝑥 = 𝑐 is a solution of the equation 𝑃(𝑥) = 0
𝑥 − 𝑐 is a factor of 𝑃(𝑥).
𝑐 is an 𝑥-intercept of the graph of 𝑃
To find the zeros of a polynomial 𝑃, we factor and then use the Zero-Product Property. For
example, to find the zeros of 𝑃(𝑥) = 𝑥 2 + 𝑥 − 6, we expand 𝑃 to get 𝑃(𝑥) = 𝑥 2 + 3𝑥 − 2𝑥 − 6
which when simplified and factored will yield 𝑃(𝑥) = (𝑥 − 2)(𝑥 + 3)
From the factored form, we can easily see that,
1.
2.
3.
4.
2 is a zero of 𝑃
𝑥 = 2 is a solution of the equation 𝑥 2 + 𝑥 − 6 = 0.
𝑥 − 2 is a factor of 𝑥 2 + 𝑥 − 6.
2 is an 𝑥-intercept of the graph of 𝑃
The same facts above are true of the other zero, −3.
88
3.3.4. Intermediate Value Theorem for Polynomials
If 𝑃 is a polynomial function and 𝑃(𝑎) and 𝑃(𝑏) have opposite signs, then there exists at least
one value 𝑐 between 𝑎 and 𝑏 for which 𝑃(𝑐) = 0.
3.3.5. Finding Zeros for Graphing a Polynomial Function
Example 3.3.5.1.
Let 𝑃(𝑥) = 𝑥 3 − 2𝑥 2 − 3𝑥.
(i)
(ii)
Find the zeros of 𝑃
Sketch a graph of 𝑃
Solution 3.3.5.1.
(i)
To find the zeros, we factor 𝑃 completely.
𝑃(𝑥) = 𝑥 3 − 2𝑥 2 − 3𝑥
Factor 𝑥 out
𝑃(𝑥) = 𝑥(𝑥 2 − 2𝑥 − 3)
𝑃(𝑥) = 𝑥(𝑥 2 + 𝑥 − 3𝑥 − 3)
𝑃(𝑥) = 𝑥[(𝑥(𝑥 + 1) − 3(𝑥 + 1))]
𝑃(𝑥) = 𝑥[(𝑥 + 1)(𝑥 − 3)]
Factor the quadratic
= 𝑥(𝑥 2 + 𝑥 − 3𝑥 − 3)
= 𝑥(𝑥 + 1)(𝑥 − 3)
The zeros are 𝑥 = 0, 𝑥 = 3 and 𝑥 = −1.
(ii)
(iii)
The 𝑥-intercepts are 𝑥 = 0, 𝑥 = 3 and 𝑥 = −1. The 𝑦-intercept is 𝑃(0) = 0. We
make a table of values of 𝑃(𝑥), making sure that we choose test points between (and
to the right and left of) the successive zeros.
We leave the sketch for the learners as an exercise.
89
Exercises
(i)
(ii)
Find the zeros of 𝑃 if 𝑃(𝑥) = −2𝑥 4 − 𝑥 3 + 3𝑥 2, and sketch the graph of 𝑃
Find the zeros of 𝑃 if 𝑃(𝑥) = 𝑥 3 − 2𝑥 2 − 4𝑥 + 8, and sketch the graph of 𝑃
3.3.6. Dividing Polynomials
In this section we begin to study the algebra on polynomials. Most of the work will be concerned
factorising polynomials, and to factorise, we need to know how to divide polynomials.
3.3.7.
Long Division of Polynomials
Dividing polynomials is much like the familiar process of dividing numbers. For example, when
we divide 47 by 5, the quotient is 9, and the remainder is 2. We write it as follows:
47
2
= 9+
5
5
The number 47 is the dividend, the number 5 is the divisor, the number 9 is the quotient, the
remainder is 2.
To divide polynomial, we use long division as follows:
3.3.8. The Polynomial Division Algorithm (Division Method):
The word algorithm means method or formula. It is used to describe the procedure for solving
problems. It is also the systematic or the step-by-step way of solving a problem.
If 𝑃(𝑥) and 𝐷(𝑥) are polynomials, with 𝐷(𝑥) ≠ 0, then there exists, unique polynomials 𝑄(𝑥)
and 𝑅(𝑥), where 𝑅(𝑥) is either 0 or of degree less than the degree of 𝐷(𝑥), such that:
𝑃(𝑥) = 𝐷(𝑥)𝑄(𝑥) + 𝑅(𝑥)
90
The polynomials 𝑃(𝑥) and 𝐷(𝑥) are called the dividend and divisor, respectively, 𝑄(𝑥) is the
quotient, and 𝑅(𝑥) is the remainder.
Example 3.3.8.1.
Divide 6𝑥 2 − 26𝑥 + 12 by 𝑥 − 4
Solution 3.3.8.1.
We identify 6𝑥 2 − 26𝑥 + 12 as the dividend and 𝑥 − 4 as the divisor.
We begin by arranging them as follows:
𝑥 − 4 ⁄ 6𝑥 2 − 26𝑥 + 12
Next, we divide the leading term (first term) in the dividend by the leading term (first term) in
the divisor to get the first term of the quotient: i.e.,
6𝑥 2
𝑥
= 6𝑥. Then we multiply the divisor by the
first term of the quotient obtained so far and subtract the result from the dividend.
Step 1. Divide leading terms
6𝑥 2
𝑥
= 6𝑥
6𝑥
𝑥 − 4 ⁄ 6𝑥 2 − 26𝑥 + 12
Step 2. Multiply 6𝑥 by 𝑥 − 4 : 6𝑥(𝑥 − 4) = 6𝑥 2 − 24𝑥
6𝑥 2 − 24𝑥
Step 3. Subtract 6𝑥 2 − 24𝑥 from 6𝑥 2 − 26𝑥 and bring the 12 down
−2𝑥 + 12
Step 4. Repeat the process using the last line involving the term −2𝑥 + 12 as the dividend.
6𝑥 − 2
𝑥 − 4 ⁄ 6𝑥 2 − 26𝑥 + 12
6𝑥 2 − 24𝑥
−2𝑥 + 12
Step 5. Multiply −2 by 𝑥 − 4 as in −2(𝑥 − 4)
91
−2𝑥 + 8
Step 6. Subtract −2𝑥 + 8 from −2𝑥 + 12
4 is the remainder.
The division process ends when the last line is of a lesser degree than the divisor. The last
line then contains the remainder, and the top line contains the quotient. The result of the
division can be interpreted in either of two ways.
6𝑥 2 −26𝑥+12
𝑥−4
= 6𝑥 − 2 +
4
, where 6𝑥 − 2 is the quotient, and
𝑥−4
4
𝑥−4
is the remainder, OR
6𝑥 2 − 26𝑥 + 12 = (𝑥 − 4)(6𝑥 − 2) + 4, where 6𝑥 2 − 26𝑥 + 12 is the dividend, (𝑥 − 4) is the
divisor, 6𝑥 − 2 is the Quotient, and 4 is the remainder.
3.3.9. The Factor Theorem
𝑐 is a zero of 𝑃 if and only if 𝑥 − 𝑐 is a factor of 𝑃(𝑥).
Proof:
If 𝑃(𝑥) factors as 𝑃(𝑥) = (𝑥 − 𝑐). 𝑄(𝑥), then 𝑃(𝑐) = (𝑐 − 𝑐). 𝑄(𝑐) = 0 . 𝑄(𝑐) = 0
Conversely, if 𝑃(𝑐) = 0, then by the Remainder Theorem
𝑃(𝑥) = (𝑥 − 𝑐). 𝑄(𝑥) + 0 = (𝑥 − 𝑐). 𝑄(𝑥), so 𝑥 − 𝑐 is a factor of 𝑃(𝑥).
Example 3.3.9.1.
Factoring a Polynomial Using the Factor Theorem
Let 𝑃(𝑥) = 𝑥 3 − 7𝑥 + 6. Show that 𝑃(1) = 0, and use this fact to factor 𝑃(𝑥) completely.
Solution 3.3.9.1.
Given that 𝑃(𝑥) = 𝑥 3 − 7𝑥 + 6, substitute 1 in 𝑃(𝑥) = 𝑃(1) = 13 − (7). (1) + 6 = 0. By the
Factor Theorem this means that 𝑥 − 1 is a factor of 𝑃(𝑥).
92
We can use the long division to factor 𝑃(𝑥) completely, but we would like to introduce another
method known as the synthetic method of dividing polynomials. We hope you would like this
method.
3.3.10. The Synthetic Division of Polynomials.
The synthetic division can be used when the divisor is of the form 𝑥 − 𝑐. In the synthetic
division, only the coefficients of the polynomial and the divisor of the long division are written,
without the variable, 𝑥.
We begin by writing the appropriate coefficients to represent the divisor and the dividend.
Divisor 𝑥 − 1
1
Bring the 1 in the dividend down and
multiply by the divisor 1, and write
the result in the middle row, below the
zero, and add 0 and 1
1
1
−7
6
1
0
1
1
1
0
6
1
−7
1
−6
0
−7
−6
6
−6
0
−6
0
We repeat this process of multiplying
and adding until the table is complete
1
We end up with
1
1
1
Dividend 𝑥 3 − 7𝑥 + 6
This corresponds with 1 in the 𝑥 2 coefficient, the next 1 is the coefficient of 𝑥, and −6 is the
constant, thus we have 𝑥 2 + 𝑥 − 6 is a factor of 𝑥 3 − 7𝑥 + 6. The quadratic factor 𝑥 2 + 𝑥 − 6
reduces to (𝑥 − 2)(𝑥 + 3)
93
3.4.0. Session 4. - Rational Expressions
A quotient of two algebraic expressions is called a fractional expression. Here are some
examples of fractional expressions.
2𝑥
𝑥−1
√𝑥+3
𝑥+1
𝑦−2
𝑦 2 +4
A rational expression is a fractional expression where both the numerator and the denominator
are polynomials. Rational expressions are also functions. Here are some examples of rational
expressions:
𝑥
2𝑥
𝑥 3 −𝑥
𝑥−1
𝑥 2 +1
𝑥 2 −5𝑥+6
We can perform standard algebraic operations such as, addition, subtraction, multiplication, and
division on rational expressions. We can determine the domain as well as simplify and factorize
them just as we do with expressions involving real numbers. The domain of an algebraic
expression is the set of real numbers that the variable is permitted to have.
Example 3.4.0.1.
Find the domain of the expressions
(a) 2𝑥 2 + 3𝑥 − 1
𝑥
(b) 𝑥 2−5𝑥+6
(c)
√𝑥
𝑥−5
Solution 3.4.0.1.
(a) The polynomial 2𝑥 2 + 3𝑥 − 1, is defined for every 𝑥. Therefore, the domain is the set ℝ
of real numbers
𝑥
(b) For the expression 2
, we first factor the denominator
𝑥 −5𝑥+6
𝑥2
𝑥
𝑥
=
− 5𝑥 + 6 (𝑥 − 2)(𝑥 − 3)
The denominator will be zero when 𝑥 = 2 or when 𝑥 = 3. Therefore, the expression is
not defined for the numbers 𝑥 = 2, and 𝑥 = 3. The domain is {𝑥: 𝑥 ≠ 2 𝑎𝑛𝑑 𝑥 ≠ 3}
94
(c) For the numerator of the expression
√𝑥
𝑥−5
to be defined, we must have 𝑥 ≥ 0. Also, we
cannot divide by zero, so 𝑥 ≠ 5. The domain is therefore, {𝑥: 𝑥 ≥ 0 𝑎𝑛𝑑 𝑥 ≠ 5}
3.4.1. Simplifying Rational Expressions
To simplify rational expressions, we factor both the numerator and the denominator and use the
following property of fractions
𝐴𝐶
𝐵𝐶
𝐴
=𝐵
This allows for cancellation of common factors and from the numerator and denominator
Example 3.4.1.1.
𝑥 2 −1
Simplify 𝑥 2+𝑥−2
Solution 3.4.1.1.
Both the numerator and denominator are of the same degree, quadratic expressions. We cannot
cancel out the 𝑥 2 , because it is not a common factor.
We note that the numerator is a difference of two squares, so it can be expressed or simplified as
(𝑥 − 1)(𝑥 + 1), and the denominator can be expressed as (𝑥 − 1)(𝑥 + 2)
𝑥 2 −1
𝑥 2 +𝑥−2
(𝑥−1)(𝑥+1)
𝑥+1
= (𝑥−1)(𝑥+2) = 𝑥+2 because 𝑥 − 1 is now a common factor that cancels out
Note that the 𝑥 2 ’s in the numerator and the denominator cannot cancel each other because the 𝑥 2
is not a common factor
95
3.4.2. Multiplying Rational Expressions
To multiply rational expressions, we use the following property of fractions.
𝐴 𝐶 𝐴𝐶
× =
𝐵 𝐷 𝐵𝐷
This says that to multiply two fractions we multiply their numerators and multiply their
denominators
Example 3.4.2.1.
Multiply and simplify the expression:
𝑥 2 +2𝑥−3
𝑥 2 +8𝑥+16
×
3𝑥+12
𝑥−1
Solution 3.4.2.1.
We proceed by factorising first the expressions:
𝑥 2 + 2𝑥 − 3
3𝑥 + 12 (𝑥 − 1)(𝑥 + 3) 3(𝑥 + 4)
×
=
×
2
(𝑥 + 4)2
𝑥 + 8𝑥 + 16
𝑥−1
𝑥−1
Next is cancelling the common factors (𝑥 − 1) and (𝑥 + 4) that appear in both the numerator
and the denominator
𝑥 2 + 2𝑥 − 3
3𝑥 + 12 3(𝑥 − 1)(𝑥 + 3)(𝑥 + 4)
×
=
2
𝑥 + 8𝑥 + 16
𝑥−1
(𝑥 − 1)(𝑥 + 4)2
After the rearrangement, the common terms (𝑥 − 1) in the numerator will cancel the (𝑥 − 1) in
the denominator, and the term (𝑥 + 4) in the numerator will cancel one term of (𝑥 + 4) in the
denominator. Finally, it reduces to
96
𝑥 2 + 2𝑥 − 3
3𝑥 + 12 3(𝑥 + 3)
×
=
𝑥 2 + 8𝑥 + 16
𝑥−1
𝑥+4
3.4.3. Dividing Rational Expressions
To divide rational expressions, we use the following property of fractions:
𝐴 𝐶 𝐴
𝐷
÷ = ×
𝐵 𝐷 𝐵
𝐶
This says that to divide a fraction by another fraction, we invert the divisor and multiply
Example 3.4.3.1.
Divide and simplify the expression
𝑥 − 4 𝑥 2 − 3𝑥 − 4
÷
𝑥 2 − 4 𝑥 2 + 5𝑥 + 6
Solution 3.4.3.1.
Invert and multiply the expressions 𝑥 2 − 3𝑥 − 4 and 𝑥 2 + 5𝑥 + 6 as done in the following
expression:
𝑥−4
𝑥 2 − 3𝑥 − 4
𝑥−4
𝑥 2 + 5𝑥 + 6
÷ 2
= 2
× 2
𝑥2 − 4
𝑥 + 5𝑥 + 6
𝑥 −4
𝑥 − 3𝑥 − 4
Next, we factorise the quadratics in both the numerator and denominator to obtain the following
expression
(𝑥 − 4)(𝑥 + 2)(𝑥 + 3)
(𝑥 − 2)(𝑥 + 2)(𝑥 − 4)(𝑥 + 1)
97
Next, we cancel the common factors in both the numerator and denominator as follows:
(𝑥 − 4)(𝑥 + 2)(𝑥 + 3)
(𝑥 − 2)(𝑥 + 2)(𝑥 − 4)(𝑥 + 1)
The final and simplified expression is below.
𝑥+3
(𝑥 − 2)(𝑥 + 1)
3.4.4. Adding and Subtracting Rational Expressions
To add or subtract rational expression, we first find a common denominator and then use the
following property of fractions:
𝐴 𝐵 𝐴+𝐵
+ =
𝐶 𝐶
𝐶
Although any common denominator will work, it is best to use the least common denominator
(LCD). The LCD is found by factorising each denominator and taking the product of the distinct
factors, using the highest power that appears in any of the factors.
Example 3.4.4.1.
Use the specified operators to simplify the given expressions.
(a)
3
𝑥
+ 𝑥+2
𝑥−1
(b)
1
𝑥 2 −1
2
− (𝑥+1)2
Solution 3.4.4.1.
(a) Here in this expression, the LCD is simply the product (𝑥 − 1)(𝑥 + 2), thus we use the
LCD to write the fractions as
3
𝑥
3(𝑥 + 2)
𝑥(𝑥 − 1)
+
=
+
(𝑥 − 1)(𝑥 + 2)
𝑥 − 1 𝑥 + 2 (𝑥 − 1)(𝑥 + 2)
98
Next, we add the fractions as follows:
3𝑥 + 6 + 𝑥 2 − 𝑥
(𝑥 − 1)(𝑥 + 2)
Next is to combine the terms in the numerator
𝑥 2 + 2𝑥 + 6
(𝑥 − 1)(𝑥 + 2)
3.4.5. Subtracting Rational Expressions
Example 3.4.5.1.
(a)
1
𝑥 2 −1
2
− (𝑥+1)2
Solution 3.4.5.1.
In this problem, the LCD of 𝑥 2 − 1 is (𝑥 − 1)(𝑥 + 1) because it is a difference of two squares,
and the LCD for (𝑥 + 1)2 is (𝑥 + 1)2
Thus, we factorise the expression as follows:
We have,
1
2
1
2
−
=
−
𝑥 2 − 1 (𝑥 + 1)2 (𝑥 − 1)(𝑥 + 1) (𝑥 + 1)2
We combine the fractions using the LCD
(𝑥 + 1) − 2(𝑥 − 1)
1
2
−
=
(𝑥 − 1)(𝑥 + 1)2
𝑥 2 − 1 (𝑥 + 1)2
99
We use the Distributive Property to obtain
𝑥 + 1 − 2𝑥 + 2
(𝑥 − 1)(𝑥 + 1)2
We combine the terms in the numerator to obtain
3−𝑥
(𝑥 − 1)(𝑥 + 1)2
Note:
We avoid making the following error:
𝐴
𝐴 𝐴
≠ +
𝐵+𝐶 𝐵 𝐶
For instance, if we let 𝐴 = 2, 𝐵 = 1, and 𝐶 = 1, then we can easily see the error:
2
2 2
≠ +
1+1 1 1
2
=? 2 + 2
2
1 =? ? 4 Wrong!!!!
100
3.4.6. Compound Fractions
A compound fraction is a fraction in which the numerator, the denominator, or both, are
themselves fractional expressions.
Example 3.4.6.1.
Simplifying Compound Fractions:
Simplify
𝑥
𝑦+1
𝑦
1−𝑥
𝑥
𝑦+1
𝑦
1−𝑥
Solution 3.4.6.1.
Here, we first convert terms in the numerator into a single fraction. We do the same for the terms
in the denominator. Next, we invert and simplify
𝑥
𝑥+𝑦
𝑥+𝑦
𝑥
𝑦+1
𝑦
𝑦= 𝑥−𝑦 = 𝑦 × 𝑥−𝑦
1−𝑥
𝑥
=
𝑥+𝑦
𝑥
×
𝑦
𝑥−𝑦
=
𝑥(𝑥 + 𝑦)
𝑦(𝑥 − 𝑦)
101
Alternative Solution
We find the LCD of all the fractions in the expression, then we multiply the numerator and
denominator by it. In this example the LCD of all the fraction is 𝑥𝑦. Thus, multiplying both
numerator and denominator by 𝑥𝑦, we have
𝑥
𝑥
𝑦+1
𝑦 + 1 𝑥𝑦
𝑦=
𝑦 ×
1−𝑥
1 − 𝑥 𝑥𝑦
We simplify as follows:
𝑥 2 + 𝑥𝑦
𝑥𝑦 − 𝑦 2
We factorise as follows:
𝑥(𝑥 + 𝑦)
𝑦(𝑥 − 𝑦)
Example 3.4.6.2.
Simplify:
1
1
−
𝑎+ℎ 𝑎
ℎ
Solution 3.4.6.2.
We begin by combining the fractions in the numerator using a common denominator.
1
1 𝑎 − (𝑎 + ℎ)
−𝑎
𝑎(𝑎 + ℎ)
𝑎+ℎ
=
ℎ
ℎ
We invert the divisor and multiply
𝑎 − (𝑎 + ℎ)
1
×
𝑎(𝑎 + ℎ)
ℎ
=
𝑎−𝑎−ℎ 1
×
𝑎(𝑎 + ℎ)
ℎ
102
We simplify further by cancelling the common factors, the ℎ’s
=
−ℎ
1
×
𝑎(𝑎 + ℎ) ℎ
=
−1
𝑎(𝑎 + ℎ)
Example 3.4.6.3.
Simplify:
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2
1 + 𝑥2
Solution 3.4.6.3.
1
We begin by factorising (1 + 𝑥 2 )−2 from the numerator
1
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2 (1 + 𝑥 2 )−2 [(1 + 𝑥 2 ) − 𝑥 2 ]
=
1 + 𝑥2
1 + 𝑥2
1
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2 (1 + 𝑥 2 )−2 [1 + 𝑥 2 − 𝑥 2 ]
=
1 + 𝑥2
1 + 𝑥2
1
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2 (1 + 𝑥 2 )−2
=
1 + 𝑥2
1 + 𝑥2
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2
1
=
3
1 + 𝑥2
(1 + 𝑥 2 )2
Alternative Solution
103
1
Since (1 + 𝑥 2 )−2 =
1
1
(1+𝑥 2 )2
is a fraction, we can clear all fractions by multiplying numerator and
1
denominator by (1 + 𝑥 2 )2 .
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2
1 + 𝑥2
1
1
1
1
1
1
1
1
1
1
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2 (1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2
(1 + 𝑥 2 )2
=
×
1
1 + 𝑥2
1 + 𝑥2
(1 + 𝑥 2 )2
(1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2 (1 + 𝑥 2 )2 − 𝑥 2 (1 + 𝑥 2 )−2
(1 + 𝑥 2 )2
=
×
1
1 + 𝑥2
1 + 𝑥2
(1 + 𝑥 2 )2
104
3.5.0. Session 5 – Exponents and Radicals
3.5.1. The Laws of Exponents (Indices).
We begin by encouraging learners to revisit the laws or rules of indices, or the law of exponents
which were taught in secondary schools or in high schools.
Here is a quick way to refresh our memories:
3.5.2. Zero and Negative Exponents.
If 𝑎 ≠ 0 is any real number and 𝑛 is a positive integer, then
1
𝑎0 = 1 and 𝑎−𝑛 = 𝑎𝑛
Example 3.4.2.1
4 0
1
1
(b) 𝑥 −1 = 𝑥 1 = 𝑥
(a) (7) = 1
(c)
(−2)−3 =
1
(−2)3
1
1
= −8 = − 8
3.5.3. Rules for Working with Exponents.
Our understanding of the following rules will be essential for working with exponents (indices)
and their bases. In the following rules, the bases 𝑎 and 𝑏 are real numbers and the exponents 𝑚
and 𝑛 are integers.
If 𝑎 > 0, and if 𝑚 and 𝑛 are any real numbers, or integers, then,
Rule
𝑎𝑚 . 𝑎𝑛 = 𝑎𝑚+𝑛
Description
(To multiply two powers of the same number, add
the exponents)
2.
𝑎𝑚
𝑎 ÷ 𝑎 = 𝑛 = 𝑎𝑚−𝑛 (𝑎 ≠ 0)
𝑎
To divide two powers of the same number, subtract
the exponents
3.
(𝑎𝑚 )𝑛 = 𝑎𝑚×𝑛 = 𝑎𝑚𝑛
To raise a power to a new power, multiply the
exponents
1.
𝑚
𝑛
4.
1
= 𝑎−𝑛 (𝑎 ≠ 0)
𝑎𝑛
5.
𝑎𝑛 = 𝑛√𝑎
1
The reciprocal of any number raised to the power 𝑛 is
the number raised to the power −𝑛
1
Any number raised to the power 𝑛 is the 𝑛th root of
that number
105
𝑚
6.
Any number 𝑎, raised to the power
𝑛
𝑎 𝑛 = √𝑎𝑚
𝑚
𝑛
is 𝑛th root of
the number raised to the power 𝑚 √𝑎𝑚
7.
(𝑎𝑏)𝑛 = 𝑎𝑛 . 𝑏𝑛
To raise a product to a power, raise each factor to the
power
8.
𝑎0 = 1
Any number raised to the power 0 is 1
(𝑎 ≠ 0)
𝑎1 = 𝑎
9.
Any number raised to the power 1 is the number
itself
Example 3.5.3.1.
1. Using the Laws of Exponents
(a) 𝑥 4 𝑥 7 = 𝑥 4+7 = 𝑥 11
using Law 1: 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
1
(b) 𝑦 4 𝑦 −7 = 𝑦 4−7 = 𝑦 −3 = 𝑦 3
(c)
𝑐9
= 𝑐 9−5 = 𝑐 4
𝑐5
using Law 1: 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
using Law 2:
𝑎𝑚
𝑎𝑛
= 𝑎𝑚−𝑛
(d) (𝑏4 )5 = 𝑏4.5 = 𝑏20
using Law 3: (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛
(e) (3𝑥)3 = 33 𝑥 3 = 27𝑥 3
using Law 4: (𝑎𝑏)𝑛 = 𝑎𝑛 𝑏𝑛
𝑥 5
𝑥5
𝑥5
(f) (2) = 25 = 32
𝑎 𝑛
Exercises 3.5.3.2
Ex 1.
Simplify the following expressions.
(a) (2𝑎3 𝑏2 )(3𝑎𝑏4 )3
4
𝑥 3 𝑦2𝑥
(b) (𝑦 ) (
(c)
)
𝑧
−4
4 )(3𝑎 3 −4 )3
(2𝑎
𝑏
𝑥
𝑏
3 𝑦 −2 𝑥 4
(d) (𝑦 2) (
𝑧
𝑎𝑛
using Law 5: (𝑏 ) = 𝑏𝑛
)
106
3.5.4. Laws of Exponents (Indices) Continued.
6.
7.
Rule
𝑎 −𝑛
𝑏 𝑛
( ) =( )
𝑏
𝑎
Description
To raise a fraction to a negative power, invert the
fraction and change the sign of the exponent
𝑎−𝑛 𝑏𝑚
=
𝑏−𝑚 𝑎𝑛
To move a number raised to a power from numerator
to denominator or from denominator to numerator,
change the sign of the exponent
Example 3.5.4.1.
Ex1.
Eliminate the negative exponents and simplify each of the following expressions.
(a)
6𝑠𝑡 −4
𝑦
(3𝑧 3 )
(b)
2𝑠−2 𝑡 2
−2
Solution 3.5.4.1.
Ex1.
(a) We apply Law 7, which allows us to move a number raised to a power from numerator to
the denominator (or vice versa) by changing the sign of the exponent.
6𝑠𝑡 −4
Given 2𝑠 −2𝑡 2, and using Law 7, the 𝑡 −4in the numerator moves to the denominator and
becomes 𝑡 4, and the 𝑠 −2in the denominator moves to the numerator and becomes 𝑠 2 . We
6𝑠𝑠2
6𝑠3
get, 2𝑡 2𝑡 4 , which becomes 2𝑡 6 =
3𝑠3
𝑡6
by applying Law 1.
(b) We apply Law 6, which allows the change of signs of the exponent of a fraction by
𝑦
inverting the fraction. We are given (3𝑧 3)
−2
107
=(
3𝑧 3
𝑦
2
) =
9𝑧 6
𝑦2
3.5.5. Radicals and Properties of Roots
We know the meaning and the value of 2𝑛 when 𝑛 is an integer. Also, we know meaning of the
symbol √𝑎 , “the positive square root of 𝑎”.
If √𝑎 = 𝑏 then it means 𝑏2 = 𝑎 and 𝑏 ≥ 0.
Since 𝑎 = 𝑏2 ≥ 0, the symbol √𝑎 makes sense only when 𝑎 ≥ 0. For instance, √9 = 3, because
32 = 9 and 3 ≥ 0.
Square roots are special cases of 𝑛th roots. The 𝑛th root of 𝑥 is the number that, when raised to
the 𝑛th power, gives 𝑥.
3.5.6. Definition of 𝑛th Root
If 𝑛 is any positive integer, then the principal 𝑛th root of 𝑎 is defined as follows:
𝑛
√𝑎 = 𝑏 means 𝑏𝑛 = 𝑎
If 𝑛 is even, we must have 𝑎 ≥ 0 and 𝑏 ≥ 0.
4
We can say √81 = 3 because 34 = 81 and 3 ≥ 0
3
√−8 = −2 because (−2)3 = −8
4
6
Note that √−8, √−8, √−8 are not defined. (For instance, √−8 is not defined because the square
of every real number is nonnegative.)
Note also that √42 = √16 = 4 but √(−4)2 = √16 = 4 = | − 4|
108
The equation √𝑎2 = 𝑎 is not always true; it is true only when 𝑎 ≥ 0. However, we can always
write √𝑎2 = |𝑎|. This last equation is true not only for square roots, but for any even root. This
and other rules used in working with 𝑛th roots are listed below. In each property we assume that
all the given roots exist.
3.5.7. Properties of 𝑛th Roots
Property
1.
2.
3.
4.
5.
𝑛
𝑛
3
𝑛
√𝑎𝑏 = √𝑎 √𝑏,
𝑛
𝑎
√𝑏 =
𝑚 𝑛
𝑛
4
√𝑎
𝑛 ,
√𝑏
√ √𝑎 =
3
3
an example √(−8)(27) = √−8 √27 = (−2)(3) = −6
4
16
an example is √81 =
√16
4
√81
3
𝑚𝑛
2
=3
6
an example is √ √729 = √729 = 3
√𝑎 ,
√𝑎𝑛 = 𝑎,
if 𝑛 is odd
an example is √(−5)3 = −5,
𝑛
if 𝑛 is even
an example is √(−3)4 = |−3| = 3
√𝑎𝑛 = |𝑎|
3
5
𝑛
√25 = 2
4
Example 3.5.7.1. Simplifying Expressions involving 𝑛th Roots
3
(a) √𝑥 4
Factor out the largest cube, and we get,
3
3
√𝑥 4 = √𝑥 3 𝑥,
3
3
3
Using Property 1: √𝑎𝑏 = 3√𝑎 √𝑏 , we get √𝑥 3 3√𝑥
3
Using Property 4: √𝑎3 = 𝑎, we get
4
(b) √81𝑥 8 𝑦 4
4
4
𝑥 3√𝑥
4
Using Property 1: √𝑎𝑏𝑐 = 4√𝑎 √𝑏 4√𝑐, we get
4
4
4
4
√81𝑥 8 𝑦 4 = √81 √𝑥 8 √𝑦 4, Using Property 5: √𝑎4 = |𝑎|, |𝑥 2 | = 𝑥 2 , we get
4
4
4
√81 √𝑥 8 √𝑦 4 = 3𝑥 2 |𝑦|
Expressions that involve roots such as √𝑎, 𝑎 + 𝑏√𝑐,
109
√𝑎
,
𝑏
etc are known as radicals.
It is frequently useful to combine like radicals in an expression such as 2√3 + 5√3. This can be
done using the Distributive Property. Thus,
2√3 + 5√3 = (2 + 5)√3
3.5.8. Combining Radicals
Example 3.5.8.1
Combine the following radicals and simplify them
(a) √32 + √200 = √(16)(2) + √(100)(2),
by factorising out the largest squares
√32 + √200 = √16√2 + √100√2
√32 + √200 = 4√2 + 10√2 = 14√2
(b) If 𝑏 > 0, then √25𝑏 − √𝑏3 = √25√𝑏 − √𝑏2 √𝑏
√25𝑏 − √𝑏3 = √25√𝑏 − √𝑏2 √𝑏
√25𝑏 − √𝑏3 = 5√𝑏 − 𝑏√𝑏
√25𝑏 − √𝑏3 = (5 − 𝑏)√𝑏
110
3.6.0. - Session 6 - Rational Exponents
𝑚
Rational exponents are also known as fractional exponents. They are of the form 𝑎 𝑛 . To define
1
rational exponents, we will use radicals. To give meaning to the symbol 𝑎𝑛 in a way that is
consistent with the Laws of Exponents, we would have to have
1 𝑛
𝑛
(𝑎 )
=
1 𝑛
( )
𝑛
𝑎
= 𝑎1 = 𝑎
By the definition of the 𝑛th root,
1
𝑎𝑛 = 𝑛√𝑎
3.6.1. Definition of Rational Exponents
In general, we define rational exponents as follows:
For any rational exponent
𝑚
𝑛
𝑚
𝑛
in lowest terms, where 𝑚 and 𝑛 are integers and 𝑛 > 0, we define
𝑚
𝑚
𝑛
𝑎 = ( 𝑛√𝑎) or equivalently 𝑎 𝑛 = √𝑎𝑚
If 𝑛 is even, then we require that 𝑎 ≥ 0.
The Laws of Exponents also holds for rational exponents.
Example 3.6.1.1.
Using the Definition of Rational Exponents
1
(a) 42 = √4 = 2
2
2
2
3
(b) 83 = ( √8) = 22 = 4
1
(c) 1253 =
(d)
1
3
√𝑥
=
4
1
1
(125)3
1
4
𝑥3
=
1
3
√125
3
3
Alternatively, 83 = √82 = √64 = 4
1
=5
4
= 𝑥 −3
111
3.6.2. Using the Laws of Exponents with Rational Exponents
1
7
8
using the Law: 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
(a) 𝑎3 𝑎3 = 𝑎3
2 7
(b)
𝑎5 𝑎5
3
𝑎5
2 7 3
6
= 𝑎5+5−5 = 𝑎5
3
3
using the Law:
3
3
(c) (2𝑎3 𝑏4 )2 = 22 (𝑎3 )2 (𝑏4 )2
3
= (√2) 𝑎3
3
2
( )
3
(𝑏4 )2
(d) (
) (
3 3
𝑦4
23 (𝑥 4)
1) =
−
𝑥 2
1 3
(𝑦 3 )
1
9
× (𝑦 4𝑥 2 ) =
8𝑥 4
𝑦
1
11
× 𝑦 4 𝑥 2 = 8𝑥 4 𝑦 3
3.6.3. Simplifying by Writing Radicals as Rational Exponents
1
(a)
(b)
1
1 1
5
(3√𝑥)(2 3√𝑥) = (3𝑥 2 ) (2𝑥 3 ) = 6𝑥 2+3 = 6𝑥 6
1
1 2
2
1
3 2
2
= 𝑎𝑚−𝑛
using the Law: (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛
9
3
𝑎𝑛
using the Law: (𝑎𝑏𝑐)𝑛 = 𝑎𝑛 𝑏𝑛 𝑐 𝑛
= 2√2 𝑎2 𝑏6
3
2𝑥 4
1
𝑦3
𝑎𝑚
3
√𝑥 √𝑥 = (𝑥𝑥 ) = (𝑥 ) = 𝑥 4
112
3.6.4. Rationalization of Denominator and the Numerator
3.6.4.1.
Rationalizing the Denominator
It is often useful to eliminate the radical from the denominator by multiplying both numerator
and denominator by an appropriate expression. This procedure is called rationalizing the
denominator. If the denominator is of the form √𝑎, we multiply numerator and denominator by
√𝑎. In doing so, we multiply the given quantity by 1, so we do not change its value.
For example, we can write
1
as follows
√𝑎
1
√𝑎
=
1
√𝑎
× 1=
1
√𝑎
×
√𝑎 √𝑎
=
𝑎
√𝑎
Note that the denominator in the last fraction contains no radical. In general, if the denominator
𝑛
𝑛
is of the form √𝑎𝑚 with 𝑚 < 𝑛, then multiplying the numerator and denominator by √𝑎𝑛−𝑚 will
rationalise the denominator, because (for 𝑎 > 0)
𝑛
𝑛
𝑛
√𝑎𝑚 √𝑎𝑚+𝑛−𝑚 = √𝑎𝑛 = 𝑎
Exercises - Rationalizing Denominators
Example. 3.6.4..1.
(a)
(b)
2
=
√3
1
2
3
=
2
7
1
√𝑥
×
√3
(c) √𝑎2 =
√𝑥 2
1
7
√𝑎
√3
=
2√3
√3
3
√3
, and noting that
𝟑
1
3
√3
×
=
2
√𝒙
𝟑
√𝒙
1
7
√𝑎2
3
=
3
√𝑥
3
√𝑥 3
=
7
√𝑎5
×7
√𝑎
=1
=
5
√𝑥
𝑥
7
√𝑎5
𝑎
113
Exercises 3.6.4.3.2
Rationalise the following expressions
(a)
(d)
1
√10
2
3
√𝑥
2
(b)
√𝑥
(e)
4
(c)
1
(f)
√𝑦 3
𝑥
√3
𝑥
2
𝑦5
114
Unit 4
Session 1.
Exponential Functions
Session 2.
The Napierian (Natural) Logarithms
Session 3.
Exponential and Logarithmic Functions
Session 4.
Logarithmic Functions
Session 5.
Common Logarithms with Base 10
Session 6.
Expanding and Combining Logarithmic Expressions
115
4.1.0. Session 1 - Exponential and Logarithmic Functions
In this section, we will be solving equations that involve exponential and logarithmic functions.
An exponential equation is one in which the variable occurs in the exponent. An example is the
equation 2𝑥 = 7 or 𝑓(𝑥) = 2 𝑥 where the independent variable is in the exponent.
The variable 𝑥 presents a difficulty because it is in the exponent. To deal with this difficulty, we
take the logarithm of each side, and we use the Laws of Logarithms to “bring down 𝑥” from the
exponent.
For example, in the above equation, we proceed as follows:
Given
2 𝑥 = 7,
Take ln (natural logarithm = ln) of each side
ln 2𝑥 = ln 7
At this moment, we will visit the Laws of Logarithms briefly before we proceed with the
solution of the problem
4.1.1. The Laws of Logarithms.
The Laws of Logarithms are:
(i)
log 𝑎 𝑚𝑛 = log 𝑎 𝑚 + log 𝑎 𝑛
(ii)
log 𝑎 ( 𝑛 ) = log 𝑎 𝑚 − log 𝑎 𝑛
𝑚
(iii)
log 𝑎 (𝑚𝑛 ) = 𝑛 log 𝑎 𝑚
(iv)
log b 𝑚 =
log𝑎 𝑚
log𝑎 𝑏
(Change of logs to a common base)
These laws apply to natural logarithms as well.
Please note:
We will look at natural logarithms shortly in Session 4.
116
We will use the Laws of Logarithms to continue with solution to the problem 2 𝑥 = 7 after taking
logarithms of both sides
ln 2𝑥 = ln 7
We use Law 3 (to bring down the exponent, 𝑥)
𝑥 𝑙𝑛 2 = ln 7
We solve for 𝑥 as follows:
𝑥=
𝑥=
ln 7
ln 2
1.945910
= 2.807355
0.693147
We want to remember that Law 3 of the Laws of Logarithms says that log 𝑎 𝐴𝐶 = 𝐶 log 𝑎 𝐴 .
The method we used to solve 2𝑥 = 7 is typical of how we would solve exponential equations in
general.
4.1.2. Properties of the Exponential Function
1. 𝑒 0 = 1
1
2. 𝑒 𝑥 = 𝑒 −𝑥
this is equivalent to 𝑎0 = 1 with 𝑎 = 𝑒.
1
which is equivalent to 𝑎𝑛 = 𝑎−𝑛 with 𝑎 = 𝑒
4.1.3. Guidelines for Solving Exponential Equations
1. Isolate the exponential expression on one side of the equation
2. Take the logarithm of each side, then use the Laws of Logarithms to “bring down the
exponent”
3. Solve for the variable
117
Example 4.1.3.1.
Find the solution of the equation 3𝑥+2 = 7, round to six decimal places
Solution 4.1.3.1.
We proceed by taking a logarithm of each (both sides) side, and use the Laws of Logarithms
We are given
3 𝑥+2 = 7
Next is, we take log of each side
log(3𝑥+2 ) = log 7
We use Law 3 to “bring down the exponent”
(𝑥 + 2) log 3 = log 7
We divide both sides by log 3
𝑥 + 2 = log 3
Subtract 2 from each side
𝑥=
log 7
log 7
log 3
−2
0.84509804
𝑥 = 0.47712125 − 2
= 1.77124377 − 2
= −0.228756
Example 4.1.3.2.
Solve the equation 8𝑒 2𝑥 = 20
Solution 4.1.3.2.
We proceed by isolating the exponential term on one side of the equation.
We do so by dividing both sides by 8
We are given
8𝑒 2𝑥 = 20
Divide both sides by 8
𝑒 2𝑥 =
118
20
8
We take natural logs of both sides (each side)
ln 𝑒 2𝑥 = ln 2.5
We “bring the exponent down” using the properties of log
2𝑥 ln𝑒 𝑒 = ln 2.5, because
ln𝑒 𝑒 = 1, we have 2𝑥 = ln 2.5
𝑥=
ln 2.5
≈ 0.458145
2
Example 4.1.3.3.
Solve the equation 𝑒 3−2𝑥 = 4 algebraically
Solution 4.1.3.3.
Since the base of the exponential term is 𝑒, the natural logarithm ( log 𝑒 𝑒) can be used to solve
this equation.
Given the equation
𝑒 3−2𝑥 = 4
We take the natural logarithm of both sides
ln(𝑒 3−2𝑥 ) = ln 4
We apply the properties of natural log
3 − 2𝑥 = ln 4
Subtract 3 from each side
−2𝑥 = ln 4 − 3
1
Divide each side by -2, or multiply each side by − ,
2
119
1
𝑥 = (3 − ln 4)
2
4.1.4. Exponential of the Quadratic Type
Example 4.1.4.1.
Solve the equation 𝑒 2𝑥 − 𝑒 𝑥 − 6 = 0
Solution 4.1.4.1.
We isolate the exponential term by factorising them
Given the expression
𝑒 2𝑥 − 𝑒 𝑥 − 6 = 0
Applying the Laws of Exponents, we can express it as
(𝑒 𝑥 )2 − 𝑒 𝑥 − 6 = 0 , this is of the
form 𝑤 2 − 𝑤 − 6 = 0, a quadratic, which we factor as
(𝑒 𝑥 − 3)(𝑒 𝑥 + 2) = 0
We use the zero-product property as follows:
𝑒𝑥 − 3 = 0
𝑒𝑥 = 3
or
𝑒𝑥 + 2 = 0
𝑒 𝑥 = −2
The equation 𝑒 𝑥 = 3 leads to 𝑥 = ln 3. On the other hand, the equation 𝑒 𝑥 = −2 has no solution
because, 𝑒 𝑥 > 0 for all 𝑥. Therefore, 𝑥 = ln 3 ≈ 1.0986 is the only solution.
4.1.5. Solving Exponential Equations
Example 4.1.5.1.
Solve the equation
3𝑥𝑒 𝑥 + 𝑥 2 𝑒 𝑥 = 0
120
Solution 4.3.5.1.
We proceed by factorising the left side of the equation
We are given
3𝑥𝑒 𝑥 + 𝑥 2 𝑒 𝑥 = 0
We factor the common factors as follows:
𝑥(3 + 𝑥)𝑒 𝑥 = 0
Divide each side by 𝑒 𝑥 (because 𝑒 𝑥 ≠ 0)
𝑥(3 + 𝑥) = 0
Applying the Zero-Product property
𝑥 = 0 or
3+𝑥 =0
𝑥= 0
or
The solutions are
121
𝑥 = −3
4.2.0. Session 2 - Logarithmic Functions
Every exponential function 𝑓(𝑥) = 𝑎 𝑥 , with 𝑎 > 0 and 𝑎 ≠ 1, is a one-to-one function, and
therefore has an inverse function. The inverse function 𝑓 −1 is called the logarithmic function
with a base 𝑎 and is denoted by log 𝑎 𝑥 .
Recall the definition of 𝑓 −1 (𝑥) = 𝑦 ⇔ 𝑓(𝑦) = 𝑥
This leads to the following definition of the logarithmic function
4.2.1. Definition of the Logarithmic Function
Let 𝑎 be a positive number with 𝑎 ≠ 1. The logarithmic function with a base 𝑎, denoted by
log 𝑎 𝑥 is defined by
log 𝑎 𝑥 = 𝑦
⟺
𝑎𝑦 = 𝑥
So log 𝑎 𝑥 is the exponent to which the base 𝑎 must be raised to give 𝑥
It is important to note that when we use the definition of logarithms to switch back and forth
between logarithmic form log 𝑎 𝑥 = 𝑦 and the exponential form 𝑎 𝑦 = 𝑥, it is helpful to notice
that, in both forms, the base is the same:
Logarithmic Form
Exponential Form
𝑦 is the Exponent
log 𝑎 𝑥 = 𝑦
𝑎 is the Base
𝑦 is the Exponent
𝑎𝑦 = 𝑥
𝑎 is the Base
4.2.2. Logarithmic and Exponential Forms
The logarithmic and exponential forms are equivalent equations: if one is true, then so is the
other. We can therefore switch from one form to the other as illustrated in the table below.
122
Example 4.2.2.1.
Logarithmic Form
log10 100000 = 5
log 2 8 = 3
1
log 2 ( ) = −3
8
log 5 𝑠 = 𝑟
Exponential Form
105 = 100000
23 = 8
1
2−3 = ( )
8
5𝑟 = 𝑠
It is important to understand that log 𝑎 𝑥 is an exponent. For example, the numbers in the right
column of the table below are logarithms (base 10) of the numbers in the left column. This the
case for all bases, as we will see in the examples that will follow in the next pages.
𝑥
104
103
102
10
1
10−1
10−2
10−3
10−4
log10 𝑥
4
3
2
1
0
−1
−2
−3
−4
4.2.3. Evaluating Logarithms
Example 4.2.3.1.
(a) log10 1000 = 3
(b) log 2 32 = 5
(c) log10 0.1 = −1
1
(d) log16 4 = 2
because
because
because
103 = 1000
25 = 32
10−1 = 0.1
because
162 = 4
1
123
Recall the Inverse Function Properties
𝑓 −1 (𝑓(𝑥)) = 𝑥
𝑓(𝑓 −1 (𝑥)) = 𝑥
When we apply the Inverse Function Property 𝑓(𝑥) = 𝑎 𝑥 and 𝑓 −1 (𝑥) = log 𝑎 𝑥 we get
log 𝑎 𝑎 𝑥 = 𝑥,
𝑥∈ℝ
𝑎loga 𝑥 = 𝑥,
𝑥>0
We list these properties of logarithms below
1.
2.
3.
4.
Property
log 𝑎 1 = 0
log 𝑎 𝑎 = 1
log 𝑎 𝑎 𝑥 = 𝑥
𝑎loga 𝑥 = 𝑥
Reason
We must raise 𝑎 to the power 0 to get 1
We must raise 𝑎 to the power 1 to get 𝑎
We must raise 𝑎 to the power 𝑥 to get 𝑎 𝑥
log 𝑎 𝑥 is the power to which 𝑎 must be raised to get 𝑥
4.2.4. Applying Properties of Logarithms
We will illustrate the properties of logarithms when the base is 5. Consider the following
examples with base 5
log 5 1 = 0
Property 1
log 5 5 = 1
Property 2
log 5 58 = 8
Property 3
5log5 12 = 12
Property 4
124
4.3.0. Session 3 – Natural Exponential Functions
4.3.1. The Natural Exponential Function
The natural exponential function is one which contains 𝑒 𝑥 , where 𝑒 is constant called the base. It
has an approximate value of 2.7183. The natural exponent arises from the natural laws of growth
and decay and is used as a base for natural or Napierian logarithms.
4.3.2. Evaluating the Natural Exponential Functions
The value of 𝑒 𝑥 may be determined by using:
(a) A calculator
(b) The power series for 𝑒 𝑥 as we will see in the next section
(c) Tables of exponential functions
The most common method of evaluating the natural exponential function is by using a scientific
notation calculator, which has now replaced the use of tables. Most scientific notation calculators
contain an 𝑒 𝑥 function which enables all practical values of 𝑒 𝑥 and 𝑒 −𝑥 to be determined correct
to 8 or 9 significant figures.
For example, the following natural exponential values are from calculators
i.
ii.
iii.
𝑒 1 = 2.7182818
𝑒 2.7 = 14.87973172
𝑒 −2.33 = 0.09729575
You can use your personal calculator or the calculator on your phone to check the following
values:
𝑒 0.12 = 1.1275, correct to 5 significant figures
𝑒 −1.54 = 0.21438, correct to 5 decimal places
𝑒 −0.431 = 0.6498589, correct to 7 decimal places
Example 4.3.2.1.
Use a calculator to evaluate, correct to 5 significant figures:
(a)
𝑒 2.731
(b)
𝑒 −3.162
(c)
125
5
3
𝑒 5.253
Solution 4.3.2.1.
(a) 𝑒 2.731 = 15.348227 … = 15.348, correct to 5 significant figures
(b) 𝑒 −3.162 = 0.04234097 … = 0.04234, correct to 5 significant figures
5
5
(c) 3 𝑒 5.253 = 3 (191.138825 … ) = 318.56, correct to 5 significant figures
4.3.3. The Power Series
The value of 𝑒 𝑥 can be calculated to any required degree of accuracy since it is defined in terms
of the following power series:
𝑒𝑥 = 1 + 𝑥 +
𝑥2
2!
+
𝑥3
3!
+
𝑥4
4!
+⋯
(4.3.3.)
Where (3! = 3 × 2 × 1 = 6) is called the ‘factorial 3’
The series is valid for all values of 𝑥.
The series is said to converge, i.e., if all terms are added, an actual value for 𝑒 𝑥 (where 𝑥 is a real
number) is obtained. The more terms that are taken, the closer will be the value of 𝑒 𝑥 to its actual
value. The value of the natural exponent 𝑒, correct to say 4 decimal places, may be determined
by substituting 𝑥 = 1 in the power series of equation (4.3.3.).
Therefore,
(1)2 (1)3 (1)4 (1)5 (1)6 (1)7 (1)8
𝑒 =1+1+
+
+
+
+
+
+
+⋯
2!
3!
4!
5!
6!
7!
8!
1
𝑒 1 = 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00139 + 0.00020 + 0.00002
𝑒 = 2.71828
i.e. 𝑒 = 2.71828 correct to 4 decimal places
126
The value of 𝑒 0.05, correct to say 8 significant figures, is found by substituting 𝑥 = 0.05 in the
power series for 𝑒 𝑥 . Thus
𝑒 0.05 = 1 + 0.05 +
(0.05)2 (0.05)3 (0.05)4 (0.05)5
+
+
+
+⋯
2!
3!
4!
5!
𝑒 0.05 = 1 + 0.05 + 0.00125 + 0.000020833 + 0.000000260 + 0.000000003 + ⋯
adding them up, yields
𝑒 0.05 = 1.0512711 correct to 8 significant figures.
We note from the example we just completed that the successive terms in the series grow smaller
very rapidly and it is relatively easy to determine the value of 𝑒 0.05 to a high degree of accuracy.
However, when 𝑥 is nearer to unity or larger than unity, a very large number of terms are
required for an accurate result.
If in the series of equation (4.3.3), we replace 𝑥 by −𝑥, then
𝑒 −𝑥 = 1 + (−𝑥) +
𝑒
−𝑥
(−𝑥)2 (−𝑥)3 (−𝑥)4
+
+
+⋯
2!
3!
4!
𝑥2 𝑥3
=1−𝑥+ − +⋯
2! 3!
In a similar manner the power series for 𝑒 𝑥 may be used to evaluate any exponential function of
the form 𝑎𝑒 𝑘𝑥 , where 𝑎 and 𝑘 are constants. In the series of equation (4.3.3), let 𝑥 be replaced by
𝑘𝑥. Then
𝑎𝑒 𝑘𝑥 = 𝑎 {1 + (𝑘𝑥) +
(𝑘𝑥)2 (𝑘𝑥)3
+
+ ⋯}
2!
3!
127
Thus
5𝑒 2𝑥 = 5 {1 + (2𝑥) +
(2𝑥)2
2!
= 5 {1 + 2𝑥 +
+
(2𝑥)3
3!
+ ⋯}
4𝑥 2 8𝑥 3
+
+⋯}
2
6
= 5 {1 + 2𝑥 + 2𝑥 2 +
4𝑥 3
+⋯}
3
Example 4.3.3.1.
Determine the value of 5𝑒 0.5, correct to 5 significant figures by using the power series for 𝑒 𝑥
Solution 4.3.3.1.
The power series is given by
𝑒𝑥 = 1 + 𝑥 +
𝑥2 𝑥3 𝑥4
+ + +⋯
2! 3! 4!
Hence
𝑒 0.5 = 1 + 𝑥 +
𝑒 0.5 = 1 + 0.5 +
𝑥2 𝑥3 𝑥4
+ + +⋯
2! 3! 4!
(0.5)2
(0.5)5
(0.5)3
(0.5)4
+
+
+
+⋯
(2)(1) (3)(2)(1) (4)(3)(2)(1) (5)(4)(3)(2)(1)
(0.5)2
(0.5)5
(0.5)3
(0.5)4
+
+
+
(2)(1) (3)(2)(1) (4)(3)(2)(1) (5)(4)(3)(2)(1)
(0.5)6
+
+⋯
(6)(5)(4)(3)(2)(1)
𝑒 0.5 = 1 + 0.5 +
𝑒 0.5 = 1 + 0.5 + 0.125 + 0.020833 + 0.0026042 + 0.0002604 + 0.0000217 + ⋯
128
𝑒 0.5 = 1.64872 correct to 6 significant figures
Hence
5𝑒 0.5 = 5(1.64872) = 8.2436, correct to 5 significant figures
Example 4.3.3.2.
Determine the value of 3𝑒 −1, correct to 4 decimal places, using the power series for 𝑒 𝑥 .
Solution 4.3.3.2.
Note that, in this example, 𝑥 = −1.
We proceed by substituting 𝑥 = −1 in the power series
𝑥2 𝑥3 𝑥4
𝑒 =1+𝑥+ + + +⋯
2! 3! 4!
𝑥
We get
𝑒
𝑒
−1
−1
= 1 + (−1) +
(−1)2 (−1)3 (−1)4
+
+
+⋯
2!
3!
4!
(−1)2 (−1)3 (−1)4 (−1)5 (−1)6
= 1 + (−1) +
+
+
+
+
+⋯
2!
3!
4!
5!
6!
𝑒 −1 = 1 − 1 + 0.5 − 0.166667 + 0.041667 − 0.008333 + 0.001389 − 0.000198 + ⋯
𝑒 −1 = 1 − 1 + 0.5 − 0.166667 + 0.041667 − 0.008333 + 0.001389 − 0.000198 + ⋯
𝑒 −1 = 0.367858
129
Hence 3𝑒 −1 = (3)(0.367858) = 1.1036, correct to 4 decimal places
Example 4.3.3.3.
Expand 𝑒 𝑥 (𝑥 2 − 1) as far as the term in 𝑥 5 .
Solution 4.3.3.3.
The power series for 𝑒 𝑥 is
𝑒𝑥 = 1 + 𝑥 +
𝑥2 𝑥3 𝑥4 𝑥5
+ + + +⋯
2! 3! 4! 5!
Hence
𝑒 𝑥 (𝑥 2 − 1) = (1 + 𝑥 +
𝑒 𝑥 (𝑥 2 − 1) = (𝑥 2 + 𝑥 3 +
𝑥2 𝑥3 𝑥4 𝑥5
+ + + + ⋯ ) (𝑥 2 − 1)
2! 3! 4! 5!
𝑥4 𝑥5 𝑥6 𝑥7
𝑥2 𝑥3 𝑥4 𝑥5
+ + + + ⋯ ) − (1 + 𝑥 + + + + + ⋯ !)
2! 3! 4! 5!
2! 3! 4! 5!
We group like terms to get
𝑒 𝑥 (𝑥 2 − 1) = −1 − 𝑥 + (𝑥 2 −
𝑥2
𝑥3
𝑥4 𝑥4
𝑥5 𝑥5
𝑥6 𝑥6
) + (𝑥 3 − ) + ( − ) + ( − ) + ( − )
2!
3
2! 4!
3! 5!
4! 6!
+⋯
𝑒 𝑥 (𝑥 2 − 1) = −1 − 𝑥 + (
2𝑥 2 − 𝑥 2
6𝑥 3 − 𝑥 3
12𝑥 4 − 𝑥 4
20𝑥 5 − 𝑥 5
)+(
)+(
)+(
)+⋯
2
6
24
120
1
5
11 4
19 5
𝑒 𝑥 (𝑥 2 − 1) = −1 − 𝑥 + 𝑥 2 + 𝑥 3 +
𝑥 +
𝑥 +⋯
2
6
24
120
After expansion as far as to the term in 𝑥 5
130
Exercises 4.3.3.4.
1. Use the power series for 𝑒 𝑥 to evaluate 5.6𝑒 −1, correct to 4 decimal places. Check your
result from a calculator.
2. Use the power series for 𝑒 𝑥 to determine, correct to 4 significant figures
(a) 𝑒 2
(b)
𝑒 −0.3
Check your result by using a calculator
4.3.5. Graphs of Exponential Functions.
The values of 𝑒 𝑥 and 𝑒 −𝑥 can be obtained from a calculator over any range of 𝑥 and plotted on
the standard 𝑥 and 𝑦-axes for any relation or exponential function given by 𝑦 = 𝑒 𝑥 or 𝑦 = 𝑒 −𝑥
Exercise 4.3.5.1.
Use a calculator to obtain the values of 𝑒 𝑥 and 𝑒 −𝑥 for −3 ≤ 𝑥 ≤ 3, and sketch the graphs of
𝑦 = 𝑒 𝑥 and 𝑦 = 𝑒 −𝑥 on the same cartesian 𝑥 and 𝑦- axes. Determine the value of 𝑦 for 𝑥 = 2.2,
and the value of 𝑥 when 𝑦 = 1.6.
131
4.4.0. Session 4 – The Napierian (Natural) Logarithms
Logarithms that have a base of 𝑒 are called hyperbolic, Napierian or natural logarithms and the
Napierian logarithm of 𝑥 is written as log 𝑒 𝑥, or more commonly as ln 𝑥.
4.4.1. The Natural Logarithms
There is also a number 𝑒, as a preferred choice for logarithms. The Natural logarithm is a
logarithm with a base of the number 𝑒 as shown in the following definition
4.4.2. Definition
The logarithm with base 𝑒 is called the natural logarithm and is denoted by 𝑙𝑛
ln 𝑥 = log 𝑒 𝑥
The natural logarithmic function 𝑦 = ln 𝑥 is the inverse function of the natural exponential
function 𝑦 = 𝑒 𝑥 . By the definition of the inverse functions, we have
ln 𝑥 = 𝑦
⟺
𝑒𝑦 = 𝑥
If we substitute 𝑎 = 𝑒 and write "ln" for ′𝑙𝑜𝑔", in the properties of logarithms we mentioned
earlier, we can obtain the following properties of natural logarithms
1.
2.
3.
4.
Property
𝑙𝑛 1 = 0
𝑙𝑛 𝑒 = 1
ln 𝑒 𝑥 = 𝑥
𝑒 ln 𝑥 = 𝑥
Reason
We must raise 𝑒 to the power 0 to get 1
We must raise 𝑒 to the power 1 to get 𝑒
We must raise 𝑒 to the power 𝑥 to get 𝑒 𝑥
𝑙𝑛 𝑥 is the power to which 𝑒 must be raised to get 𝑥
Again, calculators are equipped with an 𝐿𝑁 key that gives the values of natural logarithms
Example 4.4.2.1.
Evaluating the Natural Logarithm Function
(a) 𝑙𝑛 𝑒 8 = 8
1
Definition of natural logarithm
(b) ln (𝑒 2 ) = ln 𝑒 −2 = −2
Definition of natural logarithm
(c) ln 5 ≡ 1.609
Use of 𝐿𝑁 key on calculator
132
4.4.3. Finding the Domain of a Logarithmic Function
Example 4.4.3.1.
Find the domain of the function 𝑓(𝑥) = ln(4 − 𝑥 2 ).
Solution 4.4.3.1.
As with any logarithmic function, ln 𝑥 is defined when 𝑥 > 0. Thus the domain of 𝑓 is
{𝑥 ∶ 4 − 𝑥 2 > 0} = { 𝑥 ∶ 𝑥 2 < 4 } = { 𝑥 ∶ |𝑥| < 2} = { 𝑥 ∶ −2 < 𝑥 < 2} = (−2, 2)
4.4.4. Evaluating Napierian Logarithms
The value of a Napierian logarithm may be determined by using:
(a) a calculator
(b) a relationship between common and Napierian logarithms, or
(c) Napierian logarithm tables
The most common method of evaluating a Napierian (natural) logarithm is by a scientific
notation calculator, which has now come to replace the four-figure tables, and the relationship
between common and Napierian (natural) logarithms
log 𝑒 𝑦 = 2.3026 log10 𝑦
Most scientific notation calculators contain a ′ ln 𝑥 ′ function which displays the value of the
Napierian (natural) logarithm of a number when the appropriate key is pressed.
Using a calculator, find the following natural logarithms.
ln 4.692 = 1.5458589 … = 1.5459, correct to 4 decimal places
ln 35.78 = 3.57738907 … = 3.5774 , correct to 4 decimal places
133
Use your calculator to check the following values:
ln 1.732 = 0.54928, correct to 5 significant figures
ln 1 = 0
ln 545 = 6.30079, correct to 6 significant figures
ln 1992 = 7.59689, correct to 6 significant figures
ln 1752 = 7.4685, correct to 5 significant figures, and 4 decimal places
ln 0.17 = −1.77195684 = −1.772, correct to 4 significant figures
ln 0.00032 = −8.04719, correct to 6 significant figures, and 5 decimal places
ln 𝑒 3 = 3
ln 𝑒 1 = 1
From the last two examples, we can conclude that
log 𝑒 𝑒 𝑥 = 𝑥 log 𝑒 𝑒 = 𝑥
This is useful when we are solving exponential functions.
For example, to solve
𝑒 3𝑥 = 8,
We will take Napierian logarithms of both sides, which gives
ln 𝑒 3𝑥 = ln 8
i.e.
3𝑥 ln 𝑒 = ln 8
3𝑥 = ln 8
1
1
𝑥 = 3 ln 8 = 3 × 2.079441542 = 0.6931
𝑥 = 0.6931, correct to 4 decimal places
134
Example 4.4.4.1.
Use a calculator to evaluate correct to 5 significant figures:
(a) ln 47.291
(b)
ln 0.06213
(c)
3.2 ln 762.923
Solution 4.4.4.1.
(a) ln 47.291 = 3.8563200 … = 3.8563, correct to 5 significant figures
(b) ln 0.06213 = −2.7785263 … = −2.7785, correct to 5 significant figures
(c) 3.2 ln 762.923 = 3.2(6.6371571 … ) = 21.239, correct to 5 significant figures
Example 4.4.4.2.
Solve the equation
7 = 4𝑒 −3𝑥 to find 𝑥, correct to 3 decimal figures.
Solution 4.4.4.2.
We begin by rearranging to bring like terms together.
We have
7
7 = 4𝑒 −3𝑥 gives = 𝑒 −3𝑥
4
7
We take the reciprocal of both sides of, 4 = 𝑒 −3𝑥 we get
4
7
=
1
𝑒 −3𝑥
= 𝑒 3𝑥
Now we can take the Napierian logarithms of both sides, and that gives us
4
ln ( ) = ln (𝑒 3𝑥 ) = 3𝑥 since ln𝑒 𝑒 𝑎 = 𝑎 ln 𝑒 = 𝑎 then
7
4
ln ( ) = 3𝑥
7
135
Hence
𝑥=
𝑥=
1
4
1
1
ln ( ) = ln(0.57142857) = (−0.5596157879)
3
7
3
3
1
4
1
ln ( ) = ln(0.57142857) = 0.333333(−0.5596157879)
3
7
3
𝑥 = −0.1865367306 = −0.18654, correct to 5 significant figures
Example 4.4.4.3.
𝑡
Given that 20 = 60 (1 − 𝑒 −2 ), determine the value of 𝑡, correct to 3 significant figures
Solution 4.4.4.3.
Once again, we will rearrange to bring like terms together as follows:
𝑡
20 = 60 (1 − 𝑒 −2 )
We rearrange them to obtain
𝑡
20
= (1 − 𝑒 −2 )
60
𝑡
𝑒 −2 = 1 −
20
1 2
= 1− =
60
3 3
𝑡
𝑒 −2 =
2
3
Taking the reciprocal of both sides, we get
𝑡
𝑒2 =
3
2
We take the Napierian logarithm of both sides to get,
136
𝑡
3
ln 𝑒 2 = ln ( )
2
𝑡
3
= ln ( )
2
2
3
𝑡 = 2 ln ( )
2
𝑡 = 2 ln(1.5)
𝑡 = 2 × 0.4054651 = 0.81093 = 0.811, correct to 3 significant figures
Example 4.4.4.4.
Solve the equation
3.72 = ln (
5.14
𝑥
) , to find the value of 𝑥
Solution 4.4.4.4.
From the definition logarithms, 3.72 = ln (
5.14
𝑥
) is equivalent to 𝑒 3.72 =
Rearranging gives:
𝑥=
5.14
= 5.14 × 𝑒 −3.72
𝑒 3.72
𝑥 = 5.14 × 𝑒 −3.72
𝑥 = 5.14 × 0.02423396
𝑥 = 0.1245625847 …
𝑥 = 0.1246, correct to 4 significant figures
137
5.14
𝑥
4.5.0. Session 5 - Common Logarithms with Base 10
We will now study logarithms with base 10.
The logarithm with base 10 is called the common logarithm and is denoted by omitting the base:
log 𝑥 = log10 𝑥
From the definition of logarithms, we can easily find that
log 10 = 1
and
log 100 = 2
How do we find log 50 ?
We need to find the exponent 𝑦 such that 10𝑦 = 50
Clearly, 1 is too small and 2 is too large. So
1 < log 50 < 2
To obtain a better approximation, we can experiment to find a power of 10 closer to 50.
Fortunately, scientific calculators are equipped with functions labelled as LOG key that can
directly give values of common logarithms.
Exercise 4.5.0.1.
Use a calculator to find the logarithms for the following numbers.
(a)
(b)
(c)
(d)
log 50 =
log 99 =
log 999 =
log 15000 =
What is your observation after you have read the logs of those values?
Did you observe that all the values are numbers greater zero?
138
4.5.1. Using the Laws of Logarithms to Evaluate Expressions
We have learned that logarithms are exponents. The Laws of Exponents give rise to the Laws of
Logarithms we studied in Session 4.1.0.
Example 4.5.1.1.
Evaluate each of the following expression.
(a) log 4 2 + log 4 32
(b) log 2 80 − log 2 5
1
(c) − ( ) log 8
3
Solution 4.5.1.1.
For convenience, we have recopied the Laws of Logarithms here for the exercises
(i)
log 𝑎 𝑚𝑛 = log 𝑎 𝑚 + log 𝑎 𝑛
(ii)
log 𝑎 ( 𝑛 ) = log 𝑎 𝑚 − log 𝑎 𝑛
𝑚
(iii)
log 𝑎 (𝑚𝑛 ) = 𝑛 log 𝑎 𝑚
(iv)
log b 𝑚 =
log𝑎 𝑚
log𝑎 𝑏
(a) We have been given log 4 2 + log 4 32, we can apply Law 1 as follows:
log 4 2 + log 4 32 = log 4(2 × 32)
log 4 2 + log 4 32 = log 4 (64)
log 4 2 + log 4 32 = log 4 (43 ) = 3 log 4 4 = 3, because 64 = 43 and log 4 4 = 1
log 4 2 + log 4 32 = log 4(64) = 3
(b) Given log 2 80 − log 2 5, we will apply Law 2
80
log 2 80 − log 2 5 = log 2 ( 5 ) = log 2 16 = log 2 24 = 4 log 2 2 = 4, because log 2 2 = 1
log 2 80 − log 2 5 = 4
139
1
(c) Given − (3) log 8, we apply Law 3
1
1
1
1
− (3) log 8 = log 8−3 = log(23 )−3 = log (2) ≡ −0.301
140
4.6.0. Session 6 - Expanding and Combining Logarithmic Expressions
The Laws of Logarithms allow us to write logarithm of a product or a quotient as the sum or
difference of logarithms. This process, called expanding a logarithmic expression, is our focus in
the next set of examples
4.6.1. Expanding Logarithmic Expressions
Example 4.6.1.1.
Use the Laws of Logarithms to expand each of the expression
(a) log 2 (6𝑥)
(b)
log 5 (𝑥 3 𝑦 6)
𝑎𝑏
ln ( 3 )
(c)
√𝑐
Solution 4.6.1.1.
(a) Given log 2 (6𝑥), we apply Law 1
log 2 6𝑥 = log 2 6 + log 2 𝑥
(b) Given log 5 (𝑥 3 𝑦 6 ), we apply Law 1 followed by Law 3
log 5 (𝑥 3 𝑦 6) = log 5 𝑥 3 + log 5 𝑦 6
Law 1
log 5 (𝑥 3 𝑦 6) = 3 log 5 𝑥 + 6 log 5 𝑦
Law 3
𝑎𝑏
(c) Given ln ( 3 ), we will apply Law 2, followed by Law 1 and Law 3
√𝑐
𝑎𝑏
ln ( 3 ) = ln(𝑎𝑏) − ln 3√𝑐,
√𝑐
= ln 𝑎 + ln 𝑏 − ln 3√𝑐,
Law 2
Law 1
1
= ln 𝑎 + ln 𝑏 − ln 𝑐 3,
1
= ln 𝑎 + ln 𝑏 − 3 ln 𝑐,
141
Law 1
Law 3
4.6.2. Combining Logarithmic Expressions
The Laws of Logarithms also allow us to reverse the process of expanding that was done in
Example 4.2.8.1. That means we can write sums and differences of logarithms as a single
logarithm. This process is called combining logarithmic expressions. We shall see how to
combine logarithms in the next set of examples
Example 4.6.2.1.
Combine into a single logarithm
1
3 log 𝑥 + log(𝑥 + 1)
2
Solution 4.6.2.1.
1
We are given 3 log 𝑥 + log(𝑥 + 1)
2
We will apply the different Laws of Logarithms for this example.
We begin by applying Law 3 to get the coefficients into exponents
1
1
3 log 𝑥 + 2 log(𝑥 + 1) = log 𝑥 3 + log(𝑥 + 1)2
Law 3
Next, we apply Law 1 to turn the addition to a product of the logarithms
1
1
3 log 𝑥 + 2 log(𝑥 + 1) = log( 𝑥 3 (𝑥 + 1)2 ) Law 1
1
log( 𝑥 3 (𝑥 + 1)2 )
Example 4.6.2.1.
1
Combine 3 ln 𝑠 + ln 𝑡 − 4 ln(𝑡 2 + 1) into a single logarithm
2
142
Solution 4.6.2.1.
We have been given
1
3 ln 𝑠 + ln 𝑡 − 4 ln(𝑡 2 + 1)
2
We proceed by applying Law 3 to take the coefficients to the exponents
1
1
3 ln 𝑠 + 2 ln 𝑡 − 4 ln(𝑡 2 + 1) = ln 𝑠 3 + ln 𝑡 2 − ln(𝑡 2 + 1)4
Law 3
Next, we apply Law 1 to convert the addition to a product
1
1
3 ln 𝑠 + 2 ln 𝑡 − 4 ln(𝑡 2 + 1) = ln ( 𝑠 3 𝑡 2 ) − ln(𝑡 2 + 1)4
Law 1
Next, we apply Law 2 to convert the difference to a quotient or fraction
1
1
2
3 ln 𝑠 + 2 ln 𝑡 − 4 ln(𝑡 + 1) =
1
𝑠3 𝑡 2
ln ((𝑡 2+1)4)
Law 1
𝑠3 √𝑡
3 ln 𝑠 + 2 ln 𝑡 − 4 ln(𝑡 2 + 1) = ln ((𝑡 2+1)4)
Law 2
Special Notes
We want to be cautious to avoid certain mistakes when we are dealing with sums and differences
of logarithms
We have learned that log 𝑎 (𝑥 + 𝑦) = log 𝑎 (𝑥𝑦), we want to avoid the error or temptation to write
143
log 𝑎 (𝑥 + 𝑦) ≠ log 𝑎 𝑥 + log 𝑎 𝑦
It is wrong
Also, we want to avoid the mistake
log 𝑎
log 𝑏
𝑎
≠ log (𝑏 )
This is wrong
(log 2 𝑥)3 ≠ 3 log 2 𝑥 This is also wrong
4.6.3. Change of Base Formula
For some purposes we find it useful to change from logarithm in one base to logarithms in
another base. Suppose we are given log 𝑎 𝑥 and we want to find log 𝑏 𝑥.
Let 𝑦 = log 𝑏 𝑥
We write this in exponential form and take the logarithm, with base 𝑎, of each side.
𝑏𝑦 = 𝑥
Exponential form
log 𝑎 (𝑏 𝑦 ) = log 𝑎 𝑥
Taking log 𝑎
𝑦 log 𝑎 𝑏 = log 𝑎 𝑥
Law 3
log 𝑥
𝑦 = log𝑎 𝑏
Divide by log 𝑎 𝑏
𝑎
This is a proof of the formula
log 𝑏 𝑥 =
log 𝑎 𝑥
log 𝑎 𝑏
If we put 𝑥 = 𝑎, then log 𝑎 𝑎 = 1, and the formular becomes
log 𝑏 𝑎 =
of each side
1
log 𝑎 𝑏
144
We can now evaluate a logarithm to any base by using the Change of Base Formula to express
the logarithm in terms of common logarithms or natural logarithms and instead of using a
calculator
4.6.4. Evaluating Logarithms with the Change of Base Formula
Example 4.6.4.1.
Use the Change of Base Formular and common or natural logarithms to evaluate each of the
following logarithms, correct to five decimal places.
(a) log 8 5
(b)
log 9 20
Solution 4.6.4.1.
(a) We will use the Change of Base Formula with
log 𝑥
log 𝑏 𝑥 = log𝑎 𝑏 with 𝑏 = 8 and 𝑎 = 10, because 10 is the common base
𝑎
log
5
log 8 5 = log10 8 =
10
0.69897
0.9030
= 0.77398
log 𝑥
(b) Again, we will use the Change of Base Formula log 𝑏 𝑥 = log𝑎 𝑏 with 𝑏 = 9 and 𝑎 = 𝑒
𝑎
log 9 20 =
ln 20 2.9957322
=
= 1.36342
ln 9
2.1972246
145
4.6.5. Logarithmic Equations
A logarithmic equation is one in which a logarithm of the variable occurs. An example is,
log 2 (𝑥 + 2) = 5
To solve for 𝑥, we write the equation in exponential form.
We write log 2 (𝑥 + 2) = 5 in exponential form
𝑥 + 2 = 25
We solve for 𝑥
𝑥 = 32 − 2 = 30
Alternatively, we raise the base, 2, to each side of the equation, as follows:
2log2(𝑥+2) = 25
This reduces to 𝑥 + 2 = 25, which is a property of logarithms
𝑥 + 2 = 25
We solve for 𝑥 to obtain
𝑥 = 32 − 2 = 30
𝑥 = 30
This is a typical metho for solving these types of logarithmic problems. We now provide a
summary of steps for solving logarithmic equations
4.6.6. Guidelines for Solving Logarithmic Equations
1.
Isolate the logarithmic term on one side of the equation; you might first need to combine
the logarithmic terms
2.
Write the equation in exponential form (or raise the base to each side of the equation)
3.
Solve for the variable
146
4.6.7. Solving Logarithmic Equations
Example 4.6.7.1.
Solve for 𝑥 in each of the following equations.
(a) ln 𝑥 = 8
(b)
log 2 (25 − 𝑥) = 3
Solution 4.6.7.1.
(a) We have been given
ln 𝑥 = 8
Write the equation in exponential form
𝑥 = 𝑒 8 ≡ 2981
∴ 𝑥 ≈ 2981.
Another way for solving this problem is this:
We have been given
ln 𝑥 = 8
We raise 𝑒 to both sides of the equation as follows:
𝑒 ln 𝑥 = 𝑒 8,
By a property of logarithms
𝑥 = 𝑒 8 ≈ 2981.
(b) Our first step is to rewrite the equation, log 2 (25 − 𝑥) = 3, in exponential form as
follows:
25 − 𝑥 = 23
25 − 𝑥 = 8
𝑥 = 25 − 8 = 17
147
Alternatively, we raise both sides of the equation to the power of the base, 2
2log2(25−𝑥) = 23
This implies that
25 − 𝑥 = 23
𝑥 = 25 − 8 = 17
Example 4.6.7.2.
Solve the equation
4 + 3 log(2𝑥) = 16
Solution 4.6.7.2.
We first isolate the logarithmic term. This will allow us to write the equation in exponential form
We have been given
4 + 3 log(2𝑥) = 16
Subtract 4 from each side
4 − 4 + 3 log(2𝑥) = 16 − 4
3 log(2𝑥) = 12
Divide each side by 4
log(2𝑥 ) = 4
Write in exponential form or raise each side by 10
10log(2𝑥) = 104
2𝑥 = 104
148
𝑥 = 5000
As a check, put 𝑥 = 5000 in the original equation as follows:
4 + 3 log(2𝑥) = 4 + 3 log (2 × 5000)
4 + 3 log(2𝑥) = 4 + 3 log (10000)
4 + 3 log(2𝑥) = 4 + 3 (4) = 4 + 12 = 16
4 + 3 log(2𝑥) = 4 + 12 = 16
4.6.8. Solving Logarithmic Equation Algebraically
Example 4.6.8.1.
Solve the equation log(𝑥 + 2) + log(𝑥 − 1) = 1 algebraically.
Solution 4.6.8.1.
We begin by using the Laws of Logarithms to combine the logarithmic terms as follows:
We have been given
log(𝑥 + 2) + log(𝑥 − 1) = 1
Combine using Law 1
log[(𝑥 + 2)(𝑥 − 1)] = 1
Express in exponential form (or raise both sides to the power of 10)
(𝑥 + 2)(𝑥 − 1) = 101 = 10
Expand the left-hand side
𝑥 2 + 𝑥 − 2 = 10
149
Subtract 10 from both sides
𝑥 2 + 𝑥 − 2 − 10 = 10 − 10
𝑥 2 + 𝑥 − 12 = 0
Factorise
(𝑥 + 4)(𝑥 − 3) = 0
𝑥 = −4
or
𝑥=3
A check from the original equation shows that 𝑥 = −4 is not a solution. It confirms that
Logarithms are not defined for negative numbers. They are defined for positive numbers.
Therefore, 𝑥 = −4 is not a solution, but 𝑥 = 3 is a solution.
150
Unit 5. The Computer Numbering System
Session 1.
The Computer Numbering System
Session 2.
Conversion from Decimal to Binary
Session 3.
Conversion from Decimal to Binary Via Octal
Session 4.
Hexadecimal Numbers
Session 5.
Conversion from Decimal to Hexadecimal
Session 6.
Conversion from Binary to Hexadecimal
151
5.1.0. Session 1. -
The Computer Numbering System
5.1.1 Binary Numbers
The system of numbers in everyday use is the denary or decimal system of numbers, using the
digits 0 𝑡𝑜 9. It has ten different digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9). This system of numbering is
said to have a radix or base of 10.
The binary system of numbering has a radix of 2 and uses only the digits 0 and 1
5.1.2. Conversion from Binary to Decimal
The decimal number 234.5 is equivalent to
(2 × 102) + (3 × 101 ) + (4 × 100 ) + (5 × 10−1 )
i.e. is the sum of term comprising: (a digit) multiplied by (the base raised to some power)
In the binary system of numbers, the base is 2, so 1101.1 is equivalent to the sum of the products
of the given binary number and the appropriate powers of 2 as in the following order:
The powers of 2 begin from 0 on the last digit on the right, and increases towards the left as in
the following order:
1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 + 1 × 2−1
Thus, the decimal number equivalent to the binary number 1101.1 is
1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 + 1 × 2−1
8+4+0+1+
13.5
152
1
2
i.e. 1101.12 = 13.510, the suffixes 2 and 10 denoting binary and decimal systems of number
respectively
Example 5.1.2.1
Convert 110112 to a decimal number
Solution 5.1.2.1.
We have been given 110112. Here again, we multiply each digit by the appropriate powers of 2
starting from the power 0 on the last digit and increases towards the first digit on the leftmost
side of the given binary number
This is equivalent to
110112 = 1 × 24 + 1 × 23 × 0 × 22 + 1 × 21 + 1 × 20
= 16 + 8 + 0 + 2 + 1
= 2710
Example 5.1.2.2.
Convert 0.10112 to a decimal fraction
Solution 5.1.2.2.
We have been given 0.10112, we follow the same method to multiply by the appropriate powers
of 2, but note that the powers of 2 are negative numbers because of the fraction, the number is a
number after the decimal point
0.10112 = 1 × 2−1 + 0 × 2−2 + 1 × 2−3 + 1 × 2−4
153
We simplify further to obtain
=1×
1
1
1
1
+
0
×
+
1
×
+
1
×
21
22
23
24
=
1
1 1
+0+ +
2
8 16
= 0.5 + 0 + 0.125 + 0.0625
= 0.687510
Example 5.1.2.3
Convert 101.01012 to a decimal number
Solution 5.1.2.3.
Here we will follow the same procedure by multiplying each digit by the appropriate powers of 2
and note the decimal part of the given binary number
101.01012 = 1 × 22 + 0 × 21 + 1 × 20 + 0 × 2−1 + 1 × 2−2 + 0 × 2−3 + 1 × 2−4
We simplify to obtain their values as
= 4+0+1+0+1×
1
1
+
0
+
1
×
22
24
1
1
=5+ +0+
4
16
= 5 + 0.25 + 0 + 0.0625
154
= 5.312510
Exercises
1. Convert the following binary numbers to decimal numbers
(a) 1110
(b)
1001
(c)
1010
2. Convert the following binary numbers to the decimal numbers
(a) 0.1101
(b)
0.11001
(c)
0.01011
155
5.2.0. Session 2 - Conversion from Decimal to Binary
Any integer decimal can be converted to a corresponding binary number by repeatedly dividing
the number by 2 and noting the remainder at each stage until there is no more a remainder.
The result is obtained by arranging the remainders or writing remainders from the last remainder
as the first or most significant bit, (a bit is a binary digit and the least significant bit is the one on
the right) in that order until the first remainder becomes the last bit or the least significant bit.
The last remainder is the most significant bit, i.e., the bit on the left
Let us use the decimal number 39 as an example to convert to binary number
Example 5.2.0.1.
Convert 3910 to a binary number
Solution 5.2.0.1.
Divisor
2
2
2
2
2
2
Dividend
39
19
9
4
2
1
0
Remainder
1
1
1
0
0
1
1001112
We will arrange the remainders from the bottom to the topmost as follows:
(most significant bit → 1 0 0 1 1 1  least significant bit)
Thus 3910 = 1001112
The fractional part of a decimal number can be converted to a binary number by repeatedly
multiplying by 2, as we demonstrate with 0.625 in the example below
156
Example 5.2.0.2.
Convert 0.62510 to binary
Solution 5.2.0.2.
The Multiplication
0.625 × 2
0.250 × 2
0.500 × 2
=
=
=
The Product The Significant bits
1.250
1
0.500
0
1.000
1
(The most significant bit) . 101 (least significant bit) arranged from the first (the top bit) to the
last (the bottom bit)
For fractions, the most significant bit of the result is the top bit obtained from the integer part of
the multiplication by 2. The least significant bit of the result is the bottom bit obtained from the
integer part of the multiplication by 2
Therefore, 0.62510 = 0.1012
Example 5.2.0.3.
Convert 4710 to a binary number
Solution 5.2.0.3.
We follow the same method by repeatedly dividing by 2 and noting the remainder
Divisor
2
2
2
2
2
2
Dividend
47
23
11
5
2
1
0
Remainder
1
1
1
1
0
1
1011112
Thus 4710 = 1011112
157
Example 5.2.0.4.
Convert 0.4062510 to a binary number
Solution 5.2.0.4.
The Multiplication
0.40625 × 2
0.81250 × 2
0.62500 × 2
0.25000 × 2
0.50000 × 2
=
=
=
=
=
The Product The Significant bits
0.81250
0
1.62500
1
1.25000
1
0.50000
0
1.00000
1
0.011012
Therefore, 0.4062510 = 0.011012
Example 5.2.0.5.
Convert 58.312510 to a binary number
Solution 5.2.0.5.
The integer part is repeatedly divided by 2, giving
Divisor
2
2
2
2
2
2
Dividend
58
29
14
7
3
1
0
Remainder
0
1
0
1
1
1
1110102
1110102
158
We will follow it with repeated multiplication by 2 for the fractional part
The Multiplier
2
2
2
2
The Multiplication
0.3125 × 2
0.6250 × 2
0.2500 × 2
0.500 × 2
=
=
=
=
The Product The Significant bits
0.6250
0
1.2500
1
0.5000
0
1.0000
1
0.01012
Therefore, 58.312510 = 111010.01012
Exercises
1. Convert the following decimal numbers to binary numbers
(a) 5
(b)
15
(c)
19
2. Convert the following decimal numbers to binary numbers
(a) 0.25
(b)
0.21875
(c)
0.28125
159
5.3.0. Session 3. - Conversion from Decimal to Binary Via Octal
For decimal integers containing several digits, repeatedly dividing by 2 can be a lengthy process.
In this case, it is usually easier to convert a decimal number to a binary number via the octal
system of numbers. This system has a radix of 8, using the digits 0, 1, 2, 3, 4, 5, 6, 7.
Example 5.3.0.1
The denary number equivalent to the octal number 43178 is
Solution 5.3.0.1.
43178 = 4 × 83 + 3 × 82 + 1 × 81 + 7 × 80
43178 = 4 × 512 + 3 × 64 + 1 × 8 + 7 × 1
43178 = 2048 + 192 + 8 + 7
43178 = 225510
Any integer decimal number can be converted to a corresponding octal number by repeatedly
dividing by 8 and noting the remainder at each stage, as shown below for the number 49310
Example 5.3.0.2.
Convert 49310 to octal number
Solution 5.3.0.2.
Divisor
8
8
8
Dividend
493
61
7
0
Remainder
5
5
7
160
Therefore, 49310 = 7558
The fractional part of a decimal number can be converted to an octal number by repeatedly
multiplying by 8, as demonstrated below for the fraction 0.437510
Example 5.3.0.3.
Convert the decimal 0.437510 to an octal
Solution 5.3.0.3.
The Multiplier
8
8
The Multiplication
0.4375 × 8
0.5000 × 8
=
=
The Product
3.5000
4.0000
The Significant bits
3
4
0.348
For fractions, the most significant bit is the top integer obtained by multiplication of the decimal
fraction by 8, thus
0.437510 = 0.348
161
5.3.1. The Natural Binary Code
The natural binary code for digits 0 to 7 is shown in the Table 5.3.1. below, and an octal number
can be converted to a binary number by writing down the three bits corresponding to the octal
Octal Digit
0
1
2
3
4
5
6
7
Natural Binary Number
000
001
010
011
100
101
110
111
Table 5.3.1.
The 0 on the extreme left does not signify anything, thus 26.358 = 10 110.011 1012
We can omit the leading zero in the numbering just like we did in the previous example
26.358 = 10 110.011 1012
Conversion of decimal to binary via octal is demonstrated in the following worked problems
Example 5.3.1.1.
Convert 371410 to a binary via octal
Solution 5.3.1.1
Divisor
8
8
8
8
Dividend
3714
464
58
7
0
Remainder
2
0
2
7
72028
The octal number for 371410 = 72028
162
From the Table 5.3.1.
72028 = 111 010 000 0102
i.e.
371410 = 111 010 000 0102
Example 5.3.1.2.
Convert 0.5937510 to a binary number, via octal
Solution 5.3.1.2.
We will perform a repeated multiplication by 8 and note the integer values.
The Fraction
0.59375
0.75
The Multiplier, 8
8
8
The Result
4.75000
6.00
The Integer Part
4
6
The Octal Result
0.468
0.5937510 = 0.468
From the Table 5.3.1., 0.468 = 0.100 1102, i.e., 0.5937510 = 0.100 1102
Example 5.3.1.2.
Convert 5613.9062510 to a binary number via octal.
Solution 5.3.1.2.
Remember. The conversion is done such that the integer part is repeated divided by8, noting the
remainder, and the fraction is repeatedly multiplied by 8 and noting the integer part
We begin with the integer division:
The Divisor, 8
8
8
8
The Dividend, 5613
5613
701
87
The Remainder
5
5
163
The Octal Result
8
8
10
1
7
2
1
127558
The octal number is converted to a binary by looking up the individual values from Table 5.1.4.
From the Table 5.3.1.
127558 = 001 010 111 101 1012
561310 = 1 010 111 101 1012
We now convert the fraction part by performing the repeated multiplication by 8, and noting the
integer part
The Fraction
0.90625
0.25
The Multiplier, 8
8
8
The Result
7.25
2.0
The Integer Part
7
2
The Octal Result
0.728
From Table 5.3.1.
0.728 = 0.111 0102
0.9062510 = 0.111 012
Finally, we combine the integer and the fractional part to obtain the complete converted number
as follows:
5613.9062510 = 1 010 111 101 101.111 012
Example 5.3.1.3.
Convert 11 110 011.100 012 to a decimal number via octal
164
Solution 5.3.1.3.
For convenience, we reproduce Table 5.1.4. here to do the conversion
Octal Digit
0
1
2
3
4
5
6
7
Natural Binary Number
000
001
010
011
100
101
110
111
Table 5.3.1.
Grouping the binary number in threes from the binary point gives us:
011 110 011.100 0102
Using Table 5.3.1. to convert this binary number to an octal number will give us
363.428
Now to covert from the octal 363.428 to decimal, we multiply the octal digits by their
corresponding powers of 8 as follows:
363.428 = 3 × 82 + 6 × 81 + 3 × 80 + 4 × 8−1 + 2 × 8−2
363.428 = 192 + 48 + 3 +
1 2
+
2 64
363.428 = 192 + 48 + 3 +
1 2
+
2 64
363.428 = 243 + 0.5 + 0.03125
165
363.428 = 243.5312510
166
5.4.0. Session 4 - Hexadecimal Numbers
The complexity of computers requires higher order numbering system such as octal (base 8) and
hexadecimal (base 16), which are merely extensions of the binary system.
5.4.1. The Hexadecimal Numbering System
A hexadecimal numbering system has a radix of 16 and uses the following distinct digits:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝑎𝑛𝑑 𝐹
′𝐴′ corresponds to 10 in the denary system, 𝐵 to 11, 𝐶 to 12, 𝐷 to 13, 𝐸 to 14, and 𝐹 to 15.
5.4.2. Conversion from Hexadecimal to Decimal
To convert from hexadecimal to decimal, we multiply the hexadecimal by the powers of base 16
and find the sum to obtain the corresponding decimal numbers
Example 5.4.2.1.
Convert 1𝐴16 to decimal
Solution 5.4.2.1.
We have been given a hexadecimal 1𝐴16
We perform the base 16 multiplication as follows:
1𝐴16 = 1 × 161 + 𝐴 × 160
1𝐴16 = 1 × 16 + 10 × 1
1𝐴16 = 16 + 10 = 2610
∴ 1𝐴16 = 2610
167
Example 5.4.2.2.
Convert 2𝐸16 to a decimal number
Solution 5.4.2.2
We have been given 2𝐸16
Therefore,
2𝐸16 = 2 × 161 + 𝐸 × 160
2𝐸16 = 2 × 16 + 14 × 1
2𝐸16 = 32 + 14 = 4610
Example 5.4.2.3.
Convert 1𝐵𝐹16 to a decimal
Solution 5.4.2.3.
1𝐵𝐹16 = 1 × 162 + 𝐵 × 161 + 𝐹 × 160
1𝐵𝐹16 = 256 + 11 × 16 + 15 × 1
1𝐵𝐹16 = 256 + 176 + 15
1𝐵𝐹16 = 44710
168
Example 5.4.2.4.
Convert the following hexadecimals into their equivalent decimal numbers:
(a) 7𝐴16
(b)
3𝐹16
Solution 5.4.2.4.
(a) 7𝐴16 = 7 × 161 + 𝐴 × 160
7𝐴16 = (7 × 16) + (10 × 1) = 112 + 10 = 12210
∴ 7𝐴16 = 12210
(b) 3𝐹16 = 3 × 161 + 𝐹 × 160
3𝐹16 = (3 × 16) + (15 × 1)
3𝐹16 = (48) + (15) = 6310
∴ 3𝐹16 = 6310
Example 5.4.2.5.
Convert the following hexadecimals into their decimal equivalents:
(a) 𝐶916
(b)
𝐵𝐷16
Solution 5.4.2.5.
(a) 𝐶916 = 𝐶 × 161 + 9 × 160
𝐶916 = (12 × 16) + (9 × 1)
𝐶916 = 192 + 9 = 20110
169
Thus
𝐶916 = 20110
(b) 𝐵𝐷16 = 𝐵 × 161 + 𝐷 × 160
𝐵𝐷16 = (11 × 16) + (13 × 1)
𝐵𝐷16 = 176 + 13 = 18910
∴ 𝐵𝐷16 = 18910
170
5.4.3. The Comparison Table
Table 5.4.3. compares decimal, binary, octal and hexadecimal numbers and shows, for example,
that 2310 = 101112 = 278 = 1716
Table 5.4.3.
Decimal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
Binary
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
100000
171
Octal
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
22
23
24
25
26
27
30
31
32
33
34
35
36
37
40
Hexadecimal
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
20
Example 5.4.3.1.
Convert 1𝐴4𝐸16 into denary number.
Solution 5.4.3.1.
We proceed as usual with the multiplication of the hexadecimal number by the powers of base
16 and find the sum as follows:
1𝐴4𝐸16 = (1 × 163 ) + (𝐴 × 162) + (4 × 161) + (𝐸 × 160 )
1𝐴4𝐸16 = (1 × 4096) + (10 × 256) + (4 × 16) + (14 × 1)
1𝐴4𝐸16 = (1 × 4096) + (10 × 256) + (4 × 16) + (14 × 1)
1𝐴4𝐸16 = (4096) + (2560) + (64) + (14)
1𝐴4𝐸16 = 673410
172
5.5.0. Session 5.- Conversion from Decimal to Hexadecimal
Conversion from decimal to hexadecimal is achieved by repeatedly dividing by base 16 and
noting the remainder at each stage, as we will demonstrate in the following example.
Example 5.5.0.1.
Convert 2610 to a hexadecimal
Solution 5.5.0.1.
The Divisor, 16
16
16
The Dividend, 26
26
1
0
The Remainder
10
1
The Hexadecimal Result
1 ≡ 𝐴16
1 ≡ 116
2610 = 1𝐴16
Example 5.5.0.2.
Convert 44710 to a hexadecimal
Solution 5.5.0.2.
The Divisor, 16
16
16
16
The Dividend, 447
447
27
1
0
The Remainder
15
11
1
The Hexadecimal Result
15 ≡ 𝐹16
11 ≡ 𝐵16
1 ≡ 116
44710 = 1𝐵𝐹16
Example 5.5.0.3.
Convert the following decimal numbers into their hexadecimal equivalents:
(a) 3710
(b)
10810
173
Solution 5.5.0.3.
(a)
The Divisor, 16
16
16
The Dividend, 37
37
2
0
The Remainder
5
2
The Hexadecimal Result
5 ≡ 516
2 ≡ 216
3710 = 2516
(b)
The Divisor, 16
16
16
The Dividend, 108
108
6
0
The Remainder
12
6
The Hexadecimal Result
12 ≡ 𝐶16
6 ≡ 616
10810 = 6𝐶16
Example 5.5.0.4.
Convert the following decimal numbers into their hexadecimal equivalents:
(a) 16210
(b)
23910
Solution 5.5.0.4.
(a)
The Divisor, 16
16
16
The Dividend, 162
162
10
0
The Remainder
2
10
The Hexadecimal Result
2 ≡ 216
10 ≡ 𝐴16
16210 = 𝐴216
174
Solution 5.5.0.4. continued
(b)
The Divisor, 16
16
16
The Dividend, 239
239
14
0
The Remainder
15
14
The Hexadecimal Result
15 ≡ 𝐹16
14 ≡ 𝐸16
23910 = 𝐸𝐹16
175
5.6.0. Session 6 - Conversion from Binary to Hexadecimal
To convert from binary number to hexadecimals, the binary bits are arranged in groups of four,
starting from the right to the left. A hexadecimal symbol is assigned to each group of four binary
numbers.
As an example, consider the binary number 1110011110101001.
We will group them into fours as
1110 0111 1010
1001
From Table 5.4.3., the hexadecimals are,
𝐸
7
𝐴
176
9
For convenience, we reproduce the Table 5.4.3 here:
Decimal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
Binary
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
100000
Octal
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
22
23
24
25
26
27
30
31
32
33
34
35
36
37
40
Hexadecimal
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
20
Hence 11100111101010012 = 𝐸7𝐴916
Alternatively, we can multiply each group of 4 binary digits by the powers of 2 as in the
following demonstration
177
For the binary number 11100111101010012 we can group it into fours as follows:
10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 8 + 0 + 0 + 1 = 916
10102 = 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20 = 8 + 0 + 2 + 0 = 1016 = 𝐴16
01112 = 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20 = 0 + 4 + 2 + 1 = 716
11102 = 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 = 8 + 4 + 2 + 0 = 1416 = 𝐸16
Hence 11100111101010012 = 𝐸7𝐴916
178
5.6.1. Conversion from Hexadecimal to Binary Numbers
The conversion of hexadecimal to binary is the reverse of the conversion from binary to
hexadecimal. For any given hexadecimal, we look for the four binary numbers that represents
each hexadecimal
Example 5.6.1.1.
The hexadecimal 6𝐶𝐹316 = 0110 1100 1111 00112 from Table 5.4.3.
Example 5.6.1.2.
Convert the following binary numbers into their hexadecimal equivalents:
(a) 110101102
(b)
11001112
Solution 5.6.1.2.
(a) Grouping bits in fours from the right gives us
hexadecimal symbols to each group gives us
Therefore, 110101102 = 𝐷616
1101 0110 and assigning
D
6
(b) Also, grouping bits in fours from right to left gives us
hexadecimal symbols to each group gives us:
0110 0111 and assigning
6
7
Thus, 1100 1112 = 6716
Example 5.6.1.3.
Convert the following binary numbers into their hexadecimal equivalents:
(a) 110011112
(b)
1100111102
Solution 5.6.1.3.
(a) Grouping bits in fours from the right to the left gives us 1100 1111 and assigning
hexadecimal symbols to each group results in
𝐶 𝐹 from Table 5.4.3.
Therefore, 110011112 = 𝐶𝐹16
179
(b) Grouping bits in fours from the right gives us:
assigning hexadecimal symbols to each group gives:
as read from the Table 5.4.3.
Therefore, 1100111102 = 19𝐸16
0001 1001 1110 and
1
9
𝐸
Example 5.6.1.4.
Convert the following hexadecimal numbers into their binary equivalents:
(a) 3𝐹16
(b)
𝐴616
Solution 5.6.1.4.
(a) We space out the given hexadecimal digits to get:
each into binary to get (as read from Table 5.4.3.)
Therefore, 3𝐹16 = 1111112
3
0011
(b) Again, we space out hexadecimal digits to give us
each into binary to give us:
Table 5.4.3. Therefore, 𝐴616 = 101001102
𝐴
1010
𝐹 and convert
1111
6 we convert
0110 as read from
Example 5.6.1.5.
Convert the following hexadecimal numbers into their binary equivalents:
(a)
7𝐵16
(b)
17𝐷16
Solution 5.6.1.5.
(a) Spacing out the hexadecimal digits, we get
each into a binary number will give us:
from the Table 5.4.3.
Therefore, 7𝐵16 = 11110112
7
(b) Spacing out the hexadecimal digits again, we get
converting each digit into a binary gives us
from the Table 5.4.3.
Therefore, 17𝐷16 = 1011111012
1
7
𝐷
and
0001 0111 1101 as read
180
𝐵
and converting
0111 1011 as read
Learners, we have come to the end of this Unit. We have learned the various types of numbers
that drive computers to perform all the task and complex computations they perform. We learned
about the differences between the decimal numbers and the binary numbers, and how to covert
from decimal to binary and from binary to decimal numbers. We learned about octal numbers,
the base 8 numbers and how to use them to convert binary numbers to the decimal or denary
numbers, especially when the binary numbers become longer. We learned about hexadecimals,
the base 16 numbers. We learned how to convert from hexadecimal to decimal numbers, and
from decimal numbers to hexadecimal numbers. We learned how to convert binary numbers to
hexadecimal numbers and from binary to hexadecimal numbers. We used tables that readily give
us the values of the various types of numbers.
In the next unit and sessions, we will learn ….
Here are some exercises for you to practice and reinforce your understanding of the subjects we
have just completed
181
Unit 6
Unit 6. Trigonometry
Session 1.
The Pythagoras Theorem and Trigonometric Ratios
Session 2.
Solving Right-Angled Triangles
Session 3.
Cartesian and Polar Coordinates
Session 4.
Solving Triangles - The Sine and Cosine Rules
Session 5.
Trigonometric Identities and Equations
Session 6.
Compound Angles
182
6.1.0. Session 1 - The Pythagoras Theorem and Trigonometric Ratios
Trigonometry is the branch of mathematics that deals with the measurements of sides and angles
of triangles, and their relationship with each other.
The word trigonometry is a combination of the Greek word “trigono” which means triangle and
“metro” means to measure. It was the Greeks who laid the foundation of trigonometry. However,
the sine of an angle was developed by Aryabhata, a Hindu.
Trigonometry is one of the most widely used mathematical techniques. There are many useful
applications in engineering where knowledge of trigonometry is needed. It has applications in
other fields including surveying, navigation, GPS in cars, radio waves, digital signal processing,
etc.
We will learn to find the sines, cosines, and tangents of angles. We will learn to solve triangles,
calculate inverse trigonometric functions, and find cosecants, secants, and cotangents of angles.
6.1.1. The Theorem of Pythagoras
In addition to knowing the trigonometric ratios, it is important to know the Pythagoras theorem
about the relationship between the lengths of a right-angled triangle.
For any right-angled triangle, the Pythagoras theorem says, ‘In any right-angle triangle, the
square on the hypotenuse is equal to the sum of the squares on the other two sides.’
𝑃
ℎ𝑦𝑝2 = 𝑎𝑑𝑗 2 + 𝑜𝑝𝑝2
It is very important to remember the Pythagoras theorem, because it is an important part of this
session of the course. We can find the actual value for ℎ𝑦𝑝, by taking the square root of both
sides of the Pythagoras formula as follows.
ℎ𝑦𝑝 = √𝑎𝑑𝑗 2 + 𝑜𝑝𝑝2
183
We can use the Pythagoras formula to find any of the sides, 𝑎𝑑𝑗, or the 𝑜𝑝𝑝 of a right-angled
triangle as needed. It is a matter of transposing the Pythagoras formula to find the 𝑜𝑝𝑝 or the 𝑎𝑑𝑗
lengths when they are unknown.
Let us try some examples.
Example 6.1.1.1.
Two aircrafts leave an airfield at the same time. One travels due north at an average speed of
300km/h and the other travels due west at an average speed of 220km/h. calculate their distance
apart after 4 hours.
Solution 6.1.1.1.
B
N
4 x 300 = 1200km
W
E
C
S
A
4 x 220 = 880km
Figure 6.1.1.1.a
Figure 6.1.1.1.b
After 4 hours, the aircraft due north has travelled 4 × 300 = 1200 km, due north, and the
aircraft due west has travelled 4 × 220 = 880km due west, as shown in Figure 6.1.2.3.b.
The distance apart, represented in Figure 6.1.1.1.b is 𝐵𝐶.
Using the Pythagoras theorem, we have
𝐵𝐶 2 = 8802 + 12002
𝐵𝐶 2 = 774,400 + 1,440,000
184
𝐵𝐶 2 = 2,214,400
𝐵𝐶 = √2,214,400
𝐵𝐶 = 1,488.0860km
Hence distance apart after 4 hours = 1488km
6.1.2. Review of Trigonometric Ratios
C
c𝛽
Hypotenuse
Opposite
𝜃
B
c
A
Adjacent
Figure 6.1.2a
Consider the right-angled triangle 𝐴𝐵𝐶 (Figure 6.1.2a) above with an angle 𝜃 (theta) at 𝐴 (it is
common to use 𝜃 to denote any ‘angle’.
A right-angled triangle is a triangle that has one angle equal to 90𝑜 and this is represented by a
square, as shown in Figure 6.1.2a.
We define,
adjacent = side 𝐴𝐵, adjacent to the angle 𝜃
opposite = side 𝐵𝐶, opposite to the angle 𝜃
hypotenuse = side 𝐴𝐶, the longest side of the right-angled triangle
185
We define 𝑠𝑖𝑛 (short for sine), 𝑐𝑜𝑠 (short for cosine) and 𝑡𝑎𝑛 (short for tangent) as follows:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
(6.1.2.1.)
sin(𝜃) =
(6.1.2.2.)
cos(𝜃) = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
(6.1.2.3.)
tan(𝜃) = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
A well-known mnemonic is ‘SOHCAHTOA’ can be used to remember these common ratios.
The above ratios are called the trigonometric ratios and as each of 𝑠𝑖𝑛, 𝑐𝑜𝑠 and 𝑡𝑎𝑛 are functions
of the angle 𝜃, they are generally referred to as the trigonometric functions.
Let 𝑜𝑝𝑝 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒, 𝑎𝑑𝑗 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 and ℎ𝑦𝑝 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒, then by using Figure 4.1.0a, we
can rewrite the trigonometric functions in terms of 𝑜𝑝𝑝, 𝑎𝑑𝑗 and ℎ𝑦𝑝 (Figure 4.1.0b).
C
opp
hyp
𝜃
B
A
adj
Figure 6.1.2b
𝑜𝑝𝑝
(6.1.2.4.)
sin(𝜃) = ℎ𝑦𝑝
(6.1.2.5.)
cos(𝜃) = ℎ𝑦𝑝
(6.1.2.6.)
tan(𝜃) = 𝑎𝑑𝑗
𝑎𝑑𝑗
𝑜𝑝𝑝
We can use the above ratios to find any unknown angle or length of a side of any right-angled
triangle. Note that the triangle must be a right-angled triangle.
186
6.1.3. Fractional and Surd Forms of Trigonometric Ratios
A quantity that is not exactly expressible as a rational number is called a surd (radical). For
example, √2 and √3 are called surds because they cannot be expressed as fraction and a decimal
part may be continued indefinitely.
For example, √2 = 1.4142135 … and √3 = 1.7320508 …
Example 6.1.3.1.
(i)
(ii)
Using the triangle in Figure 6.1.3.1.a, find the length of 𝐴𝐶.
Without using calculator, evaluate sin(45𝑜 ), cos(45𝑜 ) and tan (45𝑜 ).
C
1
B
45𝑜
1
A
Figure 6.1.3.1
Solution 6.1.3.1.
(i)
How can we calculate the length of 𝐴𝐶? We have been given the length of the
adjacent side, i.e., 𝑎𝑑𝑗 = 1 and the opposite side, 𝑜𝑝𝑝 = 1. We can apply the
Pythagoras’ formula as follows:
𝐴𝐶 2 = 𝑜𝑝𝑝2 + 𝑎𝑑𝑗 2
𝐴𝐶 2 = 12 + 12 = 1 + 1 = 2
We take the square root of both sides.
𝐴𝐶 = √2
The hypotenuse, 𝐴𝐶 = √2.
(ii)
We know all the sides now. With ℎ𝑦𝑝 = √2, 𝑎𝑑𝑗 = 1, and 𝑜𝑝𝑝 = 1. We can use the
trigonometric ratios to find sin(45𝑜 ) , cos(45𝑜 ) and tan(45𝑜 ) without using a
calculator.
187
sin(45𝑜 ) =
𝑜𝑝𝑝
1
√2
=
=
ℎ𝑦𝑝 √2
2
cos(45𝑜 ) =
𝑎𝑑𝑗
1
√2
=
=
ℎ𝑦𝑝 √2
2
tan(45𝑜 ) =
𝑜𝑝𝑝 1
= =1
𝑎𝑑𝑗 1
Example 6.1.3.2.
Without using calculator, find the values of
(i)
(ii)
sin(60𝑜 ), cos (60𝑜 ) and tan(60𝑜 )
sin (30𝑜 ), cos (30𝑜 ) and tan (30𝑜 )
C
30𝑜
2
𝑥
60𝑜
B
1
A
Figure 6.1.3.2
Solution 6.1.3.2.
(i)
Considering the case for angle 60𝑜 , we have opposite 𝑜𝑝𝑝 = 𝑥, adjacent 𝑎𝑑𝑗 = 1,
hypotenuse ℎ𝑦𝑝 = 2
𝑥
a. sin(60) = , but we don’t know the value of 𝑥. We will use the Pythagoras’
2
formula to find the value of 𝑥. The Pythagoras’ formula says,
ℎ𝑦𝑝2 = 𝑜𝑝𝑝2 + 𝑎𝑑𝑗 2 .
Substituting the values in the Pythagoras’ formula, we have:
22 = 𝑥 2 + 12
188
4 = 𝑥2 + 1
Which gives
𝑥2 = 4 − 1 = 3
𝑥2 = 3
⇒ 𝑥 = √3
With the value of the 𝑜𝑝𝑝 = 𝑥 = √3, 𝑎𝑑𝑗 = 1, ℎ𝑦𝑝 = 2 we can use the appropriate values in
the trigonometric ratios for 𝜃 = 60𝑜 as follows:
(a) sin(60𝑜 ) =
(ii)
√3
,
2
(b)
1
cos(60𝑜 ) = 2
and
(c)
tan(60) =
√3
1
= √3
Considering the case for angle 𝜃 = 30𝑜 , we have
𝑜𝑝𝑝 = 1, 𝑎𝑑𝑗 = √3 and ℎ𝑦𝑝 = 2
𝑜𝑝𝑝 1
=
ℎ𝑦𝑝 2
𝑎𝑑𝑗 √3
cos(30𝑜 ) =
=
ℎ𝑦𝑝
2
sin(30𝑜 ) =
tan(30𝑜 ) =
𝑜𝑝𝑝
1
√3
=
=
𝑎𝑑𝑗 √3
3
The values we obtained in the two examples are the exact values for the corresponding
trigonometric ratios. It will be good to keep them in mind (in our memory).
We note from the above that
sin 30𝑜 = cos 60𝑜 ,
sin 45𝑜 = cos 45𝑜
and
sin 60𝑜 = cos 30𝑜
In general,
sin 𝜃 = cos (90𝑜 − 𝜃)
cos 90𝑜 = sin (90𝑜 − 𝜃)
and
For example, it can be checked by calculator that sin 25𝑜 = cos 65𝑜, sin 42𝑜 = cos 42𝑜 and
cos 84𝑜 10′ = sin 5𝑜 50′, and so on.
189
Example 6.1.3.3.
Using surd forms, evaluate
3 tan 60𝑜 − 2 cos 30𝑜
tan 30𝑜
Solution 6.1.3.3.
From triangle 6.1.3.2., we established that tan 60 = √3, cos 30 =
√3
,
2
Therefore, we can substitute in
√3
3(√3) − 2 ( 2 )
3 tan 60 − 2 cos 30
=
1
tan 30𝑜
√3
𝑜
𝑜
3 tan 60𝑜 − 2 cos 30𝑜
=
tan 30𝑜
√3
3(√3) − 2 ( 2 )
1
√3
3 tan 60𝑜 − 2 cos 30𝑜 3√3 − √3
=
1
tan 30𝑜
√3
3 tan 60𝑜 − 2 cos 30𝑜 2√3
=
1
tan 30𝑜
√3
3 tan 60𝑜 − 2 cos 30𝑜
√3
= 2√3 ( )
𝑜
tan 30
1
190
tan 30 =
1
√3
3 tan 60𝑜 − 2 cos 30𝑜
= 2(3) = 6
tan 30𝑜
191
6.1.4. Inverse Trigonometric Functions
Let the sine of angle 𝐴 be 𝑥, that is,
sin(𝐴) = 𝑥
Then the angle 𝐴 is given by,
𝐴 = sin−1(𝑥)
Where sin−1 𝑥 denotes the inverse of sine. You will remember in Section 2.2.9.2., we learned
that the inverse of functions are represented by the index −1. It is the same for the inverses of
trigonometric functions.
1
For example, what is the value of sin−1 ( ) ?
√2
You will remember in Example 3.4.1.1 that sin(45𝑜 ) =
1
√2
1
, so sin−1 ( ) = 45𝑜 .
√2
When you use a calculator to find values for inverse trigonometric functions, you will notice that
many calculators, have the SHIFT or INV or 2ndFUN buttons. You press them in the following
manner:
SHIFT − SIN − Value = or
INV − SIN − Value = or
2nd FUN − SIN − Value =
In all cases, remember to switch your calculator to the 𝐷𝐸𝐺 for degree mode.
6.1.5. Additional Trigonometric Functions
The other trigonometric functions are cosecant (cosec), secant (sec) and cotangent (cot), and are
determined by:
𝑐𝑜𝑠𝑒𝑐(𝜃) =
sec(𝜃) =
1
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
sin(𝜃)
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
1
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
cos(𝜃)
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
cot(𝜃) =
1
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
=
tan(𝜃) 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
192
(𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 sin(𝜃) ≠ 0)
(𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 cos(𝜃) ≠ 0)
(𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 tan(𝜃) ≠ 0)
6.1.6. Trigonometric Ratios of Acute Angles
c
b
𝜃
a
Figure 6.1.6.
(a) From Figure 6.1.6.
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑏
(i)
sin 𝜃 =
(ii)
cos 𝜃 =
(iii)
tan 𝜃 =
(iv)
𝑠𝑒𝑐 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 = 𝑎
(v)
𝑐𝑜𝑠𝑒𝑐 𝜃 =
(vi)
cot 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
=𝑐
=
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑐
𝑎
𝑏
𝑐
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝑎
=
𝑐
𝑏
𝑏
=𝑎
(b) Using Figure 6.1.6.
(i)
(ii)
sin 𝜃
cos 𝜃
cos 𝜃
sin 𝜃
=
𝑏
𝑐
𝑎
𝑐
= 𝑎 = tan 𝜃
=
𝑎
𝑐
𝑏
𝑐
= 𝑏 = cot 𝜃
𝑏
𝑎
sin 𝜃
i.e.
tan 𝜃 =
i.e.
cot 𝜃 = sin 𝜃
cos 𝜃
cos 𝜃
193
1
(iii)
sec 𝜃 = cos 𝜃
(iv)
𝑐𝑜𝑠𝑒𝑐 𝜃 = sec 𝜃
(v)
cot 𝜃 = tan 𝜃
1
1
Secants, Cosecants and Cotangents are known as the reciprocal ratios
Example 6.1.6.1.
9
If cos 𝑋 = 41, determine the value of the other five trigonometric ratios
Solution 6.1.6.1.
Start by drawing a right-angled triangle 𝑋𝑌𝑍, such that the side adjacent to angle 𝑋 is 𝑋𝑌 = 9,
and the side opposite the angle 𝑋 is the unknown side, 𝑌𝑍, and the side 𝑋𝑍 = 41 is the
hypotenuse.
Using the Pythagoras’ formula, it means that
412 = 92 + 𝑌𝑍 2
Which produces
1681 = 81 + 𝑌𝑍 2
𝑌𝑍 = √1681 − 81
𝑌𝑍 = √1600 = 40
Therefore, 𝑌𝑍 = 40.
Since we have obtained values for all the sides of the triangle, we can solve for all the other
trigonometric ratios
sin 𝑋 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
40
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 41
194
tan 𝑋 =
cosec 𝑋 =
sec 𝑋 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 40
4
=
=4
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
9
9
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 41
1
=
=1
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
40
40
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 41
5
=
=4
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
9
9
cot 𝑋 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
9
=
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 40
195
6.2.0. Session 2 - Solving Right-Angled Triangles
To ‘solve a right-angled triangle’ means ‘to find the unknown sides and angles.’ This is achieved
by using
(i)
(ii)
the theorem of Pythagoras, and/or
trigonometric ratios
We can transpose the standard formulae for the trigonometric ratios in various ways to evaluate
other unknown values of the right-angled triangle. By transposing, we have the following
equations:
𝑜𝑝𝑝 = ℎ𝑦𝑝 × sin (𝜃)
𝑎𝑑𝑗 = ℎ𝑦𝑝 × cos (𝜃)
𝑜𝑝𝑝 = 𝑎𝑑𝑗 × tan(𝜃)
𝑜𝑝𝑝
ℎ𝑦𝑝 = sin(𝜃)
ℎ𝑦𝑝 =
𝑎𝑑𝑗
cos(𝜃)
𝑜𝑝𝑝
𝑎𝑑𝑗 = tan(𝜃)
Example 6.2.0.1
Triangle 𝑃𝑄𝑅 whose side 𝑄𝑅 = 7.5𝑐𝑚, and the hypotenuse 𝑅𝑃 is inclined at 𝑅 at an angle 38𝑜 .
Find the lengths of the side 𝑃𝑄 and the hypotenuse, 𝑅𝑃.
Solution 6.2.0.1.
Given that, the triangle 𝑃𝑄𝑅 has hypotenuse 𝑅𝑃 inclined at an angle 𝑅 = 38𝑜, it implies that the
side opposite the angle 𝑅 is 𝑃𝑄 and the side adjacent to angle 𝑅 is 𝑄𝑅 = 7.5𝑐𝑚.
It follows that
tan 38 =
𝑃𝑄 𝑃𝑄
=
𝑄𝑅 7.5
196
𝑃𝑄 = 7.5 × tan 38
𝑃𝑄 = 7.5 × 0.7813
𝑃𝑄 = 5.8596 ≈ 5.86𝑐𝑚
We obtain the length of the hypotenuse by using the relation
cos 38 =
𝑅𝑃 =
𝑄𝑅 7.5
=
𝑅𝑃 𝑅𝑃
7.5
7.5
=
cos 38 0.7880
𝑅𝑃 = 9.5178 ≈ 9.518𝑐𝑚
We can check by using the Pythagoras’ Theorem
𝑅𝑃2 = (7.5)2 + (5.86)2
𝑅𝑃2 = 56.25 + 34.3396
𝑅𝑃2 = 90.5896
𝑅𝑃 = √90.5896
𝑅𝑃 = 9.5179 ≈ 9.518
Example 6.2.0.2.
Solve the triangle whose hypotenuse 𝐵𝐶 = 37𝑚𝑚, and the side 𝐵𝐴 = 35𝑚𝑚.
Solution 6.2.0.2.
197
To ‘solve triangle 𝐴𝐵𝐶’ means ‘to find the unknown length 𝐴𝐶 and the angles 𝐵 and 𝐶’.
We have been given the side 𝐵𝐶 = 37𝑚𝑚 and the side 𝐵𝐴 = 35𝑚𝑚, the side 𝐴𝐶 is unknown,
and the angle at 𝐵 and at 𝐶 are not given.
Given that, the side 𝐵𝐶 is the hypotenuse, whose length 𝐵𝐶 = 37 and its other side 𝐵𝐴 = 35, we
deduce that the angle 𝐵 is the angle between the side 𝐵𝐶 and the side 𝐵𝐴, and the angle at 𝐶 is
the angle between the side 𝐴𝐶 and the side 𝐶𝐵 = 𝐵𝐶, the hypotenuse.
We formulate the relation
sin 𝐶 =
𝐵𝐴 35
=
𝐵𝐶 37
sin 𝐶 = 0.94595
𝐶 = sin−1 (0.94595)
𝐶 = 71.0754 ≈ 71.08𝑜
To find angle 𝐵, we subtract the sum of angle 𝐴 = 90 and angle 𝐶 = 71.08 from 180, because
the sum of all the angles in a triangle is 180𝑜
Therefore, angle
𝐵 = 180 − (90 + 71.08)
𝐵 = 180 − (161.08) = 18.92𝑜
To find the side 𝐴𝐶, we use
sin 𝐵 =
𝐴𝐶 𝐴𝐶
=
𝐵𝐶 37
AC = 37 sin 18.92𝑜
198
AC = 37(0.32425)
𝐴𝐶 = 11.9972𝑚𝑚 ≈ 12𝑚𝑚
Again, we can verify by using the Pythagoras’ Theorem. We leave the verification as an exercise
for the learners
Special Note:
It is always advisable to make a reasonably accurate sketch of the object to visualise the expected
magnitudes of unknown sides and angles.
Evaluating Trigonometric Ratios of Any Angles
The easiest method of evaluating trigonometric functions of any angle is by using a calculator.
The following examples are read from calculators. You can use your calculators t check values ff
your self
Trigonometric Function and Angle
𝑠𝑖𝑛𝑒 18𝑜 = 0.3090
𝑠𝑖𝑛𝑒 172𝑜 = 0.1392
𝑠𝑖𝑛𝑒 241.63𝑜 = −0.8799
𝑐𝑜𝑠𝑖𝑛𝑒 115𝑜 = −0.4226
𝑐𝑜𝑠𝑖𝑛𝑒 331.78𝑜 = 0.8811
𝑡𝑎𝑛𝑔𝑒𝑛𝑡 29𝑜 = 0.5543
𝑡𝑎𝑛𝑔𝑒𝑛𝑡 178𝑜 = −0.0349
𝑡𝑎𝑛𝑔𝑒𝑛𝑡 296.42𝑜 = −2.0127
To evaluate, say, 𝑠𝑖𝑛𝑒 42𝑜 23′ using a calculating device, means finding 𝑠𝑖𝑛𝑒 42
are 60 minutes in 1 degree.
23
Since 60 = 0.3833, we have 42𝑜 23′ = 42.3833𝑜
Thus 𝑠𝑖𝑛𝑒 42𝑜 . 23′ = 𝑠𝑖𝑛𝑒 42.3833𝑜 = 0.6741, correct to 4 decimal places
199
23𝑜
60
since there
Similarly, 𝑐𝑜𝑠𝑖𝑛𝑒 72𝑜 38′ = 𝑐𝑜𝑠𝑖𝑛𝑒 72
38𝑜
60
= 𝑐𝑜𝑠𝑖𝑛𝑒 72.6333 = 0.2985 correct to 4 decimal
places
Most calculators contain only sine, cosine, and tangent functions. Thus, to evaluate secants,
cosecants, and cotangents, reciprocals are used to evaluate such functions.
Let us look at some examples. Check the values with your personal calculators.
Reciprocal Trigonometric Functions and Angles
1
1
𝑠𝑒𝑐𝑎𝑛𝑡 32𝑜 =
=
= 1.1792
𝑜
cos 32
0.848
1
1
𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡 75𝑜 =
=
= 1.0353
𝑜
sin 75
0.9659
𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 41𝑜 =
𝑠𝑒𝑐𝑎𝑛𝑡 215.12𝑜 =
1
1
=
= 1.1504
tan 41𝑜 0.8693
1
1
=
= −1.2226
𝑜
cos 215.12
−0.8179
𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡 321.62𝑜 =
1
1
=
= −1.6106
sin 321.62𝑜 −0.6209
𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 263.59𝑜 =
1
1
=
= 0.1123
𝑜
tan 263.59
8.9012
Exercises
Evaluate correct to 4 decimal places
(1) 𝑠𝑖𝑛𝑒 168𝑜 14′
(a) 0.2037
(b)
0.2045
(c)
0.2039
(d)
0.2041
(2) 𝑐𝑜𝑠𝑖𝑛𝑒 271.41𝑜
(a) 0.0241
(b)
0.0244
(c)
0.0248
(d)
0.0246
(3) 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 98𝑜 4′
(a) −7.0555
(b)
7.0558(c)
−7.0535
(4) 𝑠𝑒𝑐𝑎𝑛𝑡 161𝑜
(a) −1.0576
(b)
1.0578
(c)
(5) 𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡 279.16𝑜
200
(d)
−1.0572
−7.0558
(d)
1.0568
(a) −1.1031
1.0130
(b)
(c)
−1.0130
(d)
1.0133
6.3.0. Session 3 - Cartesian and Polar Coordinates
6.3.1. Introduction
There are two ways in which the position of a point in a plane can represented. These are;
a. by Cartesian co-ordinates, i.e., (𝑥, 𝑦) and
b. by polar co-ordinates, i.e., (𝑟, 𝜃), where 𝑟 is a ‘radius’ from a fixed point and 𝜃 is angle
from a fixed point
6.3.2. Changing from Cartesian into Polar Co-ordinates
𝑦
𝑃
𝑟
Second
Quadrant
𝑂
Third
Quadrant
First
Quadrant
𝑦
𝑄
𝑥
Figure 6.3.2.
𝑥
Fourth
Quadrant
Using the trigonometric ratios and Figure 6.3.2,
𝑦
tan 𝜃 = 𝑥
Equation 6.3.2𝑎
from which
𝑦
𝜃 = tan−1 𝑥
Equation 6.3.2𝑏
Also using the Pythagoras’ Theorem and Figure 6.3.2,
𝑟 = √𝑥 2 + 𝑦 2
Equation 6.3.2𝑐
We need the two equations Equation 6.3.2𝑏 and Equation 6.3.2𝑐 to change from Cartesian to
polar co-ordinates. The angle 𝜃, which may be expressed in degrees or radians, must always be
measured from the positive 𝑥-axis, i.e., measured from line 𝑂𝑄in Figure 6.3.2.
201
It is suggested that when changing from Cartesian to polar co-ordinates a diagram should always
be sketched.
Example 6.3.2.1.
Change the Cartesian co-ordinates (3, 4) into polar co-ordinates.
Solution 6.3.2.1.
A diagram representing the point (3, 4) is shown Figure 6.3.2.1. below:
𝑦
𝑟
𝑂
4
3
𝑄
𝑥
Figure 6.3.2.1.
Using Pythagoras’s Theorem, we can find the radius 𝑟 as follows:
𝑟 = √32 + 4 2
𝑟 = √9 + 16 = √25
𝑟 = √25 = 5
Special Note:
It is important to note that −5 has no meaning in this context
Also, by trigonometric ratios,
4
𝜃 = tan−1 (3) = tan−1 1.3333 = 53.13𝑜 or 0.9273 radians
Hence (3, 4) in Cartesian co-ordinates corresponds to (5, 53.13𝑜 ) or (5, 0.9273 𝑟𝑎𝑑) in polar
co-ordinates
202
Example 6.3.2.2.
Express in polar co-ordinates the position (−4, 3)
Solution 6.3.2.2.
A diagram representing the point (−4, 3) is shown Figure 6.3.2.2. below:
𝑦
𝑃
𝑟
3
𝜃
𝑄
𝑂
4
𝑥
Figure 6.3.2.2.
From Pythagoras’ Theorem,
𝑟 = √(−4)2 + (3)2
𝑟 = √16 + 9 = √25 = 5
By the trigonometric ratios,
3
𝛼 = tan−1 4 = tan−1 0.75 = 36.8699 ≈ 36.87𝑜 or 0.6435 𝑟𝑎𝑑
Hence
𝜃 = 180 − 36.87 = 143.13𝑜 or 𝜃 = 𝜋 − 0.6435 = 2.4981
Hence the position (−4, 3) is (5, 143.13𝑜 ) or (5, 2.4981 𝑟𝑎𝑑) in polar co-ordinates
203
Example 6.3.2.3.
Express (−5, −12) in polar co-ordinates
Solution 6.3.2.3.
We leave the sketch of the cartesian co-ordinates (−5, −12) as an exercise for the learners, but
we guide you to do the sketch.
Since (𝑥, 𝑦) = (−5, −12), for the value of 𝑥 = −5, we move 5 units in the opposite direction of
the positive 𝑥-axis, i.e., toward the left. Similarly, the value of 𝑦 = −12, means we move 12
units in the opposite direction of the positive 𝑦-axis, i.e., downwards instead of upwards.
The diagram will occupy the third quadrant, by making 5 units move to the left, and another 12
units to the bottom into the third quadrant.
We apply the Pythagoras theorem to determine the value for 𝑟
𝑟 = √(−5)2 + (−12)2
𝑟 = √25 + 144
𝑟 = √169 = 13
The acute angle 𝑟 makes with the 𝑥-axis is 𝛼
tan 𝛼 = (
𝛼 = tan−1 (
−12
12
)=
−5
5
12
) = tan−1(2.4)
5
204
𝛼 = 67.3801𝑜 or 1.1760 𝑟𝑎𝑑
Hence, the angle we desire is 𝜃 = 180𝑜 + 67.3801𝑜 = 247.3801 ≈ 247.38
Or the angle in radians is 𝜃 = 𝜋 + 1.1760 = 4.318 rad
Therefore, the point (−5, −12) in Cartesian co-ordinates corresponds to (13, 247.38𝑜 ) or
(13, 4.318 𝑟𝑎𝑑) in polar co-ordinates
Example 6.3.2.4.
Express (2, −5) in polar co-ordinates
Solution 6.3.2.4.
We leave the sketch as an exercise for the learner.
The (𝑥, 𝑦) = (2, −5), implies, we move 2 units in the positive 𝑥-axis and 5 units in the negative
𝑦-axis into the fourth quadrant
From the Pythagoras’ Theorem,
𝑟 = √22 + (−5)2 = √4 + 25 = √29 = 5.3852
5
The angle 𝛼 = tan−1 (2) = tan−1 2.5 = 68.1986 ≈ 68.20𝑜 or 1.1903 ≈ 1.190 𝑟𝑎𝑑
Therefore,
𝜃 = 360𝑜 − 68.20𝑜 = 291.80𝑜
Or
𝜃 = 2𝜋 − 1.190 = 5.0932 𝑟𝑎𝑑
205
Thus, (2, −5) in Cartesian co-ordinates corresponds to (5.385, 291.80𝑜 ) or
(5.385, 5.0932 𝑟𝑎𝑑) in polar co-ordinates
6.3.3. Changing from Polar Co-ordinates to Cartesian Co-ordinates
𝑦
𝑟
Second
Quadrant
𝑂
𝑦
𝑄
𝑥
Third
Quadrant
First
Quadrant
Figure 6.3.3.
𝑥
Fourth
Quadrant
From the right-angled triangle in Figure 6.3.3., we can deduce that
𝑥
cos 𝜃 = 𝑟
Equation 6.3.3𝑎
and
sin 𝜃 =
Hence
𝑦
Equation 6.3.3𝑏
𝑟
𝑥 = 𝑟 cos 𝜃
Equation 6.3.3𝑐
𝑦 = 𝑟 sin 𝜃
Equation 6.3.3𝑑
and
If we know the length 𝑟 and the angle 𝜃, then 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 are the two formulae
we need to change co-ordinates from polar to Cartesian co-ordinates
206
Example 6.3.3.1.
Change (4, 32𝑜 ) into Cartesian co-ordinates
Solution 6.3.3.1.
𝑦
𝑟=4
Second
Quadrant
𝑂
Third
Quadrant
𝑦
First
Quadrant
𝑥
𝑥
Figure 6.3.3.1
Fourth
Quadrant
Given the polar co-ordinates (4, 32𝑜 ),
𝑥 = 𝑟 cos 𝜃 = 4 cos 32𝑜 = 4 × 0.8480 = 3.3922
And
𝑦 = 𝑟 sin 𝜃 = 4 sin 32 = 4 × 0.5299 = 2.1197
Hence (4, 32𝑜 ) in polar co-ordinates corresponds to (3.39, 2.12) in Cartesian co-ordinates
Example 6.3.3.2.
Express (6, 137𝑜 ) in Cartesian Co-ordinates
Solution 6.3.3.3.
Given the polar co-ordinates as (6, 137𝑜 )
𝑥 = 6 cos 137𝑜 = 6 × (−0.7314) = −4.3881
𝑦 = 6 sin 137𝑜 = 6 × 0.681998 = 4.0920
207
Hence (6, 137𝑜 ) in polar co-ordinates corresponds to (−4.3881, 4.0920) in Cartesian coordinates
Special Note:
When changing from polar to Cartesian co-ordinates, it is not quite essential to draw the sketch .
the use of 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 naturally produces the correct signs.
Example 6.3.3.4.
Express (4.5, 5.16 𝑟𝑎𝑑) in Cartesian co-ordinates.
Solution 6.3.3.4.
We have been given, (4.5, 5.16 𝑟𝑎𝑑)
𝑥 = 𝑟 cos 𝜃 = 4.5 cos 5.16 = 4.5 × 0.4328 = 1.9477
𝑦 = 𝑟 sin 𝜃 = 4.5 sin 5.16 = 4.5 × (−0.901483) = −4.0567
Therefore, (4.5, 5.16 𝑟𝑎𝑑) in polar co-ordinates corresponds to (1.948, −4.057 in Cartesian coordinates
208
6.3.4. The Use of 𝑅 → 𝑃 and 𝑃 → 𝑅 Functions on Calculators
Another name for Cartesian co-ordinates is rectangular co-ordinates. Many scientific notation
calculators possess 𝑅 → 𝑃 and 𝑃 → 𝑅 functions. The 𝑅 is the first letter of the word rectangular
and the 𝑃 is the first letter of the word polar.
Check the operation manual for your calculator to determine how to use these two functions.
They make changing from Cartesian to polar co-ordinates, and vice-versa, much quicker and
easier.
209
6.4.0. Session 4 – Solving Triangles – The Sine and Cosine Rules
6.4.1. Solution of Right-Angled Triangles
To ‘solve a right-angled triangle’ means ‘to find the unknown sides and angles’. This is achieved
by using
(i)
(ii)
the theorem of Pythagoras and/or
trigonometric ratios.
Let us look at some examples.
Example 6.4.1.1.
In triangle 𝐴𝐵𝐶, shown below in Figure 6.4.1.1., find the lengths 𝐴𝐵 and 𝐵𝐶.
𝐴
𝐵
𝐶
7.5
Figure 6.4.1.1
Solution 6.4.1.1.
From the triangle in Figure 6.1.1.1.,
tan 38𝑜 =
𝐴𝐵 𝐴𝐵
=
𝐵𝐶 7.5
Hence
𝐴𝐵 = 7.5 × tan 38𝑜 = 7.5 × 0.7813
𝐴𝐵 = 5.85975 ≈ 5.860 cm
210
Also
cos 38𝑜 =
𝐵𝐶 7.5
=
𝐴𝐶 𝐴𝐶
Hence
𝐴𝐶 =
7.5
7.5
=
= 9.5178 ≈ 9.518
cos 38𝑜 0.7880
We can check to verify the result by using the Pythagoras theorem as follows:
𝐴𝐶 2 = (7.5)2 + (5.860)2 = 56.25 + 34.3396 = 90.5896
𝐴𝐶 = √90.5896 = 9.5179 ≈ 9.518
Example 6.4.1.2.
Solve the triangle 𝐴𝐵𝐶 below
𝐴
𝐶
35𝑚𝑚
𝐵
37𝑚𝑚
Figure 6.4.1.2
Solution 6.4.1.2.
In this problem, or to solve the triangle 𝐴𝐵𝐶 means ‘to find the unknowns’, i.e., ‘to find the
length 𝐴𝐶, and the angles at 𝐵 and 𝐶’
Special Note:
Please note that the symbol ∠𝑄, means ‘the angle at the point 𝑄’
211
We proceed as follows:
sin 𝐶 =
35
= 0.95459
37
Hence
∠𝐶 = sin−1 0.94595 = 71.0754 ≈ 71.08𝑜
We can calculate the angle at 𝐵 by subtracting (90𝑜 + 71.08𝑜 ) from 180𝑜, since angles in a
triangle add up to 180𝑜
Thus 𝐵 = 180𝑜 − (90𝑜 + 71.08𝑜 ) = 18.92𝑜
We can find the length of 𝐴𝐶 as follows:
sin 𝐵 =
𝐴𝐶
37
𝐴𝐶 = 37 × sin 18.92𝑜
Remember that ∠𝐵 = 18.92𝑜
𝐴𝐶 = 37 × 0.3242 = 11.9972 ≈ 12.0𝑚𝑚
We can verify using the Pythagoras theorem
𝐴𝐶 = √372 − 352 = √1369 − 1225 = √144
𝐴𝐶 = √144 = 12.0
212
Example 6.4.1.3.
Solve triangle 𝑋𝑌𝑍, given that ∠𝑋 = 90𝑜 , ∠𝑌 = 23𝑜 17′ and 𝑌𝑍 = 20.0 mm. Determine also its
area.
Solution 6.4.1.3.
It is always advisable to make a reasonably accurate sketch of the diagram to visualise the
expected magnitudes of the unknown sides and angles. However, we will leave the sketch of this
problem for the learners to make the sketch using the given sides and angles.
In this problem, two of the angles, ∠𝑋 = 90𝑜 and ∠𝑌 = 23𝑜 17′ have been given. We can
subtract their sum from 180𝑂 to obtain the angle for ∠𝑍
We have
∠𝑍 = 180𝑜 − (90𝑜 + 23𝑜 17′ ) = 66𝑜 43′
Given that ∠𝑌 = 23𝑜 17′, we need to convert the 17′ into degrees.
There are 60′ (minutes) in a degree, i.e., 60 minutes is 1𝑜 . Therefore, we have
17′ ≡
17
= 0.2833𝑜
60
Hence
23𝑜 17′ = 23𝑜 + 0.2833𝑜 = 23.2833𝑜
sin 23𝑜 17′ = sin 23.2833𝑜 =
𝑋𝑍
20.0
𝑋𝑍 = 20.0 × sin 23.2833
𝑋𝑍 = 20.0 × 0.3953 = 7.906
213
We can determine the length 𝑋𝑌 as follows:
cos 23𝑜 17′ = cos 23.2833 =
𝑋𝑍
20.0
𝑋𝑍 = 20.0 × cos 23.2833
𝑋𝑍 = 20.0 × 0.9186 = 18.3712
Again, we can use the Pythagoras theorem to verify the results. We urge the learners to use the
Pythagoras’ theorem to verify the results
Area of triangle 𝑋𝑌𝑍, like any other triangle is given by
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑋𝑌𝑍 =
1
× (𝑏𝑎𝑠𝑒) × (𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ℎ𝑒𝑖𝑔ℎ𝑡)
2
𝐴𝑟𝑒𝑎 𝑋𝑌𝑍 =
𝐴𝑟𝑒𝑎 𝑋𝑌𝑍 =
1
× (𝑋𝑌) × (𝑋𝑍)
2
1
× (18.3712) × (7.906)
2
𝐴𝑟𝑒𝑎 𝑋𝑌𝑍 = 72.6214 ≈ 72.62 𝑚𝑚2
214
6.4.2. Angles of Elevation and Depression
𝐴
𝐶
𝑩
𝑃
𝑅
𝑸
Figure 6.4.2b.
Figure 6.4.2a.
(a) Consider Figure 6.4.2𝑎, above. If 𝐵𝐶 represents the horizontal ground, and 𝐴𝐵 a vertical
pole, then the angle of elevation of the top of the pole, 𝐴, from the point 𝐶, is the angle
that the imaginary straight line 𝐴𝐶 must be raised (or elevated) from the horizontal 𝐶𝐵,
i.e., angle 𝜃.
(b) Consider Figure 6.4.2𝑏, above. If 𝑃𝑄 represents a vertical cliff and 𝑅, is a ship at sea,
then the angle of depression of the ship from point 𝑃 is the angle through which the
imaginary straight line 𝑃𝑅 must be lowered (or depressed) from the horizontal to the
ship, i.e., angle 𝜙.
Special Note:
Please, note that the angle ∠𝑃𝑅𝑄 is also 𝜙, alternate angles between parallel lines
We now consider some sample problems.
Example 6.4.2.1.
An electricity pylon stands on a horizontal ground. At a point 80 m from the base of the pylon,
the angle of elevation of the top of the pylon is 23𝑜 . Calculate the height of the pylon to the
nearest metre.
Solution 6.4.2.1.
As usual, we need to make a sketch of the pylon as a triangle with an elevation 23𝑜 from the
point 80 m from the base of the pylon.
𝐴
𝐵
23
80 𝑚
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.2.1. 𝑚
215
𝐶
From the sketch
tan 23𝑜 =
𝐴𝐵 𝐴𝐵
=
𝐵𝐶 80
The height of the pylon is
𝐴𝐵 = 80 × tan 23𝑂
𝐴𝐵 = 80 × 0.4245
𝐴𝐵 = 33.9579853 ≈ 34 𝑚 to the nearest metre
Example 6.4.2.2.
A surveyor measures the angle of elevation of the top of a perpendicular building as 19𝑜 . He
moves 120 m nearer the building and finds the angle of elevation is now 47𝑜 . Determine the
height of the building.
Solution 6.4.2.2.
As usual we will make a sketch of the given situation.
The building 𝐴𝐵 and the angles of elevations are shown here in Figure 6.4.2.2
𝐴
ℎ
47𝑜
𝐵
𝑥
𝐶
19𝑜
𝐷
120
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.2.2. 𝑚
In triangle 𝐴𝐵𝐷,
216
tan 19𝑜 =
ℎ
𝑥 + 120
Hence
ℎ = tan 19𝑜 (𝑥 + 120)
ℎ = 0.3443 × (𝑥 + 120)
Eqn 6.4.2.2𝑎
Also, in triangle 𝐴𝐵𝐶,
tan 47𝑜 =
ℎ
𝑥
Hence
ℎ = tan 47𝑜 (𝑥)
ℎ = 1.0724𝑥
Eqn 6.4.2.2𝑏
We equate Eqn 6.4.2.2𝑎 to Eqn 6.4.2.2𝑏 and solve for 𝑥
0.3443 (𝑥 + 120) = 1.0724𝑥
0.3443𝑥 + (120)(0.3343) = 1.0724𝑥
We group like terms together, so that 0.3443𝑥 moves to the other side of the equation
(120)(0.3343) = 1.0724𝑥 − 0.3443𝑥
(1.0724 − 0.3443)𝑥 = (120)(0.3343)
0.7281𝑥 = 41.316
217
𝑥=
41.316
= 56.74495 ≈ 56.7450
0.7281
𝑥 ≈ 56.7450 m
From Eqn 6.4.2.2𝑏,
ℎ = 1.0724𝑥 = 1.0724 × 56.745 = 60.853338
ℎ = 60.85 𝑚
Example 6.4.2.3.
The angle of depression of a ship viewed at a particular instant from the top of a 75 m vertical
cliff is 30𝑜 . Find the distance of the ship from the base of the cliff at this instant. The ship is
sailing away from the cliff at a constant speed and 1 minute later its angle of depression from the
top of the cliff is 20𝑜 . Determine the speed of the ship in km/h.
Solution 6.4.2.3.
We try to make a sketch of the scenario.
30𝑜
𝐴
20𝑜
75 𝑚
30𝑜
𝐵
20𝑜
𝑥
𝐶
F
𝐷
F
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.2.3. 𝑚
Figure 6.4.2.3. shows the cliff 𝐴𝐵, the initial position of the ship at 𝐶 and the final position at 𝐷.
Since the angle of depression was initially at 30𝑜 , then ∠𝐴𝐶𝐵 = 30𝑜 (alternate angles between
parallel lines).
tan 30𝑜 =
𝐴𝐵 75
=
𝐵𝐶 𝐵𝐶
218
Hence
𝐵𝐶 =
75
75
=
𝑜
tan 30
0.5774
𝐵𝐶 = 129.89 𝑚
The initial position of the ship was 129.89 m from the base of the cliff.
In triangle 𝐴𝐵𝐷,
tan 20𝑜 =
𝐴𝐵
75
75
=
=
𝐵𝐷 𝐵𝐶 + 𝐶𝐷 129.89 + 𝑥
This implies that
129.89 + 𝑥 =
75
75
=
= 206.0440 ≈ 206.0 𝑚
𝑜
tan 20
0.3640
129.89 + 𝑥 = 206.0
𝑥 = 206.0 − 129.89
𝑥 = 76.11
The ship sails 76.11 m in 1 minute, i.e., 60 seconds, hence
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠ℎ𝑖𝑝 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
=
76.11
60
𝑚/𝑠
The speed in km/h is given by converting seconds to hours and from metres to kilometres
219
1000𝑚 = 1 𝑘𝑚
There are 60 × 60 seconds in 1 hour
The speed of the ship is
𝑆𝑝𝑒𝑒𝑑 =
76.11 × 60 × 60
60 ×1000
= 4.5666 ≈ 4.57 km/h
220
6.4.3. Sine and Cosine Rules
As noted earlier in Session 6.4.1., to ‘solve a triangle’ means ‘to find the values of the unknown
sides and angles.’ If a triangle is right-angled, trigonometric ratios and the theorem of Pythagoras
may be used for its solutions, as shown in the earlier sessions in Unit 6. However, for non-rightangled triangles, trigonometric ratios and Pythagoras’ theorem cannot be used. Instead, two
rules, called the Sine Rule and the Cosine Rule, are used.
6.4.3.1.
The Sine Rule
Consider the triangle in Figure 6.4.3.1. below.
𝐴
F
𝑐
𝑏
F
F
𝐵
𝑎
F
F
𝐶
F
Figure 6.4.3.1
With reference to the triangle 𝐴𝐵𝐶 in Figure 6.4.3.1., the Sine Rules states:
𝑎
𝑏
𝑐
=
=
sin 𝐴 sin 𝐵 sin 𝐶
The Sine Rule may used only when:
(i)
(ii)
6.4.3.2.
1 side and any 2 of the angles are initially given, or
2 sides and an angle (not the included angle) are initially given
The Cosine Rule
With reference to the triangle 𝐴𝐵𝐶 in Figure 6.4.3.1., the Cosine Rules states:
𝑎2 = 𝑏2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
Or
𝑏2 = 𝑎2 + 𝑐 2 − 2𝑎𝑐 cos 𝐵
221
Or
𝑐 2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶
The Cosine Rule may used only when:
(i)
(ii)
2 sides and the included angle are initially given, or
3 sides are initially given
6.4.4. The Area of Any Triangle
The area of any triangle such as triangle 𝐴𝐵𝐶 in Figure 6.4.3.1., is given by:
(i)
(ii)
(iii)
1
2
1
2
× 𝑏𝑎𝑠𝑒 × 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ℎ𝑒𝑖𝑔ℎ𝑡, or
1
1
𝑎𝑏 sin 𝐶, or 2 𝑎𝑐 sin 𝐵, or 2 𝑏𝑐 sin 𝐴, or
√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
where 𝑠 =
𝑎+𝑏+𝑐
2
Example 6.4.4.1.
In a triangle labelled 𝑋𝑌𝑍, ∠𝑋 = 51𝑜, ∠𝑌 = 67𝑜 and 𝑌𝑍 = 15.2 cm. Solve the triangle and find
its area.
Solution 6.4.4.1.
It is advisable to make a meaningful sketch of the triangle in each problem given. We will leave
the sketch in this problem for the learners as an exercise.
We know that, in triangle 𝑋𝑌𝑍, ∠𝑋 = 51𝑜 , ∠𝑌 = 67𝑜 and 𝑌𝑍 = 15.2 cm.
We can determine angle ∠𝑍 by subtracting ∠𝑋 + ∠𝑌 = 51𝑜 + 67𝑜 from 180𝑜 as follows:
∠𝑍 = 180𝑜 − (51𝑜 + 67𝑜 ) = 180𝑜 − 118𝑜 = 62𝑜
∠𝑍 = 62𝑜
222
We know 3 of the angles and 1 of the sides 𝑌𝑍 = 15.2 cm, which is the side opposite angle
∠𝑋 = 51𝑜 , which is also the side 𝑥 = 15.2
We can apply the Sine Rule as follows:
15.2
𝑦
𝑧
=
=
𝑜
𝑜
sin 51
sin 67
sin 62𝑜
Using
15.2
𝑦
=
sin 51𝑜 sin 67𝑜
15.2 sin 67𝑜
𝑦=
sin 51𝑜
𝑦=
15.2 × 0.9205
0.7772
𝑦=
15.2 × 0.9205
0.7771
𝑦 = 18.00 = 𝑋𝑍
Using
15.2
𝑧
=
sin 51𝑜 sin 62𝑜
𝑧=
𝑧=
15.2 × sin 62𝑜
sin 51𝑜
15.2 × 0.8829
= 17.2694 ≈ 17.27
0.7771
223
𝑧 = 17.27 = 𝑋𝑌
1
Area of triangle 𝑋𝑌𝑍 = 2 𝑥𝑦 sin 𝑍
𝐴𝑟𝑒𝑎 =
1
× 15.2 × 18.00 × sin 62𝑜
2
𝐴𝑟𝑒𝑎 =
1
× 15.2 × 18.00 × 0.8829
2
𝐴𝑟𝑒𝑎 = 120.78072 ≈ 120.8 cm2
Alternatively,
1
Area of triangle 𝑋𝑌𝑍 = 2 𝑥𝑧 sin 𝑌
𝐴𝑟𝑒𝑎 =
𝐴𝑟𝑒𝑎 =
1
× 15.2 × 17.27 × sin 67𝑜
2
1
× 15.2 × 17.27 × 0.9205 = 120.817466
2
𝐴𝑟𝑒𝑎 𝑋𝑌𝑍 = 120.8 cm2
It is always worth checking with triangle problems that the longest side is opposite the largest
angle, and vice-versa.
In the problem we just solved, 𝑌 is the largest angle and 𝑋𝑍 = 𝑦 is the longest of the three sides.
224
Example 6.4.4.2.
Solve the triangle 𝐴𝐵𝐶, given that 𝐵 = 78𝑜 51′, 𝐴𝐶 = 22.31 mm and 𝐴𝐵 = 17.92 mm. Find
also its area.
Solution 6.4.4.2.
Triangle 𝐴𝐵𝐶 is shown in Figure 6.4.4.2. below.
𝐴
𝑏 = 22.31 𝑐𝑚
78𝑜 51′
𝐵
𝐶
𝑎
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.4.2
= 22.31 𝑐𝑚
We apply the sine rule on triangle 𝐴𝐵𝐶
22.31
17.92
=
sin 78𝑜 51′ sin 𝐶
17.92 sin 78𝑜 51′
sin 𝐶 =
22.31
Remember to convert 51′ to degrees
51
We have 60 = 0.85𝑜
17.92 sin 78.85𝑜
sin 𝐶 =
22.31
sin 𝐶 =
17.92 × 0.9811
22.31
225
sin 𝐶 =
17.58131292
= 0.7881
22.31
𝐶 = sin−1 0.7881 = 52𝑜 0′
We can calculate for the angle 𝐴 by subtracting the sum of the two angles 𝐵 + 𝐶 = 78𝑜 51′ +
52𝑜 0′ from 180𝑜
𝐴𝑛𝑔𝑙𝑒 𝐴 = 180𝑜 − 78𝑜 51′ − 52𝑜 0′ = 49.15𝑜 = 49𝑜 9′
We obtained the 9′ by multiplying the 0.15𝑜 by 60, since 0.15 × 60 = 9
We apply the sine rule again to find the side 𝑎 = 𝐵𝐶
We have
𝑎
22.31
=
sin 𝐴 sin 𝐵
𝑎
22.31
=
𝑜
sin 49 9′ sin 78𝑜 51′
𝑎=
22.31 × sin 49𝑜 9′
sin 78𝑜 51′
𝑎=
22.31 × sin 49.15𝑜
sin 78.85𝑜
𝑎=
22.31 × 0.7564
0.9811
226
𝑎=
16.875284
= 17.20
0.9811
𝑎 = 17.20mm
Hence
𝐴 = 49𝑜 9′ , 𝐶 = 52𝑜 0′ and 𝐵𝐶 = 17.20 mm
1
The area of the triangle 𝐴𝐵𝐶 is = 2 𝑎𝑐 sin 𝐵
=
1
2
1
1
𝑎𝑐 sin 𝐵 = × 17.20 × 17.92 × sin 78𝑜 51′
2
2
× 17.20 × 17.92 × 0.9811 = 151.2030 ≈ 151.2 mm2
Example 6.4.4.3.
Solve triangle 𝐷𝐸𝐹 and find its area given that 𝐸𝐹 = 35.0 mm, 𝐷𝐸 = 25.0 mm and ∠𝐸 = 64𝑜
Solution 6.4.4.3.
Triangle 𝐷𝐸𝐹 is shown below in Figure 6.4.4.3.
𝐷
𝑒
𝑓 = 25.0 𝑚𝑚
64𝑜
𝐸
𝑑 = 35.0 𝑚𝑚
𝐹
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.4.3
We will apply the cosine rule in this problem as follows. The side 𝑒 = 𝐷𝐹 is not known. We
know the other 2 sides and the angle between them.
227
The cosine rules says,
𝑒 2 = 𝑑 2 + 𝑓 2 − 2𝑑𝑓 cos 𝐸
𝑒 2 = 35.02 + 25.02 − [2(35.0)(25.0) cos 64𝑜 ]
𝑒 2 = 1225.0 + 625.0 − [2 × 35.0 × 25.0 × 0.4384 ]
𝑒 2 = 1850 − 767.2 = 1082.8 ≈ 1083
𝑒 = √1083 = 32.91 mm
With the value of 𝑒 = 32.91 we use it in the sine rule to evaluate the other angles.
32.91 25.0
=
sin 64 sin 𝐹
sin 𝐹 =
25.0 × sin 64𝑜
32.91
sin 𝐹 =
25.0 × 0.8988
32.91
sin 𝐹 =
22.4699
32.91
sin 𝐹 = 0.6828
𝐹 = sin−1 0.6828 = 43.06𝑜 = 43𝑜 4′
228
OR
𝐹 = sin−1 0.6828 = 136.9398𝑜 = 136𝑜 56′
𝐹 = 136𝑜 56′ is not possible in this case since 136𝑜 56′ + 64𝑜 = 200𝑜 56′ > 180𝑜
We can accept 𝐹 = 43𝑜 4′ as the valid or possible angle
We proceed to evaluate angle as ∠𝐷 = 180𝑜 − 64𝑜 − 43𝑜 4′ = 72𝑜 56′
We now evaluate the area of the triangle
1
Area of triangle 𝐷𝐸𝐹 = 2 𝑑𝑓 sin 𝐸
𝐴𝑟𝑒𝑎 =
1
(35.0)(25.0) sin 64𝑜
2
1
𝐴𝑟𝑒𝑎 = 2 × 35.0 × 25.0 × 0.8988 = 393.2 mm2
Example 6.4.4.4.
A triangle 𝐴𝐵𝐶 has sides 𝑎 = 9.0 cm, 𝑏 = 7.5 cm, and 𝑐 = 6.5 cm. Determine its three angles
and its area.
Solution 6.4.4.4.
The triangle is shown in Figure 6.4.4.4. below.
𝐴
𝑏 = 7.5 𝑐𝑚
𝑐 = 6.5 𝑐𝑚
𝐵
𝑎 = 9.0 𝑐𝑚
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.4.4.
229
𝐶
We will apply the cosine rule to find any one of the angles, say angle ∠𝐴
The cosine rule says,
𝑎2 = 𝑏2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
We rearrange as follows
2𝑏𝑐 cos 𝐴 = 𝑏2 + 𝑐 2 − 𝑎2
cos 𝐴 =
cos 𝐴 =
cos 𝐴 =
cos 𝐴 =
𝑏2 + 𝑐 2 − 𝑎2
2𝑏𝑐
7.52 + 6.52 − 9.02
2(7.5)(6.5)
56.25 + 42.25 − 81.00
97.5
98.5 − 81.00 17.5
=
= 0.1795
97.5
97.5
cos 𝐴 = 0.1795
𝐴 = cos −1 0.1795 = 79.6601 ≈ 79.66𝑜 or 360 − 79.66 = 280.34𝑜 which is impossible for an
acute angle. Since the cosine of angle, ∠𝐴 is positive.
Special Note:
If the cosine of angle 𝐴, i.e., cos 𝐴 had been negative, the angle 𝐴 would be an obtuse angle i.e.,
will lie between 90𝑜 and 180𝑜
230
Now we apply the sine rule to evaluate the other angles:
9.0
7.5
=
sin 79.66𝑜 sin 𝐵
from which we get
sin 𝐵 =
7.5 × sin 79.66𝑜
9.0
sin 𝐵 =
7.5 × 0.9838
9.0
sin 𝐵 = 0.8198
Hence 𝐵 = sin−1 0.8198 = 55.0681𝑜
and 𝐶 = 180𝑜 − 79.66𝑜 − 55.06𝑜 = 45.28𝑜.
The area of the triangle is given by
√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
where,
𝑠=
𝑎 + 𝑏 + 𝑐 9.0 + 7.5 + 6.5 23.5
=
=
= 11.5
2
2
2
Hence the area is
𝑎𝑟𝑒𝑎 = √11.5(11.5 − 9.0)(11.5 − 7.5)(11.5 − 6.5)
𝑎𝑟𝑒𝑎 = √11.5(2.5)(4.0)(5.0)
𝑎𝑟𝑒𝑎 = √575 = 23.9792 ≈ 23.98 cm2
231
Alternatively,
𝑎𝑟𝑒𝑎 =
𝑎𝑟𝑒𝑎 =
𝑎𝑟𝑒𝑎 =
1
𝑎𝑏 sin 𝐶
2
1
(9.0)(7.5) sin 45.28
2
1
× 9.0 × 7.5 × 0.7106
2
𝑎𝑟𝑒𝑎 = 23.98 cm2
Example 6.4.4.5.
Solve triangle 𝑋𝑌𝑍, shown in Figure 6.4.4.5., and find its area given that 𝑌 = 128𝑜 , the side
𝑋𝑌 = 7.2 cm and the side 𝑌𝑍 = 4.5 cm.
Solution 6.4.4.5.
Triangle 𝑋𝑌𝑍 is shown in Figure 6.4.4.5. below.
𝑋
𝑦
𝑧 = 7.2 𝑐𝑚
128𝑜
𝑍
𝑌
𝑥 = 4.5 𝑐𝑚
𝐹𝑖𝑔𝑢𝑟𝑒 6.4.4.5.
We apply the cosine rule:
𝑦 2 = 𝑥 2 + 𝑧 2 − 2𝑥𝑧 cos 𝑌
𝑦 2 = 4.52 + 7.22 − [2(4.5)(7.2) cos 128𝑜 ]
𝑦 2 = 72.09 + 39.89736
232
𝑦 2 = 111.98736
𝑦 = √111.98736 = 10.5824 ≈ 10.58 cm
We apply the sine rule:
10.58
7.2
=
𝑜
sin 128
sin 𝑍
from which,
sin 𝑍 =
sin 𝑍 =
7.2 × sin 128𝑜
10.58
7.2 × 0.7880
= 0.5363
10.58
𝑍 = sin−1 0.5363 = 32.4297𝑜
OR
𝑍 = 180𝑜 − 32.4297𝑜 = 147.5703𝑜
which is impossible in this case.
The angle ∠𝑋 = 180𝑜 − 128𝑜 − 32.4297𝑜 = 19.5703 ≈ 19.57𝑜
We calculate the area as:
𝐴𝑟𝑒𝑎 =
𝐴𝑟𝑒𝑎 =
1
𝑥𝑧 sin 𝑌
2
1
(4.5)(7.2) sin 128𝑜
2
233
𝐴𝑟𝑒𝑎 =
1
× 4.5 × 7.2 × 0.7880
2
𝐴𝑟𝑒𝑎 = 12.7658 ≈ 12.77cm2
234
6.5.0. Session 5 - Trigonometric Identities and Equations
6.5.1. Trigonometric Identities
A trigonometric identity is a relationship that is true for all values for the unknown variable. The
following are some examples of trigonometric identities:
6.5.2. The Reciprocal Identities.
sin 𝜃
1. tan 𝜃 = cos 𝜃
2. cot 𝜃 =
cos 𝜃
sin 𝜃
1
3. sec 𝜃 = cos 𝜃
4. 𝑐𝑜𝑠𝑒𝑐 𝜃 =
1
sin 𝜃
1
5. cot 𝜃 = tan 𝜃
6.5.3. The Pythagorean Identities
𝑐
𝑏
𝜃
𝑎
𝐹𝑖𝑔𝑢𝑟𝑒 6.5.3.
= 22.31 𝑐𝑚
If we apply the Pythagoras’ theorem to the right-angled triangle in Figure 6.5.3. above, it gives
us:
𝑎2 + 𝑏2 = 𝑐 2
Eqn 6.5.3.1.
Dividing each term of Eqn 6.5.1.2. by 𝑐 2 gives:
𝑎2 𝑏2 𝑐 2
+
=
𝑐2 𝑐2 𝑐2
235
𝑎2 𝑏2
+
=1
𝑐2 𝑐2
𝑎 2
𝑏 2
( ) +( ) =1
𝑐
𝑐
(cos 𝜃)2 + (sin 𝜃)2 = 1
cos 2 𝜃 + sin2 𝜃 = 1
Eqn 6.5.3.2.
Dividing each term of Eqn 6.5.3.1. by 𝑎2 gives:
𝑎2 𝑏2 𝑐 2
+
=
𝑎2 𝑎2 𝑎2
𝑏 2
𝑐 2
1+( ) =( )
𝑎
𝑎
1 + tan2 𝜃 = sec 2 𝜃
Eqn 6.5.3.3.
Dividing each term of Eqn 6.5.3.1. by 𝑏2 gives:
𝑎2 𝑏2 𝑐 2
+
=
𝑏2 𝑏2 𝑏2
𝑎 2
𝑐 2
( ) +1=( )
𝑏
𝑏
cot 2 𝜃 + 1 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃
Eqn 6.5.3.4.
Equations (6.5.3.2.), (6.5.3.3.) and (6.5.3.4.) are three further examples of trigonometric
identities. They are known as the Pythagorean Identities.
236
6.5.4. The Even-Odd Identities
sin(−𝑥) = − sin 𝑥
cos(−𝑥) = cos 𝑥
tan(−𝑥) = −tan (𝑥)
6.5.5. The Cofunction Identities
𝜋
sin ( − 𝑢) = cos 𝑢
2
𝜋
cos ( − 𝑢) = sin 𝑢
2
𝜋
tan ( − 𝑢) = cot 𝑢
2
𝜋
cot ( − 𝑢) = tan 𝑢
2
𝜋
sec ( − 𝑢) = csc 𝑢
2
𝜋
csc ( − 𝑢) = sec 𝑢
2
6.5.6. Simplifying Trigonometric Expressions
Identities enable us to write the same expression in different ways. It is often possible to rewrite
a complicated-looking expression as a much simpler one. To simplify algebraic expressions, we
used factoring, common denominators, and the Special Product Formulas. To simplify
trigonometric expressions, we use these same techniques together with the fundamental
trigonometric identities.
Example 6.5.6.1.
Simplify the expression,
cos 𝑡 + tan 𝑡 sin 𝑡
237
Solution 6.5.6.1.
We start by rewriting the expression in terms of sine and cosine:
We are given
cos 𝑡 + tan 𝑡 sin 𝑡
Which turns out to be a reciprocal identity as in the following
cos 𝑡 + tan 𝑡 sin 𝑡 = cos 𝑡 + (
sin 𝑡
) sin 𝑡
cos 𝑡
We find a common denominator
=
cos 2 𝑡 + sin2 𝑡
cos 𝑡
=
1
cos 𝑡
Because cos 2 𝑡 + sin2 𝑡 = 1
6.5.6.2.
Simplify the expression
sin 𝜃
cos 𝜃
+
cos 𝜃 1 + sin 𝜃
Solution 6.5.6.2.
We combine the fractions by using a common denominator:
sin 𝜃
cos 𝜃
sin 𝜃 (1 + sin 𝜃) + cos 2 𝜃
+
=
cos 𝜃 1 + sin 𝜃
cos 𝜃 (1 + sin 𝜃)
Next, we distribute sin 𝜃
=
sin 𝜃 + sin2 𝜃 + cos 2 𝜃
cos 𝜃 (1 + sin 𝜃)
By the Pythagoras identity, sin2 𝜃 + cos 2 𝜃 = 1
We cancel out the common terms 1 + sin 𝜃
238
=
sin 𝜃 + 1
cos 𝜃 (1 + sin 𝜃)
=
1
= sec 𝜃
cos 𝜃
6.5.7. Proving Trigonometric Identities
Many identities follow from the fundamental identities. In the examples that follow, we will
learn how to prove that a given trigonometric equation is an identity, and in the process we will
see how to discover new identities.
First, it is easy to decide when a given equation is not an identity. All we need to do is show that
the equation does not hold for some value of the variable (or variables). Thus, the equation
sin 𝑥 + cos 𝑥 = 1
𝜋
is not an identity because when 𝑥 = 4 , we have,
sin
𝜋
𝜋 √2 √2
+ cos =
+
= √2 ≠ 1
4
4
2
2
To verify that a trigonometric equation is an identity, we transform one side of the equation into
the other side by a series of steps, each of which is itself an identity.
6.5.8. Guidelines for Proving Trigonometric Identities.
1.
Start with one side: - Pick one side of the equation and write it down. The goal is to
transform it into the other side. It is usually easier to start with the more complicated side.
2.
Use known Identities: Use algebra and the identities you know to change the side you
started with. Bring fractional expressions to a common denominator, factor, and use the
fundamental identities to simplify expressions.
3.
Convert to sines and cosines: if you are stuck, you may find it helpful to rewrite all
functions in terms of sines and cosines.
239
Warning:
To prove an identity, we do not just perform the same operations on both sides of the equation.
Example 6.5.8.1.
Verify algebraically that the equation cos 𝜃(sec 𝜃 − cos 𝜃) = sin2 𝜃
Solution 6.5.8.1.
The left-hand side looks more complicated. We start with the left-hand side and try to transform
it into the right-hand side:
𝐿𝐻𝑆 = cos 𝜃(sec 𝜃 − cos 𝜃)
Use the reciprocal identity for sec 𝜃
= cos 𝜃 (
1
− cos 𝜃)
cos 𝜃
Expanding, we get
=
𝑐𝑜𝑠𝜃
− cos 2 𝜃
𝑐𝑜𝑠𝜃
= 1 − cos 2 𝜃 = sin2 𝜃 = 𝑅𝐻𝑆
Example 6.5.8.2.
Prove the identity sin2 𝜃 cot 𝜃 sec 𝜃 = sin 𝜃
Solution 6.5.8.2.
With trigonometric identities it is necessary to start with the left-hand side (LHS) and attempt to
make it equal to the right-hand side (RHS) or vice-versa.
It is often useful to change all the trigonometric ratios into sines and cosines where possible.
240
Thus
𝐿𝐻𝑆 = sin2 𝜃 𝑐𝑜𝑡𝜃 𝑠𝑒𝑐𝜃
𝐿𝐻𝑆 = sin2 𝜃 (
cos 𝜃
1
)(
)
sin 𝜃 cos 𝜃
By cancelling, we have,
𝐿𝐻𝑆 = sin 𝜃 = 𝑅𝐻𝑆
Example 6.5.8.3.
Prove that:
tan 𝑥 + sec 𝑥
tan 𝑥 = 1
sec 𝑥 (1 + sec 𝑥 )
Solution 6.5.8.3.
The LHS
=
tan 𝑥 + sec 𝑥
tan 𝑥
sec 𝑥 (1 + sec 𝑥 )
We begin by converting each term into sines and cosines where possible
sin 𝑥
1
+
cos 𝑥 cos 𝑥
=
sin 𝑥
1
cos
(cos 𝑥 ) (1 + ( 1 𝑥 ))
cos 𝑥
Please note that the division
sin 𝑥
cos 𝑥
1
cos 𝑥
produces or results in the following by turning the denominator
upside down as follows:
sin 𝑥
cos 𝑥 = ( 𝑠𝑖𝑛𝑥 ) (𝑐𝑜𝑠𝑥 )
1
𝑐𝑜𝑠𝑥
1
cos 𝑥
241
sin 𝑥 + 1
cos 𝑥
=
1
sin 𝑥 cos 𝑥
(
) [1 + (
)(
)]
cos 𝑥
cos 𝑥
1
sin 𝑥 + 1
cos 𝑥
=
(
=(
1
) [1 + sin 𝑥]
cos 𝑥
sin 𝑥 + 1
cos 𝑥
)(
)
cos 𝑥
1 + sin 𝑥
= 1 = 𝑅𝐻𝑆
Example 6.5.8.4.
Prove that:
1 + cot 𝜃
= cot 𝜃
1 + tan 𝜃
Solution 6.5.8.4.
We begin by converting trigonometric ratios cot 𝜃 and tan 𝜃 into possible sines and cosines
cos 𝜃
1+
1 + cot 𝜃
sin 𝜃
𝐿𝐻𝑆 =
=
1 + tan 𝜃 1 + sin 𝜃
cos 𝜃
sin 𝜃 + cos 𝜃
sin 𝜃
𝐿𝐻𝑆 =
cos 𝜃 + sin 𝜃
cos 𝜃
Applying the division rule of two quotients, and cancelling out the (sin 𝜃 + cos 𝜃), we have
242
𝐿𝐻𝑆 = (
sin 𝜃 + cos 𝜃
cos 𝜃
)(
)
sin 𝜃
cos 𝜃 + sin 𝜃
𝐿𝐻𝑆 =
cos 𝜃
= cot 𝜃
sin 𝜃
Example 6.5.8.5.
Show that:
cos 2 𝜃 − sin2 𝜃 = 1 − 2 sin2 𝜃
Solution 6.5.8.5.
From Eqn 6.5.3.2.
cos 2 𝜃 + sin2 𝜃 = 1
From which we can express
cos 2 𝜃 = 1 − sin2 𝜃
Hence the LHS
𝐿𝐻𝑆 = cos 2 𝜃 − sin2 𝜃
𝐿𝐻𝑆 = (1 − sin2 𝜃) − sin2 𝜃
= 1 − sin2 𝜃 − sin2 𝜃
= 1 − 2 sin2 𝜃 = 𝑅𝐻𝑆
Example 6.5.8.6.
Prove that:
√
1 − sin 𝑥
= sec 𝑥 − tan 𝑥
1 + sin 𝑥
243
Solution 6.5.8.6.
𝐿𝐻𝑆 = √
1 − sin 𝑥
1 + sin 𝑥
We multiply both numerator and denominator by (1 − sin 𝑥)
=√
(1 − sin 𝑥)(1 − sin 𝑥)
(1 + sin 𝑥)(1 − sin 𝑥)
=√
(1 − sin 𝑥)2
(1 − sin2 𝑥)
Since cos 2 𝑥 + sin2 𝑥 = 1, then 1 − sin2 𝑥 = cos 2 𝑥, we have
√
(1 − sin 𝑥)2
(1 − sin 𝑥)2
√
=
(1 − sin2 𝑥)
cos 2 𝑥
Taking the square root, we have
√
(1 − sin 𝑥)2 1 − sin 𝑥
=
cos 2 𝑥
cos 𝑥
1 − sin 𝑥
1
sin 𝑥
=
−
cos 𝑥
cos 𝑥 cos 𝑥
1
sin 𝑥
−
= sec 𝑥 − tan 𝑥
cos 𝑥 cos 𝑥
244
sec 𝑥 − tan 𝑥 = 𝑅𝐻𝑆
245
6.5.9. Trigonometric Equations
Equations which contain trigonometric ratios are called trigonometric equations. There are
usually an infinite number of solutions to such equations; however, solutions are often restricted
to those between 0𝑜 and 360𝑜. A knowledge of angles of any magnitude is essential in the
solution of trigonometric equations and calculators cannot be relied upon to give all the
solutions.
Figure 6.5.9. shows a summary for angles of any magnitude.
90𝑜
Sine (and
cosecant positive)
All Positive
Second Quadrant
First Quadrant
180𝑜
0𝑜
360𝑜
Third Quadrant
Fourth Quadrant
Tangent (and
cotangent positive)
Cosine (and
secant positive)
270𝑜
Figure 6.5.9.
246
6.5.10. Equations of the Type 𝑎 sin2 𝐴 + 𝑏 sin 𝐴 + 𝑐 = 0
(i)
When 𝑎 = 0,
𝑐
𝑏 sin 𝐴 + 𝑐 = 0,
hence sin 𝐴 = − 𝑏
𝑐
and 𝐴 = sin−1 (− 𝑏 ).
There are two values of 𝐴 between 0𝑜 and 360𝑜 that satisfy such an equation
𝑐
provided −1 ≤ 𝑏 ≤ 1
(ii)
𝑐
When 𝑏 = 0, 𝑎 sin2 𝐴 + 𝑐 = 0, hence sin2 𝐴 = − 𝑎 ,
and
𝑐
sin 𝐴 = √− 𝑎
𝑐
𝐴 = sin−1 √− 𝑎
If either 𝑎 or 𝑐 is a negative number, then the value within the square root sign is
positive. Since when a square root is taken there is a positive and negative answer,
there are four values of 𝐴 between 0𝑜 and 360𝑜 which satisfy such an equation,
𝑐
provided −1 ≤ 𝑏 ≤ 1.
(iii)
When 𝑎, 𝑏 and 𝑐 are all non-zero:
𝑎 sin2 𝐴 + 𝑏 sin 𝐴 + 𝑐 = 0 is a quadratic equation in which the unknown is sin 𝐴.
The solution of a quadratic equation is obtained either by factorising (if possible) or
by using the quadratic formula:
−𝑏 ± √𝑏2 − 4𝑎𝑐
sin 𝐴 =
2𝑎
(iv)
Often the trigonometric identities cos 2 𝐴 + sin2 𝑎 = 1, 1 + tan2 𝐴 = sec 2 𝐴, and
cot 2 𝐴 + 1 = 𝑐𝑜𝑠𝑒𝑐 2 𝐴 need to be used to reduce equations to one of the above forms
247
6.5.11. Worked Examples of Trigonometric Equations
Type I
Example 6.5.11.1.
Solve the trigonometric equation: 5 sin 𝜃 + 3 = 0 for values of 𝜃 from 0𝑜 to 360𝑜
Solution 6.5.11.1.
We have been given
5 sin 𝜃 + 3 = 0
from which we get
5 sin 𝜃 = −3
sin 𝜃 = −
3
= −0.6000
5
𝜃 = sin−1(−0.6000) = −36.8699
From Figure 6.5.9., we can see that Sine is negative in the third and fourth quadrants. The angle
𝜃 is the acute angle 36.8699𝑜 that is measured from the horizontal axis along the 180𝑜 and
along the 360𝑂 in the third and fourth quadrants respectively.
The angle we are seeking lie in both the third and fourth quadrants. The angle in the third
quadrant is obtained by adding the acute angle 36.8699𝑜 to 180𝑜. Hence, we calculate the angle
180𝑜 + 36.86990 = 216.8699𝑜
Also in the fourth quadrant, we obtain the sought after angle by subtracting the acute angle
36.8699𝑜 from 360𝑜 to obtain 360𝑜 − 36.8699𝑜 = 323.1301𝑜 .
Hence, the solution of the equation 5 sin 𝜃 + 3 = 0 for values of 𝜃 between 0𝑜 and 360𝑜 is 𝜃 =
216.8699𝑜 and 𝜃 = 323.1301𝑜
248
Example 6.5.11.2.
Solve: 1.5 tan 𝑥 − 1.8 = 0 for values of 𝑥 within the interval 0𝑜 ≤ 𝑥 ≤ 360𝑜 .
Solution 6.5.11.2.
From the given equation, 1.5 tan 𝑥 − 1.8 = 0, we have
tan 𝑥 =
1.8
= 1.2000
1.5
Hence,
𝑥 = tan−1 (1.2) = 50.1944
We note that, Tangent is positive in the first and third quadrants (see Figure 6.5.3). The acute
angle is 50.1944𝑜 .
Hence, 𝑥 = 50.1944𝑜 in the first quadrant and 𝑥 = 180𝑜 + 50.1944 = 230.1944 in the third
quadrant.
Example 6.5.11.3.
Solve: 4 sec 𝑡 = 5 for values of 𝑡 between 0𝑜 and 360𝑜.
Solution 6.5.11.3.
We have been given
4 sec 𝑡 = 5
Divide both sides by 4.
sec 𝑡 =
Note that 𝑠𝑒𝑐𝑎𝑛𝑡 =
1
𝑐𝑜𝑠𝑖𝑛𝑒
5
= 1.2500
4
, and it is positive in the first and fourth quadrants. The acute angle is
given by:
249
sec 𝑡 =
1
5
=
cos 𝑡 4
Taking reciprocals on each side,
cos 𝑡 =
4
= 0.8000
5
𝑡 = cos −1 (0.8000)
𝑡 = 36.8699𝑜
The acute angle is 𝑡 = 36.8699𝑜 and 𝑡 = 360𝑜 − 36.8699𝑜 = 323.1301𝑜
250
Type II
Example 6.5.11.4.
Solve: 2 − 4 cos 2 𝐴 = 0 for values of 𝐴 in the range 0𝑜 < 𝐴 < 360𝑜.
Solution 6.5.11.4.
From the given equation
2 − 4 cos 2 𝐴 = 0
We have
cos 2 𝐴 =
2
= 0.5
4
Hence
cos 𝐴 = √0.5000 = ±0.7071
𝐴 = cos −1(±0.7071)
Cosine is positive in quadrants one and four, and it is negative in quadrants two and three.
Therefore, in this situation there are four possible solutions, one in each quadrant.
The acute angle is 𝐴 = 45𝑜 in the first quadrant
The second value for 𝐴 is in the second quadrant, and it is given by 𝐴 = 180𝑜 − 45𝑜 = 135𝑜
The third value for 𝐴 is in the third quadrant, and it is given by 𝐴 = 180𝑜 + 45𝑜 = 225𝑜
The fourth value for 𝐴 is in the fourth quadrant, and it is given by 𝐴 = 360𝑜 − 45𝑜 = 315𝑜
Hence 𝐴 = 45𝑜 , 135𝑜 , 225𝑜 , or 315𝑜 .
251
Example 6.5.11.5.
1
Solve: 2 cot 2 𝑦 = 1.3 for 0𝑜 < 𝑦 < 360𝑜 .
Solution 6.5.11.5.
1
From the given equation, 2 cot 2 𝑦 = 1.3 we have,
cot 2 𝑦 = 2 × 1.3 = 2.6
Hence
cot 𝑦 = ±√2.6 = ±1.6125
We note that
1
cot 𝑦 = tan 𝑦 = ±1.6125
Taking the reciprocals on each side
1
1
= tan 𝑦 =
= ±0.6202
𝑐𝑜𝑡𝑦
±1.6125
𝑦 = tan−1(±0.6202)
The acute angle is 𝑦 = 31.8053𝑜
Again, we have four solutions, one in each quadrant.
The acute angle in the first quadrant is 𝑦 = 31.8053𝑜
The angle in the second quadrant is 𝑦 = 180𝑜 − 31.8053𝑜 = 148.1947𝑜
The acute angle in the third quadrant is 𝑦 = 180𝑜 + 31.8053𝑜 = 211.8053𝑜
252
The acute angle in the fourth quadrant is 𝑦 = 360𝑜 − 31.8053𝑜 = 328.1947𝑜
Hence
𝑦 = 31.81𝑜 , 148.19𝑜 , 211.81𝑜 and 328.19𝑜
253
Type III
Example 6.5.11.6.
Solve the equation:
8 sin2 𝜃 + 2 sin 𝜃 − 1 = 0, for all values of 𝜃 between 0𝑜 and 360𝑜 .
Solution 6.5.11.6.
We have been given,
8 sin2 𝜃 + 2 sin 𝜃 − 1 = 0
We begin by factorising the quadratic equation to give
8 sin2 𝜃 − 2 sin 𝜃 + 4 sin 𝜃 − 1 = 0
2 sin 𝜃 (4 sin 𝜃 − 1) + (4 sin 𝜃 − 1) = 0
(2 sin 𝜃 + 1) (4 sin 𝜃 − 1) = 0
Which means
2 sin 𝜃 + 1 = 0
OR
4 sin 𝜃 − 1 = 0
Considering
2 sin 𝜃 + 1 = 0
We get
2 sin 𝜃 = −1
sin 𝜃 = −
1
= −0.5
2
𝜃 = sin−1 (−0.5) = − 30𝑜
254
Since sine is negative in the third and fourth quadrants, the acute angle in the third quadrant will
be added to 180𝑜 to give 180𝑜 + 30𝑜 = 210𝑜 , and the acute angle in the fourth quadrant will be
subtracted from 360𝑜 to give 360𝑜 − 300 = 330𝑜
We consider
4 sin 𝜃 − 1 = 0
We have
sin 𝜃 =
1
= 0.25
4
𝜃 = sin−1(0.25) = 14.4775𝑜
Since sine is positive in the first and second quadrants, the value of 𝜃 in the first quadrant is
14.4775𝑜 ≈ 14.48𝑜
Also in the second quadrant, the value of 𝜃 = 180𝑜 − 14.48𝑜 = 165.62𝑜 .
The values of 𝜃 are 𝜃 = 14.48𝑜 , 165.52𝑜 , 210𝑜 , 310𝑜
Note that, instead of factorising, we could have used the quadratic formula
−𝑏 ± √𝑏2 − 4𝑎𝑐
θ=
2𝑎
Example 6.5.11.7.
Solve:
6 cos 2 𝜃 + 5 cos 𝜃 − 6 = 0
for values of 𝜃 from 0𝑜 to 360𝑜.
255
Solution 6.5.11.7.
We begin by factorising the quadratic
6 cos 2 𝜃 + 5 cos 𝜃 − 6 = 0
to give
6 cos 2 𝜃 + 9 cos 𝜃 − 4 cos 𝜃 − 6 = 0
Which simplifies as
3 cos 𝜃 (2 cos 𝜃 + 3) − 2(2 cos 𝜃 + 3) = 0
(3 cos 𝜃 − 2) (2 cos 𝜃 + 3) = 0
Hence
3 cos 𝜃 − 2 = 0
OR
2 cos 𝜃 + 3 = 0
For
3 cos 𝜃 − 2 = 0
cos 𝜃 =
2
= 0.6667
3
𝜃 = cos −1(0.6667) = 48.1871𝑜
For
2 cos 𝜃 + 3 = 0
We get
cos 𝜃 = −
3
= −1.5
2
256
The minimum of cosine is −1, hence the expression 2 cos 𝜃 + 3 = 0 has no solution, so we
neglect it.
Hence the angle in the first quadrant is
𝜃 = cos −1(0.6667) = 48.1871𝑜
and the angle in the fourth quadrant
𝜃 = 360𝑜 − 48.1871𝑜 = 311.8129𝑜
𝜃 = 48.19𝑜 , 311.81𝑜
257
Type IV
Example 6.5.11.8.
Solve:
5 cos 2 𝑡 + 3 sin 𝑡 − 3 = 0
for values of 𝑡 from 0𝑜 to 360𝑜 .
Solution 6.5.11.8.
We have been given
5 cos 2 𝑡 + 3 sin 𝑡 − 3 = 0
We know
cos 2 𝑡 + sin2 𝑡 = 1
And
cos 2 𝑡 = 1 − sin2 𝑡
We can substitute for cos 2 𝑡 in
5 cos 2 𝑡 + 3 sin 𝑡 − 3 = 0
To give
5 (1 − sin2 𝑡) + 3 sin 𝑡 − 3 = 0
We expand to get
5 − 5 sin2 𝑡 + 3 sin 𝑡 − 3 = 0
−5 sin2 𝑡 + 3 sin 𝑡 + 2 = 0
5 sin2 𝑡 − 3 sin 𝑡 − 2 = 0
We factorise the quadratic in sin 𝑡 to get
5 sin2 𝑡 − 5 sin 𝑡 + 2 sin 𝑡 − 2 = 0
5 sin 𝑡(sin 𝑡 − 1) + 2(sin 𝑡 − 1) = 0
258
(5 sin 𝑡 + 2)(sin 𝑡 − 1) = 0
Hence
5 sin 𝑡 + 2 = 0
OR
sin 𝑡 − 1 = 0
For
5 sin 𝑡 + 2 = 0
sin 𝑡 = −
2
= −0.4
5
𝑡 = sin−1(−0.4) = −23.5781𝑜
Sine is negative in the third and fourth quadrants. The angle = −23.5781𝑜 corresponds to =
180𝑜 + 23.5781𝑜 = 203.5782𝑜 ≈ 203.58𝑜 in the third quadrant
Also, the angle = −23.5781𝑜 corresponds to = 360𝑜 − 23.5781𝑜 = 336.4219𝑜 ≈ 336.42𝑜 in
the fourth quadrant
Also, for
sin 𝑡 − 1 = 0
sin 𝑡 = 1
𝑡 = sin−1(1) = 90𝑜
Hence
𝜃 = 90𝑜 , 203.58𝑜 , 𝑜𝑟 336.42𝑜
259
Example 6.5.11.9.
Solve: 18 sec 2 𝐴 − 3 tan 𝐴 = 21 for values of 𝐴 between 0𝑜 and 360𝑜.
Solution 6.5.11.9.
We know that
1 + tan2 𝐴 = sec 2 𝐴
We substitute for sec 2 𝐴 in 18 sec 2 𝐴 − 3 tan 𝐴 = 21, which gives
18(1 + tan2 𝐴) − 3 tan 𝐴 = 21
We expand to get
18 + 18 tan2 𝐴 − 3 tan 𝐴 − 21 = 0
18 tan2 𝐴 − 3 tan 𝐴 − 3 = 0
We factorise the quadratic as follows
18 tan2 𝐴 − 9 tan 𝐴 + 6 tan 𝐴 − 3 = 0
We simplify as
9 tan 𝐴( 2 tan 𝐴 − 1) + 3(2 tan 𝐴 − 1) = 0
(9 tan 𝐴 + 3) ( 2 tan 𝐴 − 1) = 0
Hence
9 tan 𝐴 + 3 = 0
OR
2 tan 𝐴 − 1 = 0
260
For
9 tan 𝐴 + 3 = 0
3
1
tan 𝐴 = − = − = −0.3333
9
3
𝐴 = tan−1 (−0.3333) = −18.4332𝑜
Tangent is negative in the second and fourth quadrants
The angle in the second quadrant is 180𝑜 − 18.4332𝑜 = 161.5668𝑜
The angle in the fourth quadrant is 360𝑜 − 18.4332𝑜 = 341.5668𝑜
Also, for
2 tan 𝐴 − 1 = 0
tan 𝐴 =
1
= 0.5000
2
𝐴 = tan−1 (0.5000) = 26.5651𝑜
Tangent is positive in the first and third quadrants.
The angle 26.5651𝑜 obviously lies in the first quadrant.
The angle in the third quadrant is 180𝑜 + 26.5651𝑜 = 206.5651𝑜
Hence angle 𝐴 = 26.57𝑜 , 161.57𝑜 , 206.57𝑜 𝑜𝑟 341.57𝑜
261
Example 6.5.11.10.
Solve: 3 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 5 = 4 cot 𝜃 in the range 0 < 𝜃 < 360𝑜 .
Solution 6.5.11.10.
We know that
cot 2 𝜃 + 1 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃
We substitute for 𝑐𝑜𝑠𝑒𝑐 2 𝜃 in 3 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 5 = 4 cot 𝜃 to give us
3 (cot 2 𝜃 + 1) − 5 = 4 cot 𝜃
3 cot 2 𝜃 + 3 − 5 = 4 cot 𝜃
3 cot 2 𝜃 − 4 cot 𝜃 − 2 = 0
Since we are not able to factorise the left-hand side, we will use the quadratic formula
cot 𝜃 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
With 𝑎 = 3, 𝑏 = −4, 𝑐 = −2
cot 𝜃 =
−(−4) ± √(−4)2 − 4(3)(−2)
2 (3 )
cot 𝜃 =
4 ± √16 + 24
6
cot 𝜃 =
cot 𝜃 =
4 ± √40
6
10.3246
= 1.7208
6
262
OR
cot 𝜃 =
−2.3246
= −0.3874
6
We know,
cot 𝜃 =
1
= −0.3874
tan 𝜃
cot 𝜃 =
1
= 1.7208
tan 𝜃
From
cot 𝜃 =
tan 𝜃 =
1
= −0.3874
tan 𝜃
1
= −2.5813
−0.3874
𝜃 = tan−1(−2.5813) = −68.8235𝑜
Tangent is negative in the second and fourth quadrants
Hence in the second quadrant, the angle is
180𝑜 − 68.8235𝑜 = 111.1765𝑜 ≈ 111.18𝑜
And in the fourth quadrant, the angle is
360𝑜 − 68.8235𝑜 = 291.1765𝑜 ≈ 291.18𝑜
263
Also, from
cot 𝜃 =
tan 𝜃 =
1
= 1.7208
tan 𝜃
1
= 0.5811
1.7208
𝜃 = tan−1(0.5811) = 30.1609𝑜
Tangent is positive in the first and third quadrants.
In the first quadrant, the angle is 30.1609𝑜
In the third quadrant, the angle is 180𝑜 + 30.1609𝑜 = 210.1609𝑜
Hence the angle 𝜃 = 30.16𝑜 , 111.18𝑜 , 291.18𝑜 𝑜𝑟 210.16𝑜
264
6.6.0. Session 6 - Compound Angles
6.6.1. Compound Angle Formulae
An electric current 𝑖 may be expressed as 𝑖 = 5 sin (𝜔𝑡 − 0.33) amperes. Similarly, the
displacement 𝑥 of a body from a fixed point can be expressed as 𝑥 = 10 sin(2𝑡 + 0.67) metres.
The angles (𝜔𝑡 − 0.33) and (2𝑡 + 0.67) are called compound angles because they are the sum
or difference of two angles.
The compound angle formulae for sines and cosines of the sum and difference of two angles 𝐴
and 𝐵 are:
sin (𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
sin (𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵
cos (𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
Please Note:
sin(𝐴 + 𝐵) is not equal to (sin 𝐴 + sin 𝐵) and so on.
The formulae stated above can be used to derive two further compound angle formulae:
tan(𝐴 + 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
tan(𝐴 − 𝐵) =
tan 𝐴 − tan 𝐵
1 + tan 𝐴 tan 𝐵
The compound-angle formulae are true for all values of 𝐴 and 𝐵, and by substituting values of 𝐴
and 𝐵, and by substituting values of 𝐴 and 𝐵 into the formulae they may be shown to be true.
265
Example 6.6.1.1.
Expand and simplify the following expressions:
i. sin(𝜋 + 𝛼)
ii. − cos(90𝑜 + 𝛽)
iii. sin(𝐴 − 𝐵) − sin(𝐴 + 𝐵)
Solution 6.6.1.1.
i.
sin(𝜋 + 𝛼) = sin 𝜋 cos α + cos 𝜋 sin 𝛼
= (0) cos 𝛼 + (−1) sin 𝛼
= − sin 𝛼
ii.
−cos (90𝑜 + 𝛽)
−[cos 90𝑜 cos 𝛽 − sin 90𝑜 sin 𝛽]
−[(0) cos 𝛽 − (1) sin 𝛽] = sin 𝛽
iii.
sin(𝐴 − 𝐵) − sin (𝐴 + 𝐵)
[sin 𝐴 cos 𝐵 − sin 𝐵 cos 𝐴] − [sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴]
−2 cos 𝐴 sin 𝐵
Exercise 6.6.1.2.
Prove that:
𝜋
cos(𝑦 − 𝜋) + sin (𝑦 + ) = 0
2
Solution 6.6.1.2.
𝜋
We expand cos(𝑦 − 𝜋) separately and expand sin (𝑦 + 2 ) separately and bring them together
later
cos(𝑦 − 𝜋) = cos 𝑦 cos 𝜋 + sin 𝑦 sin 𝜋
= cos 𝑦 (−1) + sin 𝑦 (0)
266
cos(𝑦 − 𝜋) = −cos 𝑦
Also,
𝜋
𝜋
𝜋
sin (𝑦 + ) = sin 𝑦 cos + cos 𝑦 sin
2
2
2
= sin 𝑦 (0) + cos 𝑦 (1)
𝜋
sin (𝑦 + ) = cos 𝑦
2
Bringing them together, we get
𝜋
cos(𝑦 − 𝜋) + sin (𝑦 + ) = − cos 𝑦 + cos 𝑦 = 0
2
Example 6.6.1.3.
Show that
𝜋
𝜋
tan (𝑥 + ) tan (𝑥 − ) = −1
4
4
Solution 6.6.1.3.
𝜋
𝜋
tan (𝑥 + ) tan (𝑥 − ) = −1
4
4
We will do the expansion separately for the addition and for the subtraction.
Therefore,
𝜋
tan 𝑥 + tan
𝜋
4
tan (𝑥 + ) =
𝜋
4
1 − tan 𝑥 tan
4
267
𝜋
We note that tan 4 = 1
𝜋
tan 𝑥 + 1
tan (𝑥 + ) =
4
1 − tan 𝑥(1)
𝜋
tan 𝑥 + 1
tan (𝑥 + ) =
4
1 − tan 𝑥
Also, for the subtraction part, we get
𝜋
tan 𝑥 − tan 4
𝜋
tan 𝑥 − 1
tan (𝑥 − ) =
=
𝜋
4
1 + tan 𝑥
1 + tan 𝑥 tan
4
Bringing them together, we have
𝜋
𝜋
tan 𝑥 + 1
tan 𝑥 − 1
tan (𝑥 + ) tan (𝑥 − ) = (
) × (
)
4
4
1 − tan 𝑥
1 + tan 𝑥
𝜋
𝜋
tan 𝑥 − 1 −(1 − 𝑡𝑎𝑛𝑥)
tan (𝑥 + ) tan (𝑥 − ) =
=
4
4
1 − tan 𝑥
1 − 𝑡𝑎𝑛𝑥
𝜋
𝜋
−(1 − 𝑡𝑎𝑛𝑥)
tan (𝑥 + ) tan (𝑥 − ) =
= −1
4
4
1 − 𝑡𝑎𝑛𝑥
𝜋
𝜋
∴ tan (𝑥 + ) tan (𝑥 − ) = −1
4
4
268
Example 6.6.1.4.
If sin 𝑃 = 0.8142 and cos 𝑄 = 0.4432, evaluate, correct to 3 decimal places:
(a) sin (𝑃 − 𝑄)
(b) cos (𝑃 + 𝑄)
(c) tan (𝑃 + 𝑄)
Solution 6.6.1.4.
We have been given values for sin 𝑃 = 0.8142, cos 𝑄 = 0.4432, we do not know sin 𝑄, and we
do not know cos 𝑃.
We need to use the inverses of sine and cosine to determine the actual angles for 𝑃 and for angle
𝑄
We proceed as follows:
Since sin 𝑃 = 0.8142, then 𝑃 = sin−1 0.8142 = 54.5081 ≈ 54.51𝑜
Therefore, cos 𝑃 = cos 54.51𝑜 = 0.5806, and
tan 𝑃 = tan 54.51𝑜 = 1.4025.
Also, from cos 𝑄 = 0.4432, then 𝑄 = cos −1 0.4432 = 63.6918𝑜 ≈ 63.69𝑜
Therefore, sin 63.69𝑜 = 0.8964, and
tan 63.69𝑜 = 2.0225.
We can substitute the values into the compound angle formulas for sums and differences for sine
as follows:
(a) sin(𝑃 − 𝑄) = sin 𝑃 cos 𝑄 − cos 𝑃 sin 𝑄
sin(54.51𝑜 − 63.69𝑜 ) = sin(54.51𝑜 ) cos(63.69𝑜 ) − cos(54.51𝑜 ) sin(63.69𝑜 )
sin(54.51𝑜 − 63.69𝑜 ) = (0.8142)(0.4432) − (0.5806)(0.8964)
269
0.3609 − 0.5204 = −0.1596 ≈ −0.160 to 3 decimal places
(b) cos(𝑃 + 𝑄) = cos 𝑃 cos 𝑄 − sin 𝑃 sin 𝑄
cos(𝑃 + 𝑄) = (0.5806)(0.4432) − (0.8142)(0.8964)
cos(𝑃 + 𝑄) = 0.2573 − 0.7298 = −0.4725 ≈ 0.473
tan 𝑃+tan 𝑄
(c) tan(𝑃 + 𝑄) = 1−tan 𝑃 tan 𝑄
tan(𝑃 + 𝑄) =
(1.4025) + (2.0225)
tan 𝑃 + tan 𝑄
=
1 − tan 𝑃 tan 𝑄 1 − (1.4025)(2.0225)
tan(𝑃 + 𝑄) =
(1.4025) + (2.0225)
3.425
=
1 − (1.4025)(2.0225) 1 − 2.8366
tan(𝑃 + 𝑄) =
3.425
= −1.8649 ≈ −1.865
−1.8366
Example 6.6.1.5.
Solve the equation: 4 sin(𝑥 − 20) = 5 cos 𝑥 for values of 𝑥 between 0𝑜 and 90𝑜
Solution 6.6.1.5.
From the compound angle formula for sums and differences for sine,
4 sin(𝑥 − 20) = 4[sin 𝑥 cos 20𝑜 − cos 𝑥 sin 20𝑜 ] = 5 cos 𝑥
We have
cos 20𝑜 = 0.9397 and sin 20𝑜 = 0.3420
4 sin(𝑥 − 20) = 4[0.9397 sin 𝑥 − 0.3420 cos 𝑥] = 5 cos 𝑥
4 sin(𝑥 − 20) = 3.7588 sin 𝑥 − 1.3680 cos 𝑥 = 5 cos 𝑥
270
Rearrange to group like terms together,
3.7588 sin 𝑥 = 5 cos 𝑥 + 1.3680 cos 𝑥
3.7588 sin 𝑥 = 6.3680 cos 𝑥
sin 𝑥 6.3680
=
= 1.6942
cos 𝑥 3.7588
tan 𝑥 = 1.6942
𝑥 = tan−1 1.6942 = 59.4482𝑜 ≈ 59.442𝑜
𝑥 ≈ 59.442𝑜
OR
𝑥 ≈ 59𝑜 27′
[Check: 𝐿𝐻𝑆 = 4 sin(59.442𝑜 − 20𝑜 ) = 4 sin 39.4420 = 4 × 0.6353 = 2.5412 ≈ 2.54]
[Check: 𝑅𝐻𝑆 = 5 cos 𝑥 = 5 cos 59.442𝑜 = 5 × 0.5084 = 2.5421 ≈ 2.54]
271
6.6.2. Double Angles.
(i.) If, in the compound angle formula for sin (𝐴 + 𝐵), we let 𝐵 = 𝐴 then
sin 2𝐴 = 2 sin 𝐴 cos 𝐴
Also, for example,
sin 4𝐴 = 2 sin 2𝐴 cos 2𝐴
And
sin 8𝐴 = 2 sin 4𝐴 cos 4𝐴, and so on.
(ii.) If, in the compound angle formula for 𝑐𝑜s (𝐴 + 𝐵), we let 𝐵 = 𝐴 then
cos 2𝐴 = cos 2 𝐴 − sin2 𝐴
Since cos 2 𝐴 + sin2 𝐴 = 1, then cos 2 𝐴 = 1 − sin2 𝐴 and sin2 𝐴 = 1 − cos 2 𝐴, and two
further formula for cos 2𝐴 can be produced.
Thus,
cos 2𝐴 = cos 2 𝐴 − sin2 𝐴
cos 2𝐴 = (1 − sin2 𝐴) − sin2 𝐴
cos 2𝐴 = 1 − 2 sin2 𝐴
And
cos 2𝐴 = cos 2 𝐴 − sin2 𝐴
cos 2𝐴 = cos 2 𝐴 − (1 − cos 2 𝐴)
cos 2𝐴 = 2 cos 2 𝐴 − 1
Also, for example,
cos 4𝐴 = cos 2 2𝐴 − sin2 2𝐴
OR
cos 4𝐴 = 1 − 2 sin2 2𝐴
272
OR
cos 4𝐴 = 2 cos 2 2𝐴 − 1
AND
cos 6𝐴 = cos 2 3𝐴 − sin2 3𝐴
OR
cos 6𝐴 = 1 − 2 sin2 3𝐴
OR
cos 6𝐴 = 2 cos 2 3𝐴 − 1
And so on.
(iii.) If, in the compound-angle formula for tan (𝐴 + 𝐵), we let 𝐵 = 𝐴, then
tan 2𝐴 =
2 tan 𝐴
1 − tan2 𝐴
tan 4𝐴 =
2 tan 2𝐴
1 − tan2 2𝐴
Also, for example,
And
tan 5𝐴 =
5
2 tan 2 𝐴
5
1 − tan2 𝐴
2
And so on.
Example 6.6.2.1.
𝐼3 sin 3𝜃 is the third harmonic of a waveform. Express the third harmonic in terms of the first
harmonic sin 𝜃, when 𝐼3 = 1
273
Solution 6.6.2.1.
When 𝐼3 = 1,
𝐼3 sin 3𝜃 = sin 3𝜃 = sin (2𝜃 + 𝜃)
From the compound-angle formula for sin (𝐴 + 𝐵)
𝐼3 sin 3𝜃 = sin 3𝜃 = sin (2𝜃 + 𝜃) = sin 2𝜃 cos 𝜃 + cos 2𝜃 sin 𝜃
= (2 sin θ cos θ) cos 𝜃 + (1 − 2 sin2θ) sin 𝜃
From the double angle expansions
= 2 sin θ cos 2θ + sin θ − 2 sin3 𝜃
= 2 sin θ (1 − sin2 𝜃) + sin θ − 2 sin3 𝜃
= 2 sin θ − 2 sin3 𝜃 + sin θ − 2 sin3 𝜃
sin 3𝜃 = 3 sin θ − 4 sin3 𝜃
Example 6.6.2.2.
Prove that:
1 − cos 2𝜃
= tan 𝜃
sin 2𝜃
274
Solution 6.6.2.2.
1 − cos 2𝜃 1 − (1 − 2 sin2 θ)
=
sin 2𝜃
2 sin 𝜃 cos 𝜃
1 − cos 2𝜃
2 sin2 θ
=
sin 2𝜃
2 sin 𝜃 cos 𝜃
1 − cos 2𝜃 sin 𝜃
=
= tan 𝜃 = 𝑅𝐻𝑆
sin 2𝜃
cos 𝜃
Example 6.6.2.3.
Prove that
cot 2𝑥 + 𝑐𝑜𝑠𝑒𝑐 2𝑥 = cot 𝑥
Solution 6.6.2.3.
𝐿𝐻𝑆 = cot 2𝑥 + 𝑐𝑜𝑠𝑒𝑐 2𝑥
cos 2𝑥
1
cos 2𝑥 + 1
+
=
sin 2𝑥 sin 2𝑥
sin 2𝑥
cos 2𝑥 + 1 (2 cos 2 𝑥 − 1) + 1
=
sin 2𝑥
𝑠𝑖𝑛2𝑥
=
(2 cos 2 𝑥)
sin 2𝑥
2 cos 2 𝑥
cos 𝑥
=
=
= cot 𝑥 = 𝑅𝐻𝑆
2 sin 𝑥 cos 𝑥 sin 𝑥
275
6.6.3. Changing Products of Sines and Cosines into Sums and Differences
i.
sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵) = 2 sin 𝐴 cos 𝐵
(from the formulas in 6.6.1.) i.e.,
1
sin 𝐴 cos 𝐵 = 2 [sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)]
ii. sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵) = 2 cos 𝐴 sin 𝐵
Equation (6.6.3.1.)
i.e.,
1
cos 𝐴 sin 𝐵 = 2 [sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵)]
Equation (6.6.3.2.)
iii. cos(A + B) + cos (𝐴 − 𝐵) = 2 cos 𝐴 cos 𝐵 i.e.,
1
cos 𝐴 cos 𝐵 = 2 [𝑐𝑜𝑠(𝐴 + 𝐵) + cos(𝐴 − 𝐵)]
Equation (6.6.3.3.)
iv. cos(𝐴 + 𝐵) − cos(𝐴 − 𝐵) = −2 sin 𝐴 sin 𝐵 i.e.,
1
sin 𝐴 sin 𝐵 = − 2 [cos(𝐴 + 𝐵) − cos(𝐴 − 𝐵)]
Equation (6.6.3.4.)
Example 6.6.3.1.
Express sin 4𝑥 cos 3𝑥 as a sum or difference of sines and cosines.
Solution 6.6.3.1.
From Equation (6.6.3.1.),
sin 4𝑥 cos 3𝑥 =
1
[sin(4𝑥 + 3𝑥) + sin (4𝑥 − 3𝑥)]
2
sin 4𝑥 cos 3𝑥 =
1
[sin(7𝑥) + sin (𝑥)]
2
276
Example 6.6.3.2.
Express 2 cos 5𝜃 sin 2𝜃 as a sum or difference of sines or cosines.
Solution 6.6.3.2.
From Equation (6.6.3.2.),
1
2 cos 5𝜃 sin 2𝜃 = 2 { [sin(5𝜃 + 2𝜃) − sin(5𝜃 − 2𝜃)]}
2
2 cos 5𝜃 sin 2𝜃 = sin 7𝜃 − sin 3𝜃
Example 6.6.3.3.
Express 3 cos 4𝑡 cos 𝑡 as a sum or difference of sines or cosines.
Solution 6.6.3.3.
From Equation (6.6.3.3.),
1
3 cos 4𝑡 cos 𝑡 = 3 { [cos(4𝑡 + 𝑡) + cos(4𝑡 − 𝑡)]}
2
3 cos 4𝑡 cos 𝑡 =
3
(cos 5𝑡 + cos 3𝑡)
2
Assuming the integral of 3 cos 4𝑡 cos 𝑡 was required, then
3
∫ 3𝑐𝑜𝑠4𝑡𝑐𝑜𝑠𝑡 = ∫ ( (cos 5𝑡 + cos 3𝑡)𝑑𝑡)
2
∫ 3𝑐𝑜𝑠4𝑡𝑐𝑜𝑠𝑡 =
3
∫ (cos 5𝑡 + cos 3𝑡)𝑑𝑡
2
∫ 3𝑐𝑜𝑠4𝑡𝑐𝑜𝑠𝑡 =
3 sin 5𝑡 sin 3𝑡
[(
+
)] + 𝑐
2
5
3
277
∫ 3𝑐𝑜𝑠4𝑡𝑐𝑜𝑠𝑡 =
3 sin 5𝑡 sin 3𝑡
[
+
]+𝑐
2
5
3
Example 6.6.3.4.
𝜋
In an alternating current circuit, voltage 𝑣 = 5 sin 𝜔𝑡 and current 𝑖 = 10 sin (𝜔𝑡 − ). Find an
6
expression for the instantaneous power 𝑝 at time 𝑡 given that 𝑝 = 𝑣𝑖, expressing the answer as a
sum or difference of sines and cosines.
Solution 6.6.3.4.
Given that
𝑣 = 5 sin 𝜔𝑡
𝜋
𝑖 = 10 sin (𝜔𝑡 − )
6
And
𝑝 = 𝑣𝑖
We get
𝜋
𝑝 = 𝑣𝑖 = (5 sin 𝜔𝑡) [10 sin (𝜔𝑡 + )]
6
𝜋
𝑝 = 𝑣𝑖 = 50 sin 𝜔𝑡 sin (𝜔𝑡 − )
6
From Equation (6.6.3.4.)
𝜋
1
𝜋
𝜋
50 sin 𝜔𝑡 sin (𝜔𝑡 − ) = (50) [− {cos (𝜔𝑡 + 𝜔𝑡 − ) − cos [𝜔𝑡 − (𝜔𝑡 − )]}]
6
2
6
6
𝜋
𝜋
𝜋
50 sin 𝜔𝑡 sin (𝜔𝑡 − ) = (−25) [cos (2𝜔𝑡 − ) − cos ]
6
6
6
The instantaneous power 𝑝 in sum or difference of sine or cosine is given by
278
𝑝 = 𝑣𝑖 = 25 [cos
𝜋
𝜋
− cos (2𝜔𝑡 − )]
6
6
279
6.6.4. Changing Sums and Differences of Sines and Cosines into Products
In the compound-angle formula let (𝐴 + 𝐵) = 𝑋 and (𝐴 − 𝐵) = 𝑌
Solving the simultaneous equations gives
𝐴=
𝑋+𝑌
2
𝐵=
𝑋−𝑌
2
And
Thus
sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵) = 2 sin 𝐴 cos 𝐵
Becomes
sin 𝑋 + sin 𝑌 = 2 sin (
𝑋+𝑌
2
) cos (
𝑋−𝑌
2
)
Equation (6.6.4.1.)
)
Equation (6.6.4.2.)
Similarly,
sin 𝑋 − sin 𝑌 = 2 cos (
𝑋+𝑌
cos 𝑋 + cos 𝑌 = 2 cos (
2
) sin (
𝑋+𝑌
2
cos 𝑋 − cos 𝑌 = −2 sin (
𝑋−𝑌
2
) cos (
𝑋+𝑌
2
𝑋−𝑌
2
) sin (
)
𝑋−𝑌
Example 6.6.4.1.
Express sin 5𝜃 + sin 3𝜃 as a product
280
2
Equation (6.6.4.3.)
)
Equation (6.6.4.4.)
Solution 6.6.4.1.
From Equation (6.6.4.1.),
sin 5𝜃 + sin 3𝜃 = 2 sin (
5𝜃 + 3𝜃
5𝜃 − 3𝜃
) cos (
)
2
2
sin 5𝜃 + sin 3𝜃 = 2 sin (
8𝜃
2𝜃
) cos ( )
2
2
sin 5𝜃 + sin 3𝜃 = 2 sin(4𝜃) cos(𝜃) = 2 sin 4𝜃 cos 𝜃
Example 6.6.3.2.
Express sin 7𝑥 − sin 𝑥 as a product
Solution 6.6.3.2.
From Equation (6.6.4.2.),
sin 7𝑥 − sin 𝑥 = 2 cos (
7𝑥 + 𝑥
7𝑥 − 𝑥
) sin (
)
2
2
sin 7𝑥 − sin 𝑥 = 2 cos 4𝑥 sin 3𝑥
Example 6.6.4.3.
Express cos 2𝑡 − cos 5𝑡 as a product
Solution 6.6.4.3.
From Equation (6.6.4.3.),
cos 2𝑡 − cos 5𝑡 = −2 sin (
281
2𝑡 + 5𝑡
2𝑡 − 5𝑡
) sin (
)
2
2
7𝑡
−3𝑡
cos 2𝑡 − cos 5𝑡 = −2 sin ( ) sin (
)
2
2
7
3
cos 2𝑡 − cos 5𝑡 = −2 sin 𝑡 sin 𝑡
2
2
3
3
Since [sin (− 2 𝑡) = − sin (2 𝑡)]
7
3
cos 2𝑡 − cos 5𝑡 = 2 sin 𝑡 sin 𝑡
2
2
Example 6.6.4.4.
Show that
cos 6𝑥 + cos 2𝑥
= cot 4𝑥
sin 6𝑥 + sin 2𝑥
Solution 6.6.4.4.
From Equation (6.6.4.3.),
cos 6𝑥 + cos 2𝑥 = 2 cos 4𝑥 cos 2𝑥
Also, from Equation (6.6.4.1.)
sin 6𝑥 + sin 2𝑥 = 2 sin 4𝑥 cos 2𝑥
Hence,
cos 6𝑥 + cos 2𝑥 2 cos 4𝑥 cos 2𝑥
=
sin 6𝑥 + sin 2𝑥
2 sin 4𝑥 cos 2𝑥
cos 6𝑥 + cos 2𝑥
= cot 4𝑥
sin 6𝑥 + sin 2𝑥
282
Self -Assessment
283