Download MAT 110 Module 2A - West East University

FOUNDATION MATHEMATICS
MAT 110
MODULE 2 A
COPPERBELT UNIVERSITY COLLEGE
Mathematics Department
Copyright
© COPPERBELT UNIVERSITY COLLEGE 2010
Acknowledgements
The COPPERBELT UNIVERSITY COLLEGE wishes to thank the following for their contribution to
this module:
EDITORS
Directorate of Distance Education (DODE)
Ministry of Education, Lusaka
AUTHORS
1. Mr. Chikunduzi C. Lyson
Lecturer
2. Mr. Kambilombilo Dennis
Lecturer
3. Mr. Likando M.Kenneth.
Lecturer
4. Mr. Malambo Priestly.
Lecturer
COPPERBELT UNIVERSITY COLLEGE
BOX 20382
KITWE
ZAMBIA
E-mail: [email protected]
Website: www.cce.co.zm
Contents
About this module 2
1
How this module is structured .......................................................................................... 1
Getting around this module
1
Margin icons ..................................................................................................................... 1
Module 2 overview
3
Is this module 2 for you? .................................................................................................. 3
Module outcomes .............................................................................................................. 3
Timeframe ........................................................................................................................ 3
Study skills ........................................................................................................................ 3
Need help? ........................................................................................................................ 5
Assignments ...................................................................................................................... 6
Assessments ...................................................................................................................... 6
Unit 1 FUNCTIONS
1.0 Introduction ................................................................................................................ 7
1.1 Definition of product set. ............................................................................................ 8
1.2 Relations ..................................................................................................................... 8
1.3 Concept of functions. ............................................................................................... 9
1.3.1 Domain and range of a function ..................................................................... 9
1.3.2 Types of functions ........................................................................................ 11
1.3.2 Even and odd functions ................................................................................ 11
1.3.3 Inverse functions .......................................................................................... 13
1.3.4 Composite functions ..................................................................................... 14
1.3.5 Graphs of quadratic functions ...................................................................... 16
1.3.6 Sketching graphs of quadratic ...................................................................... 17
Unit summary ................................................................................................................. 24
Unit 2 TRIGONOMETRY
25
2.1 Introduction ............................................................................................................... 25
2.2 Trigonometric functions ........................................................................................... 25
2.2.2 Special angles ............................................................................................... 28
2.2.3 Graphs of trigonometric functions ............................................................... 30
2.3 Trigonometric identities ............................................................................................ 35
2.4 Further trigonometric identities ............................................................................. 40
2.4.1 Trigonometric equations............................................................................... 45
2.4.2 Trigonometric equations of the form: .......................................................... 48
Unit summary ................................................................................................................. 53
Unit 3 EXPONENTIAL AND LOGARITHMIC FUNCTIONS ............................................. 53
3.0 Introduction ............................................................................................................... 53
3.1 Exponents.................................................................................................................. 56
3.2 Logarithms ................................................................................................................ 58
3.2.1 Definition...................................................................................................... 58
3.2.2 Laws of logarithms ....................................................................................... 59
3.2.2 The quotient law ........................................................................................... 60
3.2.3 The power law .............................................................................................. 61
3.3 Exponential equations ............................................................................................... 68
3.3.1 Solving exponential equations with the same base. ..................................... 68
3.3.2 Solving quadratic exponential equations ...................................................... 69
3.3.3 Solving exponential equations using the change of base method ................ 70
3.4 logarithmic equations ................................................................................................ 72
log x b  c
3.4.1 Solving equations of the form log a b  x and
........................... 72
3.4.2 Equations involving natural exponents and natural logarithms ............................. 81
3.5.0 Application of exponential and logarithmic function ............................................ 84
3.5.1 Decay and growth problems ......................................................................... 84
3.5.2 Compound interest........................................................................................ 85
3.5.3 Graphs of exponential and Logarithmic functions ................................................ 87
Graphs of exponential functions............................................................................ 87
3.5.4 The hyperbolic functions ....................................................................................... 95
Answers to activities ..................................................................................................... 101
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
About this module 2
Module 2 of Foundation Mathematics has been produced by Copperbelt
University College and is structured in the same way, as outlined below.
How this module is structured
The course content
The course is broken down into units. Each unit comprises:
 An introduction to the unit content.
 Unit outcomes.
 Core content of the unit with a variety of learning activities.
 A unit summary.
Resources
For those interested in learning more on this subject, we provide you with
a list of additional resources at the end of this module
Getting around this module
Margin icons
While working through this module you will notice the frequent use of
margin icons. These icons serve to “signpost” a particular piece of text, a
new task or change in activity; they have been included to help you to
find your way around.
Page 1
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
A complete icon set is shown below. We suggest that you familiarize
yourself with the icons and their meaning before starting your study.
Activity
Assessment
Assignment
Outcomes
Summary
Time
Help
Note it!
Page 2
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Module 2 overview
Welcome to module 2 MAT 110
This module is based on introductory concepts in Mathematics is part of the foundation
mathematics course.
Is this module 2 for you?
This module 2 is intended for Diploma holder teachers of Mathematics and Science that
aim to further teaching skills .
Module outcomes
During and upon completion of this module you will be able to:
 Understand and apply knowledge of functions
 Understand and apply knowledge of Trigonometry.
Timeframe
The expected study time is 100 hours
Study skills
As an adult learner your approach to learning will be different to that
from your school days: you will choose what you want to study, you will
have professional and/or personal motivation for doing so and you will
most likely be fitting your study activities around other professional or
domestic responsibilities.
Essentially you will be taking control of your learning environment. As a
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
consequence, you will need to consider performance issues related to
time management, goal setting, stress management, etc. Perhaps you will
also need to reacquaint yourself in areas such as essay planning, coping
with exams and using the web as a learning resource.
Your most significant considerations will be time and space i.e. the time
you dedicate to your learning and the environment in which you engage
in that learning.
We recommend that you take time now—before starting your selfstudy—to familiarize yourself with these issues. There are a number of
excellent resources on the web. A few suggested links are:
 http://www.how-to-study.com/
The “How to study” web site is dedicated to study skills resources.
You will find links to study preparation (a list of nine essentials for a
good study place), taking notes, strategies for reading text books,
using reference sources, test anxiety.
 http://www.ucc.vt.edu/stdysk/stdyhlp.html
This is the web site of the Virginia Tech, Division of Student Affairs.
You will find links to time scheduling (including a “where does time
go?” link), a study skill checklist, basic concentration techniques,
control of the study environment, note taking, how to read essays for
analysis, memory skills (“remembering”).
 http://www.howtostudy.org/resources.php
Another “How to study” web site with useful links to time
management, efficient reading, questioning/listening/observing skills,
getting the most out of doing (“hands-on” learning), memory building,
tips for staying motivated, developing a learning plan.
The above links are our suggestions to start you on your way. At the time
of writing these web links were active. If you want to look for more go to
www.google.com and type “self-study basics”, “self-study tips”, “selfstudy skills” or similar.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Need help?
For any help contact the University College Through:
1. University College Secretary – phone +260 212293003
2. Email: [email protected]
3. The Mathematics HOD: +260 955 906714
4. The District Resource Centre Coordinator as advised.
All contacts during office hours.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Assignments
Tutor marked assignments will be considered for assessment and to be
submitted at stipulated times through the District Resource centres as
advised.
Assessments
There will be a three hour test administered at the University College
during a residential school session.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Unit 1 FUNCTIONS
1.0 Introduction
This Unit is aimed at acquainting you with functions which you certainly have met
in your earlier study of Mathematics. Without doubt, this is one of the most
important concepts in Mathematics and because of that; we shall carry you from the
basics and build up the idea so as to prepare you for further Mathematics courses.
You will finally learn how to sketch graphs of quadratic function.
During and upon completion of this unit you will be able to:

Describe a product set

Discuss the concept of a relation

Discuss the concept of a function

Determine domains and ranges of functions

Distinguish different types of functions

Solve simple problems on linear functions
(including their
Work with composite functions

Determine maximum and minimum turning
points of quadra
points

Sketches graphs of quadratic functions.

Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
1.1 Definition of a product set.
With your knowledge of sets acquired in Module one, consider a situation
where you have two sets A and B. We can define the product set of A and B
denoted A  B as follows:
A  B  (a, b) : a  A, b  B.
Example 1
Let’s suppose that sets A and B are such that A  {3,4,7} and B  {1,3} .
Using our definition above, the product set of A and B is:
A  B  {( 3,1), (3,3), (4,1), (4,3), (7,1), (7,3)}
A product set has ordered pairs for its elements.
Activity 1
Given two sets A and B as described in example 1 above,
determine the product set B  A .What relationship have you
noticed between B  A and A  B ?
1.2 Relations.
In simple terms, we can define a relation as a subset of ordered pairs. Let’s
now use our knowledge of product sets to amplify and clarify the concept of
a relation as follows:
Consider the product set A  B  {( 3,1), (3,3), (4,1), (4,3), (7,1), (7,3)} . Several
subsets of this set of ordered pairs A  B can be derived such as:
{( 3,1), (4,1)} , {( 7,1), (3,3), (4,3)} , {( 7,3), (7,1)}
These subsets of ordered pairs are what are referred to as
relations and usually the letter R is used to denote a relation.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 2
(i) Derive relations from the following product set:
A  B  {( 3,1), (3,3), (4,1), (4,3), (7,1), (7,3)}
(ii) Construct product sets and determine some of the relations which can be
derived from such product sets.
1.3 Concept of functions.
You are now ready to be introduced to the concept of a function.
Definition:
A function (mapping) can be defined as a special relation in which
no ordered pairs have the same first entry. What this entails is that a
function is a rule which assigns every starting element (input or
object), say x  X a unique ending element (output or image) say
y  Y . You could be aware that we usually use the letter f to
denote a function, although any other letter can be used. You may
equally recall that, functions can be expressed using formulas,
diagrammatically or in functional notation. For instance:
y  f (x) represents a function in formula form where the object is x
and the image is f(x).
f : X  Y represents a function f in functional notation. The
meaning here is that ‘the function f maps the elements of set X
(objects) onto elements of set Y (images). We shall
illustrate the diagrammatic representation of quadratic functions
latter in this unit.
1.3.1 DOMAIN AND RANGE OF A FUNCTION
A domain of a function f denoted D (f), is a set composed of all
possible objects of such a function, whereas a range of f denoted
R (f) could be defined as a collection of all images or out puts of f.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 3
(i)
Determine which of the following relations are
functions and those which are not:
R1  (3,10), (4,10), R2  (7,10), (3,10), (7,13),
R3  (10,7), (13, )
(ii)
For the functions in part (i), state the domain and
range.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
1.3.2 TYPES OF FUNCTIONS
In this section, we are going to study different types of
functions.
(a) Surjective (onto) function:
Let f be a function where y  Y , x  X and that Y=R (f),
X = D (f). Then f is said to be surjective (onto) if every
y  Y is an image of some x  X . Thus, when R (f) is equal
to the entire co-domain of f, we say that such a function f is
surjective (onto).
(b) Injective (one-to-one) function:
Suppose that y  f (x) is a function and that x1 , x2  D( f ) .
The function f can only be said to be injective (one-to-one) if:
f ( x1 )  f ( x2 )  x1  x2 . The implication here is that a oneto-one function is one where every image has only one object
assigned to it.
(c) Many-to-one function:
Any function which is not one-to-one is called a many to one.
(d) Bijective function:
This is a function which is both Surjective (onto) and
injective (one-to-one).
1.3.2 EVEN AND ODD FUNCTIONS
An even function say y  f (x) is one which is such that
for all x  D( f ) , f ( x)  f ( x) . This implies that for a
function f to be even,
f ( x)  f ( x)  0 . On the other
hand, a function f (x) can be said to be odd
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
if f ( x)   f ( x) for all values of x. Alternatively,
f (x) is odd if for all values of x, f ( x)  f ( x)  0 .
Even functions are symmetrical along the y-axis ( x  0) ,
whereas odd functions are symmetrical about the origin
(0, 0) through 180 0 .
Example 2
Show that g ( x)  5 x  6 is a one-to-one function.
Solution:
Let x1 , x2  D( g ) . g ( x1 )  g ( x2 )  5x1  6  5x2  6
 5 x1  5 x 2
 x1  x 2
Since g ( x1 )  g ( x2 )  x1  x2 , it has been shown that
g (x ) is a one-to-one function.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 4
Show that the following functions are one-to-one:
f ( x) 
(i)
x 1
,
2x
h( x ) 
2
x3
Determine which one of the following functions is even and
which one is odd:
y  x 2 , y  x3
(ii)
Determine whether f ( x)  x  1 is odd, even or neither of
the two.
1.3.3 INVERSE FUNCTIONS
By now you should be familiar that a function maps
objects from the domain onto unique images in the range.
Nevertheless, let’s now
suppose that we require a rule which would map images
from the range onto their respective objects in the
domain. Such a rule when it exists, and as long as the
characteristic of a function is met, is called an inverse
function. You may already know that, for functions say
f ( x), g ( x), h( x) , their respective inverse functions are
denoted by f 1 ( x), g 1 ( x), h 1 ( x) .
(i)
You should never regard the superscript -1 in an inverse
function as an exponent. Thus, f 1 ( x) 
1
.
f ( x)
(ii) Only one-to-one functions are invertible (have inverses).
The following example is meant to familiarise you
algebraically with a process of finding an inverse of a
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
given one-to-one function.
Example 3
Compute the inverse function of y  6 x  3
Solution:
For you to find the inverse of this function, you ought to
start by making the object variable x the subject of the
formula in y  6 x  3 as follows:
y  6x  3
y  3  6 x  3  3 (subtracti ng 3 on both sides)
y  3  6x
y 3
 x (dividing by 6 on both sides)
6
y 3
x
6
You should then replace the variable y by x on the right
hand side of the preceding equation and use the notation
of inverse functions as follows:
f 1 ( x) 
x3
(which is the required inverse function).
6
1.3.4 COMPOSITE FUNCTIONS
There are instances when you confront single functions
which are a combination of two or more functions. Such
functions are called composite functions.
Let f (x) and g (x ) be two functions. The composition of
f (x) and g (x ) , denoted ( f  g )( x) is given by:
( f  g )( x) = f [ g ( x )] and the composition of g (x ) and f (x)
denoted ( g  f )( x) is given by:
( g  f )( x) = g[ f ( x )] .
You may be wondering what ( f  g )( x)
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
and ( g  f )( x) means. Let us therefore address this concern:
The function ( f  g )( x) implies that the
function g operates on the object x first which gives an
image upon which the function f operates on. Thus, the
image of g becomes the object of f in the composite
function ( f  g )( x) . Similarly, ( g  f )( x) means that f
operates on the object x and then g operates on the
resulting answer.
Example 4
Given that f ( x)  2 x  1 and g ( x) 
1 x
, find the
1 x
following functions:
(i) ( f  g )( x)
(ii) ( g  f )( x)
Solution:
1 x 
(i) ( f  g )( x) = f [ g ( x)]  f 

1 x 
1 x 
 2
  1 , the object replaced by g ( x)
1 x 
=
2  2x
2  2x  1  x
1 
1 x
1 x
Hence, ( f  g )( x) =
3 x
1 x
(ii) ( g  f )( x) = g[ f ( x)]  g (2 x  1) , replacing the object
by f(x).
g[ f ( x)]  g (2 x  1) =
=
1  (2 x  1) 2 x  2

1  (2 x  1)
 2x
1 x
, factoring out 2 and dividing by 2.
x
Hence, ( g  f )( x) =
1 x
x
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
(i) From the preceding example, you may have observed
that ( f  g )( x)  ( g  f )( x) .
(iii) Given two functions say f ( x) and g ( x) which are one-to-one, the
following hold:
(a) ( f  f 1 )( x)  ( f 1  f )( x)  x
(b) ( f  g ) 1 ( x)  ( g 1  f 1 )( x)
(c) ( g  f ) 1 ( x)  ( f 1  g 1 )( x)
Activity 5
If f ( x)  x  4 and g ( x)  5 x  2 , find the following
composite functions:
(i) ( f  f
1
)( x)
(ii) ( f
1
 f )( x)
(iii) ( f  g ) 1
(iv) ( g 1  f 1 )( x)
(v) ( g  f ) 1 ( x)
(vi) ( f 1  g 1 )( x)
1.3.5 GRAPHS OF QUADRATIC FUNCTIONS
In Module 1, you covered quadratic functions and different
methods of finding their roots. You equally studied the
concept of a discriminant of quadratic functions. However,
in this section, you will learn techniques of sketching
graphs of such functions. Additionally, you will study
aspects of minimum and maximum values of functions,
including the minimum and maximum turning points of
quadratic functions.
Are you able to remember that a quadratic function is one
of the form:
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
f ( x)  ax 2  bx  c , where a, b and c are constants
and a  0 ? If not, you will do well to master this fact. Since we
are now interested in graphs of quadratic functions, this is
the right stage, to let you know that a graph of a functions
like f(x) is called a Parabola. Further, parabolas will
either open upwards or downward as shown below:
Fig.2 Parabola Opening upwards
Fig 1. Parabola opening downward
You may be wondering how you can sketch the graph of f(x).
This is exactly what we shall study in the following section.
1.3.6 SKETCHING GRAPHS OF QUADRATIC UNCTIONS
For you to draw a sketch of the graph of f ( x)  ax 2  bx  c ,
where a, b and c are constants and a  0 , the following points
must be taken into account:
1. Firstly, determine whether the parabola opens upwards or downwards by
establishing the characteristic of the coefficient of x 2 in f ( x) . If a  0 then
the parabola opens upwards. However, if a  0 , then the parabola opens
downwards.
2. Determine the y-intercept by letting x  0 in f (x) .
3. Your next step should be to establish whether the graph of f (x) cuts the xaxis at two different points, exactly one point or does not touch the x-axis at
all. For you to do so, employ the discriminant: b 2  4ac .
(Refer to MAT 110).
After this, proceed to computation of the
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
x-intercept/s by solving f (x) =0.
4. Further, you ought to determine the turning point of f (x) , and this is given
 b 4ac  b 2 
 .
by   ,
2
a
4
a


5. With the information in the steps above, you can then sketch the graph
of f (x) .
Let us now use these steps above to sketch the graph of a quadratic function
in the following example:
Example 5
Sketch the graph of the function f(x) = -2x2 + 5x + 3
Solution:
From f(x), a  2  0 . This implies that the curve of function
f(x) opens downwards. You now require finding the y-intercept
by letting x=0 in f(x) as follows:
f (0)  2(0) 2  5(0)  3 =3
 (0,3) is the point at which the graph of f(x) meets the y-axis.
We shall use the discriminant to discover the behaviour of the
graph in relation to the x- axis:
From f(x), b 2  4ac = (5) 2 (4  2  3)  1  0 . Since the
discriminant is greater than Zero, the curve of f(x) touches the xaxis at two different points. You can find these points by letting
y=0 in f(x) as follows:
f ( x)  2 x 2  5 x  3 =0
 2 x 2  5 x  3  0
 (2 x  1)( x  3)  0
(Factorising the left hand side)
either  2x  1  0 or x  3  0
x
1
 0.5, x  3
2
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Therefore, the curve touches the x-axis at ( 
1
, 0) and (3,0)
2
According to our earlier considered five steps, the turning point is
given by:
  b 4ac  b 2 

 . Now since a  2, b  5, c  3 from f(x), the
,
2
a
4
a


required turning point
  5 4(2)3  52   5 49 
,
is 
   ,  . This




2

2
4

2

 4 8 
information should help you to sketch the graph of
f(x) = -2x2 + 5x + 3 as shown below:
Fig 3. Graph of f(x) = -2x2 + 5x + 3
Let us consider another example relating to graphs of quadratic
functions:
Example 6
Sketch the graph of the quadratic function f(x) = x2 + 2x − 3.
Solution:
By now it should be clear to you that a  1, b  2 and c  3 , from the function f(x).
What do you think about the whether the curve opens downwards or upwards? You
may have noticed that the curve shall open upwards, since a  0 .
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The discriminant b 2  4ac  2  [4  1  3]  2  12  14  0 . Do you remember
the implication of this value of the discriminant? A positive value implies that the
curve will cut the x-axis at two different points.
2
Let us now find those two x-values at which the graph shall cut the horizontal axis.
This shall be done by letting y  0 in the function as follows:
x 2  2x  3  0
x 2  3x  x  3  0
x( x  3)  1( x  3)  0
(Factorising the left hand side)
( x  3)( x  1)  0
 x  3  0 or x  1  0
 x  3 or x  1 are the two values at which f(x) cuts the x-axis.
Before sketching, you also need to ascertain the y-intercept and the turning point of
f(x).
We shall begin with computing the y-value at which the curve crosses the vertical
axis and to do so we let x=0 in f(x) as follows:
f (0)  (0) 2  2(0)  3  3  y  3 is the sought intercept.
  b 4a  b 2 
 ? If not,
Do you recall that the turning point of f(x ) is given by 
,
4a 
 2a
revisit the section which deals with this aspect before proceeding. Assuming that
you are familiar with turning points, you will notice that the turning point of f(x)
  2 (4  1  3)  (2) 2 
   1,4 .
is 
,
(
2

1
)
(
4

1
)


With all the above information, try to sketch the graph on a piece of paper. Did
your graph look like the one below? If not, where are the variations and what could
be the justification? If yes, then you have acquired the concepts and skill expected.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Fig 4. Graph of f(x) = x2 + 2x − 3:
(i) A quadratic function whose parabola opens upwards (u-shaped
and where a > 0), is said to have a minimum value at the turning
point and such a turning point is called a minimum turning point.
The turning point of f(x) in example 6, is a minimum turning
4ac  b 2
point. Further, the range of f(x) in this case is y 
.
4a
(ii) A quadratic function whose parabola opens downwards (nshaped and where a < 0), has a maximum value at the
turning point. A turning point of such a parabola (opening
downwards) is called a maximum turning point.
Further, the range of f(x) in this respect is y 
4ac  b 2
.
4a
  b 4a  b 2 
 is found by completing the
(iv) The turning point expression 
,
4a 
 2a
square for f ( x)  ax 2  bx  c , where a, b and c are constants
and a  0 .
In order for you to strengthen your understanding of
sketching graphs of quadratic functions, the following
questions are provided for that purpose.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 6
1. Sketch the graphs of the following quadratic functions:
(a) f(x) = 2x2 − 8x + 6 (b) f(x) = -x2 + x + 6 (c) f ( x)   x 2  5x  2
(d) f ( x)  4 x 2  x  1 (e) f ( x)  2 x 2  5 x  3
2. Determine the minimum and maximum values of the
following functions where applicable: Further, indicate the
ranges of each given function.
f ( x)  2 x 2  5x  3,
f ( x)  3x 2  7 x  1, f ( x)  9 x 2  30 x  25
f ( x)  36  48x  9 x 2
Unit Summary
In this unit you learnt concepts of product sets, relations and functions. You have
also learned how to calculate inverses of linear functions. The idea of composite
functions was tackled as well as how to find composite functions. The last aspects
which were considered are the sketching of graphs of quadratic functions, minimum
and maximum values of quadratic functions and minimum and maximum turning
points of graphs of quadratic functions.
Page 12
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Unit 2 TRIGONOMETRY
2.1 Introduction
In this unit, you will learn of the six trigonometric functions. Curves of
sin  , cos  , tan  shall be covered. You will further learn trigonometric identities
and how to prove given trigonometric identities. The last section shall be based on
how to solve trigonometric equations.
During and upon completion of this unit you will be able to:
 Work with six basic trigonometric functions.
 Prove trigonometric identities
 Solve problems involving difference and sum
trigonometric formulas.
 Work with double and half angle formulas.
 Solve trigonometric equations.
2.2 TRIGONOMETRIC FUNCTIONS
To start with, you shall be reminded of the three basic trigonometric
functions considered during your earlier study of mathematics. However,
you will study three other additional trigonometric functions which shall act
as a basis for proofs of identities. You shall in addition learn how to find
solutions of trigonometric equations.
Let us consider the right angled triangle ABC below in which angle ABC= 
and angle ACB= 90 0 .
Page 25
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Figure 5
We define the three basic trigonometric ratios as follows:
opposite side AC
adjacent side BC

, cos  

,
hypotenuse
AB
hypotenuse
AB
opposite side AC
and tan  

adjacent
BC
sin  
tan  
AC
sin 
sin 
AB  AC  tan 
, since

cos 
cos BC
BC
AB
You should now extend your knowledge to the other three trigonometric
functions, and these are explained below:
1. The secant of an angle written secant θ. We shall however, be using short
form: sec . By definition, you should know that:
sec  
1
. What identity can you derive from this definition of sec ?
cos 
You may have deduced that sec cos  1.
2. The second of the other three trigonometric functions is the cosecant of
an angle written as cosecant θ. You should get acquainted with the short
form: csc , where csc  
1
. We can derive an identity form this
sin 
definition and this is: csc sin   1 .
Page 26
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3. Lastly, one other trigonometric function is the cotangent of an angle
written as cotangent cotangent θ. You require a short form for this function
as well,
which is cot  . In terms of the basic trigonometric functions,
cot  
1
cos 

.
tan  sin 
consequently, one of the identities which could be derived from this
definition is: cot  tan   1 .
The last three trigonometric functions (cos ecant , cot angent , sec ant ) are
reciprocals of the basic three ( sin  , cos  , tan  ) .
2.2.1 ANGLE MEASUREMENT
Let us start by stating that angles can either be measured in degrees or
radians. It is usually easy to use degrees when talking about angles.
However, at this level, you should start using radians as well. What follows
therefore, is information meant to acquaint you with the usage of radians.
Suppose that the diagram AOBCA below represents a sector of a circle in
which the radius is r and the arc ACB is of length k  r  kr where k is a
constant and O is the centre of the circle to which the sector belongs.
Figure 6
It follows that the angle which the arc ACB formulates at the centre O is
worth k radians. Similarly, when the length of an arc of a circle with radius
Page 27
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
r units is r units, such an arc forms an angle of size one (1) radian at the
centre.
Can you now consider a circle whose radius is r units. You may recall that
the circumference of such a circle is given by 2  r  2 r . What do you
think then will be the size of the angle, in terms of radians, formed by the
circumference of the circle at the centre? Arising from our earlier
discussion, the angle will basically be 2 radians.
Taking into account that, in terms of degrees, a circle is worth 360 0 , we can
consequently conclude that:
360 0  2 Radians………………………………….. *1 .
Using *1 , we can always derive relationships between angles in degrees and
those measured in radians. The following are some of the several
relationships:
180 0   Radians
90 0 
45 0 

2
Radians

Radians etc
4
Can you now derive other equivalences between angles in degrees and
those in radians?
2.2.2 SPECIAL ANGLES
As you work with trigonometry, you shall meet situations where it will be
important for you to make use of angles deemed ‘special’. These angles are
special because you naturally do not require to use a calculator or any
other technological aid to evaluate a trigonometric functions involving
them. Thus, you can easily evaluate trigonometric expressions involving the
said ‘special angles’. For our study, we shall only consider three angles:
30 0 or

6
, 45 0 or

4
, and 60 0 or

3
, and these are the ones that are referred to
as the ‘special angles’ .
Page 28
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 7
Consider the isosceles triangle ABC and the equilateral triangle XYZ
below in which AB=1, BC=1, AC= 2 , angle ABC= 90 0 and XY=2,
XZ=2, YM=1, MZ=1, XM= 3 , angle XYZ= 60 0 , angle YXM= 30 0 and
angle XMZ= 90 0 .
Figure 7
Figure 8
Use the two triangles to find the values of the following:
sin 450 , cos 450 , tan 450 , sin 60 0 , cos 60 0 , tan 60 0 , sin 30 0 , cos 30 0 tan 30 .
Compare your answers with the following:
According to the isosceles and equilateral triangles given above,
sin 45 
1
sin 60 
3
1
3
, cos 60  , tan 60 
 3 and
2
2
1
2
, cos 45 
1
1
, tan 45   1
1
2
Page 29
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
sin 30 
1
3
1
, cos 30 
, tan 30 
2
2
3
Where your answers different from the ones given above? If yes, how different
and why were they different?
You shall be required to use the three angles latter on, but for now, it suffices for
you to know how to evaluate trigonometric functions using them. It should be
mentioned that the easiest way to remember the values of trigonometric functions
involving ‘special angles’ is to master the two triangles given above (isosceles and
equilateral).
2.2.3 GRAPHS OF TRIGONOMETRIC FUNCTIONS
Have you ever thought of how you can draw the graphs of trigonometric
functions?
Anyway, this section is intended to give you the knowledge of the graphs
of trigonometric functions. However, we shall only consider graphs of the
three basic trigonometric functions (i.e. sin  , cos  , tan  ).
Zeros of the function y=sin
Before we draw the graph of sin  , it is important for you to learn certain
characteristics of this function. Firstly, you need to know of angles which
would give a value of zero for the sine function. From the onset, you
should know that the sine function will have a value of zero at  radians
( 180 0 ) .
Now let us suppose that we write a term like ‘ n ’ where n belongs to the
set of integers. Upon investigation, you will notice that sin  has a value
of Zero (0) at ‘ n (i.e. 0, ±π, ±2π, ±3π, …….), for n  Z . Angles
like n ,
for n  Z and where the function has a value of zero are called the
‘zeros’ of the function. Alternatively, by the zeros of sin θ, we mean
those values of θ for which sin θ will equal 0. The next question which
Page 30
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
arises is that of the maximum or minimum value of sin  . This concern
is addressed in the next point.
The function y  sin  is continuous and will have values lying in the
interval  1  sin   1. What this means is that y  sin  has -1 as the its
minimum value whereas its maximum value is 1.
2.2.4
Graph of the function y=sin 
Taking into account the discussion in 2.4.4, we shall now draw a sketch of
the function y  sin  , and the following is the required sketch of y  sin  :
Fig 9
2.2.5
Period of the function y=sin 
You may have noticed from the graph of y  sin  that the values of this
function started repeating themselves after sometime. For instance, the
value of y  sin  at   0 , is same as its value at   2 . When the
values of a function regularly repeat themselves, we say that the function is
periodic, and the length of the repetitive pattern is what constitutes the
period of a function. What do you think is the period of y  sin  ? You
may have discovered that the graph of this function
repeated itself after 2 radians and of course this is its period.
2.2.6
Graph of the function y=cos 
Just like y  sin  , the function y  cos  has a period of 2 ( 360 0 ) and
is continuous such that  1  cos  1 . The following is a sketch of the
Page 31
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
graph of y  cos  .
Fig 10
You may have recognized that the graph of y  cos  is similar to that
of y  sin  . Apparently, when you translate the curve of y  sin  to the left
by

2
2.2.7
or 90 0 , the result is the curve of y  cos  .
Graph of the function y=tan 
Do you think that the function y  tan  is defined for all real values of  ?
If yes, how and if not, at which points is it discontinuous? Your

investigation should have revealed that y  tan  , is not defined at
or at
2

any other angle which a multiple is of . Nevertheless, the range of values
2
for y  tan  is infinite.
What follows now is the sketch of the graph of y  tan  and certainly this
curve reveals the preceding arguments.
Sketch of y  tan  :
Page 32
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Fig 11
I, II, III, and IV are referring to the first, second, third and fourth quadrants
respectively.
Page 33
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 6
1. Determine the domain and range of the function y  sin  .
Expected answers: The domain is       (any real value)
The range is:  1  y  1 .
2. Show that y  sin  odd function.
3. Determine the zeros of y  cos  .
4. What is the period of y  tan  ?
Page 34
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
2.3 TRIGONOMETRIC IDENTITIES
You may have heard of the term ‘identity’ used in mathematics. Take a
few seconds to reflect on the meaning of trigonometric identities and what
it does not mean. What ideas came into your mind upon your reflection?
We shall basically describe an identity as any relationship which is true for
all permissible values of the variables involved. However, the objective in
this unit is to acquaint you with as many identities and proofs as is
possible.
Let us start by deriving one of the important identities which shall certainly
be useful in our proofs of other identities.
Consider the diagram below in which POH   , PO=r, PH=y
and OH=x.
Fig 12
Remember that sin  
and cos  
y
,  y  r sin  .......................1
r
x
,  x  r cos  .............................................2
r
Using the triangle OPH and the Pythagoras theorem, we have:
x 2  y 2  r 2 ..........................................................................3
We are now going to use information from (1) and (2) to express (3) as
follows:
Page 35
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
r cos  2  r sin  2  r 2  r 2 cos 2   r 2 sin 2   r 2
r 2 cos 2   sin 2    r 2


r2
r2
2
2
,
cos


sin


r2
r2
(dividing by r 2 )………………….. (4)
From (4) we derive our required and significant identity:
cos 2   sin 2   1 …………………………………………………..(5)
You can derive several other identities from (5) such as the following:
sin 2   1  cos 2 
cos 2   1  sin 2 
tan 2   1  sec 2 
1  cot 2   csc 2 
etc.
With your knowledge of the above basic identities, you should now be
ready to do proofs of trigonometric identities. From the onset, you ought to
know that there is no specific method for proving identities. Does this
worry you? Be comforted that even when that is the case, you will find
proofs on identities very interesting and enriching. Let us now prove some
identities:
Example Prove the following identities:
(a)
1
1

 2 sec 2 x
1  sin x 1  sin x
(b)
1
1

 2 tan x sec x
1  sin x 1  sin x
(c)
sec x  1
sin 2 x

sec x  1
sec 2 x
Page 36
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Solutions:
(a)
1
1

 2 sec 2 x .
1  sin x 1  sin x
In order for you to prove a trigonometric identity,
you ought to start with the most complicated
side and show that, such a side is equal to the other side. For the
identity given, it would appear that the most complicated side is the
left hand side. We shall therefore start from there.
Left hand side:
1
1
1(1  sin x)  1(1  sin x)


1  sin x 1  sin x
(1  sin x)(1  sin x)

1  sin x  1  sin x
2

,
(1  sin x)(1  sin x) 1 - sin 2 x
since 1  sin 2 x  (1  sin x)(1  sin x)

2
, cos 2 x  1  sin 2 x
2
cos x

1
 2  2 sec 2 x
2
cos x
= Right hand side
(b)
1
1

 2 tan x sec x .
1  sin x 1  sin x
Again we shall start from the left hand side as follows:
Page 37
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
1
1
1  sin x  (1  sin x)


, expressing as a single fraction.
1  sin x 1  sin x (1  sin x)(1  sin x)

2 sin x
1  sin 2 x

2 sin x
2
sin x


2
cos x cos x cos x
 2 sec x tan x  R.H.S. hence proved.
(c)
sec x  1
sin 2 x
. Let us start from the right hand side this time.

sec x  1
sec 2 x
sin 2 x
sin 2 x sec x  1


, multiplyin g by a special one (1)
sec x  1 sec x  1 sec x  1

sin 2 x(sec x  1)
, because sec 2 x  1  (sec x  1)(sec x  1)
2
sec x  1

sin 2 x(sec x  1)
tan 2 x  1  1

sin 2 x(sec x  1) sin 2 x(sec x  1)

tan 2 x
sin 2 x
cos 2 x

sin 2 x(sec x  1) cos 2 x
1
 (sec x  1) 
2
sin x
sec 2 x

sec x  1
 R.H.S. hence proved
sec 2 x
Page 38
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The activity which follows is aimed at giving you an opportunity to
practice on the proofs of trigonometric identities.
Activity 7
1. Prove the following trigonometric identities:
(a)
cos x
 1  sin x
sec x  tan x
(b) sin x  cos xtan x  cot x  sec x  csc x
cos 2 x  3 sin x  1
1
(c)

2
cos x  2 sin x  2 1  csc x
(d) sec x  cos x  sin x tan x
(e) tan 2 x  sin 2 x  tan 2 x sin 2 x
(f) sec 4 x  tan 4 x  sec 2 x  tan 2 x
(g) sin x tan x  cos x  sec x
(h) tan x  cot x  sec x csc x
(i)
tan y
 sec y
sin y
(j) sin y  sin y cot 2 y  csc y
(k) sin x cos x tan x  1  cos 2 x
(l)
sin 
sin 7  sin 5

cos  cos 7  cos 5
(m) sin   sin 2  sin 3  sin 4  4 cos
(n)
cos   cos 2

 tan
sin   sin 2
2
(o)
1  sin 
2
 sec   tan  
1  sin 
(p)
tan   cot 
1

sec   csc  sin   cos 
(q) sec   csc  cot   sex csc 2 
Page 39

2
cos  sin
5
2
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
2.4 FURTHER TRIGONOMETRIC
IDENTITIES
Suppose that A and B are angles measured in either degrees or
radians, then you should know the following identity holds:
cos( B  A)  cos B cos A  sin B sin A...........................(1)
We can derive another identity using (1) as follows:
cos( B  A)  cos[ B  ( A)]
Now take a close examination of the steps below:
cos[ B  ( A)]  cos B cos(  A)  sin B sin(  A)
 cos B cos A  sin B sin A, since cos(  A)  cos A
(which is an even function ).
Additional ly, sin(  A)   sin A, ( odd function).
We have therefore derived our desired idntity, which is :
cos( B  A)  cos B cos A  sin B sin A.........................(2)
Let us now solve questions using identities (1) and (2):
Example
Evaluate the following without use of a calculator
(a) cos 15 0
(b) cos
7
12
Solutions:
(a) We can use the special angles 45 0 and 30 0 to express cos 15 0
Page 40
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
as cos( 45 0  30 0 )............................. *0 . A look at *0 shows that it is
similar to the expression in (1), and usage of identity (1) gives us
cos(45  30)  cos 45 cos 30  sin 45 sin 30............. *2 .
Can you remember the values of cos 45, cos 30, sin 45, sin 30 ?
In case you have forgotten, go back and restudy the section on
‘special angles’. However, if you understood that section, you
will discover that *2 could be written in the following form:
 1
3  1 1

cos( 45  30)  

  . We ought to rationalising the

2
2
2
2



denominators, on the right hand side of the preceding equation as
shown below:
 2 1
3  2 1 1


cos( 45  30)  


  2  2  2
2
2
2

 

=
6
2


4
4
6 2
4
Hence, cos15 0 
6 2
4
(b) In order for us to find the value of cos
of all express
7
, we should first
12
7
either as a sum or difference of ‘special
12
angles’.
For instance,
cos
7
=
12

3


4
and because of this,
7
  
= cos   .
12
3 4
  
We can now evaluate cos   using identity (2) as
3 4
follows:




  
cos   = cos cos  sin sin
3
4
3
4
3 4
Page 41
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
1   3
1 
1
 . At this stage, we ought to rationalise


  
2
2
2 
2 

=
the denominators and doing so would give us:
3 1 
     2 1 1   2

cos   = 
 



2   2 2
2 
3 4  2 2
=
2
6

.
4
4
Consequently, cos
2
6
7

=
.
4
4
12
Have you perceived how easy it is to use ‘special angles’ and
identities to evaluate trigonometric expressions? If you have not yet
mastered the skill, we advise that you go through the above examples
again before proceeding.
There are other identities which you should know and one of such is
developed for you below:
Let us suppose that we are given a real angle B. It would then
follows that:




cos  B   cos cos B  sin sin B...................(3)
2
2
2

Since we know that cos

2
=0 and sin

2
=1, equation (3) becomes:


cos  B  = 0  cos B  1 sin B  sin B
2



 cos  B   sin B ………………………….. (4)
2

The activity which follows is intended to help you to strengthen your
understanding of the process of deriving similar identities. Attempt it.
Page 42
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 8
Follow the steps we have undergone to deduce (4) to show
that:


(a) sin   B   cos B
2



(b) tan   B   cot B
2

Having already considered identities relating to the cosine of a
sum and difference of two real angles, let us now derive those for
the sine of a difference and sum of two real angles.
Suppose A and B are two real angles. Then note that:
sin B  A  cos[



 B  A]  cos[  B   A]
2
2





 cos  B  cos A  sin   B  sin A
2

2

 sin B cos A  cos B sin A.
 sin B  A  sin B cos A  cos B sin A..............(5)
In addition to the above identities, take time to master the
following as well:
1. sin B  A  sin B cos A  cos B sin A
2. tan B  A 
tan B  tan A
1  tan B tan A
3. tan B  A 
tan B  tan A
1  tan B tan A
Another category of useful formulas which you require to know,
is that of DOUBLE ANGLE FORMULAS. We now state these formulas
below for your consideration and understanding:
Page 43
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
cos 2 A  cos 2 A  sin 2 A
cos 2 A  1  2 sin 2 A
cos 2 A  2 cos 2 A  1
sin 2 A  2 sin A cos A
tan 2 A 
2 tan A
1  tan 2 A
Let us now go through some examples which make use of some
the identities we have studied above.
Example:
Evaluate the following without using a calculator:
(a) sin
11
12
(b) sin 195 0
Solutions:
(a) Before we can evaluate sin
11
11
, we should express
either as a
12
12
a sum or difference of two ‘special angles’. Upon investigation, you
will discover that
11
11 7 
 7  
 . Consequently, sin
 .
=
=sin 
12
12
6
4
4
 6
With this discovery, we can now use equation
(5) to do the evaluation as follows:
7

7

 7  
cos  cos
sin
  = sin
sin 
6
4
6
4
4
 6
 1 1    3 1 
.

= 

2   2
2 
 2
If we rationalise the denominators as we did for our similar calculations,
Page 44
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
We would end up with
Thus, sin
6
2

.
4
4
6
2
11

=
.
12
4
4
(b) sin 195 0 = sin 150  45
= sin 150 cos 45  cos150 sin 45
 1 1    3 1 


= 
  

2
2
2
2

 

=
 2
6

4
4
You have now known what double angle formulas are about, but you still need to
Learn something about HALF-ANGLE FORMULAS, and these are provided
for your masterly below. Thus, given that A is a real angle, then the following
hold:
1. cos
B
1  cos B

2
2
2. sin
B
1  cos B

2
2
3. tan
B
1  cos B

2
1  cos B
2.4.1 TRIGONOMETRIC EQUATIONS
With all the knowledge of trigonometry we have acquired so far, you are
now at a stage where you should use the same to solve trigonometric
equations. Do not be worried about the term ‘trigonometric equations’.
These are merely equations which are composed of trigonometric terms, and
the objective is to find angles which balance such equations.
Page 45
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Example
Solve the following trigonometric equations:
(a) 2 sin 2  sin   0 for  90 0    90 0
(b) 2  tan 2   3 tan   0, for 0 0    360 0
Solutions:
(a) 2 sin 2  sin   0 for  90 0    90 0 . Here we shall express the term
2 sin 2 in form where it involves sin  and cos . How do you think this
can be done? Hopefully, you have recalled the identity
sin 2  2 sin  cos  ................(1)
Let us now use (1) in our given equation as follows:
22 sin  cos   sin   0
If we expand on the left hand side of the preceding equation, we would
obtain: 4 sin  cos  sin   0 ………………(2)
Let us now factorise the left hand side of (2), and doing so gives us:
sin  4 cos  1  0 . The implication then is that
either sin   0 ………………….(3)
or 4 cos  1  0 …………………(4)
We shall begin by using (3) to solve for  as follows:
sin   0    sin 1 0 . Arising form the given interval  900    900 ,
the only value of  which satisfies sin   0 is 0 0 .
Now let us use equation (4) to find the remaining solutions:
From 4 cos 1 =0, we have:
1
cos   .........................(5)
4
1
   cos 1    75.52 0
4
You may have noticed that 75.52 0 lies within the given interval
Page 46
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
 900    900 .
If we closely examine equation (5), we would discover that the cosine
function has a positive value also in the fourth(4th ) quadrant. Therefore, the
other value of  which lies in the interval  900    900 is  75.52 0 .
Hence the required solutions are: 0 0 , 75.52 0 and  75.52 0 .
(b) 2  tan 2   3 tan   0, for 0 0    360 0
Let us factorise the left hand side of the given equation as follows:
tan 2   2 tan   tan   2  0
tan  tan   2  1tan   2  0
tan   1tan   2  0
 either tan   1  0 .........................(1)
or
tan   2  0....................................( 2)
We shall start by using equation (1) to find the required angles:
tan   1  0  tan  1 ………………..(3)
Then   tan 1 1  450
From (3), we take note that the tangent function is giving a positive
value. The angle 45 0 lies in the first quadrant, and but the tangent function
tan is also positive in the third quadrant. We therefore require to find the
angle which lies in that quadrant. What do you think this angle is?
Have you found the angle   225 0 ? If yes, how did you find it? If your
answer is different from   225 0 , what might have caused this?
Nevertheless, for you to find the value of the angle in the third quadrant
(   225 0 ), you merely need to add 180 0 and the acute angle earlier
computed ( 45 0 ).
Now that we have used equation (1), let us also use equation (2) to
find the other values of  . From tan   2  0 , we know that
Page 47
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
tan   2    tan 1 2
Thus,   63.4 0 , which lies in the interval 00    3600 . Do you
think this is the only solution here? You should take note that tan  is
has a positive value here also. It therefore follows that we should
compute the angle which lies in the third quadrant (where tan  is
positive).
In order for us to find that angle which lies in the third quadrant, we
are going to add 180 0 and the acute angle   63.4 0 . Thus, the required
angle is 243.4 0 . We have now come to the end of the calculation. Hence
solutions are: 450 ,225 0 ,63.4 0 and 243.4 0
2.4.2 TRIGONOMETRIC EQUATIONS OF THE FORM:
a cos  b sin   c
For us to find solutions of equations of the form a cos  b sin   c ,
we assume that the numerical values a, b and c are appropriately
chosen. Additionally, before proceeding to solve such equations, we
should confirm that a 2  b 2  c 2 .
The following are the steps we ought to go through to solve this type of
equations (although any other workable method can be used):
1. Multiply and divide each term on the left hand side of the equation by
a 2  b 2 as shown below:


a
b
a2  b2 
cos  
sin  ....................(i )
2
2
a2  b2
 a b

2. Secondly, we ought to introduce an auxiliary angle say  such that
cos  
a
a2  b2
................................................(ii )
By using information in ( ii ) and Pythagoras theorem
we then have:
Page 48
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
sin  
b
..............................................(iii )
a  b2
2
3. Using information from ( ii ) and ( iii ), the expression ( i ) transforms
into:
a 2  b 2 cos  cos   sin  sin  ......................(iv ).
We are now going to use our knowledge of identities to summarise
( iv ) and once summarised, it becomes:
a
2

 b 2 cos   ...............................................(v)
4. Observe that the expression ( v ) is equal to a cos  b sin  . Thus,
a cos  b sin  =
a
cos    
c
2
a  b2
2

 b 2 cos   =c. Consequently,
......................................(vi)
Alternatively, if we let sin  
a
a2  b2
in step ( ii ), where  is the
auxiliary angle, and go through a similar process, we can deduce that:
sin     
c
a2  b2
…………………………….. (vii)
5. With either equation ( vi ) or (vii) , we can then solve for the required
values of  .
(i) The general solutions of an equation cos   Z , where Z  1 , are
given by :   2n   or   360 0   0 , for all n  Z
(ii) The general solutions of
an equation of the form
sin   Z , where Z  1 are given by:
  n   1n  or   (n  180 0 )   1n  0 , for n  Z
Page 49
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Example
Solve the following trigonometric equations:
(a)
(b)
3 cos  sin   1 , for 0 0   0  360 0
3 cos  4 sin   2.5 , for  180 0   0  180 0
Solutions:
(a) Comparing the given equation with a cos  b sin   c , we
conclude that a  3 , b  1 and c  1 . You should observe that
a2  b2 
 3
2
 12  3  1  4  c 2  12  1 . Now let us use the
steps which we studied earlier to solve our equation:
Are you able to deduce that
each term of
a 2  b 2  2? We should then divide
3 cos   sin  by 2 and multiply the entire
expression by 2 as follows:
 3

1
2
cos   sin  .......................................(i )
2
 2

Equating (i) to 1, which is the right hand side of the question
equation we have:
 3

1
2
cos   sin    1...................................(ii )
2
 2

From (ii) we obtain
3
1
1
cos   sin   ........(iii )
2
2
2
At this stage, let us introduce an auxiliary angle  such that
cos  
3
1
. Consequently, sin   and then equation (iii)
2
2
transforms into
cos  cos   sin  sin  
1
1
 cos     ................(iv )
2
2
You should take note that using (iv) we would have
1
2
    cos 1    60 0 .
Let us now use the appropriate expression for finding a general
Page 50
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
angle     360 0  n  60 0............................(v)
We are looking for the value/s of  from (v ) , but we need first
of  using either cos  
3
1
or sin   . If we use our basic
2
2
knowledge of trigonometric functions, we shall discover that:
  30 0.
Replacing  by 30 0 in equation (v) and solving for  , we have:
  30 0  360 0  n  60 0..........................(vi)
For n=0 in (vi)  = 30 0  60 0 . Here we get two values. That is
either   90 0 or   30 0 . However, only   90 0 lies within
the given interval 0 0   0  360 0 . This implies that the other
value is not among the required solutions.
Suppose we now let n=1 in (vi). What do you think will be the
solution? Certainly, the only value lying within the
given interval is   330 0 . For all the other values of n  Z ,
the values of  lie outside 0 0   0  360 0 . Hence, the solutions
of
3 cos  sin   1 , for 0 0   0  360 0 are: 90 0 and 330 0 .
(b) 3 cos  4 sin   2.5 . We shall solve this equation using
the steps we studied earlier. This given equation is the same as:
4
3

5 cos   sin    2.5
5
5

3
4
2.5 1
 cos   sin  
 ...................................(i )
5
5
5
2
We shall now introduce the auxiliary angle  such that
sin  
3
   36.87 0 .
5
With the preceding information, (i) become:
sin     
1
and     30 0..............................(ii )
2
Using the expression for computing general angles for
equations such as (ii) we have:
Page 51
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
    n  180 0   1n 30 ...................................(iii )
Let us now replace  by 36.87 0 and solve for  in (iii) to
derive   36.87  n  180 0    1  30 0...............(iv )
n
for n=0, in (iv) we have   6.87 0 which lies within
 180 0   0  180 0 . Additionally, when n=1, in (iv)
We obtain   113.130 which also lies within
 180 0   0  180 0 . However, you investigate and discover
That for all other values of n  Z , the values of  would lie
outside the interval  180 0   0  180 0 . Consequently,
the solutions of 3 cos  4 sin   2.5 , for  180 0   0  180 0
are:   113.130 and   6.87 0 .
We have now completed the study of unit two. We hope you
enjoyed your discoveries. The following activity, however, is
intended to accord you an opportunity for further practice
on how to solve trigonometric equations. Enjoy your practice.
Activity 9
1. Solve the following trigonometric equations for the indicated
Intervals:
(a) 3 tan 3   3 tan 2   tan   1  0, 0 0    360 0
(b) 1+ cos 2  cos 4  cos 6  0, for  90 0    90 0
(c) 3 cos 2   5 sin   1  0, for 0 0    360 0
(d) 8 cos 2   6 sin   3  0, for -

2
 

2
(e) 2 sin   3 tan   0, for  90 0    90 0
(f) 7 cos 2  6 sin 2  5, for 0 0   0  180 0
(g) 4 sin   3 cos   2  0, for - 180 0   0  180 0
Page 52
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Unit summary
In this unit you have learnt six trigonometric
functions, and special angles. You equally studied
several kinds of identities and how to prove
trigonometric identities. Additionally, you have
learnt how to solve trigonometric equation.
Page 53
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
ANSWERS TO ACTIVITY 9
(a) 450 ,225 0 30 0 ,210 0 ,150 0 ,330 0
(b)  90 0 ,90 0 ,450 ,450  30 0 ,30 0
(c) 199.47 0 ,340.530
(d)

6
(e) 0 0 ,30 0 ,30 0
(f) 48.880 ,171.72 0
(g)  13.30 ,119.56 0
Page 54
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Unit 3 EXPONENTIAL
AND LOGARITHMIC
FUNCTIONS
3.0 Introduction
This unit will introduce you to exponents and logarithms and will
enable you to appreciate the useful role that they play in Algebra.
In order to have an easy understanding of this unit, the knowledge
of laws of indices which some of you have, will be necessary.
During and upon completion of this unit you will be able to:
(b)Use the knowledge of exponents and logarithms to simplify and
solve algebraic expressions and equations.
(c) Appreciate the application of exponents and logarithms in
solving real life problems.
(d)Sketch, draw and interpret graphs of exponential and
logarithmic functions.
(e) Identify and sketch hyperbolic graphs of hyperbolic functions
.
Page 55
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3.1 Exponents
3.1 Definition
An exponent is a power or index of an index notation or number.
Let us consider the numbers below 2 x and 73 . For the two numbers
x and 3 are called exponents, powers or indices while 2 and 7 are
called bases.
An index number to base e, that is e x is called the natural
exponent.
3.2 Laws of exponents
From your secondary or high school mathematics, you learnt about
laws of indices. We will develop this section with this assumption.
Now let’s refresh our memories with the activity below.
Activity 1
Simplify the following expressions using laws of indices:
a) 3 3 3 3 3 3 =
b) a 5  a 3 
c)  b3  
5
Solutions
(b) To simplify the expression in a) you will realize that 3 is being
multiplied by itself six times. However, this does not imply
3 3 3 3 3 3  3 6
But 3  3  3  3  3  3  31  31  31  31  31  31 in index form
And by laws of indices this can be simplified as
3  3  3  3  3  3  31  31  31  31  31  31  3111111  36
with 3 as a common base.
(c) In this question you will notice that we have been given two
simplified expressions; a5 and a3 , which we can write in
expanded form as:
Page 56
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
 a  a  a  a  a  a    a  a  a   a  a  a2 as explained in (a)
above
Now simplifying the same expression using the laws of indices,
we will work as follows:
a 5  a 3  a 5 3  a 2
This is the same answer we got using the expanded form.
Having gone through questions (a) and (b), you can now
simplify (c).
Hint: Use the law in (a) to assist you in your working. Compare
your answer with the one below.
c) b15
The three laws of indices described above can be summarized
as:
4. Multiplication law of indices
5. Division law of indices
6. power law of indices
But since an index is the same as an exponent, the laws above
are also called the laws of exponents.
Given the exponential functions
y  a b and
x  a c , then the following laws hold:
6. xy  a b  a c
xy  a b c
7.
x ab

y ac
x
 a b c
y
8. (a b ) n  a bn
These laws also hold for natural exponents. That is:
iv). e a  e b  e a b
v). e a  e b  e a b
vi). (e a ) b  e ab
Page 57
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3.2 Logarithms
Let’s start our discussion with this activity
Activity 2
Evaluate
a) 35
b) 107
To find the value of 35 , you can either multiply the base, 3, five
times or use a calculator to get 243.
Similarly to evaluate 107 , you can either multiply the base, 10,
seven times or use a calculator to get 10,000,000.
We can therefore deduce from activity 2 that the numbers 5 and 7
determine the value of the index numbers given, that is 35 and 107 .
We call such numbers logarithms.
3.2.1 Definition
The logarithm (written in short as log or lg) of a positive number to
a given base is the power to which the base must be raised to equal
the given number.
For instance, (a) 35  243 can be expressed in logarithmic form as
log3 243  5 read as, the logarithm of 243 in base 3 is equal to 5.
5 is the logarithm to base 3
Now express b) in logarithmic form in the same way.
i)
Logarithms to base 10 is written in short form as lg and
usually with no base attached.
Example:
log 10 10,000  4 can be written as lg 10,000  4 .
ii)
Logarithms to base e (2.718 to three decimal places) is
written in short form as loge x  ln x
iii)
Logarithms to base 10 are called common logarithms while
logarithms to base e are called natural logarithms.
Page 58
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
In general, given an equation a b  c , this can be expressed in
logarithmic form as
log a c  b .
In the same way eb  a can be expressed in logarithmic form as
loge a  b or simply ln a  b .
However, for this to exist
(i) a  1
(ii) a  0
Activity 3
[Self marked. Take 20 minutes on this activity]
1. Identify the logarithm and the base in each of the following
statements.
a)
1  40
b) 1728  123
1
c) 125   
5
3
2. Express the index form given into logarithmic form:
a) 63  216
3
b) 27  814
c) 0.001  103
1
d) 4   
 16 

1
2
e) 80  1
3.2.2 Laws of logarithms
Having defined the logarithm of a number, we are going to spend
some time to discuss some important laws governing logarithms.
To do this easily, we will revisit the laws of exponents we started
with at the beginning of this unit.
Page 59
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Let’s consider the statements a x  b and a y  c .
2.2.1 The product law
The statements a x  b and a y  c can be expressed in logarithmic
form as
log a 
log a 
Well done.
Now write an expression for the product, bc in index form.
bc 
But since a is a common base, we add the exponents and obtain
bc  a x  y
Converting bc  a x  y in logarithmic form we get
log a bc 
Substituting the value of x and the value y in this equation, we
obtain
log a bc  log a b  log a c
This is the product law of logarithms.
3.2.2 The quotient law
To derive this law, we shall begin by expressing the quotient
index form. I’m sure this should not be a problem.
Let’s do this together.
Expressing
b
in index form gives
c
b ax

This simplifies to
c ay
b
 a x  y . Since a is a common base.
c
Now we write this in logarithmic form to obtain
Page 60
b
in
c
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
b
lo g a   
c
Replacing the values of x and y in the equation above we get
b
log a    log a b  log a c
c
This is the quotient law of logarithms
3.2.3 The power law
The rule states that
log a b n  n log a b , where n is an integer.
Proof using the LHS
Let n be an integer
Then
bn   a x  (From above)
n
b n  a xn (by power law of exponents)
Expressing this in logarithmic form, we obtain
log a b n  xn
Substituting the value of x in the equation, we get
…………………. as required.
The three laws we have discussed above are very important in
simplifying expressions and solving equations involving logarithms
as you will soon notice.
Examples 1
1. Express as single logarithms.
a) log 2  log 3
b) log18  log 3
1
d) 3log x  log y  1
2
Page 61
1
c) log p  log q
3
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Solutions
(v) log 2  log 3
To simplify this expression we will use the product law of
logarithms explained above. That is
log 2  log 3  log(2  3)  log 6
b) log18  log 9
To write this as a single logarithm we will apply the quotient
law as follows:
log18  log 9  log(18  9)  log 2
1
c) log p  log q
3
To simplify this expression we will need to apply the quotient
and the power rules as follows:
1
1
log p  log q  log p  log q 3 using power law
3
1


= log  p  q 3  using quotient


= log
p
1
q3
= log
p
3 q
1
d) 3log x  log y  1
2
To answer the question, we need to apply the three laws of
logarithms as follows:
1
1
3log x  log y  1  log x3  log y 2  1
2
1
2
= log( x  y )  1
3
Page 62
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
 3
x
= log10  1   log10 10
 
2
y 
log10
=
10 x3
y
2. Evaluate
a) log1000
b)
1
log 3 81
2
c)  log 2
1
2
d)
log 49
log 343
Solutions
(b) To answer this question, we take the base to be 10.
Hence log1000  log10 1000
= log10 102
= 2.
b)
1
log 3 81
2
This question can easily be evaluated by first expressing 81 in
exponential form to base 3 as follows:
1
1
log 3 81  log 3 34
2
2
1
= log 3  34  2
`
= log 3 3
2
= 2 log 3 3
=2
c) We will use the power law to evaluate this question
1
1
 log 2  log 2  
2
2
 log 2
1
1
1
 log 2  21 
2
= log 2 2
=1
Page 63
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
d)
log 49
log 343
To answer this question we will first convert the given
logarithms to the same base, 7
log 49 log 7 7 2

log 343 log 7 73
=
2
3
The examples lead us to seven important identities:
For any base a, the following hold:
i) a1  a , if and only if
log a a  1
ii) a 0  1 , if and only if
log a 1  0
iii) a 1 
log a
1
, if and only if
a
1
 1
a
And from identity (i), a more general result can be formulated.
That is
For any base a, the following is true.
a x  N , if and only if (iff)
log a N  x
This gives two important results.
9. x  log a a x
10. N  a loga N
11. e ln x  x for all x  0
12. ln e x  x for all real x
Page 64
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Examples 2
Simplify
a) 3log3 4
b) log 2 27
c)
d)
e)
e ln 4.3
ln e 9
e  ln 3
f) ln e
Solutions
a) 3log3 4  4
b) log 2 27  7
c)
d)
e)
e ln 4.3
ln e 9
e  ln 3
= 4.3
=9
1
1

= e ln 3 3
1
2
f)
1
ln e ln e 
=
2
Page 65
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 4
1. Simplify the logarithms below
a) 2  3log a
2. Find the value of:
a)
log8
log 2
b) log 4 1
c) l og a
1
a
2.3 Change of base
Changing the base of a logarithmic expression involves taking the
same logarithm on both sides. In order to assist you understand this
part well. Let’s consider the general statement log a b  x , where a
is any base other than 10 or e, the natural logarithm.
Then
a x  b in exponential form
Now changing the base of the given base a to a new base, c we
obtain
log c a x  log c b
Implying that
x
log c b
log c a
Finally substituting x  log a b , we get
log a b 
log c b
log c a
In the special case where c  b
The identity
log a b 
log c b
log c b
becomes
log a b 
log c c
1

log c a log c a
Page 66
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
We usually change logarithms to base 10 or e.
Examples 3
1. Evaluate, correct to three significant figures, the following
a) log 6 7
b) log8 72.4
2. Show that log  a b 2   log  b a 3   6
Solutions
To answer the questions above, we shall employ the formula
log c b
log a b 
.
log c a
a) log 6 7 . In this logarithmic expression, a = 6 and b = 7.
Now changing the base to 10, we have
log 6 7 
log10 7 lg 7

log10 6 lg 6
To get the value of log 6 7 , we use calculators directly.
So this gives
log 6 7  1.09
Now try to answer b) on your own using the same approach. If your
working is accurate, you should be able to get 2.06 as the answer.
Otherwise, go through your work cautiously.
2. To show that the LHS = RHS, let’s work from the LHS as
follows
log( a b 2 )  log  log b a 3   2 log  a b   3log b a
2 log a b  3 
2 log a b 
1
3
 2 log a b 
log a b
log a b
3
 6 as required
log a b
Page 67
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Activity 5
[Take 25 minutes on this activity]
1. Simplify
a) 3log3 5
b)
c) log a a n
d) x log x 9
e) log 3  log 4 43 
3.0 Exponential and Logarithmic equations
Having taken you through the laws of indices and logarithms, its
now time to use these laws in solving exponential and logarithmic
equations. To do this systematically, we will begin our discussion
with exponential equations and then finish off with logarithmic
equations.
3.3 Exponential equations
An exponential equation is an equation containing a variable in an
exponent. For example, 2 x  2  3x 3 .
3.3.1 Solving exponential equations with the same base.
Examples 4
Solve for x
ii) 34 x  2  3x
i) 2 x  64
2
3
Solutions
i) To answer this question we are going to express 64 in
exponential or index form to base 2 as follows:
2 x  64  26
Now since the base is the same on both sides, it follows that even
the exponents are equal. Therefore
2 x  26
Implying that
x6
ii) 34 x  2  3x
2
3
Page 68
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
This equation can be solved directly by equating the exponents on
both sides since the base is the same. That is
4 x  2  x2  3
Implying
x 2  4 x  5  0 This is a quadratic equation.
Solving this question gives
x  1 or x  5 (Refer to module 1 on how to solve quadratic
equations)
3.3.2 Solving quadratic exponential equations
Examples 5
i) 42 x 1  7  4 x   3  0
ii) 22 x  2 x  6  0
Solutions
We shall go through the first example together and then you will
be required to answer question two on your own. So be at tentative.
i) 42 x 1  7  4 x   3  0
Step 1: Ensure that the terms expressed in exponential or index
form have the same base. If not convert them to the same base. In
the question above, the terms expressed in exponential or index
notation, that is 42 x 1 and  7(4 x ) are given to the same base, 4.
Thus, there is no need to change the base. However, we must
express 4 2 x 1 in terms of 4 x as 4  4 x  by multiplication and power
2
laws of indices.
Step 2: Let some variable, say p, represent the term in index form,
in our case 4 x .
Now letting p  4 x , we obtain
4 p2  7 p  3  0
Solving this quadratic equation gives
p  1 and p 
3
4
Page 69
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
But p  4x
Therefore for p  1 , we obtain
4 x  1  40
Implying x  0
And for p 
4x 
3
, we obtain
4
3
4
Expressing both sides in base 10, we have
x lg 4  lg
3
4
Implying that
x
lg 0.75
By use of the calculator, we get
lg 4
x  0.21 to 2 decimal places.
Now you can work out ii). Compare your answer(s) with this
answer, x  1.58 to 2 decimal places.
If you fail to get this answer, you should not be discouraged, but try
to go through your work carefully.
3.3.3 Solving exponential equations using the change of base
method
This method was introduced earlier in the unit, but we will now
demonstrate how to use it to solve exponential equations.
Examples 6
Solve for x
i) 53 x  7
ii) 2 x1  5
Solutions
Here is how to go about solving these types of equations.
i) 53 x  7
Page 70
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
To solve this equation, you must convert both sides to logarithm to
base 10.
That is
lg 53 x  lg 7
Implying that
3x 
lg 7
 1.209
lg 5
Giving
x  0.40 to 2 decimal places
ii) 2 x1  5
To solve this equation, we will use the same approach as in (i).
That is
lg 2 x 1  lg 5x , by power rule, this becomes
 x  1 lg 2 
Expanding the LHS and collecting like terms we obtain
lg 2 
Now factorizing the RHS and dividing with the LHS, we get
lg 2
 x which is the same as
lg 5  lg 3
x
lg 2
By use of calculators we obtain the value of x as
lg 5  lg 3
x  0.76 to 2 decimal places.
I’m sure you didn’t find problems in filling the missing parts. If
you didn’t have no problems then you are ready to do the activity
below.
Activity 6
1. Solve the exponential equations
a) 2 x  8
b) x 3  729
Page 71
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
c) 52 x  25
d) 83t  32 4t 1
2. Solve the equations:
a) 3x  6
b) 22 x  5
c) 42 x1  3
3. Solve the equations
a)  5 x  5 x1   10
b) 2  22 x   5  2 x   2  0
c) 32 x 1  26  3x   9  0
d) 4 x  6  2 x   16  0
3.4 logarithmic equations
A logarithmic equation is an equation which involves a logarithmic
function of some variable. For example, 7  ln x  15 .
Like exponential equations, there are several methods used to solve
logarithmic equations as you will see from the worked examples
below.
3.4.1 Solving equations of the form
log a b  x and log x b  c
where the unknown is represented by x.
Examples 7
Solve for x
a) log3 81  x
b) log x 8  2
Solutions
a) log3 81  x
Page 72
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
To solve this equation, we must express the given equation in index
form as follows:
3x  81
Now we express 81 in exponential form to base 3. So we have
3 x  34
Implying that
x4
b) log x 8  2
The approach used in solving the previous equation will be
employed to solving this equation. That is
x2  8
Implying that
x 8.
3.2.2 Solving logarithmic equation with the same base.
Examples 8
a) l og3  x  3 log3  x  3  3
b)
log2 x  log2  x  2  log2  9  2 x 
Solutions
a) l og3  x  3 log3  x  3  3
To solve this question, we will begin by simplifying the LHS by
applying the multiplication law of logarithms. That is
log3  x  3 x  3  3
Converting in index form we obtain
33   x  3 x  3
Implying that
27  x 2  9
Thus x 2  36
And
x  6 is the only answer.
Page 73
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
b) log2 x  x  2  log2  9  2 x 
Using the approach as the one above, we simplify the equation as
follows:
log2 x  x  2  log2  9  2 x 
But the LHS and RHS are both given in base two, we can cancel or
ignore the logarithms and rewrite the equation,
log2 x  x  2  log2  9  2 x  as follows:
x( x  2)  9  2 x
Simplifying the equation we obtain
x2  9  0
And
x  3 or x  3 as answers.
3.2.3 Solving logarithmic equations with different bases
Examples 9
Solve for y
a) log 2 y  log8 y  4
b) log 5 y  2 log y 5  3
Solutions
To solve equations in example 8 above, we utilize the change of
base method which we have already discussed.
a) log 2 y  log8 y  4
We will write the equation in base two as follows:
log 2 y 
log 2 y
 4 , simplifying gives
log 2 8
3log 2 y  log 2 y  12 , since ( log 2 8  log 2 23  3 )
Now applying the quotient law of logarithms and simplifying the
given equation, we obtain
y3
log 2
 12
y
Expressing in index form, we get
Page 74
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
212  y 2
2   y 
1
12 2
1
2 2
2 6  y or
y  64
b) log 5 y  2 log y 5  3
Changing the base to 5, we obtain
log5 52
log5 y 
3
log5 y
This simplifies to
 log5 y 
2
 2  3log5 y
Letting p  log5 y , we get a new equation in p as
p 2  2  3 p , giving us
p2  3 p  2  0
Solving for p, we obtain
p  1 or p  2
But
p  log5 y
Now
For p  1
Implies that
log5 y  1
This gives
y  51  5
For p  2
Implies
log5 y  2
This gives
Page 75
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
y  52
Implying that
y  25
Activity 7
[Take 40 minutes on this task]
1. Solve for x
a) log5 x  2 b) log x 125  3 c) log 5  x 2  4 x   1
d) x  log2  log4 256
e) log3 3  2 x  2
2. Solve the following equations
a) log7 x  log7 18x  61  1
b) log 5  4 x 2  1  2  log 5  2 x  1
c) log8  x 2  9   log8  x  3  2
3. Solve for x
a) log3 x  4log x 3  3  0
b) log 2 x  log x 2  2
c) log2 x  log 4  x  6 
d) log3 x  2log x 3  1
3.3 Simultaneous equations involving Exponential and Logarithmic
functions
Apart from the equations we have discussed, that is, equations in
one variable, exponential and logarithmic equation can be
presented in two variables as we shall seen.
Examples 10
Solve for x and y
a)
log x y  2 ; xy  8
b)
2 lg y  lg 2  lg x ; 2 y  4x
c)
y log 2 8  x ; 2 x  8 y  8192
d)
ln 6  ln  x  3  2ln y ; 2 y  x  3
e)
5x  2  7 y 1  3468 ; 7 y  5x  76
Page 76
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
log3 x  y  log9  2x 1
f)
Solutions
log x y  2 ; xy  8
a)
To solve this equation, we start by converting log x y  2 in index
form as follows
x2  y
Now replacing x 2  y in the equation xy  8 and solving for x we
obtain
x  x2   8
Implying that
x3  8
x2
Substituting x  2 in the equation x 2  y , we find the value of y as
y  22  4
b)
2 lg y  lg 2  lg x ; 2 y  4x
To answer this question, it’s easier to work in index form.
Therefore, we will begin by converting the equation
2 lg y  lg 2  lg x in index form as follows:
lg y 2  lg( 2 x) Can you state the laws used to simplify the equation?
Now from the equation above, you will notice that, both sides of
the equation have been expressed in the same base, hence we will
cancel the logarithms and obtain an equation in x and y only as
y 2  2x .
Expressing x in terms of y we have
x
y2
2
y2
Substituting the value of x, that is x 
in the equation 2 y  4x
2
and solving for y, we get
Page 77
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
2 4
y
y2
2
Implying that
2  2 
y
y2
2 2
, this simplifies to
2 y  2 y . Since the base is the same, then the powers also equal.
2
Implying
y  y 2 .Hence
y  1 or y  0
Replacing y in the equation, x 
y2
we obtain two values of x as
2
For x  1
x
12 1

2 2
And for x  0
x
02
0
2
y log 2 8  x ; 2 x  8 y  8192
c)
This equation is similar to the one we have just solved. As such I
will ask you to express y log 2 8  x in index form.
2x 
Now substituting x 
in the equation 2 x  8 y  8192 , we obtain
2  8 y  8192
Now expressing the LHS in index form to base 8, we get
8 y  8 y  8192
Implying that
 
8   4096 and expressing both sides in same base, 8, we have
8   8
2 8 y  8192 , dividing both sides by 2, get
y
y
4
And
Page 78
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
y4
Substituting y  4 and solving for x, we obtain
x  12
d)
ln 6  ln  x  3  2ln y 2 y  x  3
;
The same method we used to answer b) and c) above will be
utilized to solve equation d).
Again we start solving the equations by expressing
ln 6  ln  x  3  2ln y
in index form as follows
ln[ 6x  3]  ln y 2 by multiplication and power laws of logarithms
Since the LHS and RHS are expressed to the same base, then this
equation simplifies to
6x  3  y 2 . Expanding and collecting like terms gives
y 2  6 x  18  0
Now expressing x in terms of in the equation 2 y  x  3 gives
x  2y  3
And substituting this x value in the equation y 2  6 x  18  0 we
obtain a quadratic equation in y as
y 2  12 y  36  0
Solving this equation gives
y6
Implying that
x  9 .That is by substituting y  6 in the equation x  2 y  3 .
e)
5x  2  7 y 1  3468 ; 7 y  5x  76
Since the two equations have expressed in index form, we straight
away use one of the methods of solving simultaneous equations to
work out the solutions. In this case we shall apply substitution
method.
y
x
x2
y 1
We shall substitute 7  5  76 in the equation 5  7  3468
y
since 7 is the subject. This gives
Page 79
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
5 x  2  7 y 1  3468 By laws of indices this equation can written as
5 2 (5 x )  7(7 y )  3468 Replacing the value of 7 y , we obtain


5 2 (5 x )  7 5 x  76  3468
This reduces to
 
25(5 x )  7 5 x  532  3468
Implying
 
32 5 x  4000 . Dividing by 32, we get
5 x  125
Therefore
x5
Now substituting x  5 in the equation 7 y  5 x  76 , we obtain
7 y  5 3  76
Implying that
7 y  49 . Rewriting 49 in base 7 and solving for y gives
y  2.
f)
log3 x  y  log9  2x 1
The equations given above are
log 3 x  y ; y  log 9 2 x  1
Now expressing the equations in index form, we obtain
3 y  x ; 9 y  2x  1
Substituting x  3 y in the equation 9 y  2 x  1 , we get
9 y  2(3 y )  1
(32 ) y  2(3 y )  1 And from laws of indices this gives
(3 y ) 2  2(3 y )  1
This exponential equation is of the form ax 2  bx  c  0 which we
have already discussed above. Therefore
y  0 and x  1. I’m sure getting these values would not be a
problem.
Page 80
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3.4.2 Equations involving natural
exponents and natural logarithms
3.4.2.1 Solving equations involving natural exponents
Examples 11
1. Solve for x
a) e ln(2 x 1)  5
b) e 2 x 1  5
c)
e x  ex
2
2
Solutions
a) e ln(2 x 1)  5
To solve this equation, we will use the identity e ln x  x . So in our
case x  2 x  1
Implying
2x  1  5
And
x  3.
b) e 2 x 1  5
To answer this question, we will begin by writing the equation
e 2 x 1  5 in logarithmic form as
log e 5  2 x  1
But you will recall that log e x  ln x from our discussions on page
14.
Now using this, we rewrite the equation as
2x  1  ln 5
x
1
ln 5  1 . Using the calculator we get
2
x  1.305 to three decimal places.
Page 81
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
c)
e x  ex
2
2
To solve this equation, we first multiply both sides of the equation
by 2 as follows
2
e x  ex
 2  2 giving us
2
e x  ex  4
Multiplying both sides by e x we then obtain
(e x )  e x  (e e )  e  x  4e x . Simplifying gives
e 2 x  1  4e x this gives the quadratic equation
(e x ) 2  4e x  1  0
To solve this equation, we let a variable say P replace e x in the
equation. That is
p2  4 p 1  0
And solving for p gives p  2  5
But p  e x
Hence
ex  2  5
Since e x  0 for all x, we only use the positive solution for e x ;
That is e x  2  5
Now taking the natural log on both sides we obtain
ln e x  ln( 2  5)
Implying that
x  ln (2  5 )
3.4.2 Solving equations involving natural logarithms
Examples 12
Solve for x
a) ln( x  1)  1  ln x
Page 82
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
To solve this equation, we first collect like terms as follows
ln( x  1)  ln x  1
Applying the quotient law of logarithms, we express
ln( x  1)  ln x  1 as
ln
x  1  1 . Now expressing this in exponential form, we obtain
e1 
x
x 1
, using the fact that ln x  log e x
x
Implying that
xe  x  1
xe  x  1
And
x
1
 0.582 to three decimal places.
e 1
Self marked activity 8
[Take 20 minutes on this task]
Solve for x
1.
a) e 3 x 5  100
b) e 0.01x  27
3
d) e ln1 x   2 x
c) e x  e x e 4
2
e) e ln6 x
2.
2
4
  5x
a) ln x  ln 2  1
c) ln x 

b) ln( x  1)  0
1
2
ln 4  ln 8
2
3

d) ln x 2  x  2  ln x  ln x  1
Page 83
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3.5.0 Application of
exponential and logarithmic
function
Exponential and logarithmic functions are very essential in real life
situations. In this section we will look at how exponential and
logarithmic functions are applied in business and science.
3.5.1 Decay and growth problems
Decay and growth problems are solved using the formula
f ( x)  Ae kx , where A represents the initial amount of substance,
K is the constant and x represents the time.
Examples 13
1) A radioactive substance is decaying according to the
formula y  Ae 0.2 x where y is the amount of substance
remaining after x years.
2
3
If the initial amount A = 80 grams, how much is left after 3
years?
The half-life of a radioactive substance is the time it takes
for half of it to decompose. Find the half-life of this
substance in which A = 80 grams.
Solution
i. In this question, we have been given the initial amount of
substance (A = 80) and the time (x = 3)
Thus substituting these values in the general formula, that is
y  Ae kx , we obtain
y  80e 0.2(3) .This simplifies to
y  80e 0.6
Page 84
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
This gives us
y  43.9 grams to 1 decimal place
ii.
This question calls for the time x which only half of the
initial amount of substance is left. Thus half of the substance gives
us 40 grams.
Implying
1
 e 0.2 x . Now taking the natural logarithms on both sides
2
we get
 0.2 x  ln
1
2
 0.2x  ln 1  ln 2
Meaning
x  3.466 Years to 3 decimal places
3.5.2
Compound interest
The formula used in the evaluation of none compounding interest is
also the application of exponential growth. The formula for none
compound interest is given as

At  p1 

r

n
nt
where
p = initial investment
r = annual interest rate (as a decimal)
n = number of annual interest periods
t = number of years of investment
At = final value after t years
Page 85
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Examples 14
A $5000 investment earns interest at the annual rate of 8.4%
compounded monthly.
a) What is the investment worth after one year?
b) What is it worth after 10 years?
c) How much interest was earned in 10 years?
Solutions
From the data given,
p = $5000
r = 8.4% = 0.084
n = monthly
Therefore interest per month =
r
8.4% 0.084


 0.007
n
12
12
to three decimal places.
Now with this data we can find the solutions to the given question
as follows:
a) A1  50001  0.007 
112
= 50001.007 
12
= $5437 to the nearest dollar
b) A10  50001  0.007
1210
= 50001.007 
120
= $11,548 to the nearest dollar
c) To find the interest earned in 10 years, we find the difference of
the final amount and the initial investment. That is
Interest after 10 years = $11,548 - $5000 = $6548
Page 86
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3.5.3 Graphs of exponential
and Logarithmic functions
Graphs of exponential functions.
An Exponential function is a function of the form f ( x)  a x , where
a is a positive constant. That is a  0 and a  1
In other words an exponential function is one where the variable
appears as an exponent.
Let’s consider the exponential function f ( x)  3 x for  3  x  3
x
f ( x)  3 x
-3
-2
-1
0
1
2
3
0.037
0.111
0.333
1
3
9
27
From the table above, we observe that:
i) f ( x)  0 for all real values of x,
ii) As x increases f (x) increases rapidly,
iii) f (x) =1 when x = 0
iv) As x decreases (That is x  2,3,..., ) , f(x) becomes numerically
smaller. We say that as x approaches minus infinity,
f (x) approaches the value zero. This is written in short as
x  , f ( x)  0
v) The domain of f (x) is the set of positive real numbers.
Now let’s sketch the graph of f ( x)  3 x for  3  x  3
Page 87
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
From the graph of f ( x)  3 x , you will observe that the graph does
not cross the line y = 0 or the X- axis. Hence we say that the
X – axis is the horizontal asymptote.
In general, all exponential functions of the form y  a x , for a > 0,
have graphs of the same shape and all pass through the point (0 , 1)
and the X  axis or the line y = 0 is the horizontal asymptote.
Self marked Activity 9
Draw the graphs of:
a) y  2 x
b) y  1.5
x
c) y  4 x , for  3  x  3
From the basic graphs of exponential functions discussed above,
we can also draw other graphs of exponential functions which are
quite challenging.
Examples
Draw graphs of:
f ( x)  2 x3 . and f ( x)  2 x  1 for  3  x  3 .
Page 88
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The first one f ( x)  2 x3 has been done for you. Study it and then
draw the graph of f ( x)  2 x  1 .
Now the table of values of f ( x)  2 x3 are shown below:
x
-3
-2
-1
0
f(x)
0.0156 0.0313 0.0625 0.125
1
2
3
0.25
0.5
1
Graph of f ( x)  2 x3
Now try to draw the graph of f ( x)  2 x  1 on your own.
5.2 Logarithmic functions
A function of the form f ( y)  log b x , where b  0 and b  1, is
called a logarithmic function with base b.
Let’s consider the logarithmic function f ( x)  log 2 x
Page 89
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
x
0.125
0.25
0.5
1
2
4
f ( x)  log 2 x
-3
-2
-1
0
1
2
From the table, the following we can observe the following:
i)
When x  1, f (1) does not exist. Put it in a different way,
there is no real value for f (x) when x = -1.Meaning
f ( x)  log b x does not exist for negative values of x.
ii)
For x  1, f ( x)  0 and as x approaches positive infinity,
f (x) approaches positive infinity.
iii)
For x  0 , f (0) is undefined. As x decreases to zero, f (x)
approaches negative infinity. And that for 0  x  1 , f ( x)  0 .
iv)
the graph is concave down wards
v)
it is an increasing function
Now we draw the graph of f ( x)  log 2 x for  3  x  x .
Page 90
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The graph of f ( x)  log 2 x , does not cross the line x = 0 or the Yaxis. Hence we say that the Y – axis is the vertical asymptote
All graphs of the form f ( x)  log a x , a  1 and a > 0, have the
same shape as the basic function, and they all pass through the
point (1,0) and the Y – axis or the line x = 0 is the vertical
asymptote.
Self marked activity 10
Draw graphs of:
a) y  log x 3
b) y  log x 4
Now by comparing or drawing the graphs of y  4 x and y  log x 4 ,
you will observe that:
(iii)the two graphs are inverses of each other. In other words the
graph of y  4 x is the inverse of the graph of y  log x 4 .
In general the functions f ( x)  a x and g ( x)  log x a are inverse
functions.
(iv) the line y  x is the mirror line (line of reflection) of the
two graphs. That is to say you can draw the graph of y  4 x
by reflecting this graph in the line y  x and the reverse is
true.
Apart from the basic graphs demonstrated above, we also draw
higher graphs of logarithmic functions.
Example
Graph the functions y  log 2 (2 x  4) and y  log 3 ( x  2) for
1  x  2
Solution
Since logarithms can be taken only of positive numbers, the
domain of the function is the solution of 2x  4  0 ; that is , all
x  2 . Furthermore, the asymptote is x = -2.
Now to locate points on the curve, first we rewrite the logarithmic
function in exponential form. That is
Page 91
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
y  log 2 (2 x  4)  2 x  4  2 y
Now substituting the values of x in 2 y  2 x  4 , we get
x
f(x)
-1
1
0
2
1
2
3
Now graph of y  log 3 ( x  2) .
Graphs of natural exponents and logarithms behave in the same
way as those of common exponents and logarithms. For instance
the graph of the natural exponential function, y  e x passes
through the point (0, 1) and has the line y = 0 as its horizontal
asymptote.
Similarly, the graph of the natural logarithmic function,
y  ln x passes through the point (0, 1) and has the line x = 0 as its
vertical asymptote.
Refer to the graphs below.
It follows from the fore going discussion that graphs of natural
exponential functions and logarithmic functions have the same
properties as the earlier graphs discussed above.
Page 92
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The graph of y  e x
The graph of y  ln x
Self marked activity 11
Sketch the following graphs
a)
y = e power x-2
b)
y = 3e power 2x minus 1
c)
y = ln (x-2)
d)
y = ln2x
Page 93
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Unit Summary
In this unit you have learnt the laws of exponents and logarithms
and how they can be used to simplify and solve equations involving
exponential and logarithmic functions.
You have also learnt how to draw and sketch graphs of exponential
and logarithmic functions as well as hyperbolic functions.
Page 94
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
3.5.4 The hyperbolic
functions
These are functions that have been generated by the use of
(e x  e  x )
is
2
called the hyperbolic sine of x and is abbreviated sinh x . Similarly,
exponential functions. For example the function
e

 ex
is the hyperbolic cosine of x and is
2
abbreviated cosh x . The other hyperbolic functions worth
mentioning are;
the function
x
e x  ex
tanh x  x
e  e x
csc hx 
2
e  e x
sec hx 
2
e  e x
coth x 
e x  ex
e x  e x
x
x
The graph of sinh x
Page 95
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The graph of cosh x
The graph of tanh x
Page 96
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The graph of csc hx
The graph of sec hx
Page 97
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
The graph of coth x
Hyperbolic functions are not periodic.
2.
sinh x varies from   to   and cosh x varies
between  1 and  
tanh x  1
3.
0  sec hx  1
1.
However, it is not our intention to go into detailed discussions
about the hyperbolic functions in this unit, but rather to show you
the differences between these graphs and the exponential and
logarithmic functions. You will discuss the hyperbolic functions in
details in higher courses.
Assessment
1. Express as a single logarithmic
a) 2 log a 5  log a 4  log a 10
b) log 5 7  log 5 x
c) 3  2 lg 5
2. Simplify
a) log y 2  log xy
b)
log x 3  log x
log x 2  log x
Page 98
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
c) log(log y 2 )  log(log y)
3. Evaluate
a)
ln 10
ln 5
b)
ln 8
ln 0.2
c)
ln 25
3 ln 5
4. Solve for x
a) 2 x  45
c) 80e x  120
b) 100(2 x )  2000
d) 4 2 x 1  6 x e) e 2 x 1  3 x 1
5. Solve the equations
a) log 2 (2x  1)  3
b) 2 log 2 x  log 2 ( x  1)  2
c) log 3 x  2 log x 3  1
d) 4 2 x 1  5 x  2  61 x
e) 4 x  2 x 1  3  0
f) 4 32 x 1  17 3 x  7  0


 
6. Solve the simultaneous equations
a) log y x  2 , 5 y  x  12 log x y
b) log 2 x  log 2 y  3 , log y x  2
c) log x  2  log 2  2 log y , log x  3 y  3  0
d) Show that log 16  xy 
1
1
log 4 x  log 4 y . Hence or otherwise,
2
2
solve the simultaneous equations
1 log 4 x 
log 16  xy  3 ,
 8
2 (log 4 y)
7. A radioactive material is decaying according to the formula
y  Ae kx , where x is the time in years. The initial amount A = 10
grams, and 8 grams remain after 5 years.
a) Find k. Leave your answer in terms of natural logarithms.
b) Estimate the amount remaining after 10 years.
c) Find the half – life to the nearest tenth of a year.
8. Suppose that a $10,000 investment earns compound interest at
the annual interest rate of 9%. If the time of deposit of the
investment is one year (t = 1), find the value of the investment for
each of the following types of compounding.
Page 99
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
a) n = 4 (quarterly)
b) n = 52 (weekly)
c) n = 365 (daily)
9. Graph each of the following, state the domain, intercepts,
equations of the asymptotes, if:
a) f ( x)  0.2 
x
b) f ( x)  log 4 x  1
c) f ( x)  e x2
d) f ( x)  ln x  4
e) f ( x)  ln x 3
Page
100
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
Answers to activities
Activity five
a) 5 b) n
c) 9
Activity six
d) 1
1. a) 3 b) 9
d)
c) 0
5
11
2. a) 1.6309 b) 1.1610
c) -0.1038
3. a) 1.2153 c) 2
Activity seven
1. a) 25
b) 5 c) 5; -1 d) 2 e) -3
1
2. a) 9
c) 67
1
3. a) 81 ; 3
c) 3
Activity eight
1 3
4
1.
a)
c)  ;
e)
2 2
3
e
2.
a)
c) 64
2
Assessment
1. a) log a 10 b) log 5 7 x
c) log 10 40
y
2. a) log
c) 2
x
2
3. a) b) -1.292
c)
3
4. a) 5.492
c) -0.405
1
9
5. a)
c) ; 9
e)
3
2
6. a) x  4 or x  9 and y  2 or y  3
c) x  4 or 10 and y  4 or and y  2
1 4
7. a) ln
b) 6.4 grams c) 15.5 year
5 5
END OF MODULE 2
Page
101
MODULE 2
FOUNDATION MATHEMATICS – MAT 110
References
10
3