ENGINEERING MATHEMATICS 1

ENGINEERING
MATHEMATICS 1

WAN AZLIZA BINTI WAN ZAKARIA
RAHIMAH BT MOHD ZAIN @ AB. RAZAK

NIK NOOR SALISAH BINTI NIK ISMAIL

MATHEMATIC, SCIENCE & COMPUTER

ENGINEERING
MATHEMATICS

1

Wan Azliza Binti Wan Zakaria
Rahimah Binti Mohd Zain @ Ab. Razak

Nik Noor Salisah Binti Nik Ismail

Politeknik Sultan Haji Ahmad Shah

Published by
POL/TEKNIK SULTAN HAJI AHMAD SHAH
SEMAMBU
25350 KUANTAN

Copyright ©2021, by Politeknik Sultan Haji Ahmad Shah
Materials published in this book under the copyright of Politeknik Sultan Haji Ahmad Shah.
All rights reserved. No part of this publication may be reproduced or distributed in any form
or by means, electronic, mechanical, photocopying, recording, or otherwise or stored in a
database or retrieval system without the prior written permission of the publishers.

Preface

Alhamdulillah, with the blessing of Allah S.W.T., the ebook of Engineering
Mathematics 1 can be accomplished written as planned. This ebook has been prepared by
consulting the Polytechnic’s syllabus in order to facilitate the lecture session and as one of the
meaningful references to the students.

Most of the contents of the ebook have been written by referring to various text books
and course’s website. Nevertheless, because of the limited knowledge from us, we feel that
some improvement and additional contents can be made in order to enhance the ebook’s
quality in future. Thus, any comments for improvement are highly appreciated.

Engineering Mathematics 1 consists of topics such as Basic Algebra, Partial Fraction,
Trigonometry, Complex Number, Matrices and Scalar and Vector. Various type of exercise are
provided for students to improve their skill in solving problems related to topics they have
learned.

We also would like to thank to all the friend lecturers, the Head of Mathematics Unit
and also the Head of Mathematics, Science and Computer Department for their support and
encourage us in accomplishing this ebook. We really hope that this ebook will be the important
reference hence can help the students to enhance their understanding towards the excellent
achievement especially in engineering discipline. Lastly, we also hope that this ebook will
benefit to all lecturers and students.

Rahimah Bt Mohd Zain @ Ab. Razak
Wan Azliza Binti Wan Zakaria

Nik Noor Salisah Binti Nik Ismail

CONTENTS

TOPIC ITEMS PAGES

1 BASIC ALGEBRA 1
1.1 Introduction of Algebra 7
1.2 Algebraic Expression 9
1.3 Quadratic equation 9
11
1.3.1 Solving an equation by using factorization method 13
1.3.2 Solving an equation by using quadratic formula
1.3.3 Solving an equation by using completing the squares 15
21
PARTIAL FRACTION 21
2.1 Introduction of Partial Fraction 25
2.2 Proper fraction 28
32
2 2.2.1 Proper fraction with linear factor
36
2.2.2 Proper fraction with repeated linear factor 36
38
2.2.3 Proper fraction with quadratic factor 40
2.3 Improper fraction 42
46
TRIGONOMETRY 51
3.1 Introduction of trigonometry 51
54
3.1.1 Angles and their measure 57
3.1.2 Trigonometric raatios
3.2 Graph of sine, cosine and tangent 59
62
3 3.3 Positive and negative value of trigonometric 62
64
3.4 Trigonometric equations and identities 66
3.5 Sine and cosine rules 68
70
3.5.1 Sine rule 72
3.5.2 Cosine rule
3.6 Area of triangle 72
73
COMPLEX NUMBER 75
4.1 Introduction of complex number 76
4.2 The operation of complex number 77
78
4.2.1 Addition 79
4.2.2 Subtraction 80
4.2.3 Multiplication
4.2.4 Division
4.3 Conjugate of a complex numbers
4.4 Graphical representation of a complex number

4 through Argand diagram

4.4.1 Argand's diagram to represent a complex number
4.4.2 Modulus and argument
4.5 Complex number in other form
4.5.1 Complex number in polar form
4.5.2 Complex number in trigonometric form
4.5.3 Complex number in exponential form
4.5.4 Complex number in cartesian form
4.6 Multilpication and division of complex number
in polar form

MATRIX 84
5.1 Introduction of matrix 88
5.2 Operation of matrices 88
88
5.2.1 Addition 90
5.2.2 Subtracting 94
97
5 5.2.3 Multiplication 100
100
5.3 Determinant 104
5.4 Inverse matrix using minor, adjoin and cofactor
5.5 Simultaneous linear equation using matrix 107
114
5.5.1 Simultaneous equation using inverse matrix 114
5.5.2 Simultaneous equation using Cramer's rule 115

VECTOR AND SCALAR 116
6.1 Introduction of vector
6.2 The operation of vector 120
124
6.2.1 Vector addition 126
6.2.2 Addition and subtraction of vector using 129

6 parallelogram method

6.2.3 Addition and subtraction of vector using
triangle rule

6.3 Apply scalar (dot) product of two vectors
6.4 Apply vector (cross) product of two vectors

6.4.1 Application of the vector (cross) product
References

[CHAPTER 1: BASIC ALGEBRA]

BASIC ALGEBRA

OBJECTIVES:

At the end of this topic, students should be able to:

i. understand basic algebra
ii. understand the algebraic expression
iii. solve quadratic equation using:

a. factorization method
b. quadratic formula

x  b b2  4ac
2a

c. completing the square method

 x  b2  b2  c  0
 2   2

1.1 INTRODUCTION OF ALGEBRA

▪ Algebra is the use of letters and symbols to represent values and their relations, especially
for solving equations. The combination of each letters and symbols are called “Algebraic
Expressions’’.

BASIC ALGEBRA

 Basic algebra is the field of mathematics that it one step more abstract than
arithmetic.

 Remember that arithmetic is the manipulation of numbers through basic math
functions.

 Algebra introduces a variable, which stands for an unknown number or can be
substituted for an entire group of numbers.

Algebraic expression

 In mathematics, an algebraic expression is an expression built up from integer constants,

variables, and the algebraic operations (addition, subtraction, multiplication, division and

exponentiation by an exponent that is a rational number).

 Examples of algebraic expressions:

7x, x  4,5  x, x2  9, x
3

1

[CHAPTER 1: BASIC ALGEBRA]
Parts of an Algebraic Expression

i. Terms
 Every expression is separated by an operation which is called Terms.
 Like 7n and 2 are the two terms in the above figure.

ii. Factors
 Every term is formed by the product of the factors.
 7n is the product of 7 and n which are the factors of 7n .

iii. Coefficient
 The number placed before the variable or the numerical factor of the term is called
Coefficient of that variable.
 7 is the numerical factor of 7n so 7 is coefficient here.

iv. Variable
 Any letter like x , y etc. are called Variables.
 The variable in the above figure is n .

v. Operations
 Addition, subtraction etc. are the operations which separate each term.

vi. Constant
 The number without any variable is constant.
 2 is constant here.

2

[CHAPTER 1: BASIC ALGEBRA]

Examples of Algebraic Expressions
▪ An algebraic expression consists of numbers, variables, and operations. Here are a few

examples:

ALGEBRAIC EXPRESSION MEANING

8n 8 times n

3x 1 3 times x plus 1

9y 1 9 times y minus 1

n 1 N divided by 4 plus 1
4

EXAMPLE 1 EXPLANATION:
a. Move the like terms together.
1. Simplify 5x  2  2x b. Add or subtract their coefficients.

SOLUTION: EXPLANATION:
 5x  2x  2 a. Clear the parentheses.
 3x  2 b. Combine like terms by adding coefficients.
c. Combine the constants.
2. Simplify 35  x  x1  2x  7 d. Rearrange sequence of power.

SOLUTION:

 35  x  x1  2x  7

 15  3x  x  2x 2  7
 15  7  3x  x  2x 2
 22  4x  2x 2
 2x 2  4x  22

3. Simplify  2a3b  3ab2c EXPLANATION:

SOLUTION: a. Multiply the numbers (coefficients).
 2a 3b  3ab 2c b. Multiply the variables - exponents can be
 2  3  a3  a  b  b2  c
 6  a 4  b3  c combines if the base is the same.
 6a 4b3c

3

4. Simplify  8a3bc  2ab [CHAPTER 1: BASIC ALGEBRA]

SOLUTION: EXPLANATION:
a. Write the division of the algebraic terms
= −8a 3bc
2ab as a fraction.
b. Simplify the coefficient.
c. Cancel variables of the same type in the

numerator and denominator.

= - 4a2c

LET’S PRACTICE 1

Express each of the following expressions into a single algebraic fraction.

a) p3  q  qp  3

b) 3mn 1  1  Ans: 3 p  q
 m n 
Ans: 3n  m
c) 3  p
p 4 Ans: 12  p 2
4p
4

[CHAPTER 1: BASIC ALGEBRA]

 d) 2 pq  1
4p2q  2p  pq

Ans: 2 p 2 q

e) 4x2  3x 9x  6

 f) 5m 2 n3  6mn  4m 2 n3  3  6 Ans: x 23x  18

x 2 x 6 5  Ans: 3 3m 2n3  2mn  1
x3 3x 
g)  6 Ans: 5
3

5

[CHAPTER 1: BASIC ALGEBRA]

h) 6n  2a  5n  3a

Ans: n  27a

i) 7 p 4q 3 z
28 p 2 q 5

Ans: p 2 z
4q 2

j) 2  1
m3 3m

Ans: 1
m3

k) x 2 3x  3 2  2
 3x  x 1

Ans: x 1

x 1x  2

6

[CHAPTER 1: BASIC ALGEBRA]

1.2 ALGEBRAIC EXPRESSION

SOLVING LINEAR EQUATION
Linear Equation:
A mathematical expression that has an equal sign and linear expressions.

Variable:
A number that you don't know, often represented by " x " or " y " but any letter will do!

Variable(s) in linear expressions

i. Cannot have exponents (or powers).

 For example, x squared or x2
ii. Cannot multiply or divide each other.

 For example: " x " times " y " or xy ; " x " divided by " y " or x
y

iii. Cannot be found under a root sign or square root sign (sqrt).

 For example: x or the "square root x "; sqrt (x)

Linear Expression:
A mathematical statement that performs functions of addition, subtraction, multiplication,
and division.

EXAMPLE 2

1. Solve the following equation: x  12  25 EXPLANATION:
a. Isolate " x " to one side of the equation.
SOLUTION: b. Subtract 12 from each side to get
x  12  25
x  25  12 constants on the right 12.
x  13 c. Check the solution.
d. The result x equal to 13 .
2. Solve the following equation: y  6  42
EXPLANATION:
SOLUTION: a. Isolate "y" to one side of the equation.
y  6  42 b. Add 6 from each side to get constants on
y  42  6
y  48 the right (+6).
c. Check the solution.
d. The result y equal to 48.

7

[CHAPTER 1: BASIC ALGEBRA]

3. Solve the following equation: 3z  6

SOLUTION: EXPLANATION:

3z  6 1. Divide both sides by 3 to isolate the z.

3z  6 2. Check the solution
3 3 3. The result z equal to 2

z2

4. Solve the following equation: 42x  9  4x  4  6x

SOLUTION: EXPLANATION:
1. Expand the brackets and simplify.
42x  9  4x  4  6x 2. Isolate " x" to one side of the equation.
3. Add 36 from each side to get constants on
8x  36  4x  4  6x
the right  36 .
8x  4x  6x  4  36
4. Divide both sides by 10 to isolate the x .
10x  40 5. Check the solution

10x  40 6. The result x equal to 4
10 10

x4

LET’S PRACTICE 2

a) 2 x  4  30
5

b) 8  1 q  q  3 Ans: 65
4 Ans: 4

c) 2z  3  1  2z  25

Ans: 5
8

[CHAPTER 1: BASIC ALGEBRA]

d) 4 y  8  y  3  15

Ans: 4
3

e) 6  2 p  3  5 p  2

Ans: 1 6
7

1.3 QUADRATIC EQUATION

1.3.1 SOLVING AN EQUATION BY USING FACTORIZATION METHOD

EXAMPLE 3

Solve the following equations by factoring method x2  8x 12  0 .

SOLUTION EXPLANATION

x2  8x 12  0 a. This equation is already in general form of

Solution by cross multiply: ax2  bx  c  0.

8x b. To get x2 , x must be multiply with x .

c. Factor completely, x  6x  2  0 .

d. Apply the Zero Product Rule x  6  0 or
x  2  0.

e. Solve each factor that was set equal to zero
by getting the x on one side and the answer
on the other.
x  6 or x  2

x  6x  2  0 x2  0
x  2
x  6  0 or
x  6

9

[CHAPTER 1: BASIC ALGEBRA]

LET’S PRACTICE 3

Solve these equations by using factorization

a) 5x2  3x  2  0

Ans: x   2 , x 1
5

b) 4r 2  3r  10

Ans: r   5 ,r  2
4

c) p 2  7  8 p

Ans: p  7, p  1

d) 5t 2 14t  3  0 Ans: t  1 ,t  3
e) y 2  6 y  8  0 5

Ans: y  4, y  2

10

[CHAPTER 1: BASIC ALGEBRA]
1.3.2 SOLVING AN EQUATION BY USING QUADRATIC FORMULA

 When “Completing the Square” procedure is applied to a quadratic equation in general

form, ax2  bx  c  0 , then we receive the Quadratic formula.
 The expression for the solutions of a Quadratic Equation through coefficients of this

equation.

ax2  bx  c  0

x b b2  4ac
2a

Solution set for a Quadratic Equation may contain
i. Two distinct real numbers
ii. One real number (repeated root)
iii. No real solution

EXAMPLE 4

Solve the following equations using quadratic formula  3x2  6x  5  0 .

SOLUTION EXPLANATION

 3x2  6x  5  0 a. This equation is already in general form of

x   6  62  4 35 ax2  bx  c  0 .
2 3 b. Identify a , b and c , then plug them into the

x   6  36  60 quadratic formula. In this case, a  3 ,
 6 b  6 , and c  5 .
c. Use the order of operations to simplify the
 6  96  6  96 quadratic formula.
d. Then, solve the equation either positive or
6 6 negative.
e. There will usually be two answers.

x  or x 

x  0.633 x  2.633

11

[CHAPTER 1: BASIC ALGEBRA]
LET’S PRACTICE 4
Solve the equations below using quadratic formula to find the roots value for each of the following
equation.

a) 3t  5  t 2

b) 2a2  8a  4  0 Ans: t  1.19,t  4.19
c) 2 p 2  3 p  1 Ans: a  0.586, a  3.414
d) 5t 2 14t  4  0
e) y 2  6 y  28  0 Ans: p  0.28, p  1.78
f) 2x2  5x  3  0 Ans: t  3.061,t  0.261

Ans: y  9.08, y  3.08

Ans: x  1 , x  3
2

12

[CHAPTER 1: BASIC ALGEBRA]
1.3.3 SOLVING AN EQUATION BY COMPLETING THE SQUARES

▪ If u is an algebraic expression and d is a positive real number, then the equation
u 2  d has exactly two solutions:
u  d and u   d
Or

u   d 

Formula  x b 2  b 2 c 0 where a  1
 2   2 
   

Completing the square procedure
▪ Change the quadratic equation in the ax2  bx  c  0 to an equivalent equation in

the form ax  d 2  k which then can be solved using the Square Root Method.

EXAMPLE 5

Solve the following equations using completing square 5x2  4x  2  0 .

SOLUTION EXPLANATION

5x2  4x  2  0 a. Divide all terms by 5 .
x2  0.8x  0.4  0 b. Identify a, b, and c and plug them

 x   0.8  2    0.8  2  0.4  0 into the quadratic formula. In this
 2   2  case a  1 , b  0.8 , and
c  0.4 .
 x  0.8  2  0.16  0.4  0 c. Move the number term to the right
 2  side of the equation.
d. Complete the square on the left side
 0.8  2 of the equation and balance this by
 2  adding the same number to the
x   0.56 right side of the equation.
e. Take the square root on both sides
x  0.4   0.56 of the equation.

x   0.56  0.4 x2   0.56  0.4 f. Add 0.4 on right side.
x2  0.348
x1  0.56  0.4 or
x1  1.148

13

[CHAPTER 1: BASIC ALGEBRA]

LET’S PRACTICE 5

By using completing the square, find the roots value for each of the following question.

a) 3x 2 10x  2  0

Ans: x  0.19, x  3.52

b) 20r 2  40r  8  0

c) 10 p 2  6  30 p Ans: r  1.77, r  0.23
d) 5t 2  16t  3 Ans: p  0.188, p  3.188
e) y 2  6 y  7  0
Ans: t  0.178,t  3.378

Ans: y  4.414, y  1.586

14

[CHAPTER 2:PARTIAL FRACTION]

PARTIAL FRACTION

OBJECTIVES:
At the end of this topic, students should be able to understand about:

i. Proper and Improper Fraction
ii. Proper Fraction with Linear Factor
iii. Proper Fraction with Repeated Linear Fraction
iv. Proper Fraction with Quadratic Factor
v. Improper Fraction

2.1 INTRODUCTION OF PARTIAL FRACTION

DEFINE ALGEBRAIC FRACTIONS

 In the algebraic fraction a , the dividend a is called the numerator and the divisor b is
b

called the denominator.

 The numerator and denominator are called the terms of the algebraic fraction.

 The simplest way to define a numerator and a denominator is the following:

 Numerator: the top number of a fraction
 Denominator: the bottom number of a fraction

What Are Algebraic Fractions?

 Algebraic fractions are fractions using a variable in the numerator or denominator, such as
4 because division by 0 is impossible, variables in the denominator have certain
x
restrictions.

 The denominator can never equal 0.

15

[CHAPTER 2:PARTIAL FRACTION]

PARTIAL FRACTION EXPLANATION
x cannot equal 0( x ≠ 0)
2
x x cannot equal 6( x ≠ 6)
4
x6 y-z cannot equal 0( y-z ≠ 0) so y cannot equal
3 z(a ≠ b)
yz
5 Neither x or y cannot equal 0. ( x ≠ 0, ( y ≠ 0))
x2 y

PROPER AND IMPROPER FRACTIONS

 Fractions that are greater than 0 but less than 1 are called proper fractions.
 In proper fractions, the numerator is less than the denominator.
 When a fraction has a numerator that is greater than or equal to the denominator, the

fraction is an improper fraction.
 An improper fraction is always 1 or greater than 1.
 Finally, a mixed number is a combination of a whole number and a proper fraction.

IDENTIFYING PROPER AND IMPROPER FRACTIONS

 In a proper fraction, the numerator is always less than the denominator.

 Examples of proper fractions include 1 , 7 , 2
2 13 905

 In an improper fraction, the numerator is always greater than or equal to the denominator.

 Examples of improper fractions include 5 , 170 , 30
3 15

16

LET’S PRACTICE 1 [CHAPTER 2:PARTIAL FRACTION]
Fractions(Proper /Improper)
Identify the fraction below:

Equations

30
x 1

x
x 1

5
x2 1
9x3  x  6
x2 5
2x4  4x  7
x5  8

x4
x2  3x 1

6x
(x2 1)(x2  3)

CHANGING IMPROPER FRACTIONS TO MIXED NUMBERS

 An improper fraction can also be written as a mixed number.

 Mixed numbers contain both a whole number and a proper fraction.

 Examples of mixed numbers include 6 1 ,1140 ,3 6
5 18

Writing Improper Fractions as Mixed Numbers

Step 1: Divide the denominator into the numerator.
Step 2: The quotient is the whole number part of the mixed number.
Step 3: The remainder is the numerator of the fractional part of the mixed number.
Step 4: The divisor is the denominator of the fractional part of the mixed number.

17

[CHAPTER 2:PARTIAL FRACTION]

EXAMPLE 1

Write the improper fraction 47 as a mixed number.
7

SOLUTION:
a. 47  7  6 remainder 5 .Divide the denominator into the numerator.
b. The quotient, 6 , becomes the whole number.
c. The remainder, 5 , becomes the numerator.
d. The denominator, which is also used as the divisor, remains as 7 .

Answer: 47 = 65
7 7

Writing Mixed Numbers as Improper Fractions

Step 1: Multiply the denominator of the fraction by the whole number.
Step 2: Add this product to the numerator of the fraction.
Step 3: The sum is the numerator of the improper fraction.
Step 4: The denominator of the improper fraction is the same as the denominator of the fractional

part of the mixed number.

18

[CHAPTER 2:PARTIAL FRACTION]

EXAMPLE 2

Write 4 3 as an improper fraction
4

SOLUTION:

a. Multiply the denominator of the fraction by the whole number.
b. Add this result to the numerator of the fraction.
c. This answer becomes the numerator of the improper fraction.

Notice that the denominator of the improper fraction is the same as the denominator that was in
the fractional part of the mixed number.

4 3
4

4  4  16

16  3  19

19

4

Answer: 4 3 = 19
4 4

PARTIAL FRACTIONS

 To express a single rational fraction into the sum of two or more single rational fractions is
called Partial fraction resolution.

 For example,

2x  x2 1  1  1  1
x(x2 1) x x 1 x 1

2x  x2 1 is the resultant fraction and 1  x 1  x 1 1 are its partial fractions.
x(x2 1) x 1 

19

[CHAPTER 2:PARTIAL FRACTION]

PROCEDURES TO FIND PARTIAL FRACTION

PARTIAL FRACTION DECOMPOSITION
The method is called “Partial Fraction Decomposition” and goes like this:-

Step 1: Factor the bottom 6x  6 3)  (x 6x  6
(x2  2x   3)(x 1)
Step 2: Write one partial fraction for each of
those factors 6x  6 3)  (x A  B
(x2  2x   3) (x 1)
Step 3: Multiply through by the bottom so we
no longer 6x  6  A(x 1)  B(x  3)..........equation1
Step 4: Now find the Constants, A & B
Let (x  3)  0
Thus, x  3

Substituting into equation 1
6(3)  6  A(3 1)  B(3  3)

12  A(4)  B(0)

A3
Let (x 1)  0
Thus, x  1
Substituting into equation 1
6(1)  6  A(11)  B(1 3)

12  A(0)  B(4)

B3

Step 4: Rewrite the value of A and B into the 6x  6  (x 3 3)  3
partial fraction (x2  2x  3)  (x 1)

DEFINE TYPES OF PARTIAL FRACTION

20

[CHAPTER 2:PARTIAL FRACTION]

2.2 PROPER FRACTION

2.2.1 PROPER FRACTION WITH LINEAR FACTOR

FORMULA

EXAMPLE 3

Determine the partial fraction decomposition of each of the following:

a. x 8x  4 2

1x 

SOLUTION:

x 8x  4 2  A  x B 2

1x  x 1

8x  4  A(x  2)  B(x 1) …………… equation 1

if (x  2)  0

x  2
Substitute x  2 into equation 1

8 2  4  A 2  2  B 2 1

12  A0  B 3

B4

if (x 1)  0

x 1
Substitute x  1 into equation 1

81  4  A1  2  B1 1

12  A3  B0

A4

Therefore the partial fraction:- 8x  4  (x 4  4
(x 1)(x  2) 1) (x  2)

21

[CHAPTER 2:PARTIAL FRACTION]

b. x 2  11
(x 1)(x  2)(x  3)

SOLUTION

(x x2  11  3)  A  (x B 2)  C 3)
 1)( x  2)(x (x 1)  (x 

x2 11  A(x  2)(x  3)  B(x 1)(x  3)  A(x  2)(x 1) ……..equation 1

if (x 1)  0

x 1
Substitute x  1 into equation 1

(1)2 11  A(1 2)(1 3)  B(11)(1 3)  C(11)(1 2)

12  A(6)  B(0)  C(0)

A   12  2
6

if x  2  0

x  2
Substitute x  2 into equation 1

(2)2 11  A(2  2)(2  3)  B(2 1)(2  3)  C(2 1)(2  2)

15  A(0)  B(15)  C(0)

B  15 1
15

if x  3  0

x3
Substitute x  1 into equation 1

(3)2 11  A(3  2)(3  3)  B(3 1)(3  3)  C(3 1)(3  2)

20  A(0)  B(0)  C(10)

C  20  2
10

Therefore the partial fraction:- x 2  11   (x 2  (x 1 2)  ( x 2 3)
(x 1)(x  2)(x  3)  1)  

22

[CHAPTER 2:PARTIAL FRACTION]

c. 3x 1
(x2  5x  6)

SOLUTION:

(x2 3x 1 6)  (x A  B 3) * Must be factorized
 5x   2) (x 

3x 1  A(x  3)  B(x  2) ……..equation 1

According to denominator, x  3  0
x3

Substitute x  3 into equation 1

3(3) 1  A(3  3)  B(3  2)

10  A(0)  B(1)

B  1
10

According to denominator, x  2  0
x2

Substitute x  2 into equation 1

3(2) 1  A(2  3)  B(2  2)

7  A(1)  B(0)

A   1
7

Therefore the partial fraction:- 3x 1   7( 1 2)  10( 1 3)
(x2  5x  6) x x

LET’S PRACTICE 2

Determine the partial fraction decomposition of each of the following.
a. x  4
(x  1)(x  8)

Ans: 5 1)  4 8)
9(x  9(x 

23

[CHAPTER 2:PARTIAL FRACTION]

b. 6x  4
(x2  9x  8)

Ans: 44 8)  2
7(x  7(x  1)

c. 9  9x
(2x 2  7x  4)

Ans: 1 1)  (x 5 4)
(2x  

d. x2  5
x(x  3)(x  6)

Ans:  5  14 3)  41 6)
18x 27(x  54(x 

24

[CHAPTER 2:PARTIAL FRACTION]

e. 2x2  x  2
(x  3)(2x 1)(x  1)

Ans: 2( 1 3)  1 1  6( 1 1)
x x
32x 

2.2.2 PROPER FRACTION WITH REPEATED LINEAR FACTOR
FORMULA

EXAMPLE 4

Determine the partial fraction decomposition below.

3x2  2
(x  2)(x  1)2

SOLUTION:

3x2  2  A B  C
(x  2)(x 1)2 (x  2)  (x 1) (x 1)2

3x2  2  A(x 1)2  B(x 1)(x  2)  C(x  2) ……..equation 1

According to denominator, x  2  0
x2

Substitute x  2 into equation 1
3(2)2  2  A(2 1)2  B(2 1)(2  2)  C(2  2)

14  A(1)  B(0)  C(0)

A  14

25

[CHAPTER 2:PARTIAL FRACTION]

According to denominator, x  1  0
x 1

Substitute x  1 into equation 1
3(1)2  2  A(11)2  B(11)(1 2)  C(1 2)
5  A(0)  B(0)  C(1)
C  5

To find the value of B we can use the method of Equating Coefficients. We take equation 1 and
multiply-out the right-hand side, and then collect up like terms.

3x2  2  A(x 1)2  B(x 1)(x  2)  C(x  2) ……..equation 1

3x2  2  A(x 1)(x 1)  B(x 1)(x  2)  C(x  2)

Multiplying out:

3x2  2  (A  B)x2  (2A  3B  C)x  (A  2B  2C)

Left Right

Compare the same coefficients right and left side:

Use Equating coefficients Right Left
Coefficients 3
(A  B)
x2 0
x (2A  3B  C)
(A  2B  2C) 2
Constant,k

Substitute value of A,……… A  14

AB 3
14  B  3
B  11

Then; A  14, B  11, C  5

Therefore the Partial Fraction:- 3x2  2 14 11 5
(x  2)(x 1)2  (x  2)  (x 1)  (x 1)2

26

[CHAPTER 2:PARTIAL FRACTION]

LET’S PRACTICE 3

Determine the partial fraction decomposition of each of the following.

a. x2
x(x  4)2

Ans: (x 1 4)  (x 4
  4)2

b. 3 x2
x2 (x  6)

Ans: 11  13 6)
 12x  2x2 12(x 

c. (1  2x 1
x)(5x 1)2

Ans: 3 15 7
16(1  x)  16(5x  1)  4(5x  1)2

27

[CHAPTER 2:PARTIAL FRACTION]
2.2.3 PROPER FRACTION WITH QUADRATIC FACTOR
FORMULA

Note: (cx2  dx  c) cannot be factorized

UNFACTORIZED FACTORIZED

(x2 1) (x2 1)
(x2  4) (x2  4)
(x2 16) (x2 16)
(x2  64) (x2  64)

(x3 1)  (x 1)(x2  x 1)

EXAMPLE 5

Determine the partial fraction decomposition below.

6x2  8
x (x2  2)

SOLUTION:

6x2 8  A  Bx  C
x(x2  2) x (x2  2)

6x2  8  A(x2  2)  (Bx  C)(x) ……………..equation 1

According to denominator, x  0
Substitute x  0 into ……………………………equation 1

6(0)  8  A(0  2)  (B(0)  C)(0)
 8  A(2)  B(0)  C(0)
A  4

To find the value of B and C we can use the method of Equating Coefficients. We take equation 1
and multiply-out the right-hand side, and then collect up like terms.

28

[CHAPTER 2:PARTIAL FRACTION]

6x2  8  A(x2  2)  (Bx  C)(x) ………….equation 1
6x2  8  Ax2  2A  Bx2  Cx

Multiplying out:

6x2  8  (A  B)x2  Cx  2A

Left Right

Compare the same coefficients right and left side:

Use Equating coefficients

Coefficients Right Left
6
x2 (A  B)
x 0
C -8
Constant,k 2A

Substitude value of A,……… A  4

AB 6
4 B  6

B  10

C 0

Then; A  4, B  10, C  0

Therefore the Partial Fraction:- 6x2  8 4 10x
x(x2  2)   x  (x2  2)

29

[CHAPTER 2:PARTIAL FRACTION]

LET’S PRACTICE 4

Determine the partial fraction decomposition of each of the following.

a. 2x  6
(x  4)(x2  5)

Ans:  2 4)  2x  2
3(x  3(x2  5)

b. (x2 3 x
 2)(1  x)

2x  5 2
3 x2  2
 Ans:  31  x

c. 2x  3
(x2  1)x

Ans: 3x  2  3
(x2  1) x

30

[CHAPTER 2:PARTIAL FRACTION]

d. 9
x(x2  6)

Ans: 3 3x
2x  2(x2  6)

e. (x2 3x  8
 4x  1)(x  2)

Ans: x  51  1)  2 2)
13( x 2  4x 13(x 

f. 2x2  4x 10
(x2  1)x2

Ans:  4x  12  4  10
x x2
(x2  1)

31

[CHAPTER 2:PARTIAL FRACTION]

2.3 IMPROPER FRACTION

FORMULA

Note * Degree of numerator is the same or higher than the denominator n  k

EXAMPLE 6

Determine the partial fraction decomposition below.

4x3  5
x3 1

SOLUTION:

4
Use Long Division technique (x3  1) (4x3  5)

(4x3  4) 
_______

9

So, 4x3  6  4 x 9 1 …………… 9 must be state in form partial fraction
x3 1 3 x3 1

Partial fraction for 9 is….
x3 1

9  (x A  Bx  C
x3 1  1) (x 2  x  1)

9  A(x2  x 1)  (Bx  C)(x 1) ……..equation 1

According to denominator, x  1  0
x 1

32

[CHAPTER 2:PARTIAL FRACTION]

Substitute x  1 into equation 1

9  A((1)2 11)  (B(1)  C)(11)
9  A(3)  B(0)  C(0)
A3

To find the value of B and C we can use the method of Equating Coefficients. We take equation 1
and multiply-out the right-hand side, and then collect up like terms.

9  A(x2  x 1)  (Bx  C)(x 1) ……..equation 1

9  Ax2  Ax  A  Bx2  Bx  Cx  C

Multiplying out:

9  (A  B)x2  (A  B  C)x  (A  C)

Left Right

Compare the same coefficients right and left side:

Use Equating coefficients

Coefficients Right Left
A B 0
x2
x A B C 0
AC 9
Constant,k

AB 0 ABC 0
B  A 3  (3)  C  0
B  3
C  6
Then; A  3; B  3; C  6

The Partial Fraction:- 9 (x 3 1)  (x 3x  6
x3 1   2 x  1)

Therefore the Partial Fraction:- 4x3  5 4 3  (x 3x  6 1)
x3 1 (x 1) 2 x 

33

[CHAPTER 2:PARTIAL FRACTION]

LET’S PRACTICE 5

Express of the following in partial fractions.
a. 4x 3 10x  4
(2x 1)x

Ans: 2x 1 3  4
(2x 1) x

b. 2x2
(x  3)(2x 1)

Ans: 1 18 3)  1
5(x  5(2x 1)

34

[CHAPTER 2:PARTIAL FRACTION]

c. 4x3  3x  2
(x  2)(2x 1)

Ans: 2x  3  24 2)  2
5(x  5(2x 1)

d. x 3
(x  2)(x  3)

Ans: x 1 8 2)  27
5(x  5(x  3)

35

[CHAPTER 3: TRIGONOMETRY]

TRIGONOMETRY

OBJECTIVES:
At the end of this topic, students should be able to understand about:

i. Introduction of trigonometric
ii. Define sine, cosine, tangent, cosecant, secant and cotangent.
iii. Graph of sine, cosine and tangent.
iv. Positive and negative value of trigonometric.
v. Apply sine and cosine rules.

3.1 INTRODUCTION OF TRIGONOMETRY

3.1.1 ANGLES AND THEIR MEASURE

ANGLES IN UNIT DEGREES AND RADIANS

 Angles are measured in either degrees or radians (rad).

 The size of a radian is determined by the requirement that there are 2 radians in a

circle.

 Thus 2 radians equals 360 degrees.

 This mean that 1 radian  180, and 1 degree   radians.
180

EXAMPLE 1 (b) 135



1. Express in radian measure:

(a) 54

SOLUTION:

(a) 54  54   0.9425 rad
180

(b) 135  135   3  rad
180 4

36

[CHAPTER 3: TRIGONOMETRY]

2. Express each angle in degree measure:

(a)  rad (b) 5 rad
3 9

SOLUTION:

(a)  rad    180  60
3 3 

(b) 5 rad  5  180  100
9 9 

LET’S PRACTICE 1

1. Express in radian measure:

a) 330
b) 160
c) 42

2. Express each angle in degree Ans: 5.76,  2.79 & 0.733
Ans: 76.39° , 756°
measure:

a) 4 rad
3

b) 251

RIGHT ANGLE TRIANGLE

 In mathematics, the pythagorean theorem, also known as Pythagoras’ theorem, is a

fundamental relation in Euclidean geometry among the three sides of a right triangle.

 It states that the area of the square whose side is the hyotenuse is equal to the sum of

the areas of the squares on the other two sides. Formula;

 By the Pythgorean theorem that the sum of the

squares of each of the smallest sides equals the

square of the hypotenuse.

 Trigonometric functions are how the relationships

amongst the length of the sides of a right triangle c  a2 b2
vary as the other angles are changed.
Figure 3.1: Pythagoras’ Theorem

37

[CHAPTER 3: TRIGONOMETRY]

3.1.2 TRIGONOMETRIC RATIOS

 The ratios of the sides of a right triangle are called trigonometric ratios.
 Three common trigonometric ratios are the sin (sin), cosine (cos) and tangent (tan).
 The three others trigonometric ratios are the cosecant (csc), secant (sec) and

cotangent (cot).

Figure 3.2: Trigonometric ratios in right triangles

Figure 3.3: Basic and Pythagorean Identities

SOH-CAH-TOA : an easy way to remember trigonometric ratios

 The word SOHCAHTOA helps us remember the definitions of sine, cosine and
tangent.

Table 3.1: Trigonometric ratios

ACRONYM PART VERBAL DESCRIPTION MATHEMATICAL
DEFINITION

SOH Sine is Opposite over Hypotenuse sin A  opposite
hypotenuse

CAH Cosine is Adjacent over Hypotenuse cos A  adjacent
hypotenuse

TOA Tangent is Opposite over Adjacent tan A  opposite
adjacent

38

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 2

Find sin A and tan A Figure 3.4

SOLUTION:

sin A  opposite tan A  opposite
hypotenuse adjacent

sin A  3 tan A  3
5 4

LET’S PRACTICE 2

1. Find: Ans: 5 , 12 & 12
a) cosF 13 13 5
b) sin F
c) tanF

2.

a) sin H d) cscG Ans: 8 , 15 , 8 , 17 , 17 & 8
b) cos H e) secG 17 17 15 15 8 15
c) tan H f) cot G

39

[CHAPTER 3: TRIGONOMETRY]
3.

Find: d) cosecB Ans: 5 , 12 , 5 , 13 , 13 & 12
a) sin B e) sec B 13 13 12 5 12 5
b) cos B f) cot B
c) tan B

4.

Given BAC  60 0 . Find the length Ans: 22.52
of BC.

3.2 GRAPH OF SINE, COSINE AND TANGENT

GRAPH OF SINE

y  sin x

° −90° 0° 90° 180° 270° 360°
0   3 2
x rad   
2 0 22 0
y
1 1 0 1

Figure 3.5: Graph of sine

40

[CHAPTER 3: TRIGONOMETRY]

GRAPH OF COSINE

y  cos x

° −90° 0° 90° 180° 270° 360°
 3 2
x rad    0
2 2 1
y 2
0
0 1 0 1

Figure 3.6: Graph of cosine

GRAPH OF TANGENT
y  tan x

° −90° −45° 0° 45° 90° 180° 270°
  3
x rad      0 
2 4 4 22
y   1 0 1  0 

Figure 3.7: Graph of tangent

 An asymptote is an imaginary line that a curve get closer and closer to but never
touches.

41

[CHAPTER 3: TRIGONOMETRY]
3.3 POSITIVE AND NEGATIVE VALUE OF TRIGONOMETRIC
FOUR QUADRANTS

Figure 3.8a: Four quadrants

RELATED ANGLE

 Reference angles allow us to evaluate more complex angles and makes easier when
evaluating angles.

 A reference angle x and x ' , is the positive acute angle made by the terminal side of

the angle x and x-axis.
 The figure shows differing angles that lies in quadrants I, II, III and IV.
 Remember that in quadrant I, an angle and its related angle are the same measure.

= 180° − =

= 180° + = 360° −
42
Figure 3.8b: Four quadrants

where;

= Actual angle
= Related angle

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 3

Find the value of for sin   0.5 where 0° ≤ ≤ 360°

SOLUTION: Sine positive Quadrant 1:
sin  0.5 (Quadrant 1 and 2)   300

  sin 1 0.5 Quadrant 2:
  300   180 0  30 0  150 0

Therefore;
  30 0 & 150 0

LET’S PRACTICE 3

Find the value of  for the followings where 0° ≤ ≤ 360°
a. cos  0.8264

Ans: 34.27° & 325.73°

b. sin  0.8660

c. cos ec  5.461 Ans: 60° & 120°
d. sec  1.5642
Ans: 10.55° & 169.45°

Ans: 50.26° & 309.74°
43

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 4

Find the value of  for = −0.3255, where 0° ≤ ≤ 360°

SOLUTION: Cos is negative value
= −0.3255 (Quadrant 2 and 3)

= 0.3255
= 71°

Quadrant 2:
= 180° − 71° = 109°
Quadrant 3:
= 180° + 71° = 251°

Therefore;
= 109°& 251°

LET’S PRACTICE 4

Find the value of  for the followings where 0° ≤ ≤ 360°;
a. cos  0.9061

Ans: 154.97° & 205.03°

b. tan  1.364

c. sin  0.5246 Ans: 126.25° & 306.25°
d. cot  2.53
Ans: 211.64° & 328.26°

Ans: 158.43° & 338.43°
44