resource book for both teachers and grade 12 learners. It is written in a
manner that follows final Mathematics examination papers (CAPS).
It is divided by units. Units are sequenced as per the final Mathematics
Examination paper. It has important icons.
•
Unit outcomes
•
General notes
•
•
•
•
Practical examples
Common errors of the topic
End of the year questions (exam questions)
For the teacher
maths final exam with flying colours. Use it more and more daily.
Exam preparation book (Learner and Teacher)
Practically using this resource book will definitely make you pass your
Grade 12
•
Hints for the particular topics
PROBLEM SOLVED Maths
This mathematics preparation book was developed and compiled as a
Grade 12
Exam preparation book
(Learner and Teacher)
Leornard Gumani
Steven N. Muthige
Timothy M. Sibeko
Palesa T. Tsuebeane
PROBLEM SOLVED –
Maths Exam Preparation Book
Grade 12 Exam preparation book
(Learner and Teacher)
It is illegal to photocopy any pages from this book without the written
permission from Vivlia Publishers
Leonard Gumani Ntshengedzeni Steven Muthige
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PROBLEM SOLVED Maths Exam Preparation Book
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Printed and bound by:
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Acknowledgement
The publisher would like to thank all the copyright holders who permitted us to use their photographs and
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for the past exam papers.
While every effort has been made to contact and acknowledge all copyright holders, this did not prove
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photographs could be found.
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To Users
This mathematics preparation book was developed and compiled as a resource book for both teachers and
grade 12 learners. It is written in a manner that follows final Mathematics examination papers (CAPS).
It is divided by units. Units are sequenced as per the final Mathematics Examination paper. It has important
icons
•
•
•
•
•
•
•
Unit outcomes
Hints for the particular topics
General notes
Practical examples
Common errors of the topic
End of the year questions (exam questions)
For the teacher
Practically using this resource book will definitely make you pass your maths final exam with flying colours.
Use it more and more daily
Enjoy!
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Table of Content
Unit 1 Algebra
1.1 General Notes
1.2 Nature of Roots
Unit 2 Number Patterns, Sequences and Series
2.1 Quadratic Sequence
2.2 Arithmetic Sequence
2.3 Geometric Sequence
1
7
15
20
27
36
Unit 3 Functions
47
3.1
3.2
3.3
3.4
3.5
3.6
Straight Line
Parabola
Hyperbola
Exponential
Inverse Functions
Graphs
3.6.1 Straight line graph
3.6.2 Parabola
3.6.3 Hyperbola
3.6.4 Exponential graph
47
48
49
50
51
52
52
52
52
52
Unit 4 Financial Mathematics
63
4.1 Simple and Compound Growth
4.2 Simple and Compound Decay
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64
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viii | Table of Content
4.3 Nominal and Effective Interest Rates
4.4 Time Lines
4.5 Annuities
4.5.1 Future value
4.5.2 Present value
4.6 Sinking Fund
4.7 Deferred Annuity
4.8 Final Payment
Unit 5 Calculus
5.1
5.2
5.3
5.4
Average Gradient
Rate of Change (Derivative)
The Rules of Differentiation
Graphs of Cubic Polynomials
5.4.1 Concavity
5.5 Optimisation: Application of Calculus
Unit 6 Probability
6.1 Aids to Solve Probability Questions
6.1.1 Venn diagrams
6.1.2 Tree diagrams
6.1.3 Contingency table or two way table
6.1.4 Counting principles
Unit 7 Data Handling and Statistics
7.1 Types of Data
7.1.1 Ungrouped data
7.2 Five Number Summary
7.2.1 Grouped data
7.3 Skewness of the Data v/s Distribution of Data
7.4 Correlation Coefficient (r)
7.5 Regression Line or Line of Best Fit
Unit 8 Analytical Geometry
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65
67
68
68
69
71
72
73
81
81
82
85
88
89
97
105
106
106
100
110
111
117
117
117
117
118
119
120
120
129
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Table of Content | ix
Unit 9 Trigonometry
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Basic Trigonometry Ratios
Reduction Formula
Trigonometric Identities
Using Diagrams to Determine the Numerical Values of Ratios for
Angles from 0° to 360°
Trigonometric Equations
General Solution
Sine, Cosine and Area rules
Trigonometric Graphs
Unit 10 Euclidean Geometry
10.1 Revision from Lower Grades Corollaries, Theorems and Axioms
10.2 Triangles
10.2.1 Congruency axioms for triangles
10.3 Quadrilaterals
10.4 Circle
10.5 Circle Theorems
Marking Guidelines 2018 November
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148
150
153
154
162
168
175
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5
Calculus
UNIT
L E A R N E R S S H O U L D K N OW:
• How to find average gradient
• Difference between increasing and decreasing functions.
• If the gradient is positive, the function is increasing
• If the gradient is negative, the function is decreasing
• How to differentiate from first principles
• How to apply the rules of differentiation to find the first and second derivatives
• How to sketch cubic functions
• How to find the point of inflection
• How to apply calculus in optimisation
5.1 Average Gradient
You have learnt from previous grades that the gradient of a staight line is m =
y2 − y1
x2 − x1
Definition: The average gradient between two points is the gradient of a straight line drawn between
the two points.
Common Errors
1. Relating average gradient to whether a function is increasing or decreasing.
2. Failure to substitute values of x into the given function to find corresponding values of y.
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82 | Unit 5
Example
What is the average gradient of the graph of y = 3x 2 − 1 between x = 1 and x = 2.
Solution
y = 3x 2 − 1 is the same as f (x) = 3x 2 − 1.
f (1) = 3 (1) − 1 = 2 and f (2) = 3 ( 2 ) − 1 = 11
2
2
f (x2 ) − f (x1 ) 11 − 2
=
x2 − x1
2 −1
Average gradient =
=9
5.2 Rate of Change (Derivative)
Average gradient is the rate of change between two points.
The rate of change at any point on f is called the derivative, written as f / (x) .
The following are notations to represent the derivative of a function y = f (x) :
f / (x) or
dy
or y / or D x [ f (x)]
dx
From first principles, the formula to find the derivative is f / (x) = lim
h→0
f ( x + h) − f ( x )
h
Common Errors
1. Notational errors:
f (x) lim
h0
f x h f x lim f x h f x f x h f x or lim or
h
0
h0 h
h
h
2. Failing to use brackets when substituting leading to mistakes in signs of terms
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Calculus | 83
Examples
Determine f /(x) from first principles if
1. f (x) = x + 5
2. f (x) = 2 − x 2
3. f (x) = 3x 2 − 2 x + 3
Solutions
1. If f (x) = x + 5 then
f (x + h) = x + h + 5
f (x + h) − f (x)
h→0
h
x + h + 5 − (x + 5)
= lim
h→0
h
x + h + 5 − x − 5)
= lim
h→0
h
h
= lim
h→0 h
= lim 1
∴ f ′(x) = lim
h→0
=1
2. If f (x) = 2 − x 2 then
f (x + h) = 2 − (x + h)2
f (x + h) − f (x)
h
2 − (x + h)2 − (2 − x 2 )
= lim
h→0
h
2
2 − (x + 2 xh + h2 ) − 2 + x 2
= lim
h→0
h
2
2 − x − 2 xh − h2 − 2 + x 2
= lim
h→0
h
2
−2 xh − h
= lim
h→0
h
h(−2 x − h)
= lim
h→0
h
= lim(−2 x − h)
∴ f ′(x) = lim
h→0
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h→0
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2 − (x + h) − (2 − x )
h
2
2 − (x + 2 xh + h2 ) − 2 + x 2
= lim
h→0
h
2
2
2
84 | Unit
5 2 − x − 2 xh − h − 2 + x
= lim
h→0
h
2
−2 xh − h
= lim
h→0
h
h(−2 x − h)
= lim
h→0
h
= lim(−2 x − h)
= lim
h→0
h→0
= −2 x
2
3. If f (x) = 3x − 2 x + 3 then
f (x + h) = 3(x + h)2 − 2(x + h) + 3
f (x + h) − f (x)
h→0
h
3(x + h)2 − 2(x + h) + 3 − (3x 2 − 2 x + 3)
= lim
h→0
h
2
2
3(x + 2 xh + h ) − 2 x − 2h + 3 − 3x 2 − 2 x − 3
= lim
h→0
h
2
2
3x + 6 xh + 3h − 2h − 3x 2
= lim
h→0
h
6 xh + 3h2 − 2h
= lim
h→0
h
h(6 x + 3h − 2)
= lim
h→0
h
= lim(6 x + 3h − 2)
∴ f ′(x) = lim
Common Error
1. Not putting brackets
where necessary,
e.g. lim(6 x 3h 2)
h0
h→0
= 6x − 2
Hints
dy
, is the instruction to determine the derivative of y with respect to x.
dx
2. y must be the subject of the formula of an equation into x before you can determine the
derivative.
1.
3. If the function involves a quotient (fraction into x) the numerator and denominator need to
be factorised where possible.
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Calculus | 85
5.3 The Rules of Differentiation
You can differentiate using the rule.
If f (x) = ax n, f / (x) = a.nx n−1. Where a and n are real numbers.
Examples
Evaluate:
1. f / (x) if f (x) = 3x 2
2.
dy
4 x
if y = 3 +
dx
x 2
3. Dx [2 − 3 5 x ]
 x3 + 8 
4. Dx 

 x+2 
Solutions
2
1. If f ( x ) = f (x) = 3x then
f / (x) = 3 × 2 x 2−1
= 6x
4 x
1
+ = 4 x −3 + x then
3
x 2
2
1
dy


= (−3 × 4 x −3−1 ) +  × 1x1−1 
dx
2

1
= −12 x −4 + x 0
2
1
= −12 x −4 +
2
2. If y =
Unit-05.indd 85
Note: When differentiating a polynomial,
express each term in the form ax n first.
Then apply the rule for differentiation to
each term.
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86 | Unit 5
1


3. Dx 2 − 3 5 x  = Dx 2 x 0 − 3x 5 


 1 1 −1 
= ( 0 × 2 x 0−1 ) −  3 × × x 5 
 5

4
−
3
=0− x 5
5
3
=− 4
5x 5
 ( x + 2 ) ( x2 + 2 x + 4 ) 
 x3 + 8 
4. Dx 

 = Dx 

x+2
 x+2 


= Dx ( x 2 + 2 x + 4 )
= 2x + 2
Exercise 1 Nov 2011
Evaluate:
1.
dy
3 x2
if y =
−
dx
2x 2
2. f / (1) if f ( x ) = ( 7 x + 1 )
2
Nov 2016
3.
dy
5
if y = x 3 − 3
dx
x
4.John determines g / (a), the derivative of a certain function g at x = a, and arrives at the answer:
4 +h −2
lim
. Write down the equation of g and the value of a.
h→0
h
5. g ( x ) = −8 x + 20 is a tangent to f ( x ) = x 3 + ax 2 + bx + 18 at x = 1. Calculate the values of a and b.
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Calculus | 87
Solutions
1.
2.
3 x2
−
2x 2
x2
3
= x −1 −
2
2
dy 3 −2
= x −x
dx 2
3
=− 2 −x
2x
y=
f (x) = ( 7 x + 1)
2
= 49 x 2 + 14 x + 1
f / (x) = 98 x + 14
f / (1) = 98(1) + 14
= 112
3.
y = x3 −
5
x3
3
y = x 2 − 5x −3
dy 3 21
= x + 15x −4
dx 2
3 1 15
= x2 + 4
2
x
4 +h −2
h
g (x) = x
a=4
4. lim
h→0
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88 | Unit 5
5.
f (x) = x 3 + ax 2 + bx + 18
f ′(x) = 3x 2 + 2ax + b
At x = 1, mtan = −8
f ′(1) = −8
3(1)2 + 2a(1) + b = −8
3 + 2a + b = −8
2a + b = −11..............(1)
y = f (1)
= g (1)
= −8(1) + 20
= 12
1 + a + b + 18 = 12
a + b = −7...............(2)
a = −4
b = −3
5.4 Graphs of Cubic Polynomials
To draw the graph, find
1. y-intercept: This is where the graph cuts the y-axis. This is a point where x = 0 always.
2. x-intercept: This is where the graph cuts the x-axis. This is a point where y = 0 always.
3.
Coordinates of turning points: This is where the gradient of the function is always = 0. OR the derivative is equal to zero. The signs of the gradient are different before and after the turning point.
• Turning points are also called stationary points.
• Stationary points on a graph are points where the gradient of the graph is 0.
• The stationary points are at local maximum or minimum turning point.
°° Local maximum: Function changes from being increasing to decreasing. i.e gradient changes from
positive to negative
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Calculus | 89
ocal minimum: Function changes from being decreasing to increasing. i.e gradient changes from
°° L
negative to positive
• To determine the coordinates of the turning points:
°° Find the first derivative and equate it to zero
°° Solve for x
/
°° Substitute the x-values obtained into f(x) (Not f (x)) to obtain corresponding y-values.
5.4.1
Concavity
A function can either be concave up or concave down.
• Concave up corresponds to a positive second derivative
• Concave down corresponds to a negative second derivative
When a function changes from concave up to concave down (or vise versa) the second derivative must equal
zero at that point. The point of inflection is where the function changes concavity.
To find the coordinates of the point of inflection, you must find the second derivative of the function, f // (x).
Then solve for f // (x) = 0.
Examples
1.Given: f (x) = − x 3 + 6 x 2 − 9 x
1.1 Write down the coordinates of the x-intercepts of f .
1.2 Write down the y-intercept of f .
1.3 Determine the coordinates of the turning points of f .
1.4 Sketch the graph of f . Clearly show ALL the intercepts with the axes and the turning points.
1.5 Determine the values of x for which the graph of f is increasing.
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90 | Unit 5
Solutions
1.1 x-intercepts Let y = 0, then
− x3 + 6x2 − 9x = 0
x3 − 6x2 + 9x = 0
x ( x2 − 6x + 9 ) = 0
x ( x − 3) ( x − 3) = 0
x = 0 or x = 3
( 0;00 ) and ( 3; 0 )
1.2 y -intercept Let x = 0, then y = 0
∴the point is ( 0; 0 )
1.3
f (x) = − x 3 + 6 x 2 − 9 x
f ′(x) = −3x 2 + 12 x − 9
−3x 2 + 12 x − 9 = 0
x2 − 4x + 3 = 0
(x − 3)(x − 1) = 0
∴ x = 3 or x = 1
f (1) = −(1)3 + 6(1)2 − 9(1) = −4
f (3) = −(3)2 + 6(3)2 − 9(3) = 0
∴ (1; −4) and (3; 0)
1.4
y
(0;0)
(3;0)
x
f
(1;–4)
1.5 1< x < 3 or x ∈ (1; 3)
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Calculus | 91
Nov 2011
2.The function f (x) = − 2 x 3 + ax 2 + bx + c is sketched below. The turning points of the graph of f are
T(2; – 9) and S(5; 18).
y
S(5; 18)
f
x
O
T(2; –9)
2.1 Show that a = 21, b = – 60 and c = 43.
2.2 Determine an equation of the tangent to the graph of f at x = 1.
2.3 Determine the x-value at which the graph of f has a point of inflection.
Solutions
2.1
f ( x ) = −2 x 3 + ax 2 + bx + c
f / ( x ) = −6 x 2 + 2ax + b
= −6 ( x − 5 ) ( x − 2 )
= −6 ( x 2 − 7 x + 10 )
= −6 x 2 + 42 x − 60
2a = 42
a = 21
b = −60
f (5) = −2 ( 5) + 21 ( 5) − 60 ( 5) + c
18 = −25 + c
c = 43
3
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2
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92 | Unit 5
OR
f (2) = −2 ( 2 ) + 21 ( 2 ) − 60 ( 2 ) + c
−9 = −52 + c
c = 43
a = 21; b = −60; c = 43
3
2
2.2 f / (x) = −6 x 2 + 42 x − 60
mtan = −6 (1 ) + 42 (1 ) − 60
= −24
2
f (1 ) = −2 (1 ) + 21 (1 ) − 60 (1) + 43
=2
Point of contact is (1; 2)
3
2
y − 2 = −24 ( x − 1)
y = −24 x + 26
2.3
f / (x) = −6 x 2 + 4 x − 60
f // (x) = −12 x + 42
0 = −12 x + 42
x=
7
2
Common Errors
1. Forgetting to write intercepts as a coordinate.
2. Failure to factorise correctly.
3. Forgetting to equate the derivative to zero when finding the turning points.
4. Using the coefficient of x3 to determine the shape of the graph
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Calculus | 93
Exercise 2
Nov 2011
1. The graph of y = f / (x), where f is a cubic function, is sketched below.
y
–4
x
y = f/(x)
Use the graph to answer the following questions:
1.1 For which values of x is the graph of y = f / (x) decreasing?
1.2 At which value of x does the graph of f have a local minimum? Give reasons for your answer.
Nov 2009
2.Given: f (x) = − x 3 + x 2 + 8 x − 12
2.1 Calculate the x-intercepts of the graph of f .
2.2 Calculate the coordinates of the turning points of the graph of f .
2.3 Sketch the graph of f, showing clearly all the intercepts with the axes and turning points.
2.4 Write down the x-coordinate of the point of inflection of f.
2.5 Write down the coordinates of the turning points of h ( x ) = f ( x ) − 3.
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94 | Unit 5
Nov 2012
3. The graph of the function f (x) = − x 3 − x 2 + 16 x + 16 is sketched below.
y
f
0
x
3.1 Calculate the x-coordinates of the turning points of f .
3.2 Calculate the x-coordinate of the point at which f / (x) is a maximum.
4. Consider the graph of g (x) = −2 x 2 − 9 x + 5 .
4.1 Determine the equation of the tangent to the graph of g at x = –1.
4.2 For which values of q will the line y = –5x + q not intersect the parabola?
5.Given: h(x) = −4 x 3 + 5x
Explain if it is possible to draw a tangent to the graph of h that has a negative gradient. Show ALL your
calculations.
Solutions
1.1 x -value of turning point:
−4 + 1
2
3
=−
2
x=
∴ x > −
Unit-05.indd 94
3
2
or
 3 
∴x ∈ − ; ∞ 
 2 
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Calculus | 95
1.2 f / (x) < 0 for x < −4 and x > 1, so f is decreasing for x < −4 and x > 1.
f / (x) > 0 for −4 < x < 1, so f is increasing for −4 < x < 1.
∴ f has a local minimum at x = −4
2.1 x -intercept
0 = − x 3 + x 2 + 8 x − 12
x 3 − x 2 − 8 x + 12 = 0
(x − 2)(x 2 + x − 6) = 0
(x − 2)(x − 2)(x + 3) = 0
x = 2 or x = −3
∴ (2; 0) and (−3; 0)
2.2
f / ( x ) = − 3x 2 + 2 x + 8
0 = − 3x 2 + 2 x + 8
3x 2 − 2 x − 8 = 0
( x − 2 ) ( 3x + 4 ) = 0
x = 2 or x = −
4
1
= −1
3
3
∴( 2; 0 ) and ( −3; 0 )
2.3
2
–4
–3
–2
(– 4 ; – 500 )
3
27
Unit-05.indd 95
–1 –20
–4
–6
–8
–10
–12
–14
–16
–18
y
x
1
2
3
4
If candidate used
function as
f(x) = x3 – x2 – 8x+12
then max 1/3
shape
y-intercept
turning pts
(3)
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96 | Unit 5
2.4
f ′′ (x) = 0
6x − 2 = 0
1
x=
3
2.5 ( 2; −3) and ( −1, 33; −21, 52 )
3.1
f (x) = − x 3 − x 2 + 16 x + 16
f ′(x) = −3x 2 − 2 x + 16
0 = −3x 2 − 2 x + 16
3x 2 + 2 x − 16 = 0
(3x + 8)(x − 2) = 0
2
8
x = − = −2 or x = 2
3
3
3.2
f ′′(x) = 0
−6 x − 2 = 0
6x = 2
1
x=
3
4.1
g (x) = −2 x 2 − 9 x + 5
g (x) = −2(1)2 − 9(1) + 5
= 12
g ′(x) = −4 x − 9
mtan = −4 x − 9
= −5
y = −5x + c
12 = −5(1) + c
c=7
∴ y = −5x + 7
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Calculus | 97
4.2
y = −5x + q and g (x) = −2 x 2 − 9 x + 5
−5x + q = −2 x 2 − 9 x + 5
2 x 2 + 9 x − 5 − 5x + q = 0
2x2 + 4x + q − 5 = 0
x=
=
=
−b ± b2 − 4ac
2a
−4 ± 16 − 4(2)(q − 5)
2(2)
−4 ± 56 − 8q
56 − 8q < 0
q<7
4
5. h′(x) = 12 x 2 + 5
For all values of x : x 2 ≥ 0
12 x 2 ≥ 0
12 x 2 + 5 ≥ 5
12 x 2 + 5 > 0
For all values of x : h′(x) > 0
All tangents drawn to h will have a positive gardient.
It will never be possible to draw a tangent with a negative gradient to the graph of h.
5.5 Optimisation: Application of Calculus
Finding the maximum or minimum values
In the previous section, you found the local maximum or minimum points by solving f / (x) = 0. You can
apply the same principle when you find a maximum or minimum value.
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98 | Unit 5
Common Errors
Most learners do not attempt this type of question as they are not aware of the following
knowledge:
dy
is the instruction to determine the derivative of y with respect to x. in Optimisation, two
dx
variables will be given e.g. Volume and height. The rate of change of volume with height will
dV
be given by the derivative of Volume with respect to height,
.
dh
2. Contexts will differ, however it is important to pick out the two variables that have the relationship under discussion.
1.
3. It is important to know the formulae for the Perimeter, Area, Total surface area and Volume
of common shapes/objects.
Examples
March 2016
1. A soft drink can has a volume of 340 cm3, a height of h cm and a radius of r cm.
r
h
1.1 Express h in terms of r.
1.2 Show that the surface area of the can is given by A ( r ) = 2π r 2 + 680 r −1.
1.3 Determine the radius of the can that will ensure that the surface area is a minimum.
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Calculus | 99
Solutions
1.1 340 = π r 2 h
340
∴h = 2
πr
1.2 A = 2π r 2 + 2π rh
 340 
= 2π r 2 + 2π r  2 
 πr 
2
= 2π r + 680r −1
1.3 A(r ) = 2π r 2 + 680r −1
A′(r ) = 4π r − 680r −2
4π r − 680r −2 = 0
680
4π r = 2
r
680
3
r =
4π
680
r=3
cm or 3,78 cm
4π
Nov 2015
2. A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when
the radius is r cm. The angle between the cone edge and the radius is 60°, as shown in the diagram below.
Formulae for volume:
r
60°
V = π r2h
1
V = π r2h
3
V = lbh
4
V = π r3
3
h
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100 | Unit 5
2.1 Determine r in terms of h. Leave your answer in surd form.
2.2 Determine the derivative of the volume of water with respect to h when h is equal to 9 cm.
Solutions
2.1
2.2
h
= tan 60°
r
h
r=
tan 60°
h
∴r =
3
1
Vcone = π r 2 h
3
2
1  h 
= π
 h
3  3
1
= π h3
9
dV 1
= π h2
3
dh
1
dV
= π (9)2
dh h=9 3
= 27π or 84,82 cm 3 /cm
Exercise 3
Nov 2011
1.Water is flowing into a tank at a rate of 5 litres per minute. At the same time water flows out of the tank
at a rate of k litres per minute. The volume (in litres) of water in the tank at time t (in minutes) is given
by the formula V (t) = 100 − 4t.
1.1 What is the initial volume of the water in the tank?
1.2 Write down TWO different expressions for the rate of change of the volume of water in the tank.
1.3 Determine the value of k (that is, the rate at which water flows out of the tank).
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Calculus | 101
March 2010
2.A wire, 4 metres long, is cut into two pieces. One is bent into the shape of a square and the other into
the shape of a circle.
2.1If the length of wire used to make the circle is x metres, write in terms of x the length of the sides of the
square in metres.
1
 1
2.2Show that the sum of the areas of the circle and the square is given by f ( x ) =  +
 16 4π
square metres.
 2 x
x − +1
2

2.3 How should the wire be cut so that the sum of the areas of the circle and the square is a minimum?
Nov 2008
3. A drinking glass, in the shape of a cylinder, must hold 200 ml of liquid when full.
h
r
3.1 Show that the height of the glass, h, can be expressed as h =
200
.
π r2
3.2 Show that the total surface area of the glass can be expressed as S(r ) = π r 2 +
400
.
r
3.3 Hence determine the value of r for which the total surface area of the glass is a minimum.
Nov 2017
4.An aerial view of a stretch of road is shown in the diagram below. The road can be described by the
­function y = x 2 + 2 , x ≥ 0 if the coordinate axes (dotted lines) are chosen as shown in the diagram.
Benny sits at a vantage point B(0; 3) and observes a car, P, travelling along the road.
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102 | Unit 5
y
P
B(0; 3)
y=x2 + 2; x>0
O
x
Calculate the distance between Benny and the car, when the car is closest to Benny.
Solutions
1.1 V(0) = 100 − 4(0)
= 100 litres
1.2 Rate in − Rate out
= 5 − k l/min
V ′(t) = −4 l/min
1.3 5 − k = −4
k = 9 l/min
2.1 Length of sides of square =
2.2
Unit-05.indd 102
4−x
x
=1−
4
4
x = 2π r
x
r=
2π
2
2
 x 
4−x
Areas = 

 +π 
 2π 
 4 
16 − 8 x + x 2 x 2
=
+
16
4π
1
1  2
 1
=1− x + +
x
2
 16 4π 
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Calculus | 103
2.3
x=−
b
2a
1
2
=−
π +4

2

 16π 
= 1, 76m
−
3.1
V = π r2
200= π r 2 h
200
h= 2
πr
3.2
Surface Area = 2π rh + π r 2
200
S(r ) = π r 2 2 .2π r
πr
400
= π r2
r
3.3
S(r ) = π r 2 +
400
r
= π r 2 + 400r −1
S′(r ) = 2π r − 400r −2
dS
At minimum:
=0
dr
400
2π r − 2 = 0
r
3
2π r − 400 = 0
200
r3 =
π
r = 3, 99 cm
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104 | Unit 5
2
4. y = x + 2
P ( x ; x 2 + 2)
B(0; 3)
PB2 = (x − 0)2 + (x 2 + 2 − 3)2
= x2 + x 4 − 2x2 + 1
= x 4 − x2 + 1
PB will be a minimum if is a minimum PB2
d(PB2 )
=0
dx
4 x3 − 2 x = 0
2 x(2 x 2 − 1) = 0
1
x = 0 or x 2 =
2
1
x=
2
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9
Trigonometry
UNIT
L E A R N E R S S H O U L D K N OW H OW TO :
•
•
•
•
•
•
simplify trigonometric expressions
solve trigonometric equations
prove trigonometric equations (identities)
prove and apply cosine, sine and area rules
solve 2D and 3D (heights and distances)
sketch and interpret trigonometric functions
9.1 Basic Trigonometry Ratios
sin θ =
opposite
y
=
hypotenuse r
y
cos θ =
adjacent
x
=
hypotenuse r
Hypotenuse (r)
tan θ =
opposite y
=
adjacent x
Opposite (y)
?
Adjacent (x)
x
*To make it easy to remember the trig ratios, you can use anagram such as SOH-CAH-TOA
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146 | Unit 9
Note:
The longest side of the triangle is opposite the largest angle and the sum of the lengths of any two sides of
the triangle is longer than the third side.
Quadrants and Positive ratios
y
The CAST diagram mainly shows the
quadrants where the trigonometric
ratios are positive or negative.
1st quadrant
2nd quadrant
All ratios
sin
cos
tan
90°+ θ or
180°– θ
sin
tan
90°– θ or
360° + θ
x
cos
360°– θ or
–θ
180°+ θ
4th quadrant
3rd quadrant
• First Quadrant: All three
trigonometric ratios are positive
• Second Quadrant: Only sin is
positive
• Third Quadrant: Only tan is
positive
• Fourth Quadrant: Only cos is
positive
Common Errors on Reduction Formula
1. Assigning incorrect reduction rule in a given quadrant. For example using 180 in the
third quadrant.
2. Not knowing which trigonometric ratio is positive or negative in a particular quadrant
3. Not knowing the difference between an angle and the value of the trigonometric ratio, e.g.
2 cos cos 5
4. Using Negative angles. e.g. sin 45 Unit-09.indd 146
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Trigonometry | 147
9.2 Reduction Formula
1st quadrant
2nd Quadrant
3rd Quadrant
4th Quadrant
sin ( 90° − θ ) = cosθ
sin ( 90° + θ ) = cos θ
sin (180° + θ ) = − sin θ
sin ( 360° − θ ) = − sin θ
cos ( 90° − θ ) = sin θ
cos ( 90° + θ ) = − sin θ
cos (180° + θ ) = − cos θ
cos ( 360° − θ ) = cos θ
sin (180° − θ ) = sin θ
tan (180° + θ ) = tan θ
tan ( 360° − θ ) = − tan θ
cos (180° − θ ) = − cos θ
sin ( −θ ) = − sin θ
cos ( −θ ) = cos θ
tan ( −θ ) = tan θ
Special angles which their ratios should be known are:
0°
sin 0° = 0
cos 0° = 1
tan 0° = 0
90°
sin 90° = 1
cos 90° = 0
tan 90° = ∞
30°
sin 30° =
1
2
3
2
1
tan 30° =
3
cos 30° =
60°
3
2
1
cos 60° =
2
1
tan 60° =
3
sin 60° =
45°
1
2
1
cos 45° =
2
tan 45° = 1
sin 45° =
Definition: Complementary angles are two angles that add up to 90°
The sine of an angle is equal to the cosine of its complement i.e. 30° and 60° are complementary
angles. e.g.
• sin 30° = cos 60°
because 30° + 60° = 90°
⇒ sin 30° − cos 60° = 0
• cos 26° = sin 64°
because 26° + 64° = 90°
⇒ cos 26° − sin 64° = 0
*Note: A right angled triangle will always contain two complementary angles
Activity
Complete the following statement:
• sin 33° = cos .....°
Unit-09.indd 147
because 33° + ....° = 90°
⇒ sin 33° − cos ...° = 0
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148 | Unit 9
9.3 Trigonometric Identities
The following are prescribed trigonometric Identities
Identities
Compound –Angle Identities
Double-angle Identities
1. cos ( x − y ) = cos x cos y + sin x sin y
1. sin 2 x = 2 sin x cos x
∴ sin x = 1 − cos x
2. cos ( x + y ) = cos x cos y − sin x sin y
2. cos 2 x = cos2 x − sin2 x
and cos2 x = 1 − sin2 x
3. sin ( x + y ) = sin x cos y + cos x sin y
= 1 − 2 sin 2 x
4. sin ( x − y ) = sin x cos y − cos x sin y
= 2 cos2 x − 1
1. sin2 x + cos2 x = 1
2
2
sin x
2. tan x =
cos x
∴ sin x = tan x ⋅ cos x
sin x
and cos x =
tan x
*Note: These identities are given on the information sheet at the back of your exam question papers
*When identities are given as fractions, apply the basic steps when proving or solving as you would have
done in a normal fraction scenario that has numbers.
E.g. Prove the following identity:
Steps to follow
• Choose the side that you will use to work with. The
side that has more terms is usually the best side to
work with
cos x
1 + sin x
=0
−
1 − sin x
cos x
• From given example, choosing the left hand side,
first thing to do will be to find the LCM of the
denominators
• Apply the normal rules for subtraction and simplify
aster fundamental algebraic manipulation.
°° M
These skills are integral in simplifying trigonometric expressions.
*It is very important to master how to expand and contract trigonometric expressions.
e.g. cos(45° − θ )cos θ − sin(45° − θ )sin θ = cos(45° − θ + θ ) = cos 45°. This is the application
cos(x + y) = cos x cos y − sin x sin y (From identity above)
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Trigonometry | 149
Examples
Expand the following trigonometric expressions
1. sin(30° + θ )
2. cos(60° + θ )
Solutions
1. sin(30° + θ ) = sin 30° cos θ + cos 30° sin θ
2. cos(60° + θ ) = cos 60° cos θ − sin 60° sin θ
Exercise
Express the following functions as a single function
1. sin 15° cos 10° + cos 15° sin 10°
2. cos 15° cos 10° + sin 15° sin 10°
3. cos 45° cos 30° + sin 45° sin 30°
4. sin 15° cos 10° − cos 15° sin 10°
Solve for x:
5. cos 45° cos 30° + sin 45° sin 30° = 2 x
Solutions
1. sin 15° cos 10° + cos 15° sin 10° = sin(15° + 10°)
= sin 25°
2. cos 15° cos 10° + sin 15° sin 10° = cos(15° − 10°)
= cos 5°
3. cos 45° cos 30° + sin 45° sin 30° = cos(45° − 30°)
= cos 15°
4. sin 15° cos 10° − cos 15° sin 10° = sin(15° − 10°)
= sin 5°
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150 | Unit 9
5. cos 45° cos 30° + sin 45° sin 30° = 2 x
cos 15° = 2 x
0, 966 = 2 x
x = 0, 493
Common Errors on Proving Identities
1. Not applying the correct identities
2. Poor algebraic skills in manipulating fractions
3. Incorrect expansion of compound angles despite the expansions being given in the information sheet
4. Not following instructions, e.g. when a question says “Use a calculator” then you must use a
calculator as this will save you time.
5. Not easily identifying co-functions i.e cos 52 sin 38
9.4 Using Diagrams to Determine the Numerical Values
of Ratios for Angles from 0° to 360°
You can use right angled triangles to help you determine numerical values of other trigonometric ratios when
only one is given.
For example:
Given that 5 cos β − 3 = 0 and 0° < β < 270°. If α + β = 90° and 0° < α < 90°, calculate the value of
1
tan α
Steps to follow:
• Make the given trigonometric ratio the subject of the formula. In this case rearrange the equation
such that only cosine is on the left hand side
cos β =
3
5
Note: β must be placed in a reference right angled triangle drawn in the correct quadrant
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Trigonometry | 151
• To identify the correct quadrant to draw the triangle:
°° I dentify the sign of the ratio (+ or − ) as seen on the right hand side of after making the trigonometric ratio the subject of formula. This eliminates two quadrants.
°° Restrictions will guide you to make your final choice of the correct quadrant
3
ƒƒ e.g. tan x = − , sin x > 0 and cos < 0.
2
In this case the tangent is negative which implies an angle in the second or third quadrant.
However, only in the second quadrant do we find sin x > 0 and cos < 0
3
5
Cosine is positive. This implies either the first or fourth quadrant
ƒƒ From above, it is given that cos β =
In the interval 0° < β < 270° cosine is only positive in the first quadrant
• Remembering the trigonometric ratios (SOHCAHTOA) the right hand side will give the
Ratio as
Adjacent
Opposite
Opposite
or
or
depending on the ratio given
hypotenuse
hypotenuse
Adjacent
3
implies adjacent = 3 units and hypotenuse = 5 units
5
• Use the theorem of Pythagoras to calculate the third side of the triangle.
cos β =
Tip: it is important in such questions to refer to the adjacent as x, opposite as y and hypotenuse as r
2
2
2
°° r = x + y
°° After calculating y = 4
• Draw your right angled triangle in the chosen quadrant in a Cartesian plane
Examples
∧
3
and 0°< A < 90°, determine the values of the following with the aid of a sketch and
40
without using a calculator. Leave your answers in surd form, if necessary.
1.If tan A =
1.1 cos A
1.2 sin (180° + A )
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152 | Unit 9
2. P (− 7 ; 3) and S(a; b) are points on the Cartesian plane, as shown in the diagram below.
∧
∧
POR = PO S = θ and OS = 6. (March 2016)
y
P(–; 3)
O
x
R
6
S(a; b)
2.1 tan θ
2.2 sin ( −θ )
Solutions
1.1
( 40; 3)
3
7
40
r2 =
Unit-09.indd 152
(
40
A
)
2
+ 32
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Trigonometry | 153
r2 =
(
40
)
2
+ 32
= 40 + 9
= 49
r=7
40
7
cos A =
1.2
sin(180° + A) = − sin A
3
=−
7
2.1 tan θ = −
2.2
3
7
sin(−θ ) = − sin θ
(
OP 2 = − 7
)
2
+ 32
= 7 + 9 = 16
OP = 4
∴ sin (−θ ) = −
3
4
3
– 7
9.5 Trigonometric Equations
Examples
Solve the following equations
1. sin x = 0,5
2. cos2 x + cos x = 2
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154 | Unit 9
Solutions
1. sin x = 0, 5
x = sin −1 0, 5
x = 30°
2. cos2 x + cos x = 2
cos2 x + cos x − 2 = 0
( cos x + 2 ) ( cos x − 1) = 0
cos x + 2 = 0 or cos x − 1 = 0
cos x = −2 or cos x = 1
x = cos −1 (−2) or x = cos −1 1
invalid
x = 0°
9.6 General Solution
• General solution is a summary of the solutions to trigonometric equations.
Remember:
°° s in θ and cos θ have a period of 360°. This means that the functions repeat themselves after
every 360°. OR it means that the functions repeat themselves for consecutive multiples of 360°
(360°.n ; n ∈ ).
°° tan θ has a period of 180°. This means that the function repeats itself after every 180°. OR it means
that the function repeats itself for consecutive multiples of 180° (180°.n ; n ∈ )
Steps to follow when finding General solution given a ratio
1. Solve for the given ratio
2.Use the CAST diagram to determine in which two quadrants the solution lies. This will be where the
given ratio is either positive or negative.
3. Find the calculator angle by using sin‒1, cos‒1 or tan‒1 of the POSITIVE ratio.
• If the given trigonometric ratio is positive, the calculator angle will be one of the solutions in the first
quadrant. Refer to the CAST diagram to get the second solution.
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Trigonometry | 155
• If the given trigonometric ratio is negative, the calculator angle is not a solution but a reference
angle. Use this reference angle to find the solutions in the quadrants where the trigonometric ratio is
negative.
4. Solve for the unknown angle:
• Quadrant 1: x = calculator angle
• Quadrant 2: x = 180° − calculator angle
• Quadrant 3: x = 180° + calculator angle
• Quadrant 4: x = 180° − calculator angle
*Note: When a question wants specific solutions over a given interval, find the general solution first and then
get the exact angles using the general solutions by substituting values of n. Start with n = 1; n = –1; n = 2;
n = –2; and so on until your answers fall outside the given interval.
Examples
Determine the general solution of:
1. cos 2 x − 7 cos x − 3 = 0
(Nov 2015)
2. sin x = cos 2 x − 1
(March 2012)
Solutions
1.
cos 2 x − 7 cos x − 3 = 0
2 cos2 x − 1 − 7 cos x − 3 = 0
2 cos2 x − 7 cos x − 4 = 0
( 2 cos x + 1) ( cos x + 4 ) = 0
2 cos x + 1 = 0 or cos x + 4 = 0
1
cos x = − or cos x = −4 (invalid )
2
1
x = cos −1 (− )
2
x = 120°
∴ x = 120° + 360°. n , n ∈  or x = 240° + 360° .n , n ∈ 
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156 | Unit 9
2.
sin x = cos 2 x − 1
sin x = 1 − 2 sin2 x − 1
2 sin2 x + sin x = 0
sin x ( 2 sin x + 1) = 0
sin x = 0 or 2 sin x + 1 = 0
sin x = −
1
2
 1
x = sin − ’1 0 or x = sin −1  − 
 2
x = 0°
x = −30°
x = 0° + 180° n , n ∈  or x = 210° + 360° n , n ∈  or x = 330° + 360° n , n ∈ 
Exercise
Nov 2018
1.In the diagram, P(k; 1) is a point in the 2nd quadrant and is
∧
positive x-axis and obtuse R OP = θ .
5 units from the origin. R is a point on the
y
5
R
1.1 Calculate the value of k.
1.2 Without using a calculator, calculate the value of:
1.2.1
tan θ
1.2.2
cos(180° + θ )
1.2.3
sin(θ + 60°) in the form
Unit-09.indd 156
a +b
20
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Trigonometry | 157
March 2012
∧
2.In the diagram, P is the point (12; 5). OT ⊥ OP. PS and TR are perpendicular to the x-axis. α = SOP
and OR = 7,5 units.
y
T
P(12; 5)
x
α
R
O
S
Determine:
2.1 cos α
∧
2.2 T OR in terms of α
2.3 The length of OT
3. March 2012
If sin 61° = p, determine the following in terms of p:
3.1 sin 241°
3.2 cos 61°
3.3 cos 122°
3.4 cos 73° cos 15° + sin 73° sin 15°
4. Prove without using a calculator, that sin 77° − sin 43° = sin 17°
5. Simplify the following:
5.1
Unit-09.indd 157
sin(A − 360°).cos(90° + A)
cos(90° − A).tan( − A)
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158 | Unit 9
5.2
4 sin 15° cos 15°
without using a calculator. Leave your answer in simplest surd form. (March 2016)
2 sin2 15° − 1
6. Prove that
cos2 x sin 2 x + cos 4 x
= 1 + sin x (March 2013)
1 − sin x
6.1
6.2 cos 3x + 4 cos3 x − 3 cos x
7. Determine the general solution of:
7.1 tan x sin x + cos x tan x = 0 (Nov 2013)
7.2 sin x + 2 cos2 x = 1 (Nov 2009)
Solutions
k 2 = ( 5)2 − 12
= 5−1 = 4
k = −2
1.1
1.2.1
tan θ = −
1
2
1.2.2
cos(180° + θ ) = − cos θ
2
=
5
1.2.3
sin(θ + 60°) = sin θ cos 60° + cos θ sin 60°
 1   1   2  3 
=

  +−

5  2 
 5 2 
1−2 3
=
2 5
=
1−2 3
20
2.1 r = 13
cos α =
Unit-09.indd 158
12
13
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Trigonometry | 159
2.2
sin 241° = sin(180° + 61°)
= − sin 61°
=− p
2.3
TR
OT
7,5
cos(90° − α ) =
OT
∧
cos T OR =
7, 5
cos(90° − α )
7, 5
=
sin α
7, 5
=
5
13
= 19, 5
OT =
3.1
sin 241° = sin(180° + 61°)
= − sin 61°
=− p
3.2
cos 61° = 1 − sin2 61°
= 1− p
1
3.3
= 2 cos2 61° − 1
=2
(
1− p
)
2
= 2(1 − p) − 1
= 2 − 2p −1
= 1 − 2p
Unit-09.indd 159
7
cos 122° = cos 2 ( 61° )
−1
61°
p–1
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160 | Unit 9
3.4
cos 73° cos 15° + sin 73° sin 15° = cos(73° − 15°)
= cos 58° = cos(180° − 122°)
= − cos(122°)
= −(1 − 2 p)
= 1 − 2p
4. sin 77° − sin 43° = sin(60° + 17°) − sin(60° − 17°)
= sin 60° cos 17° + cos 60° sin 17° − (sin 60° cos 17° − cos 60° sin 17°)
= sin 60° cos 17° + cos 60° sin 17° − sin 60° cos 17° + cos 60° sin 17°
= 2 cos 60° sin 17°
1
= 2 × × sin 17°
2
= sin17°
5.1
sin(A − 360°).cos(90° + A) sin A( − sin A)
=
cos(90° − A).tan( − A)
sin A( − tan A)
sin A
=
tan A
sin A
cos A
=
= sin A ×
sin A
sin A
cos A
= cos A
5.2
4 sin 15° cos 15° 2(2 sin 15° cos 15°)
=
2 sin2 15° − 1
−(1 − 2 sin2 15°)
2(sin 2(15°)
=
− cos 2(15°)
= −2 tan 2(15°)
= −2 tan 30°
 1 
= −2 

 3
2
=−
3
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Trigonometry | 161
6.1
6.2
cos2 x sin 2 x + cos 4 x cos2 x(sin2 x + cos2 x)
=
1 − sin x
1 − sin x
2
cos x .(1) cos2 x
=
=
1 − sin x 1 − sin x
1 − sin 2 x
=
1 − sin x
(1 − sin x)(1 + sin x)
=
1 − sin x
= 1 + sin x
cos 3x = cos(2 x + x)
= cos 2 x cos x − sin 2 x sin x
= ( 2 cos2 x − 1 ) cos x − 2 sin x cos x . sin x
= 2 cos3 x − cos x − 2 sin 2 x cos x
= 2 cos3 x − cos x − 2 (1 − cos2 x ) cos x
= 2 cos3 x − cos x − 2 cos x + 2 cos3 x
= 4 cos3 x − 3 cos x
7.1
tan x sin x + cos x tan x = 0
tan x ( sin x + cos x ) = 0
tan x = 0 or sin x + cos x = 0
sin x = − cos x
sin x
cos x
=−
cos x
cos x
tan x = −1
x = 0° or x = 135°
x = 0° + 180°.n , n ∈  or x = 135° + 180°.n , n ∈ 
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162 | Unit 9
7.2
sin x + 2 cos2 x = 1
sin x = 1 − 2 cos2 x
sin x = −(2 cos2 x − 1)
sin x = − cos 2 x
sin x = − sin ( 90° − 2 x )
x = 180° + ( 90° − 2 x ) + 360°.n , n ∈  or x = 360° − ( 90° − 2 x ) + 360°.n , n ∈ 
or x = 360° − 90° + 2 x + 360°.n , n ∈ 
3x = 270° + 360°.n , n ∈ 
x = 90° + 120°.n , n ∈ 
or − x = 270° + 360°.n , n ∈ 
x = −270° − 360°.n , n ∈ 
9.7 Sine, Cosine and Area rules
SOH-CAH-TOA and Pythagoras theorem are used ONLY on right angled triangles. When the triangles
are not right angled, you can use either the sine or cosine rules. These rules are used to solve triangles (to
determine the lengths or size of angles which are not given). The area rule is used to find the area of a triangle
when the triangle is not a right angled triangle.
∧
∧
∧
Given ∆ABC, side opposite A is a, side opposite B is b and side opposite C is c as below
A
b
c
B
a
Sine rule
sin A sin B sin C
= =
a
b
c
Applied when given
SS∠ or ∠∠S
Two sides and an angle or
Two angles and a side
Unit-09.indd 162
C
Cosine rule
Area rule
a = b + c − 2bc cos A
Applied when given
SSS or S∠S
Three sides or
Two sides and an included angle
Used to calculate the area of a triangle
1
Area of ∆ABC = ab cos C
2
Applied when given
S∠S
Two sides and an included angle
2
2
2
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Trigonometry | 163
Examiners usually ask a question on heights and distances. See the examples below to show you what is
expected.
Common Errors on Sine, Cosine and Area Rules
1. Difficulty in seeing the different planes in the sketch.
2. Developing strategies to be used when solving right-angled triangles and triangles that are
not right-angled.
3. Using incorrect rule (refer to the formula sheet).
4. Incorrect substitution in the cosine rule.
5. Algebraic manipulation of the cosine rule.
6. Using a calculator.
Examples
March 2016
1. In the diagram below, ∆PQR is drawn with PQ = 20 – 4x, RQ = x and
P
20 4x
Q
3 = 60°.
60°
x
1.1 Show that the area of ∆PQR = 5 3x − 3x 2.
1.2 Determine the value of x for which the area of ∆PQR will be a maximum.
1.3 Calculate the length of PR if the area of ∆PQR is a maximum.
Nov 2014
∧
2.In the figure below, ACP and ADP are triangles with C = 90°, CP = 4 3, AP = 8 and DP = 4. PA bisects
∧
∧
∧
DPC. Let C A P = x and D A P = y.
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164 | Unit 9
C
4 3
x
A
8
P
y
4
D
2.1 Show, by calculation, that x = 60°.
2.2 Calculate the length of AD.
2.3 Determine y.
Solutions
1.1
1.2
∧
1
Area of ∆PQR = PQ.QR.sin Q
2
1
= x(20 − 4 x)(sin 60°)
2
 3
= 10 x − 2 x 2 

 2 
= 5 3x − 3x 2
xmax = −
=−
=2
Unit-09.indd 164
b
2a
5 3
(
2 − 3
)
=
5
2
1
2
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Trigonometry | 165
1.3
RP2 = QP2 + QR 2 − 2QP.QR.cosQ
= 102 + 2 , 52 − 2(10)(2 , 5) cos 60°
= 81, 25
∴ RP = 9, 01
CP
AP
4 3
3
sin x =
=
8
2
= 60°
∧
2.1
sin CP A=
2.2
CP A=DP A=30°
∧
∧
∧
(AP bisects D P C)
∧
AD2 = AP2 + DP2 − 2.AP.DP cos A PD
= 82 + 4 2 − 2(8)(4)cos 30°
 3
= 64 + 16 − 64 

 2 
= 24 , 57
∧
2.3
∧
sin D A P sin A PD
=
DP
AD
sin y sin 30°
=
4
4 , 96
4 sin 30°
sin y =
4 , 96
= 0, 403
y = 23, 78°
Exercise
Nov 2015
1. A corner of a rectangular block of wood is cut off and shown in the diagram below.
∧
∧
The inclined plane, that is, ∆ACD, is an isosceles triangle having ADC = A CD = θ .
∧
1
Also A CB = θ , AC = x + 3 and CD = 2x.
2
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166 | Unit 9
A
x3
D
B
1
2
2x
C
∧
1.1 Determine an expression for C AD in terms of θ.
x
1.2 Prove that cos θ =
.
x+3
1.3 If it is given that x = 2, calculate AB, the height of the piece of wood.
Nov 2017
2.AB represents a vertical netball pole. Two players are positioned on either side of the netball pole at
points D and E such that D, B and E are on the same straight line. A third player is positioned at C. The
points B, C, D and E are in the same horizontal plane. The angles of elevation from C to A and from E to
A are x and y respectively. The distance from B to E is k.
A
y
E
k
D
Unit-09.indd 166
B
x
C
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Trigonometry | 167
∧
2.1 Write down the size of A BC.
2.2 Show that AC =
k . tan y
sin x
∧
2.3If it is further given that D A C = 2x and AD = AC, show that the distance DC between the players at
D and C is 2k tan y.
Solutions
∧
1.1 C AD = 180° − 2θ
1.2
1.3
s in θ s in (180° − 2θ )
=
x+3
2x
s in θ s in 2θ
=
x+3
2x
s in θ 2s in θ cosθ
=
x+3
2x
2 x sin θ
cos θ =
2(x + 3)
x
cos θ =
x+3
2
5
∴θ = 66,42°
cos θ =
In ∆ABC:
1
AB
sin θ =
2
AC
AB
sin 33, 21° =
5
AB = 5 sin 33,,21°
= 2 , 74
∧
2.1 A BC = 90°
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168 | Unit 9
In ∆ABE:
AB
= tan y
BE
AB = k tan y
In ∆ABC:
2.2
AB
= sin x
AC
AB
AC =
sin x
k tan y
=
s in x
DC2 = AD2 + AC2 − 2AD.AC cos 2 x
2.3
= AC2 + AC2 − 2AC.AC cos 2 x
= 2AC2 (1 − cos 2 x )
(
= 2AC2 1 − (1 − 2 sin2 x )
)
= 2AC ( 2 sin x )
2
2
= 4AC2 sin2 x
DC = 2AC sin x
 k . tan y 
= 2
 sin x
 sin x 
= 2k . tan y
9.8 Trigonometric Graphs
You should be able to:
•
•
•
•
•
Unit-09.indd 168
draw trigonometric graphs
determine the period for each function
determine the range and domain
determine asymptotes where applicable
determine the equation when the graph has been drawn for you
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Trigonometry | 169
Graphs that you must be able to sketch and interpret:
y = a sin k(x + p) + q
y = a cos k(x + p) + q
y = a tan k(x + p) + q
y = a sin x
y = a sin x + q
y = sin(x + p)
y = sin kx
y = a cos x + q
y = a cos x
y = cos kx
y = cos(x + p)
y = a tan x y = tan kx y = tan(x + p)
Sine and Cosine graphs (both are like waves)
y = a sin x ; y = a sin x + q ; y = sin kx
and y = sin(x + p)
y = a cos x ;
and y = cos(x + p)
y = a cos x + q
; y = cos kx
The general form of sine and cosine functions are:
Sine: y = a sin k(x + p) + q and
Cosine: y = a cos k(x + p) + q
At most two parameters per question will be assessed.
Hints on Trigonometric Graphs
1. Master the original graphs=
of y sin
=
x , y cos x and y = tan x (you can only be an
expert at it by drawing them many, many times)
2. Identify the important features of these graphs such as turning points and intersection points with the axes.
3. Understand the effects of the parameters a, p, q and k.
4. Know how the graph changes shape when a is a negative number.
5. If you sketch graphs using a calculator, pay attention to the critical features of the
graphs. Examiners still expect a high degree of accuracy from your sketch.
6. Do a lot of questions on translating and reflecting graphs as well as questions where
you need to read off solutions from sketched graphs.
7. Note how the critical features and characteristics of the basic graphs change for each
transformation that you perform.
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170 | Unit 9
8. Pay careful attention when dealing with horizontal translations of trigonometric
graphs. The intersection points with the axes change and you need to indicate the
new intersection points after translations.
Examples
March 2016
1. Given the equation: sin(x + 60°) + 2cos x = 0
1.1 Show that the equation can be rewritten as tan x = −4 − 3.
1.2 Determine the solutions of the equation sin(x + 60°) + 2cos x = 0 in the interval –180° ≤ x ≤ 180°.
1.3 In the diagram below, the graph of f (x) = –2 cos x is drawn for –120° ≤ x ≤ 240°.
y
2
f
1
x
–120°
–60°
0°
60°
120°
180°
240°
–1
–2
1.3.1 Draw the graph of g(x) = sin(x + 60°) for –120° ≤ x ≤ 240°
1.3.2 Determine the values of x in the interval –120° ≤ x ≤ 240° for which sin(x + 60°) + 2cos x > 0.
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Trigonometry | 171
Solutions
1.1
sin(x + 60°) + 2 cos x = 0
sin xcos60°+ cos xsin 60° + 2 cos x = 0
1
3
cos x + 2 cos x = 0
sin x +
2
2
1
3
sin x = −
cos x − 2 cos x
2
2
sin x = − 3 cos x − 4 cos x
(
sin x cos x − 3 − 4
=
cos x
cos x
tan x = − 3 − 4
1.2
tan x = − 3 − 4
tan x = −
(
3+4
)
)
ref ∠ = 80, 10°
x = 80, 10° or x = 99, 9°
1.3.1
y
2
g
–120°
f
1
0°
–60°
60°
120°
180°
240°
x
–1
–2
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172 | Unit 9
1.3.2 sin(x + 60°) > −2 cos x
x ∈ (−80, 10°; 99, 9°) or − 80, 10° < x < 99, 9°
Exercise
Nov 2017
1. In the diagram, the graph of f (x) = cos 2 x is drawn for the interval x ∈[−270° ; 90°].
y
1
f
–180°
–90°
0
90°
x
–1
–2
–3
1.1Draw the graph of g (x) = 2 sin x − 1 for the interval x ∈[−270° ; 90°] . Show ALL the intercepts with
the axes, as well as the turning points.
1.2Let A be a point of intersection of the graphs of f and g. Show that the x-coordinate of A satisfies the
−1 + 5
­equation sin x =
.
2
1.3Hence, calculate the coordinates of the points of intersection of graphs of f and g for the interval
x ∈[−270° ; 90°].
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Solutions
y
1.1
1
f
–270º
–210º –180º
–90º
0
30º
90º
x
–1
g
–2
(–90º; –3)
1.2
3
cos 2 x = 2 sin x − 1
1 − sin2 x = 2 sin x − 1
sin2 x + 2 sin x − 2 = 0
sin x =
sin x =
1.3
−b ± b2 − 4ac
2a
=
−1 ± 12 − 4(1)(−1)
2(1)
−1 + 5
−1 − 5
−1 − 5
or sin x =
but
< −1
2
2
2
his has no solution
∴ th
−1 + 5
2
sin x = 0, 618
ref ∠ = 38, 17°
∴ x = 38, 17° + 360°.n , n ∈ Z or x = 141, 83° + 360°.n , n ∈ Z
x = −321, 83° or x = −218, 17°
y = 0, 24
∴ Points of intersection
sin x =
( − 321, 83°; 0, 24) and ( − 218, 17°; 0, 24)
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QR CODES
Sample Lessons Mathematics
Unit 5
Hyperbolic-Functions
Unit-09.indd 174
E
Unit 10
Euclidean Geometry
6/23/2020 6:46:26 PM
resource book for both teachers and grade 12 learners. It is written in a
manner that follows final Mathematics examination papers (CAPS).
It is divided by units. Units are sequenced as per the final Mathematics
Examination paper. It has important icons.
•
Unit outcomes
•
General notes
•
•
•
•
Practical examples
Common errors of the topic
End of the year questions (exam questions)
For the teacher
maths final exam with flying colours. Use it more and more daily.
Exam preparation book (Learner and Teacher)
Practically using this resource book will definitely make you pass your
Grade 12
•
Hints for the particular topics
PROBLEM SOLVED Maths
This mathematics preparation book was developed and compiled as a
Grade 12
Exam preparation book
(Learner and Teacher)
Leornard Gumani
Steven N. Muthige
Timothy M. Sibeko
Palesa T. Tsuebeane