Download Mathematics Teachers` Self Study Guide on the National

Mathematics
Teachers’ Self Study Guide
on the National Curriculum Statement
Grade 12
May 2012
Page 1
CONTENTS
1.
2.
3.
4.
5.
PATTERNS, SEQUENCE AND SERIES
FUNCTIONS
CALCULUS
ANALYTICAL GEOMETRY
TRIGONOMETRY
03 -18
19 -25
26 - 42
43 - 63
64 - 77
Material written by
William Buthane Chauke (GP), Anne Lindelwa Hadebe (FS), Percy Tebeila (GP),
Dumisani Mdlalose (WC) and Thandi Lomthandazo Mgwenya (MP)
Edited by Leonard Gumani Mudau
Page 2
1
PATTERNS, SEQUENCE AND SERIES
1.1
Arithmetic Sequences and Series
An arithmetic progression is a sequence in which there is a constant difference between any two
consecutive terms. For example:
100; 101; 102; 103; 104 ; ...
3; 7; 11; 15; 19; ...
The distinct feature of these sequences is that each term, after the first, is obtained by adding a
constant, d, to the previous term. In the examples above, ‘d’ is 1 and 4 respectively. In the
discussion that follows ‘a’ is used to represent the first term.
𝑑𝑑 = 𝑇𝑇2 − 𝑇𝑇1 = 𝑇𝑇3 − 𝑇𝑇2
In the above sequences if we replace the “;” by “+” the sequence becomes a series. For example,
100 + 101 + 102 + 103 + 104 +…
3 + 7 + 11 + 15 + 19 +…
There are two important formulae that can be used to solve most arithmetic sequence and series
problems:
The general or n-th term: T n = a + (n – 1)d
The sum of the first n terms: 𝑆𝑆𝑛𝑛 = 𝑛𝑛2 [2𝑎𝑎+(𝑛𝑛−1)𝑑𝑑] or 𝑆𝑆𝑛𝑛 = 𝑛𝑛2 (𝑎𝑎 +𝑇𝑇𝑛𝑛 )
Examples
1.1
1.2
Find the 32nd term of the 2nd sequence above.
Find the sum of the first 32 terms of the sequence.
Solution
1.1
We may list all 32 terms, but that will take a long time and a lot of space. The simplest or a
quickest way is to use the general term, T n = a + (n – 1)d
T n = a + (n – 1)d
T 32 = 3 + (32 – 1)4
= 3 + 31(4)
= 3 + 124 = 127
1.2
Similarly to find the sum we can add all terms together or we can use some formula:
𝑆𝑆𝑛𝑛 = 𝑛𝑛2 [2𝑎𝑎+(𝑛𝑛−1)𝑑𝑑]
or
𝑆𝑆𝑛𝑛 = 𝑛𝑛2 (𝑎𝑎+𝑇𝑇𝑛𝑛 )
𝑆𝑆32 = 322 [2 × 3 + (32 − 1)4]
𝑆𝑆32 = 322 (3 + 𝑇𝑇32 )
= 16[6 + (31)4]
= 16(3 + 127)
= 16(130)
= 16(130)
= 2 080
= 2 080
Page 3
2.1
Consider the sequence:
30 ; 22 ; 14 ; 6 ; – 2 ; . . .
Find the sum of the first 19 terms of the sequence.
2.2
The 5th term of an arithmetic sequence is 17 and a common difference of 6, determine the
first term of the sequence.
Solution
2.1
2.2
To find S 19 without first calculating T 19 we use the formula:
𝑆𝑆𝑛𝑛 = 𝑛𝑛2 [2𝑎𝑎+(𝑛𝑛−1)𝑑𝑑]
𝑆𝑆19 = 192 [2 × 30 + (19 − 1)(−8)]
= 9,5[60 − 144]
= 9,5(−84)
= −798
To find T 1 given T 5 , we may subtract d = 6 four times from 17 because T 5 = a + 4d, that is,
𝑎𝑎 = 17 − 6 − 6 − 6 − 6 = −7
OR 𝑇𝑇5 = 𝑎𝑎 + 4𝑑𝑑
17 = 𝑎𝑎 + 4 × 6
17 − 24 = 𝑎𝑎
−7 = 𝑎𝑎
Exercise 1.1
1.
2.
The cost of boring a well 300 metres deep is calculated from the following information:
The cost for the first metre is R20,00, and then the cost per metre increases by R2,00 for
every subsequent metre. Find
1.1
the cost of boring the 300th metre;
1.2
the total cost of boring the well.
A bank clerk is offered a yearly salary of R108 000 for the first year of employment. She is
also to receive R4 000 a year increase at the beginning of each subsequent year.
2.1
What is her salary at the beginning of the fifth year?
After the first five years she is to receive a yearly increase of R6 000 until she reaches a
maximum annual salary of R206 000.
2.2
What is her salary at the beginning of the sixth year?
2.3
At the beginning of which year of her employment does her salary reach the
maximum?
Page 4
3.
4.
5.
A woman deposits R100 into her son’s savings account on his first birthday. On his
second birthday she deposits R125, R150 on his third birthday, and so on.
3.1
How much money would she deposit into her son’s account on his 17th birthday?
3.2
How much in total would she have deposited after her son’s 17th birthday?
The fourth term of an arithmetic sequence is 12 and the tenth term is 42.
4.1
Given that the first term is a and the common difference is d, write down two
equations in terms of a and d that satisfy this information.
4.2
Solve the equations to find the values of a and d.
The n th term of a sequence is 3n − 2 .
5.1
Show that the sequence is an arithmetic sequence.
5.2
Determine the sum of the first 150 terms of the sequence.
5.3
Determine the value of n for which
n
∑ (3r − 2) = 70 .
r =1
6.
The sum of the first 50 terms of an arithmetic series is 1 275. Calculate the sum of the 25th and
26th terms of this series.
7.
The sum of the first n terms of an arithmetic series is: S n =
3n 2 − n
.
2
7.1
Determine S10 .
7.2
Calculate the value of ∑Tr , where T r is the rth term of the series.
10
r =5
8.
The following is an arithmetic sequence: 1 – p ; 2p – 3 ; p + 5 ; ...
8.1
Calculate the value of p.
8.2
Write down the value of:
8.2.1 The first term of the sequence
8.2.2 The common difference
8.3
9.
Explain why none of the numbers in this arithmetic sequence are perfect squares.
Given:∑99
𝑡𝑡=0(3𝑡𝑡 − 1)
9.1
Write down the first THREE terms of the series.
9.2
Calculate the sum of the series.
Page 5
10.
The sum of the first 20 terms of the following arithmetic series is 3 360:
2 x + 2.2 x + 3.2 x + ...
Calculate the value of x.
11.
If
5
∑ (r + a ) = 10b determine
a in terms of b.
r =1
12.
The book value of a company car diminishes by a fixed amount of money each year. In 10
years the value dropped by two thirds. When will it be equal to zero?
13.
The sum of 10 consecutive natural numbers is equal to 575. Find them.
14.
The formula for the nth partial sum of an arithmetic sequence is Sn = 1.5n2 + 2.5n. Find
the first three terms of the sequence.
Equally, the overview of a geometric progression would also need to be provided:
1.2
Geometric Sequences and Series
A geometric progression is one in which there is a constant ratio between any two consecutive
geometric terms. For example:
1; 10; 100; 1000; 10000;….
3; 9; 27; 81; 243;….
The distinct feature of these sequences is that each term, after the first, is obtained by multiplying
the previous term by a constant, r. In the examples above, ‘r’ is 10 and 3 respectively. Again, in
the discussion that follows ‘a’ is used to represent the first term. Each element of a sequence can
be identified by reference to its position in the sequence (or its term number). For example, in the
second sequence above, the first term, T 1 , of the sequence is a = 3 and the second term, T 2 , of the
sequence is ar = 3 x 3. The third term, T 3 , is ar2 = 3 x 32.
In the above sequences if we replace the “;” by “+” the sequence becomes a series. For example,
1 + 10 + 100 + 1000 + 10000 + ….
3 + 9 + 27 + 81 + 243 + ….
There are two important formulae that can be used to solve most arithmetic sequence and series
problems:
Page 6
The general or n-th term: T n = arn – 1
𝑛𝑛
The sum of the first n terms: 𝑆𝑆𝑛𝑛 = 𝑎𝑎 �𝑟𝑟𝑟𝑟−1−1� or 𝑆𝑆𝑛𝑛 = 𝑎𝑎 �1−𝑟𝑟
1−𝑟𝑟
𝑛𝑛 �
Examples
1.
Consider the sequence: 3 ; 12 ; 48 ; 192 ; 768; ...
1.1
Find the 12th term of the sequence
1.2
Find the sum of the first eight terms
2.
Find the sum of the first eleven terms of the following sequence:
40 ; 8 ; 1,6 ; 0,32 ; 0,064;...
3.
A geometric sequence has all its terms positive. The first term is 7 and the third term is 28.
3.1
Find the common ratio.
3.2
Find the sum of the first 14 terms.
Solutions
1.
2.
The sequence can be written as: 3 ; 3 × 4 = 12 ; 12 × 4 = 48 ; 48 × 4 = 192 ; 192 × 4 = 768
1.1
It is a geometric sequence with a = 3 and r = 4; ∴ T n = arn – 1
T 12 = 3. 412 – 1
= 3. 411
= 3 × 4 194 304 = 12 582 912
1.2
𝑆𝑆𝑛𝑛 = 𝑎𝑎 �𝑟𝑟𝑟𝑟−1−1�
𝑛𝑛
8
8
𝑟𝑟 = 40
𝑆𝑆8 = 3�44−1−1�
= 48 – 1
= 65 535
1,6
1
= 8 = 5 = 0,2 and a = 40
𝑆𝑆𝑛𝑛 = 𝑎𝑎 �1−𝑟𝑟
1−𝑟𝑟
𝑛𝑛 �
11
�
𝑆𝑆11 = 40 �1−0,2
1−0,2
= 50(1 − 0,211 )
= 50(0,9 999 999 795)
= 50,000
3.
a = 7 ; T 2 = ar2 = 28
3.1
7r2 = 28
r2 = 4
∴ r=2
Page 7
3.2
𝑛𝑛
𝑆𝑆𝑛𝑛 = 𝑎𝑎 �𝑟𝑟𝑟𝑟−1−1�
14
𝑆𝑆14 = 7�22−1−1�
= 7(214 − 1)
= 7 × 16 383
= 114 681
Page 8
Exercise 1.2
1.
2.
3.
4.
The tuition fees for the first three years of school are R2 000 ; R2 500 ; R3 125. If these
tuition fees form a geometric sequence, find:
1.1
Find the common ratio, r, for this sequence
1.2
If fees continue to rise at the same rate, calculate (to the nearest rand) the total
cost of tuition fees for the first six years of school.
A basketball is dropped vertically. It reaches a height of 2 m on the first bounce. The
height of each subsequent bounce is 90% of the previous bounce.
2.1
What height does it reach on the 8th bounce?
2.2
What is the total vertical distance travelled by the ball between the first and the
sixth time the ball hits the ground?
If 1 − 5t , 1 − t and t + 1 are the first three terms of a convergent geometric series, calculate:
3.1
The value of t.
3.2
The common ratio.
3.3
The sum to infinity of the series.
The first term of a geometric sequence is 3 and the sum of the first 4 terms is 5 times
the sum of the first 2 terms. The common ratio is greater than 1.
Calculate:
5.
4.1
The first three terms of the sequence, and
4.2
The value of n for which the sum to n terms will be 765
The first two terms of a convergent geometric series are m (m ≠ 0 ) and 6, in that order.
The sum of the infinite series is 25. Calculate the values of m. (Check that these values are
acceptable.)
Page 9
6.
Pattern 1
Pattern 2
Pattern 3
Pattern 4
In the patterns above each consecutive pattern has more shaded squares than the
1
previous one. The area of the shaded portion of the first pattern is square units.
4
Assume that the pattern behaves consistently, as shown above.
1
3
6.1
The shaded area in Pattern 2 is +
. Write down the area of the shaded
4 16
portions of Patterns 3 and 4.
6.2
Write down the area of the shaded portion of the nth pattern in sigma notation.
6.3
7.
8
If the pattern continues without end, what does the area in QUESTION 6.2
approach?
Given the geometric series: 8x2 + 4x3 + 2x4 + ...
7.1
Determine the nth term of the series.
7.2
For what value(s) of x will the series converge?
7.3
Calculate the sum of the series to infinity if 𝑥𝑥 = 32.
A sequence of squares, each having side 1, is drawn as shown below. The first square is
shaded, and the length of the side of each shaded square is half the length of the side of
the
shaded square in the previous diagram.
1
2
DIAGRAM 1
DIAGRAM 2
DIAGRAM 3
DIAGRAM 4
8.1
Determine the area of the unshaded region in DIAGRAM 3.
8.2
What is the sum of the areas of the unshaded regions on the first seven squares?
Page 10
9
Data regarding the growth of a certain tree has shown that the tree grows to a height of
150 cm after one year. The data further reveals that during the next year, the height
8
increases by 18 cm. In each successive year, the height increases by 9 of the previous year's
increase in height. The table below is a summary of the growth of the tree up to the end of
the fourth year.
Tree height (cm)
Growth (cm)
First year
150
Second year
168
18
Third year
184
16
Fourth year
19829
2
149
9.1
Determine the increase in the height of the tree during the seventeenth year.
9.2
Calculate the height of the tree after 10 years.
9.3
Show that the tree will never reach a height of more than 312 cm.
10
Healthy Living sponsors a cross country marathon and awards prize money as follows:
Every prize-winner gets half the amount of the person who finished just before him or her
and the person in 20th place gets R1. Prizes are awarded for the top 20 places. Assume
that no two people finish at exactly the same time. Determine:
11
10.1
The prize money that the winner (the person who finishes first) receives.
10.2
The position of the person who receives R2 048.
10.3
The total prize money for the event.
A tree is planted when it has a height of 1,5 metres. At the end of the first year it has
grown to a height of 2,5 metres. Thereafter the increase in height every year is 2 of the
3
increase in height during the previous year. Determine the maximum height that the tree
will reach.
Page 11
1.3
Combined Sequences and Series
Examples
1.
Consider the following sequence of numbers:
2 ; 5 ; 2 ; 9 ; 2 ; 13 ; 2 ; 17 ; …
1.1
1.2
Write down the next TWO terms of the sequence, given that the pattern continues.
Calculate the sum of the first 100 terms of the sequence.
Worked Solution
1.1
2 ; 21
(2 + 2 + ... + 2)+ (5 + 9 + 13 + ...)
for 50 terms
50
50
i =1
i =1
for 50 terms
= ∑ 2 + ∑ (4i + 1)
1.2
 50

= 2(50) +  (2(5) + 49(4))
2

= 100 + 25(10 + 196)
= 100 + 5150
= 5250
Exercise 1.3
1
Consider the sequence: 12; 4; 14 ; 7 ; 18 ; 10 ; ...
1.1
If the pattern continues in the same way, write down the next TWO terms in the
sequence.
1,2
2
Calculate the sum of the first 50 terms of the sequence.
Consider the series:
2.1
1
1×2
1
1
1
+ 2×3 + 3×4 + 4×5 + ⋯
𝑎𝑎
Express each of the following sums as a fraction of the form 𝑏𝑏 :
2.1.1 The sum of the first two terms of the series
2.1.2 The sum of the first three terms of the series
2.1.3 The sum of the first four terms of the series
2.2
Make a conjecture about the sum of the first n terms of the given series.
2.3
Use your conjecture to predict the value of the following:
1
1×2
1
1
1
1
+ 2×3 + 3×4 + 4×5 + ⋯ + 2008 ×2009
Page 12
3
3.1
Tebogo and Matthew's teacher has asked that they use their own rule to construct
a sequence of numbers, starting with 5. The sequences that they have constructed
are given below.
Matthew's sequence: 5 ; 9 ; 13 ; 17 ; 21 ; …
Tebogo's sequence: 5 ; 125 ; 3 125 ; 78 125 ; 1 953 125 ; …
3.1.1 Matthew's sequence
3.1.2 Tebogo's sequence
3.2
Nomsa generates a sequence which is both arithmetic and geometric. The first term
is 1. She claims that there is only one such sequence. Is that correct? Show ALL your
workings to justify your answer.
(The next two questions are from the Nov 2009 IEB Paper 1)
4
Alex decides to include both swimming and running in her exercise plan. On day 1, Alex
swims 100 m and runs 500 m. Each day she will increase the distance she swims by 50 m
and the distance she runs will increase by 3,5% of the distance she ran on the previous day.
4.1
Determine, in terms of n, the distance that Alex
4.1.1 swims on the nth day of her exercise plan.
4.1.2 runs on the nth day of her exercise plan.
4.2
On the same set of axes, plot points for each of the exercise types. You may join
the points to illustrate the trend.
4.3
Use your graphs to determine the first day on which the distance Alex swims will
be greater than the distance she runs.
5
5.1
5.2
2
Evaluate: ∑∞
𝑘𝑘=1 3𝑘𝑘
Refer to the figure showing the first three layers of a stack of cans. There are 30
cans in the first layer, 29 cans in the second layer and 28 cans in the third layer.
This pattern of stacking cans in layers continues.
5.2.1 Determine a formula for the number of cans in the nth layer.
5.2.2 Determine the maximum number of cans that can be stacked in this way.
Page 13
6
Three numbers summing up to 10 form a geometric sequence. If the 2nd one was increased
by 1, they would form an arithmetic sequence. Find these numbers.
7
Suppose that you have just won the Reader's Digest Sweepstakes. You have a choice of
accepting a lump sum of R200 000 now, or taking R1 000 per month for the rest of your
life. In either case, you will put the money into a savings account paying 4% per year
interest, compounded monthly, and let the interest accumulate. Assume that SARS charges
you no income tax.
7.1
If you accept the R200 000 what amount you will have
7.1.1 after 10 years?
7.1.2 after 50 years?
7.2
If you accept the R1 000 per month, what is the amount that you will have
7.2.1 after 10 years?
7.2.2 after 50 years?
7.3
How long would it be before the amount you would have from the R1 000 per
month plan would exceed the amount you would have from the R200 000 lump
sum plan?
7.4
If you can get an annual interest rate of 7%, which plan is better
7.4.1 after 30 years?
7.4.2 after 50 years?
Page 14
1.4
Quadratic Sequences
A quadratic progression is a sequence in which there is a second constant difference. That is, the
difference between any two consecutive terms is not constant, but the difference between any
two consecutive terms formed by the first difference is constant. For example:
100; 103; 110; 121; 136 ; ...
The difference between consecutive terms of the above sequence form another sequence,
obtained as follows: 103 – 100 = 3; 110 – 103 = 7; 121 – 110 = 11; etc. Thus the sequence is
3; 7; 11; 15; ...
The difference in the new sequence (the first differences of the original sequence) is constant; this
new sequence is arithmetic with d = 4.
Examples
1.
Consider the sequence: 5 ; 18 ; 37 ; 62 ; 93 ; …
1.1
If the sequence behaves consistently, determine the next TWO terms of the
sequence.
1.2
Calculate a formula for the nth term of the sequence.
1.3
Use your formula to calculate n if the nth term in the sequence is 1 278.
Worked Solution
1.1
1.2
130 ; 173
5
18
13
37
19
6
25
6
62
31
93
sequence of first difference
second difference is constant
The second difference is constant ∴ Tn is quadratic
2a = 6
a=3
∴ Tn = an2 + bn + c
∴ Tn = 3n 2 + bn + c
5 = 3(1) 2 + b(1) + c
b+c = 2
... (1)
18 = 3(2) 2 + b(2) + c
2b + c = 6
... (2)
(2) – (1): b = 4
Page 15
c = −2
∴ Tn = 3n 2 + 4n − 2
Alternative method:
a (1) 2 + b(1) + c = 5
a+b+c = 5
... (1)
a (2) 2 + b(2) + c = 18
4a + 2b + c = 18
... (2)
a (3) 2 + b(3) + c = 37
9a + 3b + c = 37
(2) – (1):
... (3)
3a + b = 13
b = 13 − 3a
Substitute b = 13 − 3a into (3)
9a + 3(13 − 3a ) + c = 37
9a + 39 − 9a + c = 37
c = −2
Substitute b = 13 − 3a and c = – 2 into (2)
4a + 2(13 − 3a ) + (− 2 ) = 18
− 2a = −6
a=3
∴b=4
1.3
3n 2 + 4n − 2 = 1278
3n 2 + 4n − 1280 = 0
(3n + 64)(n − 20) = 0
n=
− 64
or n = 20
3
n=
− 64
is not valid ∴ n = 20
3
Page 16
Exercise 1.4
1
Consider the sequence: 8 ; 18 ; 30 ; 44 ; …
1.1
Write down the next TWO terms of the sequence, if the pattern continues in the
same way.
1.2
Calculate the nth term of the sequence.
1.3
Which term of the sequence is 330?
2
Consider the sequence: 6 ; 6 ; 2 ; – 6 ; – 18 ; ...
3
2.1
Write down the next term of the sequence, if the sequence behaves consistently.
2.2
Determine an expression for the n term, T .
2.3
Show that – 6838 is in this sequence.
th
n
The following sequence of numbers forms a quadratic sequence:
– 3 ; – 2 ; – 3 ; – 6 ; – 11 ; …
3.1
The first differences of the above sequence also form a sequence. Determine an
expression for the general term of the first differences.
3.2
th
th
Calculate the first difference between the 35 and 36 terms of the quadratic
sequence.
4
th
3.3
Determine an expression for the n term of the quadratic sequence.
3.4
Explain why the sequence of numbers will never contain a positive term.
The number pattern 7; 22; 47; 82 ; 127 ; …….. is given such that it continues
consistently. Determine:
5
4.1
The first difference
4.2
The second difference, what do you notice?
4.3
The formula for the n-th term.
The sequence –5; –7; –7; –5; …. is a quadratic sequence.
5.1
Write down the next two terms.
5.2
Determine an expression for the n th term of the sequence.
5.3
What is the value of the first term of the sequence that is greater than 299?
Page 17
6
Look at the following sequence and answer the questions that follow:
10 ; 21 ; 38 ; 61 ; ..........
6.1
Determine the type of sequence.
6.2
Determine the general term.
6.3
Which term has a value of 1 245?
Page 18
2. FUNCTIONS
2.1 Challenges
2.1.1 Teachers
• Time management
Teachers should cover the work in time. They should make sure that all the aspects are
attended to.
2.1.2
•
•
•
•
•
•
•
•
Learners must know
effects of a, p and q
to differentiate between different shapes of graphs based on the equation.
to draw graphs
to analyse and interpret graphs, make deductions from the given graph.
how to recognise the domain and the range
identify the asymptote
how to find the point(s) of intersection
that hyperbola has two arms at all times, and show at least one point on the other arm.
Teaching Approach.
Know the following graphs:
•
Straight line / linear graph
 Revise number system with emphasise on real numbers.
 use table method to plot your graph as your starting point
 pay attention to the intercepts (coordinates)
 introduce the dual-intercept method.
 effects of “a” and “q” in y = ax +q
 For y=2x +0 , consider gradient intercept method: y =





•
2 (change in y over change
x
in x) or substitute x by any value and plot the coordinate with y-intercept as 0
already.
Introduce graphs like : y=4, x=-3, y=x ,etc.
Use the notation of : f(x) ,g(x) , etc.
domain and range
finding the equation
making deductions from the given sketch
N.B you may use the gradient and y-intercept method
Parabola
 Consider y=x2 , as the point of departure.
 use table method to illustrate the effect of “a” , “p” and “q”. encourage self
discovery.
Page 19
 let learners draw rough sketches (if they know the values of a, p, and q. At this
stage ignore the intercepts)
 introduce the intercepts and draw the graph considering all other characteristics
including decreasing and increasing intervals.
 domain and range.
 completing the square
 finding the equation of the graph, when given intercepts and a point and when
given the turning point and any other point.
y=a(x-x 1 )(x-x 2 ) and y=a(x+p)2+q .
 making deductions from the given sketch.
 combination of the two graphs- focus on the points of intersection and comparison
e.g. f(x)
𝑓𝑓(𝑥𝑥)
g(x) ,f(x) 2 , f(x).g(x) 0 and 𝑔𝑔(𝑥𝑥). 0.
 Consider the following Transformations : f(x) , -f(x) , f(-x), f(x+a) , a f(x), f(x)+a.
•
Hyperbola
 use table method to illustrate the effect of “a” , “p” and “q”. encourage self
discovery and explain asymptotes.
 let learners draw rough sketches (if they know the values of a, p, and q. At this
stage ignore the intercepts)
6
 introduce the intercepts ( 0 = 𝑥𝑥−2 +1 )and draw the graph







•
Write asymptotes in equation form.
Lines of symmetry : y=x+c and y=-x+c.
domain and range
finding the equation of the graph
making deductions from the given sketch
combination of the two or more graphs- focus on the points of intersection.
Hyperbola has two arms at all times if is complete and show at least one point on
the other arm with intercepts on the other. But there some cases where it has one
arm when the constrain is 𝑥𝑥 < 0 or 𝑥𝑥 > 0 .
Exponential function
 use table method to illustrate the effect of “a” , “p” and “q”. encourage self
discovery
 asymptote
 increasing/decreasing function
 let learners draw rough sketches (if they know the values of a, p, and q. At this
stage ignore the intercepts)
 introduce the intercepts and draw the graph
 asymptote
 domain and range
 finding the equation of the graph
Page 20
 making deductions from the given sketch
 combination of the two or more graphs- focus on the points of intersection
•
•
•
•
•
•
•
N.B
starting point must be the “mother function”.
introduce a, p and q gradually.
let learners participate.
encourage self discovery e.g. investigations,
make use of available technology e.g MS excel and other softwares.
always link lessons with real life examples.
assessment must be done regularly.
Page 21
EXERCISES 2.1
(September 2011)
Sketched below are the graphs of:
f(x) = ax2+c
g(x) =mx +q
and h(x) =
6
+1
x−2
y
(-6;20)
h
x
A
B
•
•
g
Identify and assemble information from the graphs.
What characteristics as reflected in the work schedule come to mind?
Determine
1.1 the coordinates of A and B
(4)
1.2 show that the values of a = 1; c = -16; m = -2 and q = 8
(8)
1.3 the equation(s) of the asymptotes of h.
(2)
1.4 the equation of the resultant functions, s, obtained when f is reflected about the x-axis
and translated 1 unit upwards.
1.5 the values of x for which g(x) – f(x) ≥ 0
(2)
(3)
Page 22
2.1 In the figure f is symmetrical to g about the line y = x. P(1 ; 3) is a point on g.
y
B
P(1;3)
C
x
2.1.1 Determine the value of a if g(x) = ax.
(2)
2.1.2 Determine the coordinates of B and C.
(2)
2.1.3 Write down the equation of f.
(2)
2.1.4 Determine the equation of h if h is symmetrical to g about the y-axis.
(1)
3.2 If t(x) = 2(2x):
3.2.1 draw a sketch graph of t, on DIAGRAM SHEET 1, indicate clearly the intercepts with
the axes;
(2)
3.2.2 write down the equation of the asymptote of t;
(1)
3.2.3 determine the equation of m if m is the reflection of t about the x-axis; and
(1)
3.2.4 Write down the range of m.
(2)
Page 23
2.2 Inverse Functions.
Work schedule and Examination Guidelines.
•
•
•
Notation : 𝑓𝑓(𝑥𝑥) =..........
𝑓𝑓 −1 (𝑥𝑥) =.........
Inter change values of x and y and make y the subject of the formula.
Reflection about the line y=x .
Challenges:
Learners:
•
•
1
Guard against :𝑓𝑓 −1 (𝑥𝑥) = 𝑓𝑓(𝑥𝑥) , it is incorrect.
Learners will want to introduce log on both sides of the equation.
Teaching Approach.
•
•
•
•
•
•
•
•
•
•
•
Definition of a function: discussing one-to-one and many-to-one mapping using sets
as examples.
Table for original graph of y= mx + c and inter change the values of x and y and lead
to the conclusion that this is reflection of original graph about the line y=x.
Use table for the original graph and then exchange the values of x and y. Draw the
graph of the inverse and find the equation of the second graph which should be the
same as the equation of the inverse when you substituted/exchanged the value of x
and y.
Use table method for inverse of exponential graph: x=3 y .
Domain /range of the original graph become the range/domain of the inverse.
Draw the graph y=x2 and its inverse ,then ask your class if the inverse is a function
or not with reasons.
Original graph y=x2 is a function by vertical line test but be careful for the inverse
to be a function you will need to restrict the domain of the original graph :i.e.
x ≥ 𝟎𝟎 or x ≤ 0 .Put condition for inverse that y=√𝑥𝑥 , x 0 .
When drawing the inverse of an exponential graph plot three points to show you
the direction of the graph.
Changing from exponential form to log form must be revised particularly for
exponential graphs.
Make learners aware of different ways to translate the graph (variations).
Use coloured chalk when discussing this topics.
Page 24
EXERCISES 2.2
The graphs of f(x) = 2x – 8 and g(x) = ax2 + bx + c are sketched below. B and C (0; 4,5) are the
y- intercepts of the graphs of f and g respectively. The two graphs intersect at A, which is the
turning point of the graph of g and the x- intercept of the graphs of f and g.
•
•
•
•
Identify and assemble information from the graphs.
Read the information given to verify your interpretation.
What characteristics as reflected in the work schedule come to mind?
Remember that -8 is not the y-intercept.
1.1.1 Determine the coordinates of A and B.
(4)
1.1.2 Write down the equation of the asymptote of the graph of f.
(1)
1.1.3 Determine the equation of h if h(x) = f(2x) + 8.
(2)
1.1.4 Determine an equation of h-1 in the form y = ....
(2)
1.1.5 Write down an equation of p, if p is the reflection of h-1 about the x-axis.
(1)
1.1.6 Calculate
3
5
k =0
k =4
∑ g(k ) − ∑ f (k ). Show your working.
(4)
Page 25
3. CALCULUS
3.1 Limits
Substitution is a very useful skill in determining the limit of f (x) as x tends to some number.
Example 1
Calculate lim x
x →1
Solution
lim x = 1
x →1
Example 2
Calculate
lim x 2 − 1
x→2
Solution
lim x 2 − 1 = 2 2 − 1
x→2
= 4 −1
=3
Example 3
A learner calculates a limit as follows:
x 0
lim =
x →0 x
0
= undefined
Argue for or against the learner’s solution
Solution
x
x/
lim = lim
x →0 x
x →0 x
/
= lim1
x →0
=1
NB. The learner needs to simplify before substituting and so his/her solution is not correct.
Exercise 3.1
Calculate the following:
1. lim 2 x
5. lim( x − 1)
x
2. lim
x→0 2
1
3. lim
x →0 x
x2 −1
4. lim
x →1 x − 1
(1 + h) 2 − 1
h →0
h
3( x + h) 2 − 3 x 2
7. lim
h →0
h
x →0
x→2
6. lim
Page 26
3.2 Average gradient
∆y
Average gradient =
∆x
y −y
= 2 1
x2 − x1
f ( x2 ) − f ( x1 )
=
x2 − x1
Example 1
Calculate the average gradient of f ( x) = x 2 between x = 1 and x = 2
Solution
f ( x2 ) − f ( x1 )
Average gradient =
x2 − x1
f ( x1 ) = 11
=1
f ( x2 ) = 2 2
=4
Average gradient =
4 −1
2 −1
3
1
=3
=
Exercise 3.2
1.Calculate the average gradient of f ( x) = 2 x 2 between x = 1 and x = 2
2. Calculate the average gradient of f ( x) = x 2 − 2 between x = 2 and x = 3
3. Calculate the average gradient of f ( x) = −3x 2 + 2 x + 1 between x = 1 and x = 2
Page 27
3.3 Derivative
The questions you may come across in the final examination are mostly of the following types:
•
•
Determine the derivative using first principles
Determine the derivative using rules
Derivative using first principles
If y = f (x) then
f ′( x) = lim
h →0
f ( x + h) − f ( x )
h
Example 1
Given f ( x) = x 2
Determine f ′(x)
Solution
f ′( x) = lim
h →0
f ( x + h) − f ( x )
h
f ( x) = x 2
f ( x + h) = ( x + h) 2
= x 2 + 2 xh + h 2
f ′( x) = lim
h →0
x/ 2 + 2 xh + h 2 − x/ 2
h
2 xh + h 2
h →0
h
= lim
h/ (2 x + h)
h →0
h/
= lim
= lim 2 x + h
h →0
= 2x + 0
= 2x
Page 28
Example 2
Given: f ( x) = −
1
x
f ( x + h) − f ( x )
.
h
2.2 What does the value in 2.1 represent?
2.1 Determine lim
h →0
Solution
2.1 f ( x) = − 1
x
f ( x + h) = −
lim
h →0
1
x+h
f ( x + h) − f ( x )
h
= lim
−
h →0
1
1
+
x+h x
h
x+h
−x
+
x ( x + h) x ( x + h)
= lim
h →0
h
−x+ x+h
x ( x + h)
= lim
h →0
h
h
x ( x + h)
= lim
h →0
h
= lim
h
xh( x + h)
= lim
1
x ( x + h)
h →0
h →0
=
1
x( x + 0)
1
= x−2 = 2
x
2.2 f ′( x)
Page 29
Exercise 3.3
Differentiate f using the first principle:
1. f ( x) = x
2. f ( x) = 2
3. f ( x) = x 2 − 2
4. f ( x) = x 2 − 2 x
5. f ( x) = x 2 − 2 x − 3
6. f ( x) =
1
x
7. f ( x) = x 3
Page 30
3.4 Derivative using rules
The derivative of f in each of the following cases can be found as follows
•
•
•
If f ( x) = x n then f ′( x) = nx n −1
If f ( x) = ax n then f ′( x) = anx n −1
If f ( x) = ag ( x) ± bh( x) then f ′( x) = ag ′( x) ± bh′( x)
Example 1
1. Given f ( x) = 1 find f ′(x)
Solution
f (x) = x 0
f ( x) = 0 x 0−1
=0
Example 2
Given f ( x) = x 2 find f ′(x)
Solution
f ′( x) = 2 x 2−1
f ( x) = 2 x
Example 3
1
2
Given f ( x) = x +
1 2
x find f ′(x)
2
Solution
1 12 −1 1
′
f ( x) = x + ⋅ 2/ x 2−1
2/
2
1
1 −
= x 2+x
2
1
= 1 +x
2x 2
1
+x
=
2 x
Note well
Given y = f (x)
dy
f ′( x) =
dx
= Dx ( y )
d
= ( y)
dx
Page 31
Exercise 4.4
1. If f ( x) = 2 find f ′(x)
2. If f ( x) =
3 2
x find f ′(x)
2
3
3. Differentiate x 2
1
4. Find the derivative of ( x − ) 2
x
5. If y = x 2 +
6. If y =
2x − 4
x
7. Determine
8. Find
1
dy
, find
2
dx
x
find Dx ( y )
dy
if y = (2 − 3x) 2
dx
dy
if 2 xy = 2 x 2 − 5 x + 6
dx
Page 32
4.5 Graphs
Often two types of graphs are asked in the examination
• A function may be given and a candidate is asked to sketch a graph.
• A graph is given and questions based on the graph are asked.
Graph sketching
The derivative of f in each of the following cases can be found as follows
•
•
•
•
Calculate the y -intercept. Here x = 0 .
Calculate the x -intercept. Here y = 0 .
Calculate the turning points. Here f ′( x) = 0 .
You may have to find the regions of increase/decrease.
Example 1
Given:
f ( x) = x3 − x 2 − 8x + 12
Draw a sketch graph of f , showing all intercepts with axes, and turning points.
Solution
y -intercept: if x = 0, then
f (0) = 03 − 0 2 − 8 ⋅ 0 + 12
= 12
y -intercept is given by (0 ;12)
x − intercept: if y = 0, then
( x − 2)( x 2 + x − 6) = 0
(x – 2)(x + 3)(x – 2) = 0
( x − 2) = 0 or ( x + 3) = 0 or ( x − 2) = 0
x = 2 or x = −3 or x = 2
x − intercepts are given by (2;0) and (−3,0)
Turning points:
f ′( x) = 0
Page 33
3x 2 − 2 x − 8 = 0
(3 x + 4)( x − 2) = 0
4
x = − or x = 2
3
4
4
4
y = (− )3 − (− ) 2 − 8(− ) + 12
3
2
3
=
500
27
= 18,52
=
500
= 18,52
27
Maximum:
1
( − 1 ; 18,52)
3
Minimum:
( 2;0)
( − 1 1 ;18,52))
3
y
(-3;0)
(0;12)
(2;0)
Calculate the point of inflection.
Solution
f ′′( x) = 6 x − 2
6x − 2 = 0
x=
1
3
Page 34
Example 2
The diagram in the accompanying sketch represents the curve of
f ( x) = x 3 + ax 2 + bx + c .
The curve has turning points at B and (1,0).The points (-1,0) and (1,0) are x-intercepts.
B
y
f
A(-1;0)
(1;0)
x
2.1. Calculate the values of a; b and c
Solution
y = 1( x + 1)( x − 1)( x − 1)
= ( x 2 − 1) x − ( x 2 − 1)
= x3 − x 2 − x + 1
a= -1; b= -1, c= 1
2.2. Write down values of x for which f ' ( x) > 0
Solution
3x 2 − 2 x − 1 > 0
(3 x + 1)( x − 1) > 0
x<−
1
or x > 1
3
2.3. Write down the x-coordinates of the turning points of the curve of f .
3x 2 − 2 x − 1 = 0
Page 35
(3x + 1)( x − 1) = 0
1
or x = 1
3
x=−
2.4. If k < 0, how many real roots will the equation x 3 − x 2 − x + 1 = k have?
Solution: one
2.5. Determine the equation of the tangent to the graph of f at x = −1
Solution
y = mx + c
m = 3(−1) 2 − 2(−1) − 1
m = 3 + 2 −1
m=4
y = 4x + c
f (−1) = (−1) 3 − (−1) 2 − (−1) + 1
f (−1) = 0
0 = 4(−1) + c
c=4
y = 4x + 4
2.6 Determine the point of inflection of f.
Solution
f ' ' ( x) = 0
f ' ( x) = 3x 2 − 2 x − 1
6x − 2 = 0
x=
1
3
Page 36
Exercise 3.5
1.
B
y
(0;3)
f
A
(1;0)
x
The diagram above represents the curve of f ( x) = x 3 + x 2 − 5 x + 3 .
The curve has a y-intercept at (0;3) and turning points at (1;0) and B.
The point A is an x-intercept.
1.1 Calculate the coordinates of A
1.2 Calculate the x-coordinate of B
1.3 Write down values of x for which f ' ( x) > 0
1.4 If k < 0, how many real roots will the equation x 3 + x 2 − 5 x + 3 = k have?
1.5 Determine the equation of the tangent to the graph of f at x = −1
1.6 Determine the point of inflection of f.
2.
y
f '(x)
1
x
1
3
2.1 Calculate the gradient of the tangent to f when x = 0
Page 37
2.2 Show that x = −
b
is the x coordinate of the point of inflection of f (x)
3a
2.3 Write down the x-coordinates of the turning points of the curve of f .
2.4 If it is further given that f ( x) = 0 when x = 0 ; x = 2 and x = 5 , sketch the graph of f.
(The y -coordinates of the turning points are not required.)
2.5 For which values of x will f (x) be decreasing?
3. Given f ( x) = − x 3 + 6 x 2 − 9 x + 4
3.1 Sketch the graph of f .Indicate the turning co-ordinates of all turning points and intercepts
with axes. Show all calculations.
3.2 Determine the equation of the tangent to the curve of f at the point (2;2).
Optimisation
Whenever the minimum or maximum value is mentioned in a problem, use f ′( x) = 0
Example
A function g ( x) = ax 2 +
b
has a minimum value at x = 4 . The function value at
x
x = 4 is 96. Calculate the values of a and b .
Solution
g ( x) = ax 2 +
b
x
g ′( x) = 2ax − bx − 2 = 2ax − b2
x
0 = 2a (4) −
8a =
b
(4) 2
b
16
b = 128a
(1)
Page 38
96 = a (4) 2 +
b
4
96 = a(4)2 + 128a
4
(2)
(1) into (2)
96 = 16a + 32a
48a = 96
a=2
b = 128a = 128(2) = 256
Exercise 3.6
1. A tank is being filled with water. The volume of the water tank(in litres) after a
time t in hours is given by the formula V = 50 − 5t + 5t 2
1.1 Calculate the initial volume.
1.2 Calculate the rate of change of volume in the tank when t =12 minutes.
1.3 Calculate the time at which the volume will be maximum.
2.
Divide 12 into two parts such that their product is a maximum.
3. A farmer has 100m of fencing and fences off a rectangular field. One of the sides of the
rectangular field is a wall as shown in the diagram. Let the length of the rectangular field be x
and the breadth be y.
WALL
3.1 Express the breadth in terms of x.
3.2 Show that the area of the rectangular field is given by: A = 100 x − 2 x 2
3.3 Hence determine the value of x such that the area enclosed by the wall and the fence will be a
Page 39
maximum.
3.4 Find the maximum area of the rectangular field.
4. The distance covered in metres by object A is given by s (t ) = 3t 2 − 5t and the distance covered
in metres by object B is given by s (t ) = t 3 − 2t 2 + 3t + 5 , where t represents seconds in both
cases.
4.1 Find an expression for the speed of object A at any time t
4.2 Determine how long it will take for both objects to reach the same speed.
4.3 Determine how long it will take before the acceleration of both objects will be exactly the
same.
5. 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.
5.1 If the length of wire used to make the circle is x in metres, write in terms of x the length of
the sides of the square in metres.
5.2 Show that the sum of the areas of the circle and the square is given by
1
1 2 x
f ( x) = ( +
) x − + 1 square metres.
16 4π
2
5.3 How should the wire be cut so that the sum of the areas of the circle and the square is a
minimum?
Page 40
Answers
Exercise 3.1
1. 0 2. 0
3. undefined 4. 2 5. 1
6. 2
7. 6 x
Exercise 3.2
1.6
2.5 3.-7
Exercise 3.3
1. 1
2. 0
3. 2 x
4. 2 x
6. −
5. 2 x − 2
Exercise 3.4
1.0
2. 3
3.
3
2
4. 2 x −
2
x3
1
x2
5. 2 x −
7. 3x 3
3
−
2
2
2
6.
−
+
2
x
x3
x2
7. − 12 + 18 x 8. 2
Exercise 3.5
1.1 -3
1.2 − 1
2
3
2
1.3 x < −1 and x > 1
3
1.4 One real root
1.5 y = −4 x + 5
1 128
)
1.6 (− ;
3 27
2.1 1
2.2 6ax + 2b = 0
2.3 1and 3
2.5 1 < x < 3
3.1 x − intercepts (1; 0), (4, 0)
y − Intercept: (0; 4)
Minimum turning point: (1;0)
Maximum turning point: (3;4)
Page 41
3.2 y = 3 x − 4
Exercise 3.6
1.1 50 litres
1.2 115
1.3
1
s
2
2. The parts are 6 and 6
3.1 100 − 2 x
3.2 100 x − 2 x 2
3.3 x = 25
3.4 Area= 1250
4.1 s′(t ) = 6t − 5
4.2 t =
5
6
4.3 t =
4
,t = 2
3
5.1 1 −
x
4
5.3
4π
π +4
Page 42
4. ANALYTICAL GEOMETRY
4.1 Formulae
You’ll find the following formulas on Analytical Geometry in the information sheet:
 x + x 2 y1 + y 2 
M 1
;
d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2

2 
 2
y − y1
m= 2
m = tan θ
y = mx + c
y − y1 = m( x − x1 )
x 2 − x1
( x − a )2 + ( y − b )2 = r 2
NB: The information sheet does not contain all Mathematical formulas.
The following table shows how to use these formulas in working out sums on Analytical Geometry
in grade 12:
Formula
Procedure for solution
Exemplar question
d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2
• Label the two points
• Substitute into the
formula.
Calculate AB if points are A(-5; 4) and B(3; -2).
is used to find distance
between two points.
Answer: d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2
AB = (3 − (−5))2 + (−2 − 4)2
AB = 64 + 36 = 100 = 10 units
 x + x 2 y1 + y 2 
M 1
;
 is used
2 
 2
to find the midpoint of a line
segment.
• Label the two points
• Substitute into the
formula.
A(-5; 3) and B(4; -1) are two points on a line
segment AB. Find the co-ordinates of M, the
midpoint of AB.
 x + x 2 y1 + y 2 
Answer: M 1
;

2 
 2
M − 5 + 4 ; 3 − 1  = M − 1 ; 1

m=
y 2 − y1
is used to find
x 2 − x1
gradient of a line.
Use m = tan θ to find an
inclination of a line.
NB: If m > 0, then an angle of
elevation θ < 90°.
P
O
θ
Acute angle
A
• Label two points on
the line as A and B.
• Substitute coordinates of A and B
in the formula to
find m.
• To find an angle
given gradient of a
line, use a scientific
calculator and key in
the following
buttons in
sequence:- shift; 2nd
function; tan −1 ; m;
and =.
2 
2
 2

Determine the gradient of a line passing through
(4; 3) and (-2; 1).
Answer: let A( x1 ; y1 ) = A(4; 3) and
B( x 2 ; y 2 ) = B(-2; 1). Therefore
y − y1
−2 1
1− 3
m= 2
=
=
=
x 2 − x1 − 2 − 4 − 6 3
Given the points A(2; 2) and B(1; -3), find the
angle which AB makes with the positive x-axis
(inclination of a line).
Answer: m =
m=
y 2 − y1
x 2 − x1
2 − (−3)
=5
2 −1
Page 43
OR
P
O
θ
• When gradient is not
given, determine it
first and then
substitute m in
[ tan −1 (m)] .
Therefore θ = tan −1 (5) = 78.69°
• Find the gradient of
the given line.
• The new line will
then have the same
gradient.
Calculate the value of x if BC // DE given coordinates of B, C, D and E as B(-1; 3), C(-5; -3),
D(4; 1) and E(x; -8).
Obtuse angle
A
If m < 0, the inclination is
90°<θ<180°. Your knowledge
of trigonometry is crucial here.
Lines l 1 and l 2 are parallel, if
and only if their gradients are
equal, i.e. ll // l 2 if m1 = m2 .
mBC =
Answer:
∴ mDE =
y2 − y1
= −3−3
x2 − x1 − 5 − (−1)
= −6 = 3
−4 2
= 11
2
y2 − y1 − 8 − 1 − 9
=
=
=3
x2 − x1
x−4 x−4 2
−9 = 3
x−4 2
⇒ 3( x − 4) = 2(−9)
⇒
3( x − 4) − 18
=
3
3
x − 4 = −6
∴ x = −2
Page 44
Lines l 1 and l 2 are
perpendicular, if the product
of their gradients is –1, i.e. if
l1 ⊥ l 2 then m1 × m2 = −1 .
NB: This formula does not
apply when lines are
horizontal (m = 0) or vertical
(m is undefined).
• Find the gradient
of the given line.
• The new line will
then have gradient
−1
m2 =
m1 .
Given the points A(-2; -1), B(1; 5) and C(− 8;2) ,
show that AB ⊥ CA .
Answer: mAB =
and mCA =
5 − (−1) 6
= =2
1 − (−2) 3
2 − (−1)
= 3 =−1
− 8 − (−2) − 6
2
∴ m AB × mCA = 2 × (− 12 ) = −1
⇔ AB ⊥ CA .
Collinear points lie on the
same line and gradients are
equal, i.e. m1 = m2
NB: You may be required to
apply your knowledge of
simultaneous equations to
solve for unknown values.
• Find gradient of
line segments
keeping one point
common.
Calculate the value of y if A(-1; 1), B(2; -3) and C(4; y) are collinear.
Answer: m1 = m2
mAB = mBC
∴
− 3 −1
y − (−3)
=
2 − (−1)
−4−2
y+3
∴ −4 =
3
−6
∴ 3( y + 3) = 24
y+3=8
y=5
To find equation of a line:
• Standard form of a line is
ax + by + c = 0
• When given m and c, use
y = mx + c
Determine the equation of a line with gradient of
• Use any form of a
straight line, unless 2 and y-intercept of -3.
otherwise
Answer: y = mx + c
specified, and
substitute the
y = 2x − 3
information given.
Determine the equation of a line whose x- and yintercepts are -3 and 2 respectively.
• When given x and y
x y
intercepts, use + = 1
a b
x y
Answer: a + b = 1
x
y
+ =1
−3 2
Page 45
2 x − 3 y = −6 or
y = 23 x + 2
Determine the equation of a line whose gradient
is -3 and passes through (2; -1)
Answer: y = m( x − x1 ) + y1
• When given m and a point,
use y = m( x − x1 ) + y1
y = −3( x − 2) − 1 = −3 x + 5
Determine the equation of a line passing through
(2; 3) and (8; 6).
• When given two points,
use
y − y1
y= 2
( x − x1 ) + y1
x 2 − x1
To find an equation of a circle
with centre at origin (0; 0), use
x2 + y2 = r 2
Answer: y =
y=
• Substitute coordinates of a
point on a circle to
find the radius.
NB: This equation works in the
same way as the Pythagoras
Theorem, and so questions
may require you to find
measurement of an unknown
side when two are given.
To find an equation of a circle
with centre (a; b) and radius r,
(x − a )
2
+ ( y − b) = r .
2
2
NB: Questions may
require you to use the
method of completing
the square to the
centre and radius of a
circle.
y 2 − y1
( x − x1 ) + y1
x 2 − x1
6−3
( x − 2) + 3 = 12 x + 2
8−2
Find the equation of the circle passing through
point (4; -5). What is the radius of this circle?
Answer: x 2 + y 2 = r 2
(4)2 + (− 5)2
= r 2 = 41
Equation is
• Substitute coordinates of the
centre and radius
when they are
given.
• When centre and
radius are not
given, Complete
the square in x and
y.
• Simplify
and radius is
41 units.
Write down the equation of the circle with the
centre (−2;3) and a radius 13 units.
Answer: ( x − a ) + ( y − b ) = r 2
2
2
( x + 2 ) 2 + ( y − 3) 2
=
( 13 )
2
x 2 + 4 x + 4 + y 2 − 6 y + 9 = 13
x 2 + y 2 + 4x − 6 y = 0
Consider the following questions:
x 2 + y 2 + 4 x − 12 y + 4 = 0 is the equation of a
circle with centre M and radius r.
1. Calculate the co-ordinates of the centre M and
the length of the radius r.
Page 46
2. Write down the co-ordinates of the point(s)
where this circle intersects the x-axis without
any further calculations.
y
Answers: 1. x 2 + y 2 + 4 x − 12 y + 4 = 0
O(a; b)
) = ( 42 ) + ( −122 ) − 4
x 2 + 4 x + ( 42 ) + y 2 − 12 y + ( −12
2
2
r
(x
2
2
) (
2
2
)
+ 4 x + 4 + y 2 − 12 y + 36 = 4 + 36 − 4
( x + 2 )2 + ( y − 6 )2
x
= 36
Therefore M(-2; 6) and
2. ( x + 2 )2 + ( y − 6 )2 = 36
(x + 2)2 + (0 − 6)2
= 36
( x + 2 )2
=0
Co-ordinates are (-2; 0).
To find an equation of a
tangent to a circle, use the
fact about perpendicular lines
that m1 × m2 = −1 because
radius ⊥ tangent
and equation of a line
y = m( x − x1 ) + y1
NB: This question lends itself
to solving simultaneous
equations.
• Find the centre of
the circle.
• Find gradient of
the radius, i.e. line
joining the centre
of the circle and a
point of contact.
• Use m1 × m2 = −1
to find the gradient
of a tangent.
• Substitute gradient
of a tangent into
the equation of a
line
y = m( x − x1 ) + y1
Determine the equation of the tangent to the
circle (x − 1)2 + ( y − 2 )2 = 20 at P(5; 0).
Answer: (x − 1)2 + ( y − 2 )2 = 20
Centre is M(1; 2)
mradius =
2−0
2
=
= − 12
1− 5 − 4
∴ m tan gent = 2
Therefore the equation of tangent is
y = m( x − x1 ) + y1
y = 2( x − 5) + 0
y = 2 x − 10
Page 47
4.2 QUADRILATERALS
It is extremely important to know definitions when you're doing sums, e.g. when asked to prove
that a particular quadrilateral is a rectangle, and you already know it is a parallelogram, all you
need to show is that one angle is a right angle. You need to recall properties and area formulas of
quadrilaterals, i.e. trapezium, parallelogram, rectangle, rhombus, square and kite, and apply
knowledge thereof in problem situations. Properties relate to sides (equal and parallel), angles
(opposite and adjacent) and diagonals (bisect each other).
A trapezium
Definition: A trapezium is a quadrilateral with at least one pair of opposite sides parallel.
Property of a trapezium:
a
• one pair of opposite sides parallel
Area of a trapezium =
h
1
h(a + b )
2
b
where a and b are parallel sides or bases and h is the
height.
A parallelogram
Definition: A parallelogram is a quadrilateral with two pairs of opposite sides parallel.
Properties of a parallelogram:
•
•
•
•
both pairs of opposite sides parallel
both pairs of opposite sides equal
both pairs of opposite angles equal
diagonals bisect each other.
Area of a parallelogram = DC × h
B
A
h
E
D
C
OR
Area of parallelogram = 2 × Area ∆ADC
A rectangle
Definition: A rectangle is a parallelogram with a right angle or
A rectangle is a quadrilateral with four right angles.
Page 48
Properties of a rectangle:
• both pairs of opposite sides parallel
• both pairs of opposite sides equal
• both pairs of opposite angles are right angles
• diagonals bisect each other
A
B
D
C
Area of a rectangle = AB × AD
A rhombus
Definition: A rhombus is a parallelogram with four equal sides in length.
Properties of a Rhombus:
• Diagonals of a rhombus bisect each other at right angles.
• Diagonals of a rhombus bisect opposite angles
NB: A rhombus is like a square in that:
1)
All of the sides are equal;
2)
Both pairs of opposite sides are parallel.
But a rhombus is very different from a square in that none of the interior angles is 90°.
B
d1
A
C
d2
D
Area of a rhombus =
1
× d1 × d 2
2
OR
Area of a rhombus = base × breadth
(as in the case of a kite)
(as in the case of a parallelogram)
Example
Determine the measure of the angles in this rhombus. It helps to know a theorem that states if
two parallel lines are cut by a transversal, then co-interior angles (i.e. angles on the same side of a
transversal) are supplementary.
Answer: 3x + x = 180° (co-interior angles)
4x = 180°
x = 45°
and 3x = 135°
x
3x
Page 49
A kite
Definition: A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides
that are adjacent to each other.
NB: A kite may either be convex or concave, but the word "kite" is often restricted to the convex variety,
i.e. a "dart" or "arrowhead".
Properties of a kite:
• One diagonal is the perpendicular bisector of the other diagonal.
• One diagonal bisects a pair of opposite angles.
• One diagonal divides a quadrilateral into two congruent triangles.
Area of a kite =
B
1
× d1 × d 2 .
2
A
C
d2
d1
D
A square
Definition: A square is a special case of a rhombus with equal sides
and equal opposite angles or
a regular quadrilateral with all sides equal and interior
angles right angles or
a square is a parallelogram with a right angle or
a square is a rectangle with all sides equal.
a
A
B
d1
a
d2
D
C
Properties of a square:
• The diagonals of a square bisect
2
each other and meet at 90°.
Area of a square = a
• The diagonals of a square bisect its angles.
• The diagonals of a square are equal.
• Opposite sides of a square are both parallel and equal in length.
• All four interior angles of a square are equal, and so every angle is a right angle.
Page 50
Exercise 4.1
1.
A(-2; 1), B(p; -4), C(5; 0) and D(3; 2) are vertices of a trapezium ABCD in a Cartesian plane
with AB // DC.
1.1
Show that p = 3 .
1.2
Calculate the ratio AB:CD in its simplest
form.
1.3
If N ( x; y ) is on AB and NBCD is a
parallelogram, determine the
coordinates of N.
y
D(3;2)
•
A(−2;1)
•
•
x
1.4
Determine the equation of the straight
line passing through B and D.
1.5
What is the inclination of line BD?
1.6
Calculate the area of parallelogram NBCD.
1.7
R(-1; q), A and C are collinear. Calculate the value of q.
2.
x 2 + 4 x + y 2 + 2 y − 8 = 0 is the equation of a circle with centre M in a Cartesian plane.
2.1
Prove that the circle passes through the point N (1;−3) .
2.2
Determine the equation of PN, the tangent to the circle at N.
2.3
Calculate θ, the angle of inclination of the tangent, rounded off to ONE decimal digit.
2.4
Determine the co-ordinates of T, the point where the tangent in QUESTION 2.2 intersect the
x-axis.
2.5
Calculate the co-ordinates of the point(s) where the circle with centre M cuts the y-axis.
3.
P(-3; 2) and Q(5; 8) are two points in a Cartesian plane.
3.1
Calculate the length of PQ.
3.2
Calculate the angle that PQ forms with the x-axis, rounded off to one decimal digit.
3.3
Determine the equation of the perpendicular bisector of PQ in the form ax + by + c = 0 .
O
C (5;0)
Page 51
4.
In the diagram A, B, C and D(3;9) are the vertices of a rhombus. The equation of AC is
x + 3 y = 13 .
4.1
Show that the equation of BD is 3 x − y = 0 .
4.2
Calculate the co-ordinates of K if the diagonals
of the rhombus intersect at point K.
4.3
Determine the co-ordinates of B.
4.4
Calculate the co-ordinates of A and C if
y
A
(3)
x
C
0
AD = 73 units.
5.
D(3;9)
B
The points A(−3;4) , B, C (−2;−1) and D are
the vertices of a square ABCD in a Cartesian
plane. D is a point on the y-axis. The
diagonals AC and BD intersect at M.
A(−3;4)
y
5.1
Show that the length of the diagonal AC is
26 units.
5.2
Determine the co-ordinates of M.
5.3
Calculate the length of any side of the
square.
5.4
Determine the co-ordinates of D.
5.5
Calculate the co-ordinates of B.
6.
The equation of a circle is x 2 + y 2 + 4 x − 2 y − 4 = 0 .
6.1
6.2
Determine the coordinates of M, the centre of the circle, as well as the length of the radius.
Calculate the value of p if N ( p; 1) with p > 0 is a point on the circle.
6.3
Write down the equation of the tangent to the circle at N.
D
B
x
C (−2;−1)
y
7.
In the accompanying diagram, AB is a straight
line with the equation y = mx + c , which
P( x; y )
B
passes through the point Q(4; 2) .
• Q ( 4;2)
Show that the equation of AB can be written
x
y
as
+
= 1.
4m − 2 2 − 4m
m
8.
O
A
x
Prove that the straight line y = x + 3 is a tangent to the circle x 2 + y 2 - 6 x - 4 y + 5 = 0 .
Page 52
9.
ABCD is a quadrilateral
with vertices A(1; 6),
B(3; 0), C(6; 1) and D(7; t
) in a Cartesian plane.
AD//BC.
9.1
Calculate the gradient of
BC.
9.2
Determine the equation
of AD in the form y = …
9.3
Show that t = 8.
9.4
Calculate the lengths of
AD, BC and AB.
9.5
Show that AB is perpendicular to BC.
9.6
Calculate the area of the quadrilateral ABCD. (Simplify your answer.)
9.7
Determine θ, the angle of inclination of BC.
10.
A(0; 5) and B(–8; 1) are two points on the circumference of the circle centre M, in a
Cartesian plane. M lies on AB. DA is a tangent to the circle at A. The coordinates of D are (3;
–1) and the coordinates of C are (–12; –1). Points C and D are joined. K is the point (0; –7).
CTD is a straight line.
Page 53
10.1 Show that the coordinates of M, the midpoint of AB, are (–4; 3).
10.2 Determine the equation of the tangent AD.
10.3 Determine the length of AM.
10.4 Determine the equation of circle centre M in the form ax 2 + by 2 + cx + dy + e = 0 .
10.5 Quadrilateral ACKD is one of the following:
parallelogram; kite; rhombus; rectangle Which one is it? Justify your answer.
Page 54
Answers and suggested solution methods
1.1
m
AB
=m
DC
m
2 − 0 − 4 −1
=
3−5
p+2
OR
2( p + 2 ) = −5 × −2
DC
=
Analytical Geometry
2−0
= −1 = m
AB
3−5
At (-2; 1), the equation of AB is y = −( x + 2) + 1
∴ y = −x −1
At B(p; -4), we have − 4 = − p − 1
∴ p=3
p+2=5 ∴p =3
1.2
AB =
CD =
(3 + 2)2 + (− 4 − 1)2
(5 − 3)2 + (0 − 2)2
= 25 + 25 = 50 = 5 2
= 4+4 = 8 = 2 2
AB 5 2 5
=
= = 21
2
CD 2 2 2
AB : CD = 5 : 2
m NB = m DC = −1
y+4
−1 =
x −3
y + 4 = 3− x
Therefore equation of NB is y = − x − 1
Therefore
1.3
m BC = m ND
0+4
y−2
=
=2=
5−3
x−3
y − 2 = 2( x − 3)
OR ….
x NB = x DC
xN − 3 = 3 − 5
xN = 1
And y N + 4 = 2 − 0
y N = −2
∴ N (1; − 2)
y = 2x − 4
Since N is the point of intersection of lines NB and ND: − x − 1 = 2 x − 4 .
3 x = 3 , and therefore x = 1
Substitute in y = 2(1) − 4 = −2
∴ N (1; − 2)
OR
Midpoint of NC = midpoint of BD
3+3 2−4
Diagonals of the parallelogram NBCD intersect at 
;
 = (3; − 1)
2 
 2
x+5
Therefore
=3
2
x = 1 and
y+0
= −1
2
y = −2
∴ N (1; − 2)
Page 55
OR
m NB = m DC = −1
y+4
x−3
y + 4 = 3− x
Therefore equation of NB is y = − x − 1 ……… (1)
BN 2 = CD 2
(x − 3)2 + ( y + 4)2 = (5 − 3)2 + (0 − 2)2
∴ −1 =
x 2 − 6 x + 9 + y 2 + 4 y + 16 = 4 + 4
x 2 − 6 x + y 2 + 8 y + 17 = 0 …………………… (2)
Subst (1) in (2): x 2 − 6 x + (− x − 1)2 + 8(− x − 1) + 17 = 0
x 2 − 6 x + x 2 + 2 x + 1 − 8 x − 8 + 17 = 0
x 2 − 6x + 5 = 0
(x − 1)(x − 5) = 0
x = 1 or x = 5
And y = −2 or y = −6
∴ N (1; − 2) note N (5; − 6) is incorrect
OR
DC 2 = NB 2
(3 − 5)2 + (2 − 0)2 = (x − 3)2 + ( y + 4)2
4 + 4 = x − 6 x + 9 + y + 8 y + 16
x 2 − 6 x + y 2 + 8 y + 17 = 0 ………………….. (1)
DN 2 = CB 2
(x − 3)2 + ( y − 2)2 = (5 − 3)2 + (0 + 4)2
x 2 − 6 x + 9 + y 2 − 4 y + 4 = 4 + 16
x 2 − 6 x + y 2 − 4 y − 7 = 0 …………………… (2)
(1) – (2): 12 y + 24 = 0
∴ y = −2
And x = 1
∴ N (1; − 2)
2
2
OR
AN = AB − CD
= 50 − 8
=5 2 −2 2 =3 2
AN : NB = 3 2 : 2 2 = 3 : 2
(m × x B ) + (n × x A )
xN =
m+n
3 × 3 + 2 × (− 2 )
=1
xN =
3+ 2
(m × y B ) + (n × y A )
and y N =
m+n
3 × (− 4 ) + 2 × 1
yN =
= −2
3+ 2
∴ N (1; − 2)
1.4
x=3
1.5
θ = 90°
1.6
1
Area of parallelogram NBCD = 2 Area∆BCD = 2  × 6 × 2 = 12 sq units
2
Page 56
OR
DN = DC = 2 2 units
Altitude from D to AB = 3 2 units
Area of //m NBCD = 2 Area∆BCD

1
= 2 × DC × h 

2
OR
= 2 2 ×3 2
≈ 12 square units
1.7
tan α = M DC = −1
∴ α = 135°
tan β = BC = 2
∴ β = 63,4°
DCˆ B = 108,4°
BC = (0 − (−4) ) + (5 − 3) = 20 = 2 5 units
Area of //m NBCD = 2 Area∆BCD

1
= 2 × DC × BC × sin DCˆ B 

2
2
2
= 2 2 × 2 5 × sin 108,4°
≈ 12 square units
m AR = m RC
m AR =
m RC
1− q
= q −1
− 2 +1
∴ q −1 =
q
−6
y − 0 0 −1
1
=
=−
x−5 5+2
7
1
5
y =− x+
7
7
1
5 6
At R(-1; q), q = − (− 1) + =
7
7 7
Equation of AC:
q−0
q
=
=
−1− 5 − 6
OR
∴ − 6(q − 1) = q
6 = 7q
q=
2.1
6
7
Subst. x = 1 in x 2 + 4 x + y 2 + 2 y − 8 = 0
(1)2 + 4(1) + y 2 + 2 y − 8 = 0
y2 + 2y − 3 = 0
( y + 3)( y − 1) = 0
y = –3
N(1; –3)
OR
Subst. y = –3 in x 2 + 4 x + y 2 + 2 y − 8 = 0
x 2 + 4 x + (− 3) + 2(− 3) − 8 = 0
2
x 2 + 4x − 5 = 0
(x + 5)(x − 1) = 0
x=1
N(1; –3)
Page 57
OR
Subst. (1; –3) in x 2 + 4 x + y 2 + 2 y − 8 = 0
LHS = (1) + 4(1) + (− 3) + 2(− 3) − 8
2
2
= 0 = RHS
Therefore N(1; -3) lies on the circle.
2.2
x 2 + 4x + 4 + y 2 + 2 y + 1 = 8 + 4 + 1
(x + 2)2 + ( y + 1)2
= 13
Therefore centre is M(-2; -1)
− 3 +1
2
m MN =
=−
1+ 2
3
3
∴ m PN =
2
3
x+c
2
3
−3= +c
2
9
c=−
2
y=
∴y =
3
(x − 1) − 3
2
∴y =
3
9
x − or 3 x − 2 y − 9 = 0 or 3 x − 2 y = 9
2
2
or
2.3
3
2
θ = 56,3°
2.4
3 x − 2(0) = 9
x=3
T(3; 0)
2.5
Subst. x = 0 in ( x + 2 )2 + ( y + 1)2 = 13
tan θ =
(2)2 + y 2 + 2 y + 1 = 13
y2 + 2y − 8 = 0
( y + 4)( y − 2) = 0
Co-ordinates are (0; –4) and (0; 2)
3.1
P(–3; 2) and Q(5; 8)
∴ PQ =
3.2
(− 3 − 5)2 + (2 − 8)2
= 10 units
2−8
3
=
−3−5 4
3
∴ tan θ =
4
θ = 36,87°
m PQ =
Page 58
3.3
 −3+5 2+8
M
;
 = M (1; 5)
2 
 2
4
From 3.2, m2 = − 3
4
x+c
3
−4
+c
5=
3
19
c=
3
y=−
4.1
∴y =−
4
(x − 1) + 5
3
∴y =−
4
19
x+
or 4 x + 3 y − 19 = 0 or 4 x + 3 y = 19
3
3
or
1
13
1
Equation of AC: y = − x +
where m AC = −
3
3
3
∴ m BD = 3
y = 3x + c
At (3; 9), y = 3(x − 3) + 9
or
c=0
∴ y = 3 x or y − 3 x = 0
y − 3x = 0
1
13
3x = − x +
3
3
9 x + x = 13
13
 13  39
∴x =
and y = 3  =
10
 10  10
 13 39 
Hence K  ; 
 10 10 
4.2
4.3
4.4
3 x − y = 0 ………………(1)
x + 3 y = 13 ………………(2)
3×(1): 9x – 3y = 0 ………………(3)
OR
(1)+(3): 10x = 13
13
 13  39
∴x =
and y = 3  =
hence
10
 10  10
 13 39 
K ; 
 10 10 
x + 3 13
y + 9 39
=
=
2
10
2
10
and
2
6
x=−
y=−
5
5
−2 −6
;
B

 5 5 
AD = (x − 3) + ( y − 9 ) = 73
2
2
x 2 − 6 x + 9 + y 2 − 18 y + 81 = 73
(13 − 3 y )2 − 6(13 − 3 y ) + y 2 − 18 y + 17 = 0
169 − 78 y + 9 y 2 − 78 + 18 y + y 2 − 18 y + 17 = 0
10 y 2 − 78 y + 108 = 0
Page 59
x=
x=
− b ± b 2 − 4ac
2a
− (− 78) ±
(− 78)2 − 4(10 × 108)
2(10 )
9
5
y A = 6 and y C =
38
5
9
38


A(-5; 6) and C  5 ; 5 


x A = −5 and y C =
5.1
AC 2 = (− 3 + 2 ) + (4 + 1) = 1 + 25 = 26
2
2
AC = 26 units
5.2
 −5 3
 − 3 − 2 4 −1 
M
;
; 
 = M
2 
 2 2
 2
5.3
2 AD 2 = 26
AD 2 = 13
AD = 13 units
5.4
[
]
2 (− 3) + (4 − y ) = 26
2
2
9 + 16 − 8 y + y 2 = 13
y 2 − 8 y + 12 = 0
( y − 6)( y − 2) = 0
∴ D(0; 2 )
5.5
xB + 0 − 5
=
2
2
x B = −5
and
∴ B(− 5; 1)
6.1
yB + 2 3
=
2
2
yB = 1
x 2 + 4x + 4 + y 2 − 2 y + 1 = 4 + 4 + 1
(x + 2)2 + ( y − 1)2
=9
Centre is M(–2; 1) and radius = 3 units
6.2
p 2 + 4 p + (1) − 2(1) − 4 = 0
2
p2 + 4 p − 5 = 0
( p + 5)( p − 1) = 0
∴ p = 1, p > 0
∴ N (1; 1)
Page 60
6.3
1−1
=0
1+ 2
m tan = undefined (a vertical line)
Equation of tangent is x = 1
mradius =
7.
From y = mx + c find the y-intercept
2 = 4m + c
c = 2 − 4m ……………………………...…… (1)
y−c
Now make x the subject of formula: x =
…… (2)
m
0 − (2 − 4m )
x=
m
Find the x-intercept, i.e. subst. (1) in (2):
4m − 2
……………………………...………… (3)
x=
m
x y
Subst. (1) and (3) in + = 1
a b
x
y
+
=1
4m − 2 2 − 4m
m
8.
x 2 − 6 x + (x + 3) − 4( x + 3) + 5 = 0
2
x 2 − 6 x + x 2 + 6 x + 9 − 4 x − 12 + 5 = 0
2x 2 − 4x + 2 = 0
x 2 − 2x + 1 = 0
∴ x = 1 (one point of contact)
The line y = x + 3 is a tangent to the circle x 2 + y 2 - 6 x - 4 y + 5 = 0
1− 0 1
=
6−3 3
9. 1
m BC =
9-2
m AD = m BC =
1
……………………….AB//BC
3
∴ Equation of AD is:
1
y = x+c
3
1
6 = (1) + c
3
OR
17
c=
3
1
17
∴y = x+
3
3
1
y = ( x − 1) + 6
3
1
1
y = x− +6
3
3
1
17
y = x+
3
3
Page 61
9.3
9.4
1
17
x+
3
3
1
17
t = (7 ) +
=8
3
3
y=
t −6 1
=
7 −1 3
OR
t −6 = 2
⇒t =8
= 40 = 2 10 units
AD = (8 − 6) 2 + (−1 − 3) 2
BC = (6 − 3) 2 + (1 − 0) 2 = 10 units
AB = (6 − 0) 2 + (1 − 3) 2 = 40 = 2 10 units
9.5
9.6
OR
6−0
= −3
1− 3
1− 0 1
m BC =
=
6−3 3
1
m AB × m BC = × −3 = −1
3
∴ AB ⊥ BC
m AB =
Area of quad ABCD = area of ∆ADC + area of ABC
1
1
= 2 10 2 10 +
10 2 10 = 20 + 10 =30 square units
2
2
(
)(
) ( )(
)
1
Area of ABCD = (sum of parallel sides) × h
2
1
= (2 10 + 10 )2 10 = 10 (3 10) = 30 square units
2
9.7
m BC = tan θ =
∴ θ =18,43º
10.1
10.2
10.3
1
3
 − 8 + 0 1+ 5 
Midpoint AB 
;
 = (−4; 3)
2 
 2
5 +1 − 2
=
0−3
1
y =m(x− x 1 ) + y 1
y = −2(x − 0) + 5
y = −2x + 5
AM 2 = (5 − 3) 2 + (0 + 4) 2 = 2 2 + 4 2
M AD =
AM = 20 units
10.4
(x + 4)2 + ( y − 3)2 = ( 20 )
(x + 4)2 + ( y − 3)2 = 20
2
x 2 + y 2 + 8x − 6 y + 5 = 0
Page 62
AT = TK = 6
CD ⊥ AK
Diagonal CD bisects diagonal AK at right angles. Therefore, ACKD is a kite since.
OR
10.5
CAˆ D = 90°
M KC × M KD =
6
6
× = −1
− 12 3
∴ CKˆ D = 90°
∆CAD & ∆CKD are right angles & congruent. Therefore, ACKD is a kite
Page 63
5. TRIGONOMETRY
5.1 Defining the trigonometric ratios
The theorem of Pythagoras is key in Trigonometry.
If we know the lengths of two sides of a right-angled triangle, then the Pythagoras’ Theorem
allows us to find the length of the third side. (But remember it only works on right-angled
triangles!)
Example 5.1
If we know the values of the hypotenuse and a
side, then we calculate the magnitude of the
other side.
In this case, the values of b and c are known.
We need to find the value of b .
a=5
b2 = c2 − a 2
c = 13
b 2 = 132 − 52
b 2 = 169 − 25
b 2 = 144
b = 144
b = 12
b
1. Using the following triangle
c
a
b
Calculate: ( if two sides are given)
1.1
a if b = 21 and c = 29
1.2
b if a = 18 and c = 40
1.3
c if a = 12 and b = 16
1.4
a if b = 13 and c = 30
Page 64
2.
A ladder is placed against a wall as shown below. Determine the length of the ladder.
A
Note that AB = 3,2m ;
and ∆ABC is a rightangled triangle since
the wall and the
horizontal surface are
perpendicular to each
other.
ladder
3,2m
2,8m
B
C
‘Trigonometry’ - means ‘triangle measurement’
Trigonometry is an important tool for evaluating measurements of height and distance. It plays an
important role in surveying, navigation, engineering, astronomy and many other branches of
physical sciences.
Trigonometry involves the ratios of the sides of right triangles.
A
Hypotenuse
Opposite
B
θ
Adjacent
C
sin θ =
opposite
hypotenuse
cos θ =
adjacent
hypotenuse
tan θ =
opposite
adjacent
PRACTICAL PROBLEMS
PRE – KNOWLEDGE NEEDED:
•
•
•
Right-angled triangle
Adjacent; Opposite and Hypotenuse sides in a right-angled triangle
Pythagoras theorem
Page 65
New concepts:
•
•
Angle of elevation
Angle of depression
C Object
Angle of elevation
θ
β
B
A
Observer

Angle of depression
HINTS ON SOLVING TRIGONOMETRY APPLICATIONS OR PROBLEMS
•
•
•
•
•
•
•
If no diagram is given, draw one yourself.
Mark the right angles in the diagram.
Show the sizes of the other angles and the lengths of any lines that are known.
Mark the angles or sides you have to calculate.
Consider whether you need to create right triangles by drawing extra lines. For example,
divide an isosceles triangle into two congruent right triangles.
Decide whether you will need Pythagoras theorem, sine, cosine or tangent.
Check that your answer is reasonable. The hypotenuse is the longest side in a
rightangled triangle.
5.2 Defining ratios in the Cartesian plane
The Cartesian plane is divided into 4 quadrants by two coordinate axes. These 4 quadrants are
labelled 1, 2, 3 and 4 respectively.
y
A
All functions are positive
S
sine is positive
2
3
T
tangent is positive
1
4
x
C
cosine is positive
Page 66
When working with ratios on the Cartesian plane we will make use of the symbols x, y and r
y
(radius).
P(x; y )
y
sin θ =
r
r
y
x
cosθ =
r
θ
x
x
y
tan θ =
x
We can find any trigonometric functions if the co-ordinates of the terminal side is given.
90 0
90 0 ≤ θ ≤ 180 0
0 0 ≤ θ ≤ 90 0
+
1800
−180°
2
Angles are measured in this direction
1
3
00
360°
4
−
270 0 ≤ θ ≤ 360 0
180 0 ≤ θ ≤ 270 0
2700
− 90°
Example 5.2
If sin θ =
1.1
−4
and 90 0 ≤ θ ≤ 270 0 . Determine
5
cos 2 θ
1.2
5 sin θ − 3 cos θ
Solution
90 0
T
180 0
y = −4
y
θ
x=?
O
x
0 0 3600
r =5
P
270 0
Page 67
You must find first the length of the other side,
x2 + y2 = r 2
x 2 + (− 4 ) = (5)
2
2
x 2 = 25 − 16
Using Pythagoras:
x2 =9
x=±3
∴ x = − 3 ( x is negative in the 3rd quadrant).
The x-coordinate is – 3 while the length is 3.
1.1
1.2
cos 2 θ =  − 3  = 9
25
 5 
2
5 sin θ − 3 cos θ = 5 − 4  − 3 − 3 
 5 
 5 
9
=− 4 +
5
11
=−
5
=− 21
5
Example 5.2
If tan 127 0 = − P , express each of the following in terms of P.
2.1
sin 127 0
2.2
cos127 0
2.3
sin 2127 0
cos 2127 0
Solution
y
tan 1270 = P = ; Since x 2 + y 2 = r 2 ;
−1
x
2.1
2.2
2.3
sin 127 0 =
i.e (−1 )2 + P 2 = r 2
1 + P2 = r 2
1 + P2 = r
P
1+ P 2
1
cos127 0 = −
1+ P 2
sin 2 127 0
= tan 2 127 0 = P 2
2
0
cos 127
Page 68
5.3
USING SPECIAL ANGLES TO GET EXACT VALUES
30 0
3
1
x
y
45 0
1
1
2
r
60 0
1
3
2
2
(0 ;1)
y
90 0
60 0
45 0
2
2
1
1
45
30 0
0
1
(1;0)
180 0
(− 1; 0)
x
00 0
360
(0 ; − 1)
270 0
3
sin 0 0 = 0
cos 0 0 = 1
tan 0 0 = 0
sin 90 0 = 1
cos 90 0 = 0
tan 90 0 = undefined
sin 180 0 = 0
cos180 0 = − 1
tan 180 0 = 0
sin 270 0 = − 1
cos 270 0 = 0
tan 270 0 = undefined
sin 360 0 = 0
cos 360 0 = 1
tan 360 0 = 0
5.4 REDUCTION FORMULAE
 0 0 ≤ θ ≤ 360 0 use:
y
90°
360° − θ
180 0 − θ
x 0°
360° − θ
180° + θ
•
•
(
cos (180
)
± θ ) = − cos θ
sin 180 0 + θ = − sin θ
0
(
)
sin 180 0 − θ = sin θ
Page 69
•
•
•
(
tan (180
tan (360
)
− θ ) = − tan θ
− θ ) = − tan θ
(
tan (180
sin 360 0 − θ = − sin θ
0
0
 Co-functions:
(
)
+θ ) =
cos 360 0 − θ = cos θ
)
0
(
sin 90 0 ± x = cos x
tan θ
)
cos 90 0 + x = − sin x
cos (90° − x) = sin x
 θ > 360 : reduce angles to less than 360 0 by subtracting multiples of 360 0 .
0
(
)
e.g tan θ + k . 360 0 = tan θ
 θ < 0 0 : sin (− x) = − sin x ; cos (− x ) = cos x
tan (− x ) = − tan x
Examples 5.4
Simplify without using a calculator:
1.
cos (90° − x). sin (360° + x) cos 2(360° + x). tan 150°
+
sin (− 90°)
tan (− 150°)
2. sin 220° . cos130° −
cos 320° . cos 220°. tan 40°
tan 220°
Solutions
1.
cos (90° − x). sin (360° + x) cos 2(360° + x). tan 150°
+
sin (− 90°)
tan (−150°)
2
(sin x)(sin x) (cos x) tan(180° − 30°)
=
+
tan(−180° + 30°)
−1
2
cos x(− tan 30°)
= − sin 2 x +
tan 30°
= − sin 2 x − cos2 x
= −(sin 2 x + cos2 x)
= −1
2.
sin 220° . cos 130° −
cos 320° . cos 220°. tan 40°
tan 220°
cos(360° − 40°). cos(180° + 40°). tan 40°
= sin(180° + 40°). cos(90° + 40°) −
tan(180° + 40°)
(cos 40°)(− cos 40°)(tan 40°)
= (− sin 40°)(− sin 40°) −
tan 40°
2
2
= sin 40° + cos 40°
=1
Page 70
5.6 TRIGONOMETRIC IDENTITIES (work with the LHS and RHS separately)
•
•
•
Choose most difficult side and use identities to simplify it.
Look for square identities
Change functions to sin x and cos x
•
•
•
If there are fractions: Get LCM
Factorise or simplify if necessary
If the expression contains cos x − 1 and cannot be solved, then multiply above and below
by cos x + 1
•
NB:
See which value(s) of x the expression is not defined.
“hence” means that the previous answer must be used.
 Identities:
tan x =
sin x
;
cos x
sin 2 x + cos 2 x = 1
 Compound angles:
o cos (α + β ) = cos α . cos β − sin α . sin β
o
o
o
cos (α − β ) = cos α . cos β + sin α . sin β
sin (α + β) = sin α . cos β + sin β . cos α
sin (α − β) = sin α . cos β − sin β . cos α
 Doubles angles
o sin 2 A = 2 sin A . cos A
o
cos 2 A = 2 cos 2 A − 1
o
cos 2 A = 1 − 2 sin 2 A
o
cos 2 A = cos 2 A − sin 2 A
Examples
Prove the identity:
1 + sin x 1 − sin x 4 tan x
−
=
1 − sin x 1 + sin x
cos x
Hints:
•
•
•
Simplify the LHS
Get an LCD
Then simplify
Page 71
Exercise 5.1
1. If sin 230 = t , find the values of the following in terms of t .
0
1.1 sin(−23 )
0
1.2 sin 67
1.3 2 − cos2 670
1.4 tan 23° −
1
2 sin 67°
1.6 cos 8330 −
1
sin(−247°)
1.5
cos 460
cos134° − cos2 113° − 1
Simplify the following:
2.
sin(180° − x). cos(− x). tan(180° − x)
sin( x + 540°). cos(− x + 450°)
3.
sin(180° + x). cos(− x + 360°)
sin(180° − x). cos(90° + x) + cos(540° + x). cos(− x)
4. Show that :
4.1 sin 2 20° + sin 2 40° + sin 2 80° = 3
2
4.2 cos 75° =
(
)
2 3 −1
4
5.4 GENERAL SOLUTIONS
•
•
•
•
•
•
Write the equation on its own on one side of the equation
Start by simplifying an equation as far as possible. Use identities, double and compound
angle formulae, and factorisation where possible. You want to have one trig ratio and one
1
angle equal to a constant, for example cos x =
2
Find the reference angle
Identify the possible quadrants in which the terminal rays of the angles could be,based on
the sign of the function
Because trig functions are periodic, there will be a number of possible solutions to an
equation. You will need to write down the general solution of the equation.
Once the solution has been solved, write down the general solution by adding the
following:
 + k 360 0 for cosine and sine, because they repeat every 360 0
Page 72
 + k 180 0 for tangent, because it repeats every 180 0
•
If an equation contain double angles, for example, sin 3θ , cos 2 x and tan 5 y , find the
general solutions for 3θ , 2 x and 5 y first. Only then divide by 3, 2 or 5 to find the final
•
solutions. If you divide first, you will lose valid solutions.
Apply restrictions
Examples
Find the general solutions of the following:
1. 2 cos x sin x −cos x = 0
2. 15 sin 2 x − 7 cos x + 15 cos 2x − 2 = 0
Solution
1. 2 cos x sin x −cos x = 0
cos x(2 sin x − 1) = 0
cos x = 0 or 2 sin x = 1
x = cos−1(0) or sin x = 1
2
x = 90° or x = sin −1 1 
2
x = 30°
x = 90° + n.360°, n ∈ Z or
x = 270° + n.360°, n ∈ Z or
x = 30° + n.360°, n ∈ Z or
x = 150° + n.360°, n ∈ Z
Hint: Factorise; equate factors to zero; look for quadrants and find solution.
2.
15 sin 2 x − 7 cos x + 15 cos 2x − 2 = 0
15 sin 2 x − 7 cos x + 15(cos2 x − sin 2 x) − 2 = 0
15 sin 2 x − 7 cos x + 15 cos2 x − 15 sin 2 x − 2 = 0
15 cos2 x − 7 cos x − 2 = 0
(5 cos x + 1)(3 cos x − 2) = 0
cos x = − 1 or cos x = 2
5
3
Page 73
x = cos−1 − 1  or x = cos−1 2 
 3
 5
x = 101,54° or x = 48,19°
x = 48,19° + n.360°, n ∈ Z or
x = 311,81° + n.360°, n ∈ Z or
x = 101,54° + n.360°, n ∈ Z or
x = 258,46° + n.360°, n ∈ Z
Exercise 5.2
Find the general solutions of the following:
1. 1 = −2
cos x
2. 5 cos2 2θ − 4 = 0
3. 4 sin x − 2 = 0
4. (2 sin x − 1)(sin x − 1) = 0
5. 2 cos x − 4 cos x − 1 + 2 = 0
sin x
cos x
6. 4 cos2 x = 1 − 2 cos x sin x
7. 5 cos2 2θ − 4 = 0
Page 74
5.5
SOLVING TRIANGLES THAT ARE NOT RIGHT – ANGLED
2D AND 3D PROBLEMS
1. PROBLEMS IN TWO DIMENSIONS
• Problems in two dimensions usually require that you solve a combination of triangles in
one plane ( a flat surface )
• It is easier to work with right angled triangles, so identify them first.
• The triangles usually share a common side, which you can then use in the second
triangle.
Sine Rule:
•
•
a
sin Aˆ
=
b
sin Bˆ
=
c
sin Cˆ
sin Aˆ sin Bˆ sin Cˆ
=
=
c
a
b
OR
The first arrangement rule is recommended to calculate a side, and,
The second arrangement is recommended to calculate an angle.
B
a
c
C
b
A
•
Use the sine rule for triangles for which you are given:
 Two angles and a side
 Two sides and the non – included angle.
Page 75
Cosine rule
•
There are different arrangements of the cosine rule depending on what you are trying to
find:
 For calculating a side, use a 2 = b 2 + c 2 − 2 b c cos Aˆ OR
b2 + c2 − a2
 For calculating the included angle, use cos Aˆ =
2b c
B
c
A
•
a
C
b
Use the cosine rule when you have been given:
 Three sides
 Two sides and the included angle.
Area rule
•
Area ∆ ABC =
1
1
1
a b sin Cˆ = b c sin Aˆ = a c sin Bˆ
2
2
2
A
b
C
c
a
B
2. PROBLEMS IN THREE DIMENSIONS
• Three-dimensional problems use the same principles as two – dimensional ones. However,
triangles are oriented in two planes (horizontal and vertical) and the difficulty can be
visualising the situation.
• Draw a sketch of the situation if you have not been given one, and shade the horizontal
plane to make it easier to see.
• Fill in all the given information and fill in every possible angle
• Look out for right –angled triangles as this is often the place you will start. Use trig ratios
and Pythagoras to solve these triangles.
Page 76
•
•
•
If there are no right-angled triangles, start with the triangle with the most given
information.
Look for common sides so that you use one triangle to help you to solve another side.
Use the correct rule for the given situation and the correct arrangement of that rule.
Exercise 5.3
A
1. In the diagram, BC is the diameter of circle BCD.
α
Dˆ = 90 0
1
B
BCˆ D = θ
θ
2
C
p
ACˆ B = α
D
AB = BC
BD = p units
1.1 Express BC in term of p and θ .
1.2 Determine, without stating reasons, the size of B̂1 in terms of α .
1.3 Hence, prove that AC =
p . sin 2α
sin θ . sin α
A
2. AB is a vertical lighthouse 100m high with B
at sea level. At a point P the captain of a
fishing boat measures the angle of the angle
100
of elevation of A from P to be 16 0 .
Travelling in a straight line from P to Q ,
B
the captain now measures the angle of
elevation of A from Q
to be 14 0 .
He also determines that PBˆ Q = 110 0 .
14
0
1100
160
P
Q
2.1 Calculate to the nearest metre, how far from B the boat is at P and Q .
2.2 If the boat took 20 minutes to travel from P to Q , calculate, to the nearest whole
number, the average speed of the boat in km per hour.
2.3 Calculate the area of triangle PBQ
Page 77