MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
EXAM PREPARATION PAPER 2 (D)
Learner Note: In this session you will be given the opportunity to work on a past
examination paper (Feb/Mar 2010 DoE Paper 2). The paper consists of 12 questions.
Question 10, 11 and 12 will be dealt with in this session.
Structure of Paper 2
TOPIC
APPROXIMATE MARK ALLOCATION
Coordinate Geometry
40
Transformations
25
Trigonometry
60
Data Handling
25
TOTAL
150
Before writing the maths examination, it is essential for you to have at least two blue pens, a
pencil, a ruler and a workable scientific calculator (non-programmable and non-graphical). You
will be given time to read through the paper prior to starting. It is essential that you read the
instructions carefully and ensure that the paper is complete.
Don’t waste time on a specific question. If you cannot do that question, rather move on to the
next question. Remember to number the questions correctly. You might wish to start with
Question 3. This is fine, provided that the question is clearly indicated.
SECTION A: TYPICAL EXAM QUESTIONS (QUESTION 10, 11 AND 12)
Important information regarding Question 10, 11 and 12
Question 10
This question tests trigonometry. Learners must have mastered the basics before attempting
this question.
Question 11
Trig graphs are tested in this question. The four main transformations must be known and
learners must be aware that at most two transformations can be applied to one given function.
Page 1 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
Question 12
Make sure learners know when to use the sine, cosine and areas rules.

The sine-rule can be used when the following is known in the triangle:
- more than 1 angle and a side
- 2 sides and an angle (not included)
sin A sin B sin C


a
b
c

The cosine-rule can be used when the following is known of the triangle:
- 3 sides
- 2 sides and an included angle
a 2  b2  c2  2bc cos A

The area of any triangle can be found when at least two sides an included angle are
known
Area of ABC 
1
ab sin C
2
QUESTION 10
10.1
10.2
10.3
If sin 36 cos12  p and cos36 sin12  q , determine in terms of p and q the value of:
10.1.1
sin 48
(3)
10.1.2
sin 24
(3)
10.1.3
cos 24
(3)
Show that:
sin 2 20  sin 2 40  sin 2 80 
3
2
10.3.1
sin 4 x  sin 2 x cos2 x
 1  cos x
Prove:
1  cos x
10.3.2
For which values of x is
(7)
(4)
sin 4 x  sin 2 x cos2 x
 1  cos x not true?
1  cos x
(2)
QUESTION 11
Given:
f ( x)  1  sin x and g ( x)  cos 2 x
11.1
Calculate the points of intersection of the graphs for x  180 ; 360
(7)
11.2
Draw sketch graphs of f and g for x  180 ; 360 on the same set of axes.
(4)
11.3
For which values of x will f ( x)  g ( x) for x  180 ; 360
(3)
Page 2 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
QUESTION 12
In the diagram below A, B and C are three points in the same horizontal plane. D is vertically
above B and E is vertically above C. The angle of elevation of E from D is θ. F is a point on EC
such that DF || BC.
ˆ  , ACB
ˆ   and AC  b metres
BAC
12.1
Prove that: DE 
b sin 
sin(  )cos 
(6)
12.2
Calculate DE if b  2000 metres,   43 and   27
(3)
Page 3 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
SECTION B: SOLUTIONS AND HINTS TO SESSION 40
QUESTION 7
7.1.1
7.1.2
P / (5 ;  2)
 answer
P / (5 ; 2)
 x-coordinate
 y-coordinate
(1)
(2)
7.2.1
7.2.2
7.2.3
Reduction by a scale factor of
1
:
2
1 1 
( x ; y)   x ; y 
2 2 
Reflection about the line y  x
1 1  1 1 
 x ; y   y ; x
2 2  2 2 
1 1 
 x ; y    y ; x 
2 2 
If the first transformation is the reflection, then:
H / (8 ;16)
If the first transformation is the reduction, then:
H / (8 ; 4)
Area of original 1
 2
Area of image
k
Area KUHLE
1


4
// // // // //
2
Area K U H L E
1
 
2
 reduction
 reflection
(4)
 H / (8 ;16)
 H / (8 ; 4)
(2)
 answer
(2)
 Area KUHLE": Area K // U // H // L// E //  4 :1
Page 4 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
QUESTION 8
8.1
A / ( x cos   y sin  ; y cos   x sin )
x /  x cos   y sin 
 x /  3cos120  2sin120
 x /  3( cos 60)  2sin 60
 formula
 simplification
 substitution
 answer
 simplification
 answer
1  3
 x  3    2 

 2   2 
(6)
/
3  2 3
2
/
y  y cos   x sin 
 x/ 
 x /  2 cos120  3sin120
 y /  2( cos 60)  3sin 60
1  3
 y /  2    3 

 2   2 
2  3 3
2
Q(x ; y )  Q / ( 2 ; 0)
 y/ 
8.2
 Q(x ; y )  Q /  x cos120  y sin120 ; y cos120  x sin120 
2  x cos120  y sin120 and 0  y cos120  x sin120
 3
 1
2  x     y 

 2
 2 
and
 3
 1
0  y     x 

 2
 2 
4   x  3 y
and
0   y  3x
4  x  3y
and
y  3x
 3
 1
 2  x     y 

 2
 2 
 3
 1
 0  y     x 

 2
 2 
 x-coordinate
 y-coordinate
(4)
 4  x  3( 3 x)
 4  x  3x
 4  4x
x 1
 y  3(1)  3
 Q(1; 3)
Page 5 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
QUESTION 9
9.1.1
3
4
and cos   
5
5
7
 sin   cos   
5
sin   
 correct quad and
values
3
5
4
 cos   
5
 sin   
 answer
(4)
9.1.2
9.2.1
sin 2
2sin  cos 

cos 2 cos 2   sin 2 
 3  4 
2     
5
5  24
  2

2
7
 4  3



 

 5  5
tan 2 
cos(360  x) tan x
sin( x  180) cos(90  x)
2
(cos x) tan 2 x

sin  (180  x)  (  sin x)
(cos x) tan 2 x

 sin(180  x)( sin x)
sin 2
cos 2
 2sin  cos 
2
2
 cos   sin 

 substitution
 answer
(5)
 (cos x)
 ( sin x)
 ( sin x)
sin 2 x

cos 2 x
 answer
(5)
 sin x 
(cos x) 

cos 2 x 


( sin x)( sin x)
2
9.2.2
sin 2 x
 cos2 x
sin x
sin 2 x
1

 2
cos x sin x
1

cos x
Let x  30

1
1
2


cos 30
3
3
2
 x  30
 answer
(2)
Page 6 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
SECTION B: SOLUTIONS AND HINTS TO SESSION 41
10.1.1
10.1.2
sin 48  sin(36  12)
 sin 36 cos12  cos 36 sin12
 36  12
 expansion
 answer
 pq
sin 24  sin(36  12)
 36  12
 expansion
 answer
 sin 36 cos12  cos 36 sin12
10.1.3
 pq
sin 48  2sin 24 cos 24
(3)
(3)
 identity
 substitution
 answer
 p  q  2( p  q ) cos 24
pq
2 p  2q
OR
cos 2 24  1  sin 2 24
(3)

 cos 24  1  sin 2 24
 cos 24  1  ( p  q) 2
10.2
sin 2 (60  20)
sin 2 20  sin 2 40  sin 2 80

 sin 2 20  sin 2 (60  20)  sin 2 (60  20)
 sin 2 (60  20)
 expansions
 substitution
 simplification
 answer
 sin 2 20  sin 60 cos 20  cos 60 sin 20
2
 sin 60 cos 20  cos 60 sin 20
2
 3 

1
 sin 20  
 cos 20    sin 20 
2
 2 

2
(7)
2
 3 

1
 
 cos 20    sin 20
2
 2 

2
 3 cos 20  sin 20 
 sin 20  

2


2
2
 3 cos 20  sin 20 


2


2
Page 7 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
3cos 2 20  2 3 cos 20 sin 20  sin 2 20
 sin 20 
4
2

3cos 2 20  2 3 cos 20 sin 20  sin 2 20
4
6 cos 2 20  2sin 2 20
 sin 20 
4
2

4sin 2 20  6 cos 2 20  2sin 2 20
4
6(sin 2 20  cos 2 20)

4
6(1) 3


4
2
10.3.1
sin 4 x  sin 2 x cos 2 x
1  cos x

factorisation

sin 2 x  cos2 x  1
sin 2 x(sin 2 x  cos 2 x)

1  cos x


1  cos2 x
(1  cos x)(1  cos x)
(4)
(sin 2 x)(1)

1  cos x
1  cos 2 x
1  cos x
(1  cos x)(1  cos x)

1  cos x
 1  cos x
1  cos x  0

10.3.2
 cos x  1


1  cos x  0
x  180  k.360
(2)
 x  180  k .360
where k 
Page 8 of 10
MATRIC REVISION
MATHEMATICS
GRADE 12
SESSION 41
(LEARNER NOTES)
QUESTION 11
11.1
1  sin x  cos 2 x





1  sin x  1  2sin 2 x
 sin x  2sin 2 x  0
 sin x(1  2sin x)  0
 sin x  0 or
sin x  
1  2sin 2 x
sin x(1  2sin x)  0
two equations
general solutions
answers
(7)
1
2
For sin x  0 :
x  0  k 360
or
x  180  k 360
1
For sin x   :
2
x  30  k 360
or
x  210  k 360
 x  180 ; 210 ; 330 ; 360
11.2
2 y
1
x
0
180
225
270
315
360
–1
–2
Page 9 of 10
MATRIC REVISION
MATHEMATICS
11.2
11.3
GRADE 12
SESSION 41
see diagram
f ( x)  g ( x)
180  x  210 or 330  x  360
(LEARNER NOTES)
For


For


f ( x)  1  sin x
max and min values
shape
g ( x)  cos 2 x
amplitude
intercepts
 180  x  210
 330  x  360
 inequality signs
correct
(4)
(3)
QUESTION 12
12.1
b
BC

sin 180  (  )  sin 
b
BC

sin(  ) sin 
b sin 
 BC 
sin(  )
But BC  DF

b sin 
sin(  )
DF
Now cos  
DE
DF
 DE 
cos 
b sin 
 DE 
sin(  ) cos 
2000sin 43
DE 
sin 79 cos 27
 DE  1559,50m
 sine rule
 180  (  )
 sin(  )
b sin 
 BC 
sin(  )
 BC  DF
 manipulation
(4)
 DF 
12.2
 substitution numerator
 substitution
denominator
 answer
(3)
Page 10 of 10