GAUTENG DEPARTMENT OF EDUCATION
MATHEMATICS
SENIOR SECONDARY INTERVENTION PROGRAMME
GRADE 12
SESSION 22
(LEARNER HOMEWORK SOLUTIONS)
SOLUTIONS TO HOMEWORK: SESSION 22
TOPIC: CONSOLIDATION – EXAM PREPARATION PAPER 2 (B)
QUESTION 10
10.1.1
sin 48  sin(36  12)
 sin 36 cos12  cos 36 sin12
 pq
 36  12
 expansion
 answer
10.1.2
sin 24  sin(36  12)
 sin 36 cos12  cos 36 sin12
 pq
 36 12
 expansion
 answer
10.1.3
sin 48  2sin 24 cos 24
(3)
cos 24  1  sin 24
2
2
 p  q  2( p  q ) cos 24 OR  cos 24  1  sin 2 24

10.2
pq
2 p  2q
(3)
 identity
 substitution
 answer
(3)
 cos 24  1  ( p  q ) 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 1 of 4
GAUTENG DEPARTMENT OF EDUCATION
MATHEMATICS
GRADE 12
 sin 2 20 

SENIOR SECONDARY INTERVENTION PROGRAMME
SESSION 22
(LEARNER HOMEWORK SOLUTIONS)
3cos 2 20  2 3 cos 20 sin 20  sin 2 20
4
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
sin 2 x(sin 2 x  cos 2 x)

1  cos x




factorisation
sin 2 x  cos 2 x  1
1  cos 2 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
 cos x  1
 x  180  k .360
where k 

10.3.2


1  cos x  0
x  180  k.360
(2)
[22]
Page 2 of 4
GAUTENG DEPARTMENT OF EDUCATION
MATHEMATICS
GRADE 12
SENIOR SECONDARY INTERVENTION PROGRAMME
SESSION 22
(LEARNER HOMEWORK SOLUTIONS)
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
[7]
11.2
2 y
1
x
0
180
225
270
315
360
–1
–2
Page 3 of 4
GAUTENG DEPARTMENT OF EDUCATION
MATHEMATICS
11.2
11.3
GRADE 12
SENIOR SECONDARY INTERVENTION PROGRAMME
SESSION 22
see diagram
(LEARNER HOMEWORK SOLUTIONS)
For


For


f ( x)  g ( x)
180  x  210 or 330  x  360
f ( x)  1  sin x
max and min values
shape
g ( x)  cos 2x
amplitude
intercepts
 180  x  210
 330  x  360
 inequality signs
correct
(4)
(3)
[7]
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)
[7]
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