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REVISING TRIGONOMETRY
24 AUGUST 2015
Section B: Exam practice questions
Question 1
If 13sin  5  0
where
 90 ; 360 and 5cos   4  0 where tan   0, calculate without the
use of a calculator and with the aid of diagrams the value of the following:
1.1
sin(  )
(9)
1.2
tan(90  )
(5)
Question 2
Simplify without using a calculator:
2.1
cos(50  x) cos(20  x)  sin(50  x) sin(20  x)
(3)
2.2
cos(140) cos 740  sin140 sin(20)
(6)
Question 3
Show that:
sin 2 20  sin 2 40  sin 2 80 
3
2
(7)
Question 4
4.1
4.2
Prove:
sin 4 x  sin 2 x cos2 x
 1  cos x
1  cos x
For which values of x is
sin 4 x  sin 2 x cos2 x
 1  cos x
1  cos x
(4)
not true?
(2)
Question 5
A, B and L are points in the same horizontal plane, HL is a vertical pole of length 3 metres,
AL = 5,2 m, the angle
5.1
5.2
5.3
AL̂B  113 ° and the angle of elevation of H from B is 40.
Calculate the length of LB.
Hence, or otherwise, calculate the length of AB.
Determine the area of ABL.
(2)
(4)
(4)
H
3
m
5,2
mm
L
113
40
B
m
A
Question 6
Right-angled triangle ABC is drawn. Equilateral triangles ACD, BCE and ABF are drawn on each side
as shown.
Prove that:
Area ADC  Area ABF  Area CBE
(7)
Section C: Solutions
1.1
(x ; 5)
 diagram for 
13
5

13sin   5  0
5
 sin  
(positive in Quad 1 or 2)
13
  90 ; 360 (Quad 2, 3 or 4)
 sin  
5
13
Common quad is 2
x 2  (5) 2  (13) 2
 x 2  144
 x  12
 x  12
4

 diagram for 
5
(  4; y)
5cos   4  0
4
 cos  
(negative in Quad 2 or 3)
5
tan   0 (Quad 1 or 3)
Common quad is 3
 cos  
(  4)2  y 2  (5)2
 y2  9
 y  3
 y  3
4
5
sin(  )
 sin  cos   cos  sin 
 5  4   12  3 
     
 
 13  5   13  5 
20 36


65 65
56

65
 sin  cos   cos  sin 
 correct substitution
 answer
(9)
1.2
tan(90  )
sin(90  )

cos(90  )
cos 

 sin 
12
 13
5

13
12

5
sin(90  )
 cos(90  )
 cos 
  sin 
 correct substitution
 answer
(5)
2.1
cos(50  x) cos(20  x)  sin(50  x) sin(20  x)
cos (50  x)  (20  x) 

 cos30
3
 2
(3)
  cos 40
 cos 20
 sin 40
  sin 20
  cos60
1

 2
(6)
 cos  (50  x)  (20  x) 
 cos 30
3
2
cos(140) cos 740  sin140 sin(20)

2.2
 cos140 cos 20  (sin 40)( sin 20)
 ( cos 40)(cos 20)  sin 40 sin 20
  cos 40 cos 20  sin 40 sin 20
 (cos 40 cos 20  sin 40 sin 20)
  cos 60
1

2
3
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
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


 sin 2 20 

2
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
 sin 2 20 
6 cos 2 20  2sin 2 20
4
4sin 2 20  6 cos 2 20  2sin 2 20

4
6(sin 2 20  cos 2 20)

4
6(1) 3


4
2
(7)
4.1
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 

4.2
5.1
5.2
1  cos x  0
x  180  k.360
(2)
3
 tan 40
LB
3
 LB 
tan 40
 LB  3,58 m
 definition
 answer
AB2  AL2  BL2  2.AL.BL.cos113
 cosine rule
 substitution
 answer
 AB2  (5, 2) 2  (3,58) 2  2(5, 2)(3,58) cos113
5.3


 AB2  54, 40410138 m 2
 AB  7,38 m
1
ˆ
Area of ABL  AL.BL.sin ALB
2
1
 (5, 2)(3,58) sin113
2
 8.568059176
 8,57 m
(3)
(4)
 area rule
 substitution
 answer
(4)
6.
Area ADC 
1
(b)(b) sin 60
2
1
3
3b 2
 Area ADC  b 2

2
2
4
Area ABF 
3c 2
4
Area CBE 
3a 2
4
3c 2
3a 2

4
4
3 2
 Area ABF  Area CBE 
(c  a 2 )
4
 Area ABF  Area CBE 
3 2
3b 2
 Area ABF  Area CBE 
(b ) 
4
4
 Area ADC  Area ABF  Area CBE
 area rule
 equal sides
 60
3b 2
4
3c 2
 Area ABF 
4
 Area ADC 
3a 2
 Area CBE 
4
 use of Pythagoras to
establish identity
(7)