Ch 14 Trigonometry I
GHS Past Paper Question Bank – Conventional Question
Page 1 of 6
Trigonometry 1
Conventional Questions
1.
[00-01 Standardized test2-5]
Solve the equation 3 cos θ – 2 sin2θ =0, where 0≤θ≤360o.
2.
[00‐01 Final Exam-3]
In an acute‐angled ABC, simplify
3.
sin( A  B ) cos( A  B )
.
tan( 270   C ) sin( 360   C )
[4 marks]
[00‐01 Final Exam-7]
Find the points of intersection of y = sin(x + 30) and y = cos(x + 30) for 0  x <360
4.
(5 mark)
.[6 marks]
[01‐02 Final Exam-7]
Given that g(  ) = 3 sin (180  )  tan  . Find  if g(  ) = 0, where 0    360 .
5.
(Correct youranswers to 1 decimal place)
[02‐03 Final Exam-6]
6.
[03-04 Standardized test 2-2]
7.
[03-04 Standardized test 2-4]
Form 4
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[5 marks]
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Ch 14 Trigonometry I
8.
GHS Past Paper Question Bank – Conventional Question
Page 2 of 6
[03-04 Final Exam-5]
If cos  
2 sin   5 cos 
5
where 180    270 . Find
.
13
2  3 sin 
(3 marks)
9. [03-04 Final Exam-10]
sin(180  x) tan(360  x) cos(90  x)
(a) Simplify
.

sin(270  x)
cos(180  x)
(b) Hence, solve
sin(180  x) tan(360  x) cos(90  x)
= 2 for 0  x  360 . (6 marks)

sin(270  x)
cos(180  x)
10. [04-05 Standardized test 2-1]
Form 4
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Ch 14 Trigonometry I
GHS Past Paper Question Bank – Conventional Question
Page 3 of 6
11. [04-05 Standardized test 2-2]
12. [04-05 Standardized test 2-5]
Solve cos 2   2 cos  sin   sin 2   0 where 0    360 .
(3 marks)
13. [04-05 Standardized test 2-11]
Let f (  )  5 sin 2   7 cos   1 for 90    180 .
(a) Rewrite f (  ) in the form a cos 2   b cos   c where a, b and c are integers.
(1 mark)
(b) If f (  )  0 , find the value of
(3 marks)
(i) cos  and
(4 marks)
(ii) cos( 180   )  tan( 270   )  sin( 360   ) .
14. [05-06 Standardized test 2-1]
The figure shows the graph of y = acosx + b sinx for 0º  x  360º.
Using the graph, solve the equation -bsinx – acosx – 3 = 0 for 0º  x  360º.
(correct your answer to nearest degree)
[4 marks]
15. [05-06 Standardized test 2-2]
tan 2   1
 2sin 2   1
tan 2   1
16. [05-06 Standardized test 2-3]
. Prove
Solve the equation 2sin2  + sin cos – cos 2  = 0 where 0    180
(correct your answer to 3 sig. fig. when necessary)
[4 marks]
[4 marks]
17. [05-06 Final Exam-5]
Solve the equation 2 cos 2   3 sin   3  0 for 0     360 .
Form 4
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(4 marks)
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Ch 14 Trigonometry I
GHS Past Paper Question Bank – Conventional Question
Page 4 of 6
18. [06-07 Final Exam-7]
In figure 5,
(a) prove that ABC is a right‐angled triangle;
(b) hence, without solving , find the value of
sin(180   ) tan(90   ) .
(3 marks)
A

(2 marks)
17
B
15
8
19. [06-07 Final Exam-10]
C
Figure 5
cos 2 x
1
(a) Express

in the form a sin 2 x  b sin x  c  0 where a, b and c are
13 sin x  4 10
integers.
(2 marks)
(b) Solve the equation in part (a) for 0  x  360 .
(c) Figure 6 shows the graph of y 
y
(3 marks)
2
cos x
for 0  x  360 .
13 sin x  4
2
y=
1.5
cos x
13sinx – 4
1
0.5
50
100
150
200
250
300
350
x (in degree)
– 0.5
– 1
– 1.5
Figure 6
Using the graph, solve the equation 10 cos 2 x  39 sin x  12 for 0  x  360 .
(3 marks)
20. [0708 Mock ‐11]
Given that tan (180   ) 
1  5 sin (270   )
,
sin 
(a) rewrite the above equation in the form a cos 2   b cos   c  0 where a, b and c are
integers ;
(4 marks)
(b) hence solve the given equation for 0    360 .
(3 marks)
Form 4
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Ch 14 Trigonometry I
GHS Past Paper Question Bank – Conventional Question
Page 5 of 6
21 [0809 Mock ‐6]
2 sin x  3 cos x 1
 , where 0  x  360 , giving the answers correct to the
sin x  2 cos x 3
Solve the equation
nearest degree.
(3 marks)
22. [0809 Mock ‐12]
Figure 5 shows the graph of y  a sin x  b cos x  1 for 0  x  90 .
y
3
y = a sinx + b cosx – 1
2.5
2
1.5
1
0.5
– 10 º 0
– 0.5
10
o
20
o
30
o
40
o
50
o
60
o
70
o
80
o
90
o
o
100
x
(a)
Find the values of a and b.
(3 marks)
(b) By adding a suitable line on the graph, solve the equation a sin x  b cos x  3 , where
0  x  90 .
(2 marks)
(c) (i) When the graph of y  a sin x  b cos x  1 is transformed to become
y  a sin( x  10)  b cos( x  10)  2 , describe the transformation(s) involved.
(ii) On Figure 5, sketch the graph for y  a sin( x  10)  b cos( x  10)  2 for
10  x  90.
(2 marks)
23. . [09-10 Standardized test 1-3]
sin(180   )
(a) Show that
 sin(360   ) cos(90   )  sin 2 (90   ) ;
(3 marks)
cos(90   )
(b) Hence, find the maximum and minimun values of

 sin(180   )
2
 sin(360   ) cos(90   )   1 .
 cos(90   )

(2 marks)
24. [09-10 Mock ‐6]
(a)
Prove that
(b)
sin  cos   cos 2 
cos   sin 
2
Hence solve
2

1
1  tan 
sin  cos   cos 2 
cos 2   sin 2 
.
2
(2 marks)
where 0    360 .
(1 mark)
25. [09-10 Mock ‐7]
Form 4
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Ch 14 Trigonometry I
GHS Past Paper Question Bank – Conventional Question
It is given that ABC is an acute‐angled triangle.
(a) Show that cos( A  B )   cos C .
cos( A  B) sin C
(b) Simplify
and express the answer in terms of C.
tan(360  C )
Form 4
GHS Past Paper Question Bank – Conventional Question
Page 6 of 6
(1 mark)
(2 marks)
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