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Published by Magdalane Lim Hwee Cheng, 2019-08-20 02:47:57

Chapter 7 Trigonometry

Chapter 7 _Trigonometry

Chapter 7 : Trigonometry Mathematics
SM015

Q1 2004 /2005 t1 @ t2
3
If tan x   t , find sin x and cos x in terms of t .
2 When t  1 ,  tan x   1 ( I , III)
3 2 3
Hence, solve cos x  7sin x  5for 0  x   .
Since 0  x  
tan x   t , 2 tan x  2t 0  x   ( I quadrant only)
2  2  1 t 22
Given tan x  tan 2  x   2
Thus, x  0.322 radian
1 2 2
x  0.644 radian
1 t 2
2t When t  2 ,  tan x   1 ( I , III)
2 3
x
1t2 Since 0  x  
0  x   ( I quadrant only)
Based on above triangle, 22
sin x  2t
1 t2 Thus, x  1.107 radian
cos x  1 t 2 2
1 t 2 x  2.214 radian

Given cos x  7sin x  5 Therefore, x  0.644 rad , 2.214 rad

1t2  7 2t 2   5 Q2 2005 /2006
1 t2  t  Show that tan(A  B)  tan A  tan B
1
1 tan Atan B
1 t 2 14t  5(1 t 2 )
From the compound angles,
6t 2 14t  4  0 sin( A  B)  sin Acos B  cos Asin B
cos(A  B)  cos Acos B  sin Asin B
3t 2  7t  2  0
(3t 1)(t  2)  0

1

Chapter 7 : Trigonometry Mathematics
SM015

sin x  2t
1 t2

tan(A  B)  sin( A  B) cos x  1 t 2
cos(A  B) 1 t 2

 sin Acos B  cos Asin B Thus, 3cos x  4sin x  5
cos Acos B  sin Asin B
3 1  t 2   4 2t 2   5
sin Acos B  cos Asin B 1  t 2 1 t 
 cos Acos B cos Acos B
3  3t 2  8t  5  5t 2
cos Acos B  sin Asin B
cos Acos B cos Acos B 8t 2  8t  2  0
 tan A  tan B shown.
1 tan Atan B 4t 2  4t 1  0

Q3 2006 /2007 (2t 1)2  0
If tan x   t , find sin x and cos x in terms of t .
2t 1  0
2
Hence, find all values of t which satisfy t  1
3cos x  4sin x  5 . 2

2 tan x  Q4 2007 /2008
Show that cos6x  cos 2x(4cos2 2x  3)

Given tan x   t , tan x  2    2t
2 tan 2  x 1 t 2
1
2
cos 6x  cos(2x  4x)
1 t 2  cos 2x cos 4x  sin 2x sin 4x
2t  cos 2x(2 cos2 2x 1)  sin 2x(2sin 2x cos 2x)
 cos 2x(2 cos2 2x 1)  2sin2 2x cos 2x
x  cos 2x(2 cos2 2x 1)  2(1 cos2 2x) cos 2x
1t2  cos 2x(2 cos2 2x 1 2  2 os2 2x)
 cos 2x(2 cos2 2x  3) shown
Based on above triangle,

2

Chapter 7 : Trigonometry Mathematics
SM015

Q5 2007 /2008 1 tan2   sec2  shown
(a) Let P(x, y) be a point on a unit circle with (b) Given cos x(sin x  cos x)  1  0

center O at the origin, such that OP make cos x(sin x  cos x)  1
an acute angle  with the positive x-axis.
Prove that sin 2   cos2  1and hence (sin x  cos x)  1
cos x
show that sec2  1  tan2  .
Divided both sides with cos x
(b) Show that the equation sin x  cos x  1
cos x(sin x  cos x)  1  0 can be reduced cos x cos x cos2 x
to tan x(1  tan x)  0 . Hence, solve for x
tan x 1  sec2 x
on the interval 0,2 .
tan x 1 1 tan2 x
(a)
P(x,y) tan x  tan2 x  0
tan x(1 tan x)  0 Shown
1y
 Hence,
x
tan x  0 or tan x  1
From the diagram, cos  x  x
1 x  0, ,2 x   , 5
44
sin  y  y
1 Thus, the solution x  0,  , , 5 ,2
44
Equation of unit circle, x2  y2  1
Q6 2008/2009
cos2   sin 2   1 shown. Find A and B if sin 2x cos3x  Asin 5x  B sin x

Divide both sides with cos2  From compound angle,
cos2   sin 2   1 sin( A  B)  sin Acos B  cos Asin B      (1)
cos2  cos2  cos2 
sin( A  B)  sin Acos B  cos Asin B      (2)

Eq(1) + Eq(2)
sin( A  B)  sin( A  B)  2sin Acos B

sin Acos B  1 sin( A  B)  sin( A  B)

2
Compare sin Acos B  sin 2xcos3x

3

Chapter 7 : Trigonometry Mathematics
SM015

A  2x and B  3x (a) cos 3x  cos(2x  x)
 cos 2x cos x  sin 2x sin x
sin 2x cos3x  1 sin(2x  3x)  sin(2x  3x)  (2 cos2 x 1) cos x  2 sin x cos x sin x
 2 cos3 x  cos x  2 sin 2 x cos x
2  2 cos3 x  cos x  2(1 cos2 x) cos x
 2 cos3 x  cos x  2 cos3 x  2 cos x
 1 sin 5x  sin( x)  4 cos3 x  3 cos x proved

2 From cos3x  4 cos3 x  3cos x

 1 sin 5x  sin x 4cos3 x  cos3x  3cos x

2 cos3 x  1 cos3x  3cos x
 1 sin 5x  1 sin x
4
22
Thus, if x = 2x
Compare with Asin 5x  Bsin x
cos3 2x  1 cos3(2)x  3cos(2x)
Thus A  1 and B  1
22 4

Q7 2009/2010  1 cos6x  3cos2x shown
Prove that for  n , where n is an integer,
4
sin  sec 4
cos (1  2sin 2   sin 4  )  tan  .

sin   sin  (b) Given 2cos 3x + cos 2x + 1 = 0
cos (1  2 sin 2   sin 4  ) cos (1  sin 2  )2 2(4cos3 x  3cos x)  cos 2x 1  0
8cos3 x  6cos x  cos 2x 1  0
 sin   1
cos cos4  8cos3 x  6cos x  2cos2 x 11  0
8cos3 x  2cos2 x  6cos x  0
 tan sec4  proved cos x(4cos x  3)(cos x 1)  0

Q8 2010/2011 cos x  0  x  90, 90

(a) Prove that cos3x  4 cos3 x  3cos x . Hence, cos x  3  x  41.4, 41.4
4
show that cos3 2x  1 cos6x  3cos2x
cos x  1  x  180
4
 x  90,  41.4, 41.4, 90,180o
(b) Use the above identity to find all the solutions
in the interval 180  x 180. 180 of the
equation 2cos3x  cos2x 1  0 .

4

Chapter 7 : Trigonometry Mathematics
SM015

Q9 2011/2012 tan 7 p  tan 2 7 p 
6  12 
(a) Given tan p  3 and tan p  1 2 tan 7 p
34  12
1   tan 7 p 2
Express tan 7 p in the form of a  b  12 
12  2(2  3)
1  (2  3)2
where a and b are integers. Hence, show
that tan 7 p   1  42 3
64 3
6 b
 2 3 2 3
(b) Find R and  such that the expression 32 3 2 3
9sin x + 12cos x can be expressed in form
of R sin (x + ) , where R > 0, 0 <  <  43
90. Hence, if 9 sin x+ 12cos x = 5, solve 63 34 36
for in interval 0  x 360.
 1 shown
(a) 3

tan 7 p  tan p  p  (b) 9 sin x +12 cos x = R sin (x+ α)
12  3 4 

tan p  tan p = R sin x cos α + R cos x sin α
3 4

1 tan p tan p By comparing

34

 3 11 3 R cos  9 R sin   12
1 3 1 3
R 2 (cos2   sin 2  )  81  144

 3 2 3 1 R 2  225
13
R  225
 42 3 R  15
2
R sin   12
 2  3 R cos 9

R tan  12
9

  tan112   53.13
9

9 sin   12 cos  15 sin(  53.13)

5

Chapter 7 : Trigonometry Mathematics
SM015

Given 1
cos 2x
9sin  12 cos  5
15sin(  53.13)  5  sec2x shown

sin(  53.13)  1 (I & II ) Q11 2012/2013
3
Given f    3sin   2cos .
19.47o 19.47o   sin 1 1  (a) Express f   in the form of Rsin  
3
where R > 0, 0    
 19.47o 2

Given 0  x 360 Hence, find the maximum and minimum

53.13o  x + 53.13o  413.13 values of f  .

x  53.13  19.47o ,180o 19.47o ,360o 19.47o (b) Solve f    13 for 0    360
x  33.66o ,160.53, 379.47
x  33.66o ,107.40, 326.34 2

Thus, x  107.40o ,326.34o (a) f    3sin  2cos  Rsin( )

Q10 2012/2013 3sin  2cos  Rsin cos  Rcos sin 
Prove that 1 tan2x tan x  sec2x
By comparing
From 1 tan 2x tan x  1  sin 2x  sin x  R cos  3      (1)
 cos 2x  cos x  R sin   2      (2)
R2 (cos2   sin 2  )  9  4
 cos2x cos x  sin 2xsin x
cos2x cos x R2  13
R  13
 cos(2x  x) Rsin   2
cos2x cos x R cos 3
tan  2
 cos x 3
cos2x cos x   tan 1 2   33.69o
3

6

Chapter 7 : Trigonometry Mathematics
SM015

 3sin   2cos  13 sin   33.69o for 0o    180o .
Given 3sin   6 cos  R sin(   )
f ( ) to be maximum when sin(  33.69o )  1
f ( )  13(1)  13 (maximum) 3sin  6cos  Rsin cos  Rcos sin 
f ( ) to be minimum when sin(  33.69o )  1
f ( )  13(1)   13 (minimum) Comparing

(b) Given f    13 R cos  3      (1)

2 R sin   6      (2)

 13 sin   33.69o  13 R2 (cos2   sin 2  )  9  36
2 R2  45
R  45
 sin   33.69o  1 ( I & II)
2 Rsin   6
R cos 3
  sin 1 1 
 2 tan  2

45o 45o  45o   tan 12  63.43o

0    360 Therefore 3sin   6cos  45 sin(  63.43o )
-33.69    33.69o  326.31
  33.69o  45o ,135o To solve
  78.69o ,168.69o 3sin   6cos  5
45 sin(  63.43o )  5
sin(  63.43o )  5
45
sin(  63.43o )  1 ( I & II)
3

Q12 2013/2014 19.47o 19.47o   sin 1 1 
3
State the values of R and  such that
3sin   6 cos  R sin(   ) where R > 0 and  19.47o
0o    90o. Hence, solve
3sin   6cos  5 0    180

7

Chapter 7 : Trigonometry Mathematics
SM015

63.43   + 63.43o  243.43 cos    cos( )  cos
 2 sin( )  sin 
  63.43o  19.47o ,180 19.47o      
  63.43o  19.47o ,160.52o  sin  2
  43.96, 97.10o
cos    
  97.10o  2 
sin   
 

Q13 2013/2014 2

(a) Show that sin   sin   cot     cot     shown
cos  cos   2  2

(b) Use trigonometric identities to verify that

 (b) (i) From half angle

2 tan 2 sin   cos  
2 sin    2   2 
(i) sin   tan2 
1
1 2

1 tan2  2sin   cos  
2  2 2
(ii) cos  
cos2     sin 2   
1 tan2 2 2 2

Hence, solve the equation 3sin  cos  2 for Divide both denominator and numerator with
0o    180o . Give your answer correct to three cos2   
decimal places.
2

(a) From Thus,

sin   sin  2sin     cos     2sin  cos  
2 2 2 2

 cos2    
cos  cos   2sin    sin    
2 2 tan 2
  sin 2   2
2 2 sin   cos2    

2 2 1  tan 2

cos     cos2   
 2  2
 sin    

 2

8

Chapter 7 : Trigonometry Mathematics
SM015

b(ii) From half angle

To solve 3sin  cos  2

cos2     sin 2     2 tan   1 tan2  
cos   2  2 3 2   2 
 tan2     1     2

1 1  2 tan 2 2

cos2     sin 2    6 tan 1 tan2   2(1 tan2  )
 2 2

cos2     sin 2    22 2

2 2 3tan2   6 tan 1  0
22
Divide both denominator and numerator with
cos2    By using quadratic formula, we get

2

Thus, tan  1.8164 or tan   0.1835
22

cos2     sin 2    Since given 0o    180o
2 2 0o    90o ( I quadrant)
2
cos2   
2
cos  cos2     sin 2   

2 2 Thus, for tan  1.8164
2
cos2   
2

sin 2      61.165
 2  2
1  cos2      122.3o
For tan   0.1835
 2 2
sin 2      10.398
1  2  2
cos2      20.8o
2

1 tan2 
2
  shown

1  tan2 2

 The solution is   20.8o ,122.3o

9

Chapter 7 : Trigonometry Mathematics
SM015

Q14 2014/2015 sin 6x  sin 2x  sin 4x

Solve the equation 2cos2 x 1  sin x , for  2cos 4x sin 2x  sin 2(2x)
0  x  2 . Give your answer in term of  .
 2cos4xsin 2x  2sin 2xcos2x

 2sin 2xcos4x  cos2x

 2 sin 2x2 cos 4x  2x  cos 4x  2x 
 2  2
2cos2 x 1  sin x
 4cos3xsin 2xcos x

2(1 sin 2 x) 1  sin x (b) sin 6x  sin 2x  sin 4x  4sin 2xcos x
4cos3xsin 2xcos x  sin 2xcos x
2sin 2 x  sin x 1  0 4cos3xsin 2xcos x  sin 2xcos x  0

2sin x 1sin x 1  0 sin 2xcos x4cos3x 1  0

sin x  1 or sin x  1 sin 2x  0 @ cos x  0 @ cos3x  1
2 4

x   ,   x  3 When sin 2x  0
66 2 2x  0o ,180o ,360o
x  0o ,90o ,180o
x   , 5
66 When cos x  0
x  90o
Thus, the solution x   , 5 , 3
66 2 When cos3x  1
4
Q15 2014/2015
3x  75.52o ,284.48o ,435.52o
(a) Express sin 6x  sin 2x in a product form. x  25.17o ,94.83o ,145.17o
Hence, show that
sin 6x  sin 2x  sin 4x  4cos3xsin 2xcos x . Therefore, the final solutions are

(b) Use the result in (a) to solve x  0o ,25.17o ,90o ,94.83o ,145.17o ,180o
sin 6x  sin 2x  sin 4x  4sin 2xcos x

for 0o  x  180o

(a) sin 6x  sin 2x  2cos 6x  2x sin 6x  2x  Q16 2015/2016
 2  2
Given cos ec 2 x  cot x  3 , show that
 2cos4xsin 2x cot 2 x  cot x  2  0. Hence, solve the equation

cos ec 2 x  cot x  3 for 0  x   .

From cos ec 2 x  cot x  3
1  cot 2 x  cot x  3  0
cot 2 x  cot x  2  0 shown

10

Chapter 7 : Trigonometry Mathematics
SM015

To solve cos ec 2 x  cot x  3 R 2 cos2   R 2 sin 2   9  16

cot 2 x  cot x  2  0  R2 cos2   sin 2   25

cot x 1cot x  2  0 R2 (1)  25
R5
cot x  1 or cot x  2
R sin   4
1  1 1 2 R cos 3
tan x tan x tan  4

tan x  1 tan x  1 3
2   tan1 4   53.13o

Given 0  x   ( I & II Quadrant only) 3

For tan x  1  3sin  4cos  5sin   53.13o

x  3 (b) To solve 3sin   4cos  2
4
 5sin   53.13o  2
tan x  1  sin   53.13o  0.4 (I & II)
For 2
  53.13o  23.58o ,156.42o
x  0.148   76.71o ,209.55o

Therefore, the solutions are Q18 2016/2017
Show that sin 2 x  1 cos x . Hence, solve
x  0.148 , 3 
4 1 cos x
sin 2 x  cos 2x for 0o    360 o .
Q17 2015/2016 1 cos x
From sin 2 x  1  cos2 x
(a) Determine the values of R and , where
R > 0 and 0o    90o so that 1  cos x 1  cos x
 (1 cos x)(1 cos x)
3sin  4cos  Rsin . 1 cos x
 1 cos x shown
(b) Hence, solve the equation
11
3sin   4cos  2 for 0o    360 o

(a)

3sin  4cos  Rsin 

 Rsin cos  Rcos sin 

By comparing,
R cos  3    (1)
R sin   4    (2)

Chapter 7 : Trigonometry Mathematics
SM015

To solve sin 2 x  cos 2x By comparing,
1 cos x R cos  3    (1)
1 cos x  cos2x R sin   2    (2)
1  cos x  2 cos2 x  1
2 cos2 x  cos x  2  0 R 2 cos2   R 2 sin 2   3  4

By using quadratic formula,  R2 cos2   sin 2   7

 (1)  (1)2  4(2)(2) R 2 (1)  7
2(2)
R 7
 1.281 @  0.781
R sin   2
Thus cos x  1.281 @ cos x  0.781 R cos 3

For 0o    360 o tan  2
3
cos x  0.781 (II & III)
x  141.35o @ x  218.65o   tan1 2   49.10 o  49o6'
3
Q19 2016/2017
Consider a function f (x)  3 cos 2x  2sin 2x .   3 cos 2x  2sin 2x  7 cos 2x  49o.6'
(a) Express f in the form of R cos(2x   ) for
f (x) to be maximum when cos(2x  49o6')  1
R > 0 and 0o    90o and  to the nearest The maximum value of f (x)  7(1)  7
minute. State the maximum and minimum R
values of f. f (x) to be minimum when cos(2x  49o6')  1
(b) Hence, solve 3 cos2x  2sin 2x   2 for The minimum value of f (x)  7(1)   7
0o  x  180o . Give your answer to the
nearest minute. (b) 3 cos2x  2sin 2x   2

 7 cos 2x  49.1o.   2

 cos 2x  49.1o   2
7

(a) 3 cos 2x  2sin 2x  R cos(2x  )
 R cos2x cos  Rsin 2xsin 

12

Chapter 7 : Trigonometry Mathematics
SM015

57.69o For cos 2  1
57.69o
2  0,2
  0,

2x  49.1o'  122.31o , 237.69o Q21 2017/2018
2x  171.41o ,286.79o
x  85.7o ,143.4o Express cos  2 sin in the form of
x  85o24' , 143o 24' R sin(   ) , where R > 0 and  is an acute angle.

(Answer in nearest minutes) Hence,
(a) Solve the equation cos  2 sin   3 by
Q20 2017/2018
2
Solve the equation cos  cos5  2cos3 for giving all the solutions between 0o and 360o.
0     . Give your answer in the term of .
(b) Show the greatest value of
1  5 3 .

cos  2 sin   5 22

cos  cos5  2cos3

2 cos  5  cos  5   2 cos3  0 cos  2 sin   Rsin   
2 2
 Rsin cos  R cos sin 

2cos3 cos2  2cos3  0 By comparing,
R cos  2    (1)
2cos3cos2 1  0 R sin   1    (2)

2cos3  0 @ cos2 1  0 R 2 cos2   R 2 sin 2   2  1

cos3  0 @ cos 2  1  R2 cos2   sin 2   3

For cos3  0 R 2 (1)  3

R 3

3   , 3 , 5 R sin   1
222 R cos 2
tan  1
    , 5
, 2
62 6   tan1 1   35.3o

 2

13

Chapter 7 : Trigonometry Mathematics
SM015

 Thus cos  2 sin   3 sin   35.3o  5 3
25  3
(a) cos  2 sin   3
2  5  3 shown.
22
 3 sin   35.3o  3
2
 sin   35.3o  1 ( I & II )
2

Given 0o  x  360 o
35.3o  x  395.3o

30o 30o

  35.3o  180o  30o , 360o  30o
  35.3o  150o ,390o
  114.7o ,354.7o

(b) 1  1
cos  2 sin   5 3 sin(  35.3o )  5

The greatest value ( maximum ) of
1 occurs when

cos  2 sin   5

 sin   35.3o  1

 1 1
3 sin(  35.3o )  5 3(1)  5
 1 5 3
5 3 5 3

14


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