Chapter 7 : Trigonometry Mathematics
SM015
Q1 2004 /2005 t1 @ t2
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
1t2 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
1t2 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
1t2 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 11 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 42 3
64 3
6 b
2 3 2 3
(b) Find R and such that the expression 32 3 2 3
9sin x + 12cos x can be expressed in form
of R sin (x + ) , where R > 0, 0 < < 43
90. Hence, if 9 sin x+ 12cos x = 5, solve 63 34 36
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 11 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
13
R 225
42 3 R 15
2
R sin 12
2 3 R cos 9
R tan 12
9
tan112 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 12 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 2xcos4x cos2x
2 sin 2x2 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 1sin x 1 0 sin 2xcos x4cos3x 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 1cot x 2 0 R2 (1) 25
R5
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 tan1 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 tan1 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)
2cos3cos2 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 tan1 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