Chapter (5)
Trigonometry
73
Trigonometric Identities
a
b
a
 cos    tan  
c
c
b
sin a b a c a

      tan 
cos  c c c b b
sin

 tan   (1)
cos 
 sin 
 a 2  b 2  c 2   c 2 
c
a

b
 a 2   b2 
a 2 b2 c 2
 2    2   1  sin 2  cos 2  1



c2 c2 c2
c  c 
 sin 2   cos 2   1  ( 2)
 sin 2   1  cos 2   ( 2  1)
,
 cos 2   1  sin 2   ( 2  2)
Examples:
1] Prove that (sin  cos ) 2  (sin  cos ) 2  2
sin 2   2 sin cos  cos 2   sin 2   2 sin cos  cos 2  
2 sin 2   2 cos 2   2(sin 2   cos 2  )  2  1  2  proved
1
2] If x  2 sin , y  cos prove that x 2  36 y 2  4
3
2
 1
 
1

x  36 y  ( 2 sin )  36  cos  )    4 sin 2   36 cos 2  
 
9

 3
2
2
2
 4 sin 2   4 cos 2   4(sin 2   cos 2  )  4  1  4  proved
3] Show that the equation sin  cos  2(sin  cos ) can be written as tan  3
sin  cos   2 sin  2 cos   cos
sin cos  2 sin 2 cos 



 tan   1  2 tan   2  2 tan   tan   1  2
cos  cos 
cos 
cos 
 tan   3  proved
74
 1

4] Prove that 1  sin  
 tan    cos
 cos

L. H . S 
1
sin
1
sin
 tan  
 sin tan  
 tan   tan   sin 

cos
cos
cos
cos
1
sin 2  1  sin 2  cos 2 



 cos  R. H . S
cos cos
cos
cos
5] Show that sin 2   cos  7  0 can be written as a quadratic equation in cos only.
sin 2   cos  7  0  1  cos 2   cos  7  0   cos 2  cos  8  0
 cos 2   cos  8  0
6] If a  2sin  cos and b  2 cos  sin , show that a 2  b 2 is constant for all values of 
a 2  ( 2 sin  cos ) 2  4 sin 2   4 sin cos  cos 2 
b 2  ( 2 cos  sin ) 2  4 cos 2   4 sin cos  sin 2 
a 2  b 2  4 sin 2   4 sin cos  cos 2   4 cos 2   4 sin cos  sin 2 
a 2  b 2  5 sin 2   5 cos 2   5(sin 2   cos 2  )  5  1  5  constant
7] Prove the identity
1  tan 2 
 1  2 sin 2 
1  tan 2 
 sin 
sin 2 
cos 2   sin 2 
1 

1
cos 2   sin 2  1  sin 2   sin 2 
 cos  
cos 2  
cos 2 
L. H . S 



2
1
sin 2 
cos 2   sin 2  cos 2   sin 2 
 sin 
1
1 

cos 2 
cos 2 
 cos 
2
1  2 sin 2   R. H . s
8] Prove that
L. H . S 
sin x
sin x

 2 tan 2 x
1  sin x 1  sin x
sin x(1  sin x )  sin x(1  sin x ) sin x  sin 2 x  sin x  sin 2 x 2 sin 2 x


 2 tan 2 x  R. H . S
2
2
(1  sin x )(1  sin x )
1  sin x
cos x
75
 2
9] Given that x  sin 1   , find the exact values of:
 5
i) cos 2 x
ii) tan 2 x
2
2
4 21
 2
 2
 x  sin    sin x   cos 2 x  1  sin 2 x  cos 2  1     1 

5
25 25
 5
 5
4
2
2
sin
x
sin
x
4
4


 tan 2 x  
 25 
 tan 2 x 
 
2
21
cos x 21 21
 cos x 
25
1
Exercise
1] Prove that
1  cos 2 
 sin tan 
cos
 1

2] Prove the identity 1  sin 
 tan    cos
 cos

3] If a  2sin  cos and b  2 cos  sin ,
i) show that a 2  b 2 is constant for all values of 
ii) Given that 2a = 3b , show that tan 
4
7
4] Show that the equation 3 tan  2 cos can be written as 2 sin 2   3 sin  2  0
1  tan 2 
 1  2 sin 2 
5] Prove the identity
2
1  tan 
6] Show that the equation 3 sin x tan x  8 can be written as 3 cos 2 x  8 cos x  3  0
7] Prove the identity
1  sin x
cos x
2


cos x
1  sin x cos x
8] Show that the equation 3( 2 sin x  cos x )  2(sin x  3 cos x ) can be written in the form tan x  
9] Show that the equation sin  cos  2(sin  cos ) can be written as tan  3
10] Prove the following identity:
a) (1  tan 2 x )(1  sin 2 x )  1
b) (1  tan 2 x ) cos 2 x  1
76
3
4
Trigonometric Graphs
* y  sin   It is periodic, with a period of   360 (2 rad )
,
 1  sin  1
* y  cos   It is periodic, with a period of   360 (2 rad )
,
 1  cos   1
* y  tan   It is periodic, with a period of   180 ( rad )
77
,
   tan   
Examples:
1] Sketch the graph of y  2  sin  for 0    2
4
3

0

2

y
2
3
2
3
2
1
2
2
1
2
0
2] Sketch the graph of y  sin 2  for 0    180
4
 2  90   45
3

0
45
90
1
y
0
1
0

3
2
2
45
90
135
180
180
360
540
720
90
180
270
360
2
135 180
-1

2
0
-1
-2
1
3] Sketch the graph of y  2  sin   for 0    720
2
4
1
   90   180
2
2

0
-1
y
2
1
180 360 540 720
1
2
3
3
-2
2
7
4] Sketch the graph of y  5  2 cos   for 0    360
6
5

0
90
y
3
5
180 270 360
7
5
4
3
3
2
1
-1
78
5] i) Sketch and label on the same diagram the graphs of y  2 sin x and
y  cos 2 x for
0 x
ii) Hence state the number of solutions of the equation 2 sin x  cos 2 x  in the interval 0  x  
i)
x
0
90
180
y
0
2
0
4
y  2 sin x
3
2
1
x
0
45
90
y
1
0
-1
y  cos 2 x
135 180
0
-1

4

2
3
4

1
ii) 2 sin x  cos 2 x has two solution.

6] The function f is defined by f ( x )  a  b cos 2 x for 0  x   . It is given that f ( 0)  1 and f ( )  7 . i)
2
Find the value of a and b.
ii) Sketch the graph of y  f ( x )
i)  f (0)  1  1  a  b cos(2  0)  1  a  b  (1)

7

 f ( )  7  7  a  b cos(2  )  7  a  b cos( )  7  a  b  ( 2)
2
2
Add (1) and (2)  a  3 and b  4
6
5
4
3
ii) y  3  4 cos 2 x
x
0
45
90
y
-1
3
7
2
135 180
3
1
-1
-1
7] The diagram shows the graph of
y  a sin( bx )  c for 0  x  2
Find the values of a , b and c
9
6
3

-3
-6
79
2

4

2
3
4

 We have two periods of sine in 2  b  2
Use point (0 , 3) in y  a sin( 2 x )  c  3  a sin( 2  0)  c  0  c  3  c  3
Use point (

4
, 9) in y  a sin( 2 x )  3  9  a sin( 2 

4
)  3  a  3  9  a  6
8] The function f is defined by f ( x )  a  b cos x for 0  x  360 , where a and b are positive constants.
The maximum value of f(x) is 10 and the minimum value is -2.
ii) Sketch the graph of y  f ( x )
i) Find the value of a and b.
i) The maximum value of f ( x ) when cos x is minimum  minimum value of cos x  1
 10  a  b( 1)  a  b  10  (1)
The minimum value of f ( x ) when cos x is maximum  maximum value of cos x  1
 2  a  b(1)  a  b  2  (1)
Add (1) and (2)  a  4 and
b6
ii) y  4  6 cos x
x
0
90
180 270 360
y
-2
4
10
4
10
8
-2
6
4
2
-2
90
180
270
360
-4
Exercise
Sketch each of the following graph:
1] y  3 cos x for 0  x  360
2] y  sin 2 x for 0  x  360
3] y  2 sin x  1 for 0  x  360
4] y  3  cos 2 x for 0  x  360
5] i) Sketch and label on the same diagram the graphs of y  2 cos x and
y  sin 3 x for
0  x  2
ii) Hence state the number of solutions of the equation 2 cos x  sin 3 x  in the interval 0  x  2
80
Solving Trigonometric Equations
ASTC Rule:
+ ve
sin ve
180
90
90   
  180  
  360  
  180  
All  ve
A
S
0 , 360
T
180
C
0 , 360
T
cos ve
tan  ve
A
S
  180  
  180  
270
C
270
360  

- ve
Examples:
1] If sin 
1
, find the possible value of  in the interval 0    360
2
  180  
1
1
Let sin      sin 1  30
2
2
  30 and   180  30  150
S
A
T
C
 
2] If tan   3 , find the values of  for 0    360
Let tan   3    tan
1
3  60
  60 and   180  60  240
3] If cos  
 
S
A
T
C
  180  
1
, find the possible value of  for 0    360
2
  180  
1
1
Let cos      cos 1  60
2
2
  180  60  120 and   180  60  240
S
A
T
C
  180  
Note:  is an acute angle, so all its trigonometry functions are positive.
81
4] If tan  1 , find the possible value of  for 0    2
Let tan   1    tan 1 1  45
  180  
   180  45  135 and   360  45  315
3
7
    rad and    rad
4
4
5] Find the values of x for which sin x  
Let sin  
T
C
2
in the interval  720  x  720
2
+ ve
x  360  45  315
S
A
T
C
x  180  
We add 360
 x  225  360  585 and
A
  360  
2
2
   sin 1
 45
2
2
 x  180  45  225 and
S
x  360  
x  360  315  675
 x  180  45  135 and
A
T
C
- ve
x  45
We subtract 360
 x  135  360  495 and
S
x  180  
x  
x  45  360  405
6] Solve the equation sin   cos for 0    2
sin
cos 
 sin   cos    cos  

 tan   1
cos 
cos 
Let tan   1    tan 1 1  45
S
A
T
C
  360  
   180  45  135 and   360  45  315
3
7
    rad and    rad
4
4
7] Solve the equation 2 sin 2   3 sin  1  0 for 0    360
(2 sin  1)(sin   1)  0 sin 
  180  
1
and sin  1
2
82
Let sin  
1
1
   sin 1  30
2
2
 
  180  
  30 and   180  30  150
S
A
Let sin   1    sin 1 1  90
T
C
  90
8] i) Prove that the equation sin 2   3 sin  cos   4 cos 2  can be written as a quadratic equation
in tan  .
ii) Hence, solve the equation sin 2   3 sin  cos   4 cos 2  for 0    180
i)  sin 2   3 sin cos   4 cos 2    cos 2  
sin 2  3 sin cos  4 cos 2 


cos 2 
cos 2 
cos 2 
 tan 2   3 tan   4  tan 2   3 tan   4  0
ii) tan 2   3 tan   4  0  (tan   4)(tan   1)  0
 tan   4 and tan   1
  180  
Let tan   4    tan 1 4  76
S
A
T
C
  360  
   180  76  104 and   360  76  284  rejected
 
Let tan   1    tan 1  45
S
A
   45 and   180  45  225  rejected
T
C
  180  
9] i) Show that the equation tan xsin x can be written in the form
1  cos 2 x
cos x
ii) Hence, solve the equation 2 tan x sin x  3 for 0  x  360
i) tan x sin x 
sin x
sin 2 x 1  cos 2 x
(sin x ) 

cos x
cos x
cos x
 1  cos 2 x 
  3  2  2 cos 2 x  3 cos x  2 cos 2 x  3 cos x  2  0
ii)  2 tan x sin x  3  2
 cos x 
83
 (2 cos x  1)(cos x  2)  0 cos x 
Let cos  
1
and cos x  2  rejected
2
x 
1
1
   cos 1  60
2
2
 x  60 and
S
A
T
C
x  360  
x  360  60  300
10] Solve the equation sin 2 
1
for 0    360
2
Let 0  2  720 and sin  
1
1
   sin 1  30
2
2
 
  180  
 2  30 , 2  180  30  150
S
A
We add 360
T
C
 2  360  30  390 , 2  360  150  510
   15 ,   75 ,   195 ,   255
11] Solve cos(  45)  
2
for 0    360
2
  180  
Let 45    45  405 and cos  
2
2
   cos 1
 45
2
2
   45  180  45  135 ,   45  180  45  225
   135  45  90 ,   225  45  180
sin 2   sin  0  sin (sin   1)  0 sin  0 and sin  1
and
 
A
T
C
  180  
2
12] Solve sin 2   sin  , 0    2
  0 ,  , 2
S
1
-1

-2
2
84

2

3
2
2
13] Solve 8 sin 2   1  3 , for 0    360
8 sin 2   4  sin 2  
Let sin  
1
2
4 1
  sin 
8 2
   sin 1
1
2
1
2
and sin  
1
2
  180  
 45
S
T
  45 and   180  45  135
  180  
  360  45  225 and   360  45  315
Exercise
1] Solve the following equations:
a) sin  0.5 , for 0    360
b)
2 cos  1 , for      
c) tan 2   3 , for 0    180
2] Solve the equations:
a) sin x  0.2 , for 0  x  360
b) cos 2 x  0.7 , for 0  x  360
3] Solve the following equations for 0  x  360
a)
3 sin x  cos x  0
b) 2 cos 2 x  3 sin x  3
4] a) Given that sin  
3
and  is obtuse. Find the values of tan  and cos 
5
b) Given that cos A  
5
and 180  A  270 . Find the values of sin A and tan A
13
………………………………………………………
85
 
A
C
  360  
General Exercise
86
87
88
89
26.
27.
………………………………………………………
90