Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Page 1Page 1 of 42

Trigonometric
Functions
Learning Outcome
 Sketch and use graphs of the sine, cosine and tangent functions
 Use the exact values of the sine, cosine and tangent of 30°, 45°, 60°
 Use the notations sin−1x, cos−1x, tan−1x to denote the principal values of the inverse
trigonometric relations
 Find all the solutions of a trigonometry equation lying in a specified interval
 Find the general solution of a trigonometry equation
 Use properties and graphs of all six trigonometric functions for angles of any
magnitude
 Use trigonometry identities and formulas for the simplification and exact evaluation of
expressions and in the course of solving equations, and select an identity or identities
appropriate to the context, showing familiarity in particular with the use of:
sin 

≡ tan θ and sin2θ + cos2θ ≡ 1
cos 


sec2 θ ≡ 1 + tan2 θ and cosec2 θ ≡ 1 + cot2 θ



the expansions of sin(A ± B), cos(A ± B) and tan(A ± B)



the formulae for sin 2A, cos 2A and tan 2A



the expressions of a sin θ + b cos θ in the forms R sin(θ ± α) and R cos(θ ± α)

,

Contents
 Concepts & Formulas
 Examples
 Practice Problems
 Suggested Solution to All Practice Problems

1
...
1 INTRODUCTION
Radian

The angle subtended at centre of a circle by an arc of length equal to the radius of the circle is 1
radian
1o=

r
r

r



radian

180 o

180 o

1 radian =



1o = 60’

Positive &
Negative
Angles

A positive angle is measured in an
anticlockwise direction

A negative angle is measured
clockwise direction

in a

1
...
3 GRAPHING TRIGONOMETRY FUNCTIONS
Terminology

The graphs of sin, cosine and tangent are known as periodic since they repeat
themselves over and over again
...
The fixed interval is
called the period
...
The
period of the graph is 3600
...

2
2
y R

3
,
...
0

Examples

EXAMPLE 1
Convert the following angles in degree to radians in terms of π:
a)
300
b)
600
c)
1200
d)
1750
e)
2000
Suggested Solution
a)

30 0 

30   
 radians
180
6

b)

600 

60   
 radians
180
3

c)

1200 

120   2

radians
180
3

d)

1750 

175   35

radians
180
36

e)

2000 

200   10

radians
180
9

EXAMPLE 2
Convert the following angles in radians to degrees ( Take π=3
...
5 radians
b)
2
...
82π radians
d)
1
...
19π radians
Suggested Solution
1
...
93 0
3
...
93  60 '  56 ' )

a)

1
...
82 radians 

e)

2
...
05 0
3
...
19 radians 

1
...
26 0
3
...
34 radians 

d)

1
...
67 180
 95
...
142
 95 o 40 '

2
...
46 0
3
...
5)

c)

d)

 7(0
...
3090)
  6
...
5

EXAMPLE 5

EXAMPLE 4
If A is an acute angle, and sin A 

 7 cos 223 0  5 sin 198 0

1
, find the
2

exact values of sin 2A

Given that sin x 

12
4
, sin y  and both the
13
5

angles x and y are in the first quadrant, find the
values of cos(y-x)

Suggested Solution
Suggested Solution

If sin A 

1
3
1
 cos A 
and tan A 
2
2
3

 sin 2 A
 2 sin A cos A
 1  3 

 2 
 2  2 




3
2

Trigonometry Functions

If sin x 

12
5
12
 cos x  and tan x 
13
13
5

If sin y 

4
3
4
 cos y  and tan y 
5
5
3

 cos( y  x)
 cos y cos x  sin y sin x
3 5 4 12

...
5 for 0°≤ x ≤360°},
B = {x : cos (x – 30°) = – 0
...

Find the elements of
(i) A,
(ii) A B

EXAMPLE 7
Given that tan  6 , find the exact value of
tan  45
Suggested Solution

tan   tan 45
1  tan  tan 45
6 1

1 6
5

7

tan  45 

Suggested Solution

cos( x  30)  0
...
5
x  30, 150

x  150, 270
(i)

A = {300, 1500}

(ii) A B={300, 1500, 2700 }

EXAMPLE 8

EXAMPLE 9

Given that sin x = p and cos x = 2p, where x is
acute, find the exact value of p and the exact value
of cosec x
...


Method 1

Method 2

sin x
p 1
 tan x 

cos x
2p 2

sin x  cos x  1

Suggested Solution
2

2

p 2  2 p   1
2

5 p2  1
1
5
1
p
5

p2 

From the triangle,
1
sin x  p 
5
1
p
5

cosec x 

1
1
1
 
1
sin x p

sin  

1
3

2
3

sin 
cos   sin 
1
3

2
 1
3
3
1
3


3
2 3 3





 5 ( Answer )
5

 cos  



3
( 3 )( 6  3 )

( 3 )( 6  3 )
3( 6  3)
3( 6  3)
18  3

3
3
9 2 3
 2 1
3

 1  2 ( Answer )

Trigonometry Functions

Page 8

EXAMPLE 10

EXAMPLE 11
If sin x  t , and   x   radians, express in

Solve the equation sin 2 x  2 cos 2 x , for
0 0  x  180 0
...
4

o

Suggested Solution
a)

or 243
...
7 or 121
...

b) 3 sin x tan x  8 can be written as 3 cos 2 x  8 cos x  3  0
Suggested Solution
a)  2 tan 2  cos   3
2

sin 2 
cos   3
cos 2 
 2(1  cos 2  )  3 cos 

 3 sin x tan x  8
sin x
 3 sin x
8
cos x
 3 sin 2 x  8 cos x

 2  2 cos 2   3 cos 

 3(1  cos 2 x)  8 cos x

 2 cos 2   3 cos   2  0

 3  3 cos 2 x  8 cos x

b)

 3 cos 2 x  8 cos x  3  0
Trigonometry Functions

Page 9

EXAMPLE 15
Solve each of the following equations giving all solutions in the given interval
a)
2 tan 2 x  2, for 0 0  x  180 0
...


Suggested Solution
a)

b)

2 tan 2 x  2
tan 2 x  1

3 cos x  1
1
cos x 
3
1
x  cos 1
3
0
x  54
...
5 0 , 112
...
2 sin A cos A
cos A
 2 sin 2 A
 1  cos 2 A ( Shown)
Note : cos 2 A  1  2 sin 2 A

c)

b)

1  sin x
cos 2 x
1  sin x

1  sin 2 x
1  sin x

(1  sin x)(1  sin x)
1

( Shown)
1  sin x


1  tan 2 x
1  tan 2 x
sin 2 x
1
cos 2 x

sin 2 x
1
cos 2 x
cos 2 x  sin 2 x
cos 2 x


...


Solve the equation 2 cos 2 x  sin x  1  0 , giving
all solutions in the interval 0 0  x  360 0

 

Suggested Solution

Suggested Solution

3 3 sin x  cos x
sin x
1

cos x 3 3
1
tan x 
3 3

Since range of x :
 180 0  x  180 0
00  x  1800

 1800  x  00

x  48
...
6)
x  48
...
4

x = 48
...
40, 2700

EXAMPLE 19

EXAMPLE 20

Solve the equation







3 sin x  70  8 cos x  70  7  0 ,
for 0 0  x  180 0
o



Show that

tan  





4t 1  t 2
1  6t 2  t 4

, where t  tan
...
75

x  10
...
10



 4 sin 2 x  2  sin x  1  0

4 sin x  3

x  10
...
9)

0

2(1  2 sin 2 x)  sin x  1  0
4 sin 2 x  sin x  3  0
(4 sin x  3)(sin x  1)  0

 1 

x  tan 1 


3 3
x  10
...

1  t 2  1  t 2 1  t 2  4t 2 


2
4t 1  t

1  2t 2  t 4  4t 2
4t 1  t 2

( Shown)
1  6t 2  t 4





3 sin 2 x  70 0  8 cos x  70 o  7  0
Let   x  70 0
 3 sin 2   8 cos   7  0
 3(1  cos 2  )  8 cos   7  0
 3  3 cos 2   8 cos   7  0
 3 cos 2   8 cos   4  0
 3 cos 2   8 cos   4  0
 (3 cos   2)(cos   2)  0
2
 cos    OR cos   2 (N/A)
3
   131
...
2





Hence,
x    70
x  131
...
2-70





 
 



x  61
...
2 0 ( Answer )

Note:

Use Compound-Angle Formula
tan( A  B) 

Trigonometry Functions

tan A  tan B
1  tan A tan B

Page 11

EXAMPLE 21

EXAMPLE 22

Solve the following equation giving all solutions
in the given interval , cos 3x  cos x  0 , for
0 0  x  180 0
...

Suggested Solution

Suggested Solution

cos 3x  cos x  0
2 cos

3x  x
3x  1
cos
0
2
2
2 cos 2 x cos x  0

2 sin 3 x  2 sin 2 x  sin x  1  0
2 sin 2 xsin x  1  sin x  1  0
(sin x  1)(2 sin 2 x  1)  0

2 cos x(2 cos 2 x  1)  0
2 cos x  0
2 cos 2 x  1  0
cos 2 x  0

x  90 0

2 x  90 0 ,270 0

sin x  1
x  90

sin 2 x  0
...
5
x  45, 135, 225, 315

x  45 o ,135 o

 x  45 o , 90 0 , 135 0

2 sin 2 x  1  0

 x  45 o , 90 o , 135 o , 225 o , 315 o

Note

cos A  cos B  2 cos

A B
A B
cos
2
2

EXAMPLE 23

EXAMPLE 24

Solve the following equation giving all solutions
in the given interval

Solve the following equation, giving all solutions
in the given interval
2 sin x 1  2 cos x

 5, 0  x  360 0
1  cos x
sin x

sin 6 x  sin 2 x  2 sin 4 x, 0  x  1800
Suggested Solution

Suggested Solution
 sin 6 x  sin 2 x  2 sin 4 x  0
1
1
 2 sin (6 x  2 x) cos (6 x  2 x)  2 sin 4 x  0
2
2
 2 sin 4 x cos 2 x  2 sin 4 x  0
 2 sin 4 x(cos 2 x  1)  0

2 sin 4 x  0
sin 4 x  0
4 x  0,180,360,540,720
x  0,45,90,135,180

Note:

sin A  sin B  2 sin

A B
A B
cos
2
2

cos 2 x  1
2 x  0,360
x  0,180

2 sin x 1  2 cos x

5
1  cos x
sin x
2 sin x
...
6)
x  36
...
10

Trigonometry Functions

Page 12

EXAMPLE 25
Express 7 cos   24 sin  in the form
R cos(   ) , where R > 0 and 0 0    90 0 ,
giving the exact value of R and the value of 
correct to 2 decimal places
Suggested Solution

EXAMPLE 26
Express cos   3 sin  in the form

 

1
R cos    , where R  0 and 0     ,
2
giving the exact values of R and 
...
(1)
 R sin   24
...
(1)

 R 2 cos 2   1  R 2 sin 2   3

 R  576  49  625  R  25
2

 cos 

cos   3 sin   R cos  cos   R sin  sin 

 (1) 2  (2) 2 :



2

 3 sin   R cos(   )

 R sin   3
...
74
25

1
2

   cos 1

1
1
 60 0 or 
2
3

EXAMPLE 27
Find all acute angles  for which cos 4  2 cos 2   0
...

cos(2  2 )  2 cos 2   0
(cos 2 cos 2  sin 2 sin 2 )  2 cos 2   0

(2 cos   1)   (2 sin  cos ) )  2 cos   0
2

2

2

2

(4 cos 4   4 cos 2   1)  4 sin 2  cos 2   2 cos 2   0
(4 cos 4   4 cos 2   1)  4(1  cos 2  ) cos 2   2 cos 2   0
(4 cos 4   4 cos 2   1)  4 cos 2   4 cos 4   2 cos 2   0
8 cos 4   6 cos 2   1  0

Let cos 2   u

8u 2  6u  1  0
1
1

2
0
4u  1  cos x  4  cos x   2  60

(4u  1)((2u  1)  0  
2u  1  cos 2 x  1  cos x   1  45 0

2
2

x  45 0 , 60 0

Trigonometry Functions

Page 13

EXAMPLE 28

If 2 x  y 

4

,show that tan y 



1  2 tan x  tan 2 x

...


Suggested Solution
 tan y


 tan  2 x 
4



tan


4

1  tan

 tan 2 x



tan 2 x
4
1  tan 2 x

1  tan 2 x
2 tan x
1
1  tan 2 x

2 tan x
1
1  tan 2 x
1  tan 2 x  2 tan x
1  tan 2 x


1  tan 2 x
1  tan 2 x  2 tan x
1  2 tan x  tan 2 x

( Shown)
1  2 tan x  tan 2 x

tan y 

1  2 tan x  tan 2 x
1  2 tan x  tan 2 x


 
 tan 2  

8
  
8
tan  2   
4

 8   1  2 tan   tan 2   

 
8
8

 
1  2 tan  tan 2  
8
8
tan 0 

 
1  2 tan  tan 2  
8
8
tan 0  0  Denominator  0  Numerator  0
1  2 tan



tan

8


8

 tan

 2 tan



1  0
8
  2  4  4(10(1)
tan 
8
2
tan 2

 2  1 or - 2  1 (Reject)


8

 2  1 ( Shown)

EXAMPLE 29
Find the general solution, in degrees of the following equations
a) tan 3x  tan 30 0
b) 5 cos 2  3
Suggested Solution
a)

tan 3x  tan 300
3x  (180(0)  30), (180(1)  30), (180(2)  30), (180(3)  30),
...

x  600 n  10 where n  Z ( Answer )

b)

5 cos 2  3
cos 2  0
...
6
2  53
...
130), (360  53
...

2  360n  53
...
6 0 ( Answer )
Trigonometry Functions

Page 14

EXAMPLE 30
a)

c)

It is given that f(x) = a sin(bx) + c, where a, b and c are integers
...
Find the value of a, of b and of c
...
The maximum value of f is 11, the
minimum value of f is 3 and the period of f is 72 °
...

Find the greatest and least possible values of 26(1  cos  )  8

d)

The function f is such that f ( x)  3  4 cos 2 x , for 0 ≤ x ≤ π
...
( 1)
Minimum : 3  a  c
...
(1)
Minimum : 3  a  c
...
of cos2x = 1

 Least : 34  26(1)  8

Min
...
Given that the amplitude of f is 2 and the period of f is 120°
...
Find the value of p and q
...
Substitute the x-coordinate and the y-coordinate into the equation

1  2  3 tan 3q
3 tan 3q  3
tan 3q  1
3q 


4

q


12

( Answer )

Trigonometry Functions

Page 16

EXAMPLE 33
Sketch the graph y  6 sin 2 x  3 , for the domain 0  x  2
...

Suggested Solution

6 sin 2 x  3  0
sin 2 x  0
...


6  5 sin(2 x   ) where tan  

a)

Show that f(x) can be expressed as

b)
c)

4
3

Find the maximum and minimum values of f(x)
Sketch the graph of y = f(x) for 0  x  2

Suggested Solution
a)

f ( x)  10 cos 2 x  2 sin 2 x  6 sin x cos x
...
(1)
 R sin   4
...





a)

Show that f(x)can be expressed as 10  4 sin x 

b)
c)



Find the maximum and minimum values of f(x)
Sketch the graph of y = f(x) for 0  x  360


3

Suggested Solution
a)

Let 2 sin x  (2 3 ) cos x  R sin( x   )
2 sin x  (2 3 ) cos x  R sin x cos   R cos x sin 
Equate the coefficien ts of cos x and sin x
 R cos   2
...
(2)
 (1) 2  (2) 2 :



 



 R 2 cos 2   4  R 2 sin 2   12
 R  4  12  16  R  4
2



(2) sin 2 3



 tan  3 
(1) cos
2
3



 10  2 sin x  2 3 cos x  10  4 sin  x   ( Shown)
3

b)

Maximum of f(x)
Happens when sin(x+π/3)=1  f(x) =10 +4(1) = 14
Minimum of f(x)
Happens when sin(x+π/3)=-1  f(x) =10 +4(-1) = 6

c)

Trigonometry Functions

Page 18

3
...


Convert the following angles in degree to radians in terms of π:
a)
2500
b)
3300
c)
5500
d)
7800
e)
10000

2
...
142)
7  radians
a)
12

b)
c)
d)
e)
3
...


0
...
03π radians
7  radians
0
...


Answer:

-0
...
8658

Answer:

6
...

Find the elements of the set A

16/65

Answer

-16/63

Answer : A  00 , 600 , 1200 , 1800 

 3 
1
9
and x  sin 1 
 , show that tan 3x 


2
13
 10 

7
...


Given that tan  6 , find the exact value of :
a) tan 2
b) tan 3

9
...


5

Answer :  12
Answer : 198

35

107

Show that the equation
b)

3 tan  2 cos  can be expressed as 2 sin 2   3 sin   2  0
sin   cos   2(sin   cos  ) can be expressed as tan  3

c)

sin x  30 0  2 cos x  60 0 can be written in the form 3 3 sin x  cos x

a)

d)


tan30

0



    2 tan60

Trigonometry Functions

0


 
   can be written in the form tan   6 3 tan  5  0
2

Page 19

10

Solve each of the following equations giving all solutions in the given interval
a)
sin 2 x  3 cos 2 x  0 for 0 0  x  180 0
...
2 0 ;144
...

3 tan x  2 cos x , for 00    3600
...
Solve each of the following equations giving all solutions in the given interval
(a)
25 cos  73
...

(b)
(c)
(d)

Answer : 20
...
90

17 sin(  61
...

13 sin( x  67
...
40, 186
...
40, 175
...
31)  4 , for 0 0    360 0

Answer : 27
...
40

12
...
Find without the use of a calculator, the exact value of tan ,given that
a) tan(   )  4

b) tan(   )  3 cos(   )

2

Answer : a) /9 b)

13
...


/5

Prove the following trigonometry identities
a)
1  sin x
cos x
2
b)

14
...
Prove the identity

sin 3x  3 sin x  4 sin 3 x
18
...
Prove the identity

sin 2 x cos 2 x 

1
1  cos 4 x 
8

20
...
Solve the following equation, giving all solutions in the given interval
2 cos x 1  sin x

 4 cos x, 0  x  360 0
1  sin x
cos x
Answer : 194
...
5

Trigonometry Functions

Page 20

22
...
Prove the identity,

tan x  450  tan 450  x  2 tan 2 x
24
...
correct to 2 decimal places
Answer R  17;   61
...
Express 5 sin x  12 cos x in the form R sin( x   ) , where R > 0 and 0 0    90 0 , giving the value
of  correct to 2 decimal places
...
380 ;

26
...

Answer : R 

26 ;  11
...
Express 9 sin   12 cos  in the form R sin     , where R > 0 and 0 0    90 0 , giving the
exact value of R and the value of  correct to 2 decimal places
...
130 ;

28
...


Prove the identity

30
...

1  4 cos 4 x

Show that

1  cos 2 x  sin 2 x
 cot x
...


32
...


Solve the equation 5 sec 2 2 x  tan 2 x  9 , for 0  x  180 0
Answer : 22
...
7o, 112
...
7o

34
...


Show that cos 5x  16 cos 5 x  20 cos 3 x  5 cos x
a)
b)





Prove that sin 2 cosec   sec   4 cos 2
i) Solve for 0 0    180 0 the equation sin 2 2 cosec 2  sec 2   3
2

2

2





ii) Find the exact value of cosec 2 150  sec 2 150
Answer : b) i)20
...


i)

Show that the equation tan(60 0   )  tan(60 0   )  k can be written in the form

2 3 (1  tan  )  k (1  3 tan  )
2

2

ii) Hence solve the equation tan(60 0   )  tan(60 0   )  3 3
giving all solutions in the interval 0 0    180 0
Answer : ii) 16
...
2

37
...
Prove that tan A  tan B  tan C  tan A tan B tan C,

38
...
6

39
...
1o , the general solution of the equation

4 sin 2 x cos x  tan 2 x

40
...
9 nZ

Find, correct to 0
...


or 180(n)  10
...
34) , n  Z

42
...


Find the general solution, in degrees, of the equation (7 sin x  3)(3 sin x  2)(4 sin x  1)  0
Answer : 360n  25
...
6 or 360n  41
...
2 or 360n  138
...
5

44
...
Find
a)
The amplitude of f
b)
The period of f

Trigonometry Functions

Page 22

45
...
Write down the
equation of the resulting graph
46
...


The function f is defined for 0 0  x  180 0 , by f ( x)  3 cos 4 x  1
a) State the amplitude and period of f
b) State the maximum and minimum values of f
...
: -4

48
...

Answer : ii )

49
...
Express f (x ) in the form R cos( x   ) , where R > 0 and

1
0     , giving the value of R and the value of  in terms of b
...
State the
relationships between these graphs in terms of geometrical transformations
...
Sketch the curve y  cos x  sin x for  2  x  2

Trigonometry Functions

Page 23

4
...
8391)  3(0
...
8660)  2(1
...
5)

d)

  1
...
6428)  3(0
...
6643)
 6
...


Given that sin x  12 , sin y  4 and both the angles x

13

2

a)
b)

12

 1

  0
...


7 180
 105 0
12
0
...
37 radians 
 21
...
142
1
...
03 radians 
 59
...
142
7  radians  7 180  252 0
5
5
0
...
78 radians 
 44
...
142
7

b)

Find the values of each of the following
a)
 4 tan 3200  3 cos 1420

c)

Convert the following angles in radians to degrees
( Take π=3
...


2

cos 2A
tan 2A

5

and y are in the first quadrant, find the values of:
a)
sin(x – y)
b)
tan(y – x)

Suggested Solution
Suggested Solution

If sin A 
a)

1
3
1
 cos A 
and tan A 
2
2
3
b)

 cos 2 A
 2 cos A  1
2

2

 3

 2
 2  1


3
 1
2
1

2

6
...

Find the elements of the set A

12
5
12
 cos x  and tan x 
13
13
5
4
3
4
If sin y   cos y  and tan y 
5
5
3
If sin x 

 tan 2 A
2 tan A

1  tan 2 A
2 1 


3
 
2
1  1 


3

2
3  3  3

2
3
3
3
 3

a)

 sin x cos y  cos x sin y
12 3 5 4

...


Given that 0  x  1 and x  sin 1  3  , show that





2

tan 3x 

Given that tan  6 , find the exact value of :
a)
tan 2
b)
tan 3

8
...
3
1 9 
9
 4
13
4
9

( Shown)
13


9
...

 2 cos 
cos 
 3 sin   2 cos 2 
 3 sin   2(1  sin 2  )
 3 sin   2  2 sin 2 
 2 sin 2   3 sin   2  0 ( Shown)

Trigonometry Functions

b)

 sin   cos   2(sin   cos  )
 sin   cos   2 sin   2 cos 
 sin   3 cos 
sin 

3
cos 
 tan   3 ( Shown)

Page 25

c)







sin x  30 0  2 cos x  60 0





 







tan 30 0    2 tan 60 0   can be written in the form

d)

 

can be written in the form 3 3 sin x  cos x

tan 2   6 3 tan  5  0

Suggested Solution

Suggested Solution







 sin x  30 0  2 cos x  60 0



 tan300     2 tan600   
tan 30  tan 
tan 60  tan 

2
1  tan 30 tan 
1  tan 60 tan 
1
 tan 
 3  tan  

 3 1
 2
 1  3 tan  
1  3 tan 



 sin x cos 30  cos x sin 30  2(cos x cos 60  sin x sin 60)


 cos x

3
1
3

sin x  cos x  2
 2  2 sin x 
2
2



3
1
sin x  cos x  cos x  3 sin x
2
2
 3

 1

 sin x
 2  3   cos x1  2 










 1  3 tan 

 32 3 1
  cos x
 sin x

 2
2





2





 2 3  tan 



2

 1  2 3 tan   3 tan 2   6  4 3 tan   2 tan 2 
 tan 2   6 3 tan   5  0 ( Shown)

 3 3 sin x  cos x ( Shown)

10
...


b)

Suggested Solution

0
0
tan x tan 2 x  1, for 0  x  180

Suggested Solution

sin 2 x  3 cos 2 x  0

tan x tan 2 x  1
2 tan x
tan x
1
1  tan 2 x
2 tan 2 x  1  tan 2 x

sin 2 x  3 cos 2 x
sin 2 x
 3
cos 2 x
tan 2 x  3

3 tan 2 x  1
1
tan 2 x 
3
1
1
tan x 
or 
3
3
x  30 0 , 150 0 ( Answer )

2 x  (180  71
...
6) 0
x  54
...
2 0 ( Answer )

(in the interval 0  x  180)
c)

2 sin x tan x  3 , for 00    3600
...

Suggested Solution

Suggested Solution

3 tan x  2 cos x
sin x
3
 2 cos x
cos x
3 sin x  2 cos 2 x

2 sin x tan x  3
sin x
2 sin x
...
5
x  60 0 , 300 0

Trigonometry Functions

cos x  2 (Reject )

3 sin x  2  2 sin 2 x
2 sin 2 x  3 sin x  2  0
(2 sin x  1)(sin x  2)  0
2 sin x  1
sin x  0
...


Solve each of the following equations giving all solutions in the given interval

Suggested Solution
0
0
25 cos  73
...


a)

b)

Suggested Solution

0
0
17 sin(  61
...


Suggested Solution

25 cos   73
...
93)  14
14
sin(  61
...
93  55
...
44)

cos   73
...
6

  73
...
130 , 306
...
130  73
...
870  73
...
44  61
...
56  61
...
60 , 126
...
4 0 , 186
...
38)  11 , for 0 0    180 0

c)

0
0
26 cos(  11
...
38)  11
11
sin( x  67
...
38  57
...
2
x  9
...
82

26 cos(  11
...
31  38
...
67

cos(  11
...
00 , 310
...
82, (360  9
...
82,350
...
4 0 , 175
...


Given that tan   2
...


Prove the following trigonometry identities

Suggested Solution
a)
1  sin x
cos x
2


cos x
1  sin x cos x

b)

Suggested Solution

Suggested Solution

1  sin x
cos x


cos x
1  sin x
(1  sin x)(1  sin x)  cos 2 x

cos x(1  sin x)
1  2 sin x  sin 2 x  cos 2 x
cos x(1  sin x)
2  2 sin x

cos(1  sin x)
2(1  sin x)

cos x(1  sin x)
2

( Shown)
cos x


14
...

2

Use this result to explain why

0 0    90 0

 tan x(1  cos 2 x)
 tan x  tan x cos 2 x
sin x sin x


2 cos 2 x  1
cos x cos x
sin x
sin x

 2 sin x cos x 
cos x
cos x
 2 sin x cos x





 sin 2 x ( Shown)

a)

2

tan x(1  cos 2 x)  sin 2 x

2

Show that 2 sin x  cos x 2 can be written in the form

5
3
 2 sin 2 x  cos 2 x
2
2
Suggested Solution

Suggested Solution

 2 sin x  cos x 
 2 sin x  cos x)(2 sin x  cos x 
2

a)

 tan 2   sin 2 
sin 
 sin 2 
cos 2 
sin 2   sin 2  cos 2 

cos 2 
2
sin 

1  cos 2 
cos 2 
 tan 2  sin 2  ( Shown)


2





b)

RHS > 0 → tan2θ > sin2θ
tan θ > sin θ if θ acute
...


cos 3x  4 cos 3 x  3 cos x
Suggested Solution

Prove the identity

sin 3x  3 sin x  4 sin 3 x
Suggested Solution

 cos 3 x

 sin 3 x

 cos(2 x  x)

 sin( 2 x  x)

 cos 2 x cos  sin 2 x sin x

 sin 2 x cos x  cos 2 x sin x (Note)

(Note)

 (2 cos x  1) cos x  (2 sin x cos x)(sin x)

 2 sin x cos x cos x  (1  2 sin 2 x) sin x

 2 cos 3 x  cos x  2 sin 2 x cos x

 2 sin x cos 2 x  sin x  2 sin 3 x

2

 2 cos 3 x  cos x  2 cos x(1  cos 2 x)

 2 sin x(1  sin 2 x)  sin x  2 sin 3 x

 2 cos x  cos x  2 cos x  2 cos x

 2 sin x  2 sin 3 x  sin x  2 sin 3 x

3

3

 4 cos x  3 cos x ( Shown)
3

Note:

18
...
2 sin x cos x)  8 cos 2 x  4
 4 cos 4 x  4 cos 2 x  1  4 sin 2 x cos 2 x  8 cos 2 x  4
 4 cos 4 x  4 cos 2 x  1  4(1  cos 2 x)(cos 2 x)  8 cos 2 x  4
 4 cos 4 x  4 cos 2 x  1  4 cos 2 x  4 cos 4 x  8 cos 2 x  4
 8 cos 4 x  3 ( Shown)

19
...


Solve the following equation, giving all solutions in the
interval 0  x  3600

21
...
5
sin x   0
...


Prove the identity,





  3 cos x





sin x  0
...
5 0 , 345
...
50 , 345
...


Prove the identity,









tan x  450  tan 450  x  2 tan 2 x

cos x  30 0  sin x  60 0 
Suggested Solution

 1 or - 0
...


25
...
(1)

 (1) 2  (2) 2 :



 



 R 2  64  225  289  R  17 ( Answer )

Express



 



 R 2 cos 2   25  R 2 sin 2   144

 R  25  144  169  R  13 ( Answer )
2

 Replace R in (1) :

 Replace R in (1) :

26
...
(1)

 (1) 2  (2) 2 :

 R 2 cos 2   64  R 2 sin 2   225

 cos 

8
17

   cos 1

5 sin x  12 cos x  R sin x cos   R cos x sin 
Equate the coefficien ts of cosx and sinx
 R sin   12
...
(2)

 cos 

5 sin x  12 cos x in the form R sin( x   ) ,

Express

where R > 0 and 0 0    90 0 , giving the value of  correct
to 2 decimal places
...
correct to 2 decimal places
0

   cos 1

8
 61
...


27
...
380 ( Answer )
13

Express 9 sin   12 cos  in the form R sin     , where
R > 0 and 0 0    90 0 , giving the exact value of R and the
value of  correct to 2 decimal places
...
(1)
 R sin   1
...
(1)

 (1) 2  (2) 2 :

 (1) 2  (2) 2 :



 

 R sin   1 2
...
310 ( Answer )
26

Trigonometry Functions



 



 R 2 cos 2   81  R 2 sin 2   144

 R  81  144  225  R  15 ( Answer )
2

 Replace R in (1) :
 cos 

9
15

   cos 1

9
 53
...


Prove the identity

29
...

2
cos x cos x cos x
1
sin x


2
cos x cos 2 x
1  sin x

cos 2 x
1  sinx

1 - sin 2 x
1  sin x

(1  sin x)(1  sin x)
1

( Shown)
1  sin x

30
...


4
Show that, 1  4 sin x is negative for all values of x
...


Show that 1  cos 2 x  sin 2 x  cot x
...


33
...
8 or 1
 2 x  141
...
3,45,225
 x  22
...
7 o , 112
...
7 o ( Answer )

cos 5x  16 cos 5 x  20 cos 3 x  5 cos x

Suggested Solution

 cos 5 x
 cos(3x  2 x)
 cos 3x cos 2 x  sin 3x sin 2 x
Expand:

cos 3x cos 2 x

Expand:

sin 3x sin 2 x

 (cos 2 x cos x  sin 2 x sin x)(2 cos 2 x  1)

 sin 3 x sin 2 x
 sin(2 x  x)2 sin x cos
 (sin 2 x cos x  cos 2 x sin x)2 sin x cos x

 ((2 cos 2 x  1) cos x  (2 sin x cos x sin x))(2 cos 2 x  1)

 (2 sin x cos x cos x  (2 cos 2 x  1) sin x)2 sin x cos x

 (2 cos 3  cos x  (2 sin 2 x cos x))(2 cos 2 x  1)

 (2 sin x cos 2 x  2 sin x cos 2 x  sin x)2 sin x cos x

 (2 cos 3 x  cos x  (2(1  cos 2 x) cos x))(2 cos 2 x  1)

 (4 sin x cos 2 x  sin x)2 sin x cos x

 (2 cos 3 x  cos x  (2 cos x  2 cos 3 x))(2 cos 2 x  1)

 8 sin 2 x cos 3 x  2 sin 2 x cos x

 (4 cos 3 x  3 cos x)(2 cos 2 x  1)

 8(1  cos 2 x) cos 3 x  2(1  cos 2 x) cos x

 8 cos 5 x  4 cos 3 x  6 cos 3 x  3 cos x

 8 cos 3 x  8 cos 5 x  2 cos x  2 cos 3 x

 8 cos 5 x  10 cos 3 x  3 cos x

 8 cos 5 x  10 cos 3 x  2 cos x

 cos 3 x cos 2 x
 cos(2 x  x)(2 cos x  1)
2

 cos 3x cos 2 x  sin 3x sin 2 x
 8 cos 5 x  10 cos 3 x  3 cos x  (8 cos 5 x  10 cos 3 x  2 cos x)
 8 cos 5 x  10 cos 3 x  3 cos x  8 cos 5 x  10 cos 3 x  2 cos x
 16 cos 5 x  20 cos 3 x  5 cos x ( Shown)

Trigonometry Functions

Page 33

35
...




i)

tan(60 0   )  tan(60 0   )  k can be
written in the form

2 3 (1  tan  )  k (1  3 tan  )



2

ii)

Find the exact value of
0

2

2

Hence solve the equation

tan(60 0   )  tan(60 0   )  3 3

cosec 15  sec 15
2

Show that the equation

0

giving all solutions in the interval 0 0    180 0

Suggested Solution
a)



 sin 2 2 cos ec 2  sec 2 

Suggested Solution



i)
 tan(600   )  tan(600   )  k

1 
 1
 sin 2
...
75
2  41
...
6

37
...
7 0 , 159
...
Prove that
tan A  tan B  tan C  tan A tan B tan C,

From part(i), k  3 3

 2 3 (1  tan 2  )  3 3 (1  3 tan 2  )
 2(1  tan 2  )  3(1  3 tan 2  )
 2  2 tan 2   3  9 tan 2 
 11 tan 2   1
 tan 2  

1
11

1
11
   16
...
2 0 ( Answer )
 tan   

Suggested Solution

 A  B  C  180
 A  180  ( B  C )
tan 180  tan(B  C )
1  tan 180 tan(B  C )
0  tan(B  C )
 tan A 
1
 tan A   tan(B  C )
 tan A 

 tan B  tan C 
 tan A   

1  tan B tan C 
 tan A(1  tan B tan C )   tan B  tan C
 tan A  tan A tan B tan C   tan B  tan C
 tan A  tan B  tan C  tan A tan B tan C ( Shown)

Trigonometry Functions

Page 34

38
...


Find, correct to 0
...
6, (180  71
...
6),
...


sin 2 x  0

 The general solution : 180n  45 or 180n  71
...
25

sin x  0
x  0,180,360,540
...
63
x  50
...
9) o , (720  50
...

x  360 o n  50
...


Find, correct to 0
...


Find the general solution, in radians, of the equation

sin 2 ( A  30 o )  4 sin 2 ( A  30 o )

sec x  5 tan x  3 cos x

Suggested Solution

Suggested Solution

 sec x  5 tan x  3 cos x

sin 2 ( A  30 o )  4 sin 2 ( A  30 o )
sin( A  30)  2 sin( A  30)

1
sin x
5
 3 cos x
cos x
cos x
1  5 sin x

 3 cos x
cos x
 1  5 sin x  3 cos 2 x  1  5 sin x


Case 1

 sin( A  30)  2 sin( A  30)
 sin A cos 30  cos A sin 30  2(sin A cos 30  cos A sin 30)
 sin A cos 30  cos A sin 30  2 sin A cos 30  2 cos A sin 30
 3 cos A sin 30  sin A cos 30
sin A 3 sin 30


cos A cos 30
1
3 3
 tan A  3 tan 30  3 

 3
3
3 3
 A  60,180  60,2(180)  60 
...
34 radians or
3

 sin x  2 (N/A)

Case 2

 sin( A  30)  2 sin( A  30)
 sin A cos 30  cos A sin 30  2(sin A cos 30  cos A sin 30)
 sin A cos 30  cos A sin 30  2 sin A cos 30  2 cos A sin 30
  cos A sin 30  3 sin A cos 30
sin A
sin 30


cos A 3 cos 30
1
1 1
1 3
3
 tan A  tan 30  


3
3
9
3 3 3 3
 A  10
...
9o ,2(180)  10
...


 General Solution : 180(n)  10
...
9 o

Trigonometry Functions

where n Z

Positve angles:

   0
...
34, 3  0
...
34,
...
34, -   (0
...
34), - 3  (0
...


The π (ignoring negative) moves in an arithmetic series: π, 2π,
3π, 4π, …
...
34’ alternate between a positive and negative
...
34)
The general solution : n   1n (0
...


Find the general solution, in radians, of the equation

sin 2 x  sin 3x

43

Find the general solution, in degrees, of the equation

(7 sin x  3)(3 sin x  2)(4 sin x  1)  0

Suggested Solution

(7 sin x  3)(3 sin x  2)(4 sin x  1)  0

 sin 2 x  sin 3x
 2 sin x cos x  sin(2 x  x)

7 sin x  3  0

 2 sin x cos x  sin 2 x cos x  cos 2 x sin x

sin x 

 2 sin x cos x  2 sin x cos x cos x  (1  2 sin 2 x) sin x
 2 sin x cos x  2 cos 2 x sin x  sin x  2 sin 2 x

3
7

 2 sin x cos x  2 sin x  2 sin 2 x  sin x  2 sin 2 x

3
x  sin 1  
7
x  25
...
4)

 2 sin x cos x  3 sin x  4 sin 3 x

x  25
...
6

 2 sin x cos x  2 sin x(1  sin 2 x)  sin x  2 sin 2 x

 2 sin x cos x  3 sin x  4 sin 3 x  0
 sin x(2 cos x  3  4 sin 2 x)  0

3 sin x  2  0

 sin x(4(1  cos 2 x)  2 cos x  3)  0

sin x 

 sin x(4 cos x  2 cos x  1)  0
2

2
x  sin 1  
3
x  41
...
8)

Solution 1

sin x  0  x  0,180,360,540,  x  0,  ,2 ,3 ,
...
: x  n where n  0, 1, 2, 3
...
8, 138
...
309 or 0
...
5, 345
...

2  
2  
2  
2 

x   
,   
,  2   

...

5 
5 
5 
5 

3
x  2n 
where n  0, 1, 2, 3,
...
4



 2  ,
5


x  2n 


5



 2  ,
5




 2  2  
...


360n  41
...
6

Or



 
x   ,
5

The period of sin x is 360, hence each of these solutions will repeat
every 3600

Or

cos x  0
...
5), (360  14
...
309
x

2
3

360n  138
...
5

Or

360n  345
...


General Solution :


3 

 x  n OR x  2n  OR x  2n  
5
5

where n  0
...
2
...
4
...


The function f is given by f(x) = 5sin 3x
...


Amplitude : 5

b)

Period 

2
3

The diagram shows the graph of

a)
b)

y  f (x) , where f ( x)  sin 2 x

State the period of the function f
The graph of y  f (x) is stretched parallel to the y-axis with scale factor 3
...


The function f is defined for 0 0  x  180 0 , by

The function f is defined for 0 0  x  180 0 , by

47
...

Sketch the graph of f

Suggested Solution

a)

Amplitude of

f = 3 (Answer)

Suggested Solution
a)

3600
4
period  90 0 ( Answer )

The maximum value for cos ax  1
So,

period 

5(1)  b  3

b  2 ( Answer )
b)
c)

d)

48
...

Suggested Solution
(i)

(iii)

(ii)



From the sketch above, the other coordinate of intersection is  

Trigonometry Functions

2



,3

Page 38

49
...
Express f (x ) in the form R cos( x   ) , where R > 0 and 0    1  , giving the value of

2

R and the value of  in terms of b
...
State the relationships between these graphs in terms of geometrical transformations
...
(1)

 R  1  b  26  R  1  b

 R sin   b
...

The graph y = cos(x – α) has amplitude
‘1’
...


Sketch the curve

y  cos x  sin x for  2  x  2

Suggested Solution

cos x  sin x  R cos(   )

 Replace R in (1) :

cos x  sin x  R cos x cos   R sin x sin 
Equat the coefficine ts of cos x and sin

 cos 

 R cos   1
...
(2)
 (1) 2  (2) 2 :



2

 

2

2



 R  11  2  R  2
2 cos( x 


4

4

)


4

)|

)

Sketch of the curve y  cos x  sin 

Trigonometry Functions



y | cos x  sin x || 2 cos( x 

2

Sketch the curve

1


2 4

cos x  sin x  2 cos( x 

 R cos   1  R sin   1
2

1
2

 
 
2  cos x   

4 
 


Page 40

Trigonometry Functions

Page 41

Trigonometry Functions

Page 42



---