Download form 5 – trigonometric functions

CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
PAPER 1
1.
Given  is an acute angle and sin   p . Express each of the following in
terms of p.
[3 marks]
a) tan 
b) cos ec  90o   
2.
Given cos   p and 270o    360o .Express each of the following in
terms of p.
[3 marks]
a) sec
b) cot  90o   
3.
Given tan   r , where r is a constant and 180o    270o . Find in
terms of r.
[3 marks]
a) cot 
b) tan 2
4.
Solve the equation 6 cos ec 2 x  13cot x  0 for 0o  x  360o
5.
Solve the equation 2sin 2 A  cos 2 A  sin A  1 for 0o  A  360o
6.
Solve the equation 2 cos 2 y  7 sin y  2 for 0o  y  360o
7.
Solve the equation 15cos 2 x  cos x  4 cos 600 for 0o  A  360o
8.
Solve the equation 3cot x  2sin x  0 for 0o  x  360o
[4 marks]
[4 marks]
[4 marks]
[4 marks]
[4 marks]
106
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
PAPER 2
1.
(a) Sketch the graph of y  sin 2 x for 0o  x  180o
[4 marks]
(b)
(b) Hence, by drawing suitable straight line on the same axes, find the
1
x
number of solution to the equation sin x cos x  
for
2 360o
0o  x  180o
[3 marks]
(a) Sketch the graph of y  3cos x for 0  x  2
[4 marks]
2.
(b) (b) Hence, by using the same axes, sketch a suitable graph to find the
2
 3cos x  0 for 0  x  2 .
number of solution to the equation
x
State number of solutions.
3. (a) Sketch the graph of y  3sin 2 x for 0  x  2
[3 marks]
(b) Hence, by using the same axes, sketch a suitable straight line to find
x
the number of solution to the equation 2  3sin 2 x 
for
2
0  x  2
cot x  tan x
 cos ec 2 x
(a) Prove that
2
3
(b) (i) Sketch the graph of y  2sin x for 0  x  2
2
(ii) Find the equation of a suitable straight line to solve the
3
3
1
x .
equation sin x 
2
2
2
Hence, on the same axes, sketch the straight line and state the
3
3
1
x  for
number of solutions to the equation sin x 
2
2
2
0  x  2 .
(a) Prove that sec2 x  2 cos 2 x  tan 2 x   cos 2 x
[3 marks]
(b) (i) Sketch the graph of y  cos 2 x for 0  x  2
[6 marks]
4.
5.
[4 marks]
[2 marks]
[6 marks]
[2 marks]
(ii) Hence, using the same axes, draw a suitable straight line to find
x
the number of solutions to the equation 2 cos 2 x  1  for

0  x  2 .
107
CHAPTER 5
6.
7.
TRIGONOMETRIC FUNCTIONS
FORM 5
(a) Prove that 2  2sin 2 x  2 cos 2 x
[2 marks]
(b) Sketch the graph of y  tan 2 x  1 for 0  x  2 . By using the same
9
x and state the number of solution to
axes, draw the straight line y  3 
2
9
x  2 for 0  x  2
equation tan 2 x 
2
[6 marks]
(a)
Prove that cot 2 x  cos ec 2 x  tan 2 x  sec2 x
3
(b) Sketch the graph y  cos x and y  2sin x for 0  x  2 . State
2
1
3 
the number of solution to equation sin x   cos x  for
2
2 
0  x  2
[2 marks]
[6 marks]
108
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
ANSWERS (PAPER 1)
1 a)
b)
tan  
1
p
1  p2
cos ec  90o    =
1
sin  90o   
=
2 a)
b)
1
cos 
1
=
p
b)
4
1
sec =
1
cot  90o    = tan 
=
3 a)
1  p2
1
1
1
1  p2
p
1
tan 
1
=
r
2 tan 
tan 2 =
1  tan 2 
2r
=
1 r2
6 1  cot 2 x   13cot x  0
cot  =
1
1
1
1
6 cot 2 x  13cot x  6  0
3cot x  2 2cot x  3  0
3cot x  2  0 OR
1
2cot x  3  0
3
2
OR tan x 
2
3
o
'
0
x  56 19 or 56.31 and x  33o 41' or 33.69o
tan x 
x  56o19' , 236o19' 33o 41' , 213o 41'
Or 56.31o , 236.31o ,33.69o , 213.69o
5
2sin 2 A  1  sin 2 A   sin A  1
1
1
1
2sin 2 A  1  sin 2 A  sin A  1
3sin 2 A  sin A  2  0
3sin A  2sin A 1  0
1
109
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
3sin A  2  0
sin A 1  0
OR
2
OR
3
A  90o and 41.81o
sin A  
A  90o , 221.81o ,
6
FORM 5
sin A  1
1
318.19o
1
2cos2 y  7sin y  2  0
2(1  sin 2 y)  7sin y  2  0
1
2sin 2 y  7sin y  4  0
 2sin y 1sin y  4  0
1
1
2
sin y  4 (not accepted)
sin y 
7
y  30o
1
y  30o , 150o
1
15cos 2 x  cos x  4 cos 600
15cos 2 x  cos x  4 cos 600  0
15cos2 x  cos x  4(0.5)  0
15cos 2 x  cos x  2  0
1
5cos x  23cos x 1  0
1
5cos x  2  0 OR
cos x 
8
2
5
3cos x 1  0
OR cos x 
1
3
x  66.42  or 66 25' and 70.53 or 70  31'
1
x  66.42 ,293.58 ,70.53 ,289.47
x  66o 25' , 70o31' , 289o 28' , 293o 35'
cos x
3
 2sin x  0
sin x
3cos x  2sin 2 x  0
1
3cos x  2 1  cos 2 x   0
1
3cos x  2  2 cos 2 x  0
2 cos 2 x  3cos x  2  0
110
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
 2cos x 1 cos x  2  0
 2cos x  1  0
cos x  
1
2
1
 cos x  2  0
OR
OR cos x  2 (unaccepted)
x  120o
1
x  120o , 240o
1
(ANSWERS)PAPER 2
1
y

y  1
x
180o

x
45o
90o 
135o
180o

1 ( shape)
1(max/min)
1(one
period)
1(complete
from 0 to
180o)
1 (straight
line)

y  sin 2 x
sin x cos x 
1
x

2 360o
y 1
x
 
2 2 360o
x
y  1
180o
Number of solutions= 3
1
1
111
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
2 a)
y
y
2
x
x
1(for line
2
y
x
y  3cos x
b)
(b)
1 ( shape)
1(max/min)
1(one
period)
1(complete
from 0 to
2 or
360 o)
2
 3cos x  0
x
2
y0
x
2
y
x
1
1
Number of solution =2
3
y
x
1 ( shape)
1(max/min)
1(one
period)
1(complete
from 0o
to 2 )
1( for the
straight
line)
2  3sin 2 x 
x
2
x
2
x
y  2
2
2 y 
1
1
Numbers of solutions= 8
112
CHAPTER 5
4
TRIGONOMETRIC FUNCTIONS
(a) Prove that
FORM 5
cot x  tan x
 cos ec 2 x
2
LHS
1  cos x sin x 
 


2  sin x cos x 
1  cos2 x  sin 2 x 
 

2  sin x cos x 
1
1
1

 

2  sin x cos x 
1

2sin x cos x
1

sin 2x
 cos ec2x
1
b)

1(shape)
1(max/min)
1(one
period)
y
y
3
x 1

1(for the
straight
2 line)
x


2
3
2

3
y  2sin x
2

sin
3
3
1
x
x
2
2
2
1
 3
y  2
x 
2
 2
3
y  x 1
1

Number of solution = 1
5
1
(a) LHS  sec2 x  2 cos 2 x  tan 2 x
 1  tan 2 x  2 cos 2 x  tan 2 x
1
 1  2 cos 2 x
  cos 2 x
(proved)
1
113
CHAPTER 5
TRIGONOMETRIC FUNCTIONS
FORM 5
b)
1(shape)
1(max/min)
1(one
period)
y
y  cos 2 x
1(for the
straight
line )


2
y
x
3
2
x

-
2 cos 2 x  1  
y
x

1
x

Number of solutions = 2
6
1
2  2sin 2 x
2 1  sin 2 x 
1
2 cos 2 x
1
(proved)
b)
1(shape)
1(max/min)
1(one
period)
y
y  tan 2 x  1
1(complete
cycle
from 0 to
2 )
2

2

x
3
2
2
-2
9
y  3
x
2
1(for the
straight
line)
1
Number of solution = 3
114
CHAPTER 5
7
(a)
TRIGONOMETRIC FUNCTIONS
FORM 5
RHS  cos ec 2 x  tan 2 x  sec2 x
 1  cot 2 x  tan 2 x  sec 2 x
1
 cot 2 x  1  tan 2 x  sec 2 x
 cot 2 x  sec2 x  sec2 x
 cot 2 x
1
(proved)
y

3
y  cos x
2


2

x
3
2
2
1,1(shapes)
1(max/min)
1(one
period)
1(complete
cycle
from 0 to
2 )


Number of solutions = 3
y=2sinx
1
115