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C2 Chapter 10
Solutions
Resource sheet 1
1. i. a. sin 90  1 
c. sin 45 
 a  4
1
2

2
2
e. sin 0  0 
2. i.
4
2
0
2
Special angles:
b. sin 60 
 a  2
d. sin 30 
 a  0
3
2
 a  3
1
1

 a  1
2 2
(5)
opp a
adj a

cos  

hyp c
hyp c
Hence cos   sin  , as required.
sin  
[5]
(1)
(1)
ii. Using the triangle (which exists for 0    90 )     90    90  
Since cos   sin  , it follows that cos  90     sin 
(1)
cos  90     sin  is also true when   0 or 90 since cos 90  sin 0  0
and cos 0  sin 90  1
(2)
a
b
, tan  90     tan  
b
a
b a
Hence tan  90    tan     1 as required.
a b
for all angles 0    90 .
iii. Using the triangle, tan  
(2)
(1)
When   0 or   90 , one of the terms in the product tan  90    tan  is
undefined.
(2)
3. i.
sin 30  4 cos 60 
1
1 5
 4 
2
2 2
(3)
2
 3 1
ii. 2sin 60  cos 60  2 
 
 2  2
3 1
  1
2 2
2
iii. 5 tan 60 cos 30  5 3 
3 15

2
2
(1)
(2)
(3)
[10]
1 
 1
iv.  sin 45  cos 45   


2
 2
2
2
(1)
2
 2 

 2
 2
v.
 sin 60  3cos 30 
2
 3
3
 
 3

2 
 2

 2 3
vi.
(2)

2
2
(1)
 12
(2)
3
3
1
 2 sin 45  1  2 
tan 30
2
3
(1)
 3 1  2
4. i.
a tan 30  tan 45  a 
(2)
tan 45
1

1
tan 30
 
[18]
(1)
3
 a 3
ii. a 2 tan 30 

 3 
2
1
3
3
 3
3
tan 60  3
Hence a 2 tan 30  tan 60 , as required.
 3 
1
3
1
 3 3
3
3
iii. a3 tan 30 
(1)
(1)
(1)
(1)
3
(1)
(1)
[7]
C2 Chapter 10
graphs:
Resource sheet 5
More trigonometrical
Solutions
1. i.
y
y  sin 3 x
1
O
60 
360 
120 
x
-1
(2)
ii. a. sin3x  0 for 0  x  360 has solution x  60 , 120 ,180, 240, 300
(2)
b. sin 3x  1 for 0  x  360 has solution x  30 , 150 , 270
[Accept any method e.g. by a substitution]
2. i. Stretch along the x-axis scale factor
1
2
0.5
(2)
(2)
ii.
y
y  cos x
O
360 
x
y  cos  0.5x 
iii.
3. i.
x  240  cos x  0.5 , cos  0.5x   0.5
(2)
Hence the graphs intersect at  240 ,  0.5 , as required.
(1)
From the sketch over the range 0  x  360 ,
cos x  cos0.5x  240  x  360
Hence the solution to the inequality is 240  x  360
(2)
sin 
cos 
 2sin  cos   sin   0
2sin   tan   2sin  
 sin   2cos 1  0
(1)
[6]
 sin   0 or cos  
Hence   0 , 60 ,180
1
2
(2)
(2)
ii.
y
 60 , 3 
y  2sin 
2
O
90 
x
180 ,0
y  tan 
(3)
4. i.
y
y  sin 4 x
1
O
90 
x
-1
(2)
ii. The line y  k intersects the graph y  sin 4 x in exactly two places for
only.
k  1
(2)
C2 Chapter 10
equations
Resource sheet 2
Solution of trigonometric
70
solutions
1. Using appropriate graphs or any other method:
i.
sin x  0  x  0, 180 , 360
(2)
ii. cos x  0  x  90 , 270
iii. 2sin x  1  sin x 
(2)
1
 x  30 , 150
2
iv. 3tan x  3  tan x 
3
1

3
3
 x  30 , 210
(2)
12
3

4
2
v. 4 cos x  12  cos x 
 x  30 , 330
vi. 2 3 sin x  3  sin x 
(2)
3
2 3

(2)
3
2
 x  60 , 120
2. Answers are given to one decimal place.
i. tan x  4  x  104.0 , 76.0
3
4
 x  48.6 , 131.4
(2)
[12]
(2)
ii. 4sin x  3  sin x 
4
5
x  36.9
(2)
iii. 5cos x  1  3  cos x 
iv.
1
tan x  1  0  tan x  2
2
 x  63.4 , 116.6
(2)
(2)
[8]
3. i.
sin 2   cos2   1  sin    1  cos2 
Hence sin    1    54 

9
25

(1)
2
3
5
Since 90    180 , sin   0 and so sin  
(2)
3
, as required.
5
3

sin 
3
5
ii. tan  
 4  .
cos    5 
4
4. i.
1
2
   30 ,  150
(1)
(2)
[6]
sin 2   0.25  sin   
1
3
   30 ,  150
(4)
ii. 3tan 2   1  tan   
iii.
(4)
1
1
 2  cos   
2
cos 
2
   45 ,  135
5. 2cos2   cos 1  0   2cos 1 cos  1  0
1
, 1
2
   60 ,180 ,300
 cos  
6. i. Replacing cos 2  with 1  sin 2  results in 2sin 2   3sin   1  0
2sin 2   3sin   1  0   2sin  1sin  1  0
1
 sin   , 1
2
   30 , 90 , 150
ii. Replacing sin 2  with 1  cos 2  results in 5cos 2   8cos   4  0
(4)
[12]
(1)
(1)
(2)
(2)
(1)
(1)
(2)
(2)
[4]
5cos2   8cos  4  0  5cos  2 cos  2  0
2
,  2   no real values 
5
   66.4 , 293.6 (1 decimal place)
sin  3
3
  tan  
iii. 2sin   3cos  
cos  2
2
 cos  
   56.3 , 236.3 (1 decimal place)
iv.
2 tan  
1
1
 0  2 tan   
2 cos 
2 cos 
 cos  
sin 
1

cos 
4
 sin   
1
4
   194.5 , 345.5
v. 2 tan  
3
sin 
3
0  2

sin 
cos 
sin 
 2sin 2   3cos   0
(1)
(1)
(2)
(2)
(1)
(1)
(1)
(1)
(2)
(1)
(1)
Replacing sin 2  with 1  cos 2  results in 2 cos 2   3cos   2  0
(2)
2cos2   3cos  2  0   2cos   1 cos   2   0
(1)
1
, 2   no real values 
2
   120 , 240 (1 decimal place)
 cos   
(1)
(2)
[28]
C2 Chapter 10
angles:
Resource sheet 3
Triangles without right
20
Solutions
1. i.
x
12

sin130 sin 25
12
 sin130  21.75...
sin 25
(2)
 x  21.8 cm (1 decimal place)
(1)
 x
ii. The missing angle  180   78  37  65
Hence
x
19
19

 x
 sin 65  17.60..
sin 65 sin 78
sin 78
 x  17.6 mm (1 decimal place)
(1)
(2)
(1)
[7]
2. i. sin120  sin 180 120
 sin 60 
ii. Area 
3
2
1
1
ab sin C  16  5  sin120
2
2
3
 40 
 20 3 cm2 , as required.
2
iii. a. AB 2  52  162  2  5 16  cos120
 361
Hence AB  361  19 cm, as required.
b. 52  192  162  2 19 16  cos 
192  162  52
 cos  
 0.9736....
2 19 16
   cos1  0.9736...  13.17...
 13 (nearest whole degree)
iv. a. The angle of CB against the North line is 120  90  30
Hence the bearing of B from C is 030
b. The angle of BA against the North line is 180  30  13  223
Hence the bearing of A from B is 223 (nearest whole degree)
(1)
(1)
(1)
(1)
(1)
(1)
(2)
(1)
(2)
(2)
[13]
C2 Chapter 10
Resource sheet 4
2
 120
3
Circular measure: Solutions
c
1. i.
ii.
5 c
  112.5
8
iv.
1
 28.6 (1 decimal place)
2
c
iii. 2.3c  131.8 (1 decimal place)
2. i.
2
40   c
9
iii. 220 
(4)
3
ii. 135   c
4
11 c

9
iv. 12.5 
5 c

72
3. i. arc length AB  r  0.8  8  6.4 cm, as required.
ii. Perimeter = (arc length AB) + BC + CA
(4)
(1)
(1)
 6.4  2  8  36.10...
 36.1 cm (1 decimal place)
(1)
iii. Triangle ABC is isosceles with equal angles  at A and B
A

C
0.8

B
Hence 2  0.8     
0.2
 0.1
2
(2)
20
4. i.
A
1 2
1
r    92    56.7
2
2
   1.4c , as required.
Area of ABC 
Hence
(2)
1 2
1
r sin    92  sin1.4
2
2
 39.910... cm2
(1)
Area ABC 39.910...

 0.70389...  70% , as required.
Sector area
56.7
(1)
ii. AB 2  92  92  2  9  9  cos1.4
 134.46...
Hence AB  134.46...  11.59
 11.6 cm (1 decimal place)
[Accept alternative method e.g. using the fact triangle ABC is isosceles]
(1)
(1)
(1)