Worked solutions
11
Relationships in space: geometry and
trigonometry in 2D and 3D
Skills check
1 a
5 10 ≈ 15.8
2 a i 12600 cm
2
b
b
2 6 ≈ 4.9
ii
1.26 m2
b
4 π ≈ 12.6 m3 ≈ 12566 l
c
Exercise 11A
1 a
c
(3, 0, 0)
(3, 0,2)
(3, 4, 0)
(3, 4,2)
b
d
d
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
0 + 3 0 + 4 0 + 2
,
,
= =
 (1.5,2,1)
2
2 
 2
2 a
4 a
( x2 − x1 )
2
+ ( y2 − y1 ) + ( z2 − z1 )
=
(3 − 0 )
=
9 + 16 + 4=
2
2
+ ( 4 − 0 ) + (2 − 0 )
2
2
29 ≈ 5.4
d
c
( −3,3,7)
d
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
(0.5, −0.5, −6 )
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
 −5.1 + 1.4 −2 + 1.7 9 + 11 
=
,
,

2
2
2 

d=
1 a
=
( 4 − 2)
=
4 + 0 + 16=
2
25 + 9 + =
9
d=
( x2 − x1 )
=
(1 + 1)
2
20 ≈ 4.47
2
2
43 ≈ 6.56
+ ( y2 − y1 ) + ( z2 − z1 )
2
4 + 36 + 64 =
2
( −2 − 2)
2
2
( x2 − x1 )
+ ( y2 − y1 ) + ( z2 − z1 )
2
2
+ ( −6 − 2 ) + ( −7 − 3)
( −5 − 1)
=
36 + 64 + 100=
2
2
( x2 − x1 )
2
2
200 ≈ 14.1
+ ( y2 − y1 ) + ( z2 − z1 )
2
2
+ ( 0 + 4 ) + (5 − 2 )
2
16 + 16 + =
9
( x2 − x1 )
2
41 ≈ 6.4
2
(1 + 1)
=
4 + 9 + 16=
2
2
+ ( y2 − y1 ) + ( z2 − z1 )
=
d=
2
20 ≈ 4.47
=
2
2
+ (1 + 1) + (3 − 3)
d=
( 4 − 0)
104 ≈ 10.2
2
16 + 4 + =
0
2
2
+ ( y2 − y1 ) + ( z2 − z1 )
=
d=
2
2
( x2 − x1 )
d=
2
+ ( −3 − 3) + ( 4 + 4 )
2
2
+ (2 + 1) + (3 + 1)
2
( x2 − x1 )
2
(1 − 4)
=
9 + 0 + 16=
2
2
29 ≈ 5.39
+ ( y2 − y1 ) + ( z2 − z1 )
=
2
2
+ (1 − 1) + (1 + 3)
2
2
25 = 5
SA = x 2 + 2 xl = 202 + 2 × 20 × 26
2
b
SA = x 2 + 2 xl = 42 + 2 × 4 × 6.3 = 66.4 cm2
c
SA = x 2 + 2 xl = 52 + 2 × 5 × 13 = 155 cm2
d
SA = π r 2 + π rl = π × 52 + π × 5 × 13
+ (3 − 3) + (1 − 5)
2
+ ( 4 − 7 ) + ( −1 − 2 )
2
d=
2
= 1440 cm2
( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
2
=
2
Exercise 11B
=
( −1.85, −0.15,10)
3 a
(2 + 3)
=
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
 5 − 4 2 − 3 −4 − 8 
= 
,
,
=
2
2 
 2
=
=
(0.5,1.5,3)
 −4 − 2 4 + 2 5 + 9 
= 
,
,
=
2
2 
 2
c
b
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
 −4 + 5 4 − 1 3 + 3 
= =
,
,

2
2 
 2
b
2
+ ( y2 − y1 ) + ( z2 − z1 )
2
=
Distance of OF
d=
( x2 − x1 )
=
e Midpoint of OF
f
d=
2
= 283 cm2
e
SA = π r 2 + π rl = π × 62 + π × 6 × 14
= 377 cm2
f
SA = π r 2 + π rl = π × 42 + π × 4 × 12
≈ 201cm2
© Oxford University Press 2019
Worked solutions
1
Worked solutions
2 a
7 Volume of the water tank, VT
= 4π =
SA
r 2 4 π × 52 ≈ 314 cm2
4 3 4
V =
πr =
π × 53 = 524 cm3
3
3
1 2
=
VT π r 2hcyl + π r hcone
3
2
b
3
SA =4 π r 2 =4 π ×   =
28.3 cm2
2
= π × 12 × (13 − 2 ) +
= 36.7 m3
3
4
4
3
V = π r 3 = π ×   =14.1cm3
3
3
2
Conversion to litres
1000000 cm3
1
L
×
1m3
1000 cm3
= 36700L
1
3 a V
=
(base area × height )
3
36.7 m3 ×
1
( 4 × 4 × 12=) 64 cm3
3
=
8 Volume of each ball,
3
4 3 4
Vball =
πr =
π × (3.35) =
157.5 cm3 .
3
3
1
(base area × height )
3
1  10 × 13.1

=
=
× 11  240 cm3

3
2

b V
=
hcyl =×
3 6.7 =
20.1cm
Vcyl =
π r 2h =
π × 3.352 × 20.1 =
708.7 cm3
total free space is
VT = Vcyl − 3VB ≈ 708.7 − 3 × 157.5 =
236 cm3
1
(base area × height )
3
1
=
(9 × 7 × 5=) 105 cm3
3
c V
=
4
Exercise 11C
SA c =curved surface area,
1 a
SA T = total surface area
a
b
l = 52 + 122 = 25 + 144 = 169 = 13
SA c = π rl = π × 5 × 13 ≈ 204cm2
b
c
SA T =π r + π rl =π × 5 + 204.2
2
2
d
≈ 283 cm2
c
1 2
1
πr h =
π × 52 × 12 ≈ 314 cm2
3
3
V=
5 a =
SA c
b
SA=
T
2 a
4π r 2
4 π × 32
=
= 56.5 cm2
2
2
b
4π r 2
2
+ π r=
56.5 + π × 32
2
c
= 84.8 cm2
c
6 a
d
4 3 2
π r = π × 33 = 56.5 cm3
V = 6
3
3
1 2
14

VT =VC + VHS = π r h +  π r3 
3
23

=
b Sum of the curved surface area of the
cone, SA cc , and curved surface area of
the hemisphere, SA CHS
l=
19
19
, θ sin−1= 44.7°
=
27
27
33
33
, θ tan−1= 30.5°
tan θ = =
56
56
sin θ =
12
12
, θ tan−1= 67.4°
=
5
5
11
−1 11
, θ cos = 56.6°
cos θ = =
20
20
tan θ =
x
, x 27 =
cos 22° =
=
cos 22° 25.0
27
44
44
,x
= 42.5
tan 46° ==
tan 46°
x
7
7
tan 46° =
= ,x
= 6.76
x
tan 46°
x
, x 22 =
sin 43° ==
sin 43° 15.0
22
h
, h 12 =
tan25° =
=
tan25° 5.60
12
tan a =
1
14
3
3
π × 42 × 10 +  π × 4  = 302 cm
3
23

2
1
π × 12 × 2
3
5.6
5.6
, a tan−1 = 29.2°
=
10
10
q
,
55
55
=
q 55 tan30
=
°
= 31.75... ≈ 31.8
3
4
tan30° =
tan50
=
°
2
10 + 4 ≈ 10.8 cm
31.75
55
( − p)
(55 − p ) tan50° =31.75
SA T =
SA CC + SA CHS =
π rl + 2π r 2
= π × 4 × 10.8 + 2 π × 4 = 236 cm
2
q
=
55
( − p)
2
31.75
(55 − p ) =
tan50°
© Oxford University Press 2019
2
Worked solutions
p=
55 −
=
h 50 =
sin55° 41m
31.75
≈ 28.4
tan50°
5
7
7
cos 20° =
= ,x
= 7.45
x
cos 20°
6
6
6
, h = 10.4
tan30° =
=
h
tan30°
7 a
3 a
0.8
=
θ sin−1 = 15.5°
3
b
1
× 20 × 15 × 30 =
3000 cm3
3
4
b AC = 2NC = 202 + 152 = 25 . Then
NC = 12.5 cm
TC = 302 + NC 2 = 302 + 12.52
Then
1056.25 32.5 cm
= =
c
sin θ =
30
30
, θ sin−1 = 67.4°
=
32.5
32.5
d
TM =
TC2 − MC2 =
e
8 a
32.52 − 7.52 = 31.6
30
30
,
sin TMN
= =
TM 31.6
30
= sin−1 = 71.7°
TMN
31.6
c
d
EO 146.5
,
tan EMO
= =
OM 115.2
146.5
EMO
= tan−1
= 51.8° ≈ 52°
115.2
1
AT =4 × AF + AB , AF = × AB × EM
2
= 139000 cm2
EM + BM
EO 146.5
sin EBO
= =
EB 219.1
sin−1
146.5
= 42.0°
219.1
Exercise 11D
h
tan25° =
12
=
h 12 =
tan25° 5.60 m
2
=
H 3 tan70
=
° 8.24 m
6 a
sin55° =
h
50
cos 36° =
N
25
sin36° =
W
25
sin68° =
W
51
8 =
H HE + HD
tan23° =
HE
300
tan23° 127.3m
=
HE 300
=
tan30° =
HD
300
=
HD 300
=
tan30° 173.2 m
H = HE + HD = 301m
9 Let C be the bottom of the Eiffel tower.
Then
2
186.42 + 115.22 = 219.1
=
H
3
=
W 51 =
sin68° 47.3km
AT =
2 × 230.4 × 186.4 + 230.4 × 230.4
1
tan70° =
7
146.52 + 115.22 ≈ 186 cm
e =
EB
81.5
= 97.1m
tan 40°
=
W 25 =
sin36° 14.7 km
146.52 + OM 2
2
81.5
d
5 R and J are the same height, so it cancels
out. Then the calculation is
b
AB = 2 OM ,
=
tan 40° =
=
d
1
V=
× 230.4 × 230.4 × 146.5
3
=
EM
h = 32 − 0.82 =2.89 m
=
N 25 =
cos 36° 20.2km
= 2592276.5 cm3
b
0.8
3
sin θ =
=
AB ABC − BC
40° + 32° + θ= 90°
θ= 18°
tan (32° + 18° ) =
ABC
300
=
ABC 300
=
tan50° 357.53
tan18° =
BC
300
=
BC 300
=
tan18° 97.48
AB = ABC − BC = 260 m
10 tan75° =
© Oxford University Press 2019
D + 1.5
498
3
Worked solutions
=
D 498 tan75
=
° − 1.5 1857 m
Area = 173cm2
5
Exercise 11 E
1
Area =
a
1
ab sin C
2
Area =
1
× 8 × 6 × sin80°
2
6
=
Area 24 =
sin 80° 23.6 cm2
b
1
Area = × 2.5 × 3.9 × sin34°
2
Area =
Area = 5 ×
1
× 4 × 4 × sin72° = 38.0 m2
2
1
( x + 2) (2x + 1) sin60° =5 3
2
2 x 2 + 5x − 18 =
0
Either factorise, or use the quadratic
formula:
1
× 4 × 7 × sin96°
2
x1,2 =
Area=
1
× 12 × 20 × sin (180° − 80° − 40° )
2
=
Area 120
=
sin 60° 104cm2
f
Area=
1
× 14 × 18 × sin(180° − 78° − 60°)
2
Area 126
sin 42° 84.3 cm2
=
=
g
Area=
1
× 12 × 8 × sin(180° − 30° − 67°)
2
2
=
Area 48=
sin83° 47.6 cm
2
Area =
a
Then x = 2
Exercise 11F
sin θ
sin35°
=
23
45
1 a
1
× 8 × 5 × sin39°
2
Area= 2 ×
b
1
2
1
ab sin C
2
6 sin75°
18
=
θ sin−1
4 4 faces, so area is multiplied by 4
3 angles in an equilateral triangle are 60°
d
4 sin66°
= 27.2°
8
sin θ
sin75°
=
6
18
sin θ =
=
Area 240
=
sin 60° 208 cm2
1
ab sin C
2
4 sin 66°
8
=
θ sin−1
c
23 sin35°
= 17.0°
45
sin θ
sin66°
=
4
8
sin θ =
Area = 20 × 12 × sin60°
Area= 4 ×
23 sin35°
45
θ sin−1
=
1
C
= sin−1 = 30°
2
3
We take the positive value, as distances
cannot be negative.
1
× 8 × 8 × sin C
2
sin C =
2×2
−5 ± 13
4
sin θ =
Area 20 =
sin39° 12.6 cm2
=
b 16 =
x1,2 =
1
ab sin C
2
Area =
−5 ± 52 − 4 × 2 × ( −18 )
−5 ± 25 + 144
=
4
=
Area 14 =
sin 96° 13.9 cm2
e
10 3
= 20
3
2
2x 2 + x + 4x + 2 =
20
=
=
Area 4.875
sin34° 2.73 cm2
d
1
ab sin C
2
1)
( x + 2 ) (2 x + =
1
Area = × 10 × 15 × sin125°
2
Area 75 =
sin125° 61.4 cm2
=
c
Area= 5 ×
6 sin75°
= 18.8°
18
sin θ
sin 48°
=
22
63
sin θ =
Area =2 × 10 × 10 × sin60°
© Oxford University Press 2019
22 sin 48°
63
4
Worked solutions
=
θ sin−1
We substitute back into the first equation
22 sin 48°
= 15.0°
63
h=
sin θ
sin82°
=
20
29
e
sin θ =

sin70°  sin50°  
sin70°
−15 ×
1 −

 h =
sin20°  sin 40°  
sin20°

20 sin 82°
= 43.1°
29
sin70°
−15 ×
sin20°
18.1
h =
=
sin70°  sin50° 
1−
sin20°  sin 40° 
sin θ
sin78°
=
18
34
f
sin θ =
2 a
4
18 sin78°
34
θ sin−1
=
sin70°  sin50°

h − 15 

sin20°  sin 40°

so h
=
20 sin 82°
29
=
θ sin−1
sin70°
x
sin20°
sin74° sin 64°
=
10
x
=
x
18 sin78°
= 31.2°
34
sin99° sin (180° − 99° − 18° )
=
37
x
10 sin 64°
= 9.35km
sin74°
5 We have
sin 49° sin 41°
=
d
20 − b
sin 63°
=
= 33.4 cm
x 37
sin 99°
20 − b =
d
sin 41°
sin 49°
sin53° sin 44°
=
x
7
=
b 20 − d
sin 41°
sin 49°
b
and
sin53°
=
x 7= 8.05
sin 44°
c
b=d
sin23° sin77°
=
x
15
20 − d
sin33° sin108°
=
x
24
sin (180° − 100° − 22° )
x
=
d
=
sin22°
10
sin58°
=
x 10
= 22.6
sin22°
f
sin (180° − 52° − 56° )
x
sin52°
=
6
sin72°
=
x 6= 7.24
sin52°
3 base = 15 + x
We have that
and
sin70° sin20°
=
h
x
sin 40° sin50°
=
h
15 + x
sin50°
Then 15 + x =
h
sin 40°
=
x
sin50°
h − 15
sin 40°
sin 41°
sin52°
=
d
sin 49°
sin38°
 sin 41° sin52° 
20 d 
=
+

 sin 49° sin38° 
sin33°
=
x 24
= 13.7
sin108°
e
sin52°
sin38°
We equate both expressions for b
sin23°
x 15
=
= 6.02
sin77°
d
sin38° sin52°
=
d
b
20
= 9.31km
sin 41° sin52°
+
sin 49° sin38°
6 We have
sin74°
sin16°
=
16 + AD
DC
sin74°
DC
16 + AD =
sin16°
=
AD DC
and
sin74°
− 16
sin16°
sin 62° sin28°
=
AD
DC
AD = DC
sin 62°
sin28°
We equate both expressions for AD
DC
sin74°
sin 62°
− 16 =
DC
sin16°
sin28°
 sin74° sin62° 
DC 
−
16
=
 sin16° sin28° 
© Oxford University Press 2019
5
Worked solutions
16
= 9.96km
sin74° sin62°
−
sin16° sin28°
=
DC
7 We have
 10

=
B sin−1 
sin 45°=
 62.1°
8


and 180° − 62.1
=
° 117.9°
sin15° sin75°
=
h
10 + d
4
sin75°
=
+ d h = 3.73h
10
sin15°
=
sin C
=
d 3.73h − 10
and
and 180° − 53.5
=
° 126.5°
sin72°
=
d h= 3.08h
sin18°
5
Then 3.08
=
h 3.73h − 10
1
Area = × AB × BC × sin B =20
2
1
× 8 × 10 × sin B =
20
2
3.73h − 3.08h =
10
0.65h = 10
h = 15.3
sin B =
sin55° sin 90°
=
h
4
The obtuse angle is 180° − 30
=
° 150°
sin78° sin 47°
and
=
4
b
Exercise 11H
sin 47°
=
b 4= 2.99077..mm
sin78°
1 a
1
1
Area =
bh =
× 3.27660 × 2.99077 =
4.90 mm2
2
2
a = 11.1cm
b
8
sin64°
10
 8

=
A sin−1 
sin 64=
°  46.0°
10


and 180° − 46.0
=
° 134°
2
sin20° sin A
=
3
5
sin A
=
5
sin20°
3
b = 36.4 cm
c
3
sin 45° sin B
=
8
10
=
sin B
10
sin 45°
8
a2 = b2 + c 2 − 2bc cos A
a2= 142 + 222 − 2 × 14 × 22 × cos 80°
a = 23.9 m
d
c 2 = a2 + b2 − 2ab cos C
c 2= 102 + 92 − 2 × 10 × 9 × cos 66°
c = 10.4 m
e
a2 = b2 + c 2 − 2bc cos A
a2= 402 + 252 − 2 × 40 × 25 × cos 20°
a = 18.6 cm
f
5

=
C sin−1  sin20°=
 34.8°
3


and 180° − 34.8
=
° 145.2°
b2 = a2 + c 2 − 2ac cos B
b2= 152 + 282 − 2 × 15 × 28 × cos112°
Exercise 11G
=
sin A
a2 = b2 + c 2 − 2bc cos A
a2= 122 + 92 − 2 × 12 × 9 × cos 62°
Then
sin 64° sin A
=
10
8
20
40
20
= sin−1 = 30°
B
40
=
h 4 sin55
=
° 3.27660..mm
1
30
sin 40°
24
 30

=
C sin−1 
sin 40°=
 53.5°
24


sin18° sin72°
=
h
d
8 We have
sin 40° sin C
=
24
30
a2 = b2 + c 2 − 2bc cos A
a2= 212 + 302 − 2 × 21 × 30 × cos123°
a = 45.0 cm
2 a
cos θ =
10.42 + 182 − 21.92
2 × 10.4 × 18
cos−1 ( −0.1267 ) =
97.3°
θ =
b
cos θ =
© Oxford University Press 2019
8.62 + 3.12 − 9.72
2 × 8.6 × 3.1
6
Worked solutions
c
θ =−
cos−1 ( 0.197299 ) =
101°
Exercise 11I
652 + 552 − 1182
cos θ =
2 × 65 × 55
1
=
θ cos
d
cos θ =
−1
Area =
52 + 52 − 32
2×5×5
=
θ cos−1 0.82
=
= 34.9°
e
cos θ =
2
2
2
24 + 22 − 20
2 × 24 × 22
θ cos 0.625
=
= 51.3°
cos θ =
3.82 + 72 − 42
2 × 3.8 × 7
150 × 106
sin7° =
d
3 a
=
θ cos 0.891729
= 26.9°
sin55°
sin PRS
=
11.84..
14
1
ab sin C
2
sin PRS
= 14 ×
1
A=
× 12 × 9 × sin29.0°= 26.1cm2
2
4
sin PSR sin PRS
=
PR
PS
b
=
θ cos−1 0.875
= 29.0°
A=
PR2= PS 2 + RS 2 − 2 × PS × RS × cos S
PR = 11.8 m
92 + 122 − 62
cos θ =
2 × 9 × 12
b
150 × 106
= 1230.8 million km
sin7°
PR2= 142 + 112 − 2 × 14 × 11 × cos 55°
−1
3 a
1
bc sin A
2
1
× 20 × 12 × sin 43.5°= 82.6 cm2
2
=
d
−1
f
202 + 122 − 142
2 × 20 × 12
=
A cos−1 0.725
= 43.53°
−3337
= 159°
3575
2
cos A =
sin55°
11.84..
c 2 = a2 + b2 − 2ab cos C
ˆ = sin−1 14 × sin55°  = 75.4809..°
PRS
11.84.. 

c 2= 602 + 302 − 2 × 60 × 30 × cos160°
ˆ= 180° − 75.4809..
=
° 104.519...°
PRQ
c = 89km
5 a
tan33° =
ˆ= 180° − 50° − 104.519..
QPR
=
° 25.4809...°
h1
46
ˆ
ˆ
sin PQR
sin QPR
=
PR
QR
=
h1 46=
tan33° 29.9 m
and tan17° =
h2
28
sin50°
sin25.4809..°
=
11.84...
QR
=
h2 28 =
tan17° 8.56 m
QR =
11.84.. ×
b =
A 180° − 33° − 17
=
° 130°
b = 462 + 29.872 =
54.9
1
A = × QS × PS × sin S
2
c
c = 282 + 8.562 =29.3
=
a2 = b2 + c 2 − 2bc cos A
=
a2 54.92 + 29.32
− 2 × 54.9 × 29.3 × cos130°
4 a
a = 77.0 m
cos ADB =
DB2 + DA2 − BA2
2 × DB × DA
122 + 202 − 282
2 × 12 × 20
ADB = cos−1 − 0.5 = 120°
c 2 = a2 + b2 − 2ab cos C
c 2 = 92 + 152 − 2 × 9 × 15 × cos140°
1
× (11 + 6.68 ) × 14 × sin50°= 94.81m2
2
cos ADB =
6 =
C 210° − 70
=
° 140°
sin25.4809..°
=
6.65m
sin50°
b
c = 22.6 km
1
Area = × BD × DA × sin ADB
2
Area=
c
1
× 12 × 20 × sin120°= 104 m2 .
2
sin DCB sin BDC
=
BD
BC
© Oxford University Press 2019
7
Worked solutions
sin DCB sin60o
=
12
13
sin DCB =
12
sin60o
13
 5

=
°  48.3°
A sin−1 
sin 95=
6.67


−1 12
=
=
DCB sin
sin60o 53.1o
13
d
c
ACD
= 180° − 32° − 48.3
=
° 99.7°
d Let C=AD. Then
sin D
sin C
=
d
c
CBD = 180o − BCD − BDC
= 180o − 53.1o − 60o = 66.9o
sin 48.3.. ° sin 99.7..°
=
6.67..
c
sin BAD sin ADB
=
BD
AB
sin99.7..°
c 6.67..
=
= 8.8039..
sin 48.3..°
sin BAD sin120o
=
12
28
sin BAD =
5
sin 95°
6.67
=
sin A
12
sin120o
28
e
−1 12
sin120o 21.79o
BAD sin
=
=
28
1
× 4 × 5 × sin 95°= 9.96 cm2
2
AABC =
1
× 6.67.. × 8.80.. × sin32° =
AACD =
15.55.. cm2
2
Then
ABD = 180o − ADB − BAD
= 180o − 120o − 21.79o = 38.2o
and so
ABC = CBD + ABD = 66.9o + 38.2o
(3 s.f.)
= 105.1o ≠ 90o
5 a
7
sin ABC
sin ACB
=
AC
AB
22.5
sin 46o
48
h
10 − x =
tan60°
= 180 − 46 − 19.71 = 114.3
o
o
o
2
BC
= AC 2 + AB2
− 2 × AC × AB × cos BAC
=
BC 2 22.52 + 482
− 2 × 22.5 × 48 × cos114.3°
BC = 60.8 m
6 a Let AC=b
=
x 10 −
b2 = 52 + 42 − 2 × 5 × 4 × cos 95°
h
h
= 10 −
tan50°
tan 60°
1
1


h
+
10
=
tan50
°
tan
60
°


=
h
10
= 7.06 m
1
1


 tan50° + tan60° 


8 a 180° − 67
=
° 113°
b = 6.67 cm
113° + 123° + ABC
= 360°
b Let BAC = A. Then
sin 95° sin A
=
6.67
5
h
tan60°
We equate both expressions for x and get
b2 = a2 + c 2 − 2ac cos B
sin B sin A =
b
a
h
tan50°
(10 − x ) tan60° =h
BAC = 180o − ABC − ACB
o
h
x
h
and tan60° =
10 − x
−1 22.5
=
=
ACB sin
sin 46o 19.7o
48
b
tan50° =
x =
sin 46o sin ACB
=
48
22.5
sin ACB =
AABCD = AABC + AACD = 9.96.. + 15.55.. = 25.5cm2
ABC
= 124°
b
b2 = a2 + c 2 − 2ac cos B
b2= 802 + 1202 − 2 × 80 × 120 × cos124°
b = 178km
© Oxford University Press 2019
8
Worked solutions
c
cos C =
a2 + b2 − c 2
2ab
=
VD
=
1202 + 1782 − 802
= = 0.9289
2 × 120 × 178
b
= 21.7°
cos−1 0.9289
OD2 + VO2
2
7.072 + 20=
21.2 cm
tan α =
VO
OM
20
=
α tan−1= 76.0°
5
complement to 123° is 57° , then
360° − 57° − 21.73
=
° 281°
c Let K be the point that connects A with
OA perpendicularly. Let M denote the
midpoint of BA
9 We use lowercase letters for sides opposite
capital letter angles. Find p
Complementary angle to 84° :
OA2 + MA2 − OM 2
2 × OA × MA
21.22 + 52 − 20.62
=
2 × 21.2 × 5
cos OAM =
180° − 84=
° 96° .
ˆ = 360° − 210° − 96° = 54°
Then HPQ
We have two sides and one angle
a
OAM
= cos−1 0.236226
= 76.3 °
ˆ
p = q + h − 2hq cos HPQ
2
2
2
Then sin BAK =
2
3402 + 1602 − 2 × 340 × 160 × cos 54°
p
=
BK= AB sin BAK= 10 × sin76.3=
° 9.72
p = 278km
The angle between two sloping edges,
β is formed by two sides of length BK
and the diagonal of the base
b We find H as
cos H =
2
BK
AB
2
2
p +q −h
2 pq
2782 + 3402 − 1602
=
2 × 278 × 340
= 0.884913
cos β =
9.722 + 9.722 − 14.42
2 × 9.72 × 9.72
β cos−1
=
=
= 27.8°
H cos−1 0.884913
Then 84° + 27.8
=
° 111.8° is Bc , the
complement of the angle
complementary to the bearing B .
12
a cos B
=
9.722 + 9.722 − 14.42
= 95.6 °
2 × 9.72 × 9.72
−11
a2 + c 2 − b2 62 + 82 − 122
=
=
2ac
2×6×8
24
b The cosine of the angle is negative, so
ABC > 90° , i.e. we have an obtuse
angle.
c
B
=
=
° 68.2°
180° − 111.8
Then =
B 360° − 68.2
=
° 292°
10 Triangle ABC with sides a, b, c
=
B 360° − (180° − 30° ) − 100
=
° 110°
b2 = a2 + c 2 − 2ac cos B
Chapter Review
2
b
=
3202 + 5002 − 2 × 320 × 500 × cos110°
=
b = 680km
Then
l =
=
= 26.2°
A cos 0.897
Bearing: 360° − (180° − 30° − =
A ) 236°
11 a Let M be the midpoint of AD. Then
triangle OMD is right-angled with
OM=5, MD=5, then
Then
32 + 42 = 5
b = x 2 + 2 xl = 82 + 2 × 8 × 5 = 144 m2
−1
OM 2 + MD2 = 7.07 cm
1
(8 × 8 × 3=) 64 m3
3
Slant height l is the hypotenuse of the
triangle formed by the pyramid height and
the distance from the origin O to the
midpoint of a side of the base.
b2 + c 2 − a2
cos A =
2bc
6802 + 5002 − 3202
=
2 × 680 × 500
= 0.897
OD =
1
( base area × height )
3
1 a
=
2
a=
1 2
1
3
π r h=
π × 62 × 8= 96 cm
π
3
3
slant height l is the hypotenuse of the
triangle formed by the cone height and the
cone radius
l =
82 + 62 = 10
2
b = π r 2 + π rl = π × 62 + π × 6 × 10 = 96 cm
π
© Oxford University Press 2019
9
Worked solutions
4 3 32
=
πr
π
3
3
3
=
V
4 3 32
r =
3
3
=
r3 = 8
c
r =2
Then SA = 4π r 2 = 4π × 22 = 16 m
π 2
Vcone =
OC =
1 2
1
πr h =
π × 42 × 10
3
3
Then
VO2 + OC 2
=
VC
=32 + 4.242 =
5.20 cm
1 2
1
π rcut hcut =
π × 22 × (10 − 6 )
3
3
16 π
= = 16.8 cm3 3
as required.
Vcut =
d We split the triangle BVC into two rightangles triangles, BVM and MVC, M is the
midpoint of BC.
Vtr =168 − 16.8 =151cm3
Then
d = 2r
sin BMV
sin BVM
=
VB
BM
sin90° sin BVM
=
5.20
3
65 = 2r
r =
65
2
sin BVM = 3
=
r 32.5
=
mm 3.25 cm
b Each ball has a diameter of
2 × 3.25 =
6.5 cm
2BMV = BVC
e Slant height
=
VM
6 tennis balls fit in the cylinder
=
V=
Vcyl − 6Vball
air
8 a
1m3
1000 000 cm3
4
=
VT π r h + π r 3
3
2
=
π × 1.52 × 8.5 +
b
d=
( x2 − x1 )
d=
(1 − 1)
2
2
+ ( y2 − y1 ) + ( z2 − z1 )
2
2
+ (5 − 0 ) + ( 3 − 3 )
2
2
d =5
b Midpoint
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
6 a =
VT Vcyl + Vsph
d 3
= = 1.5m
2 2
5.202 − =
32 4.25 cm
= 62 + 2 × 6 × 4.25 = 87.0 cm2
= 0.431 × 10−3 m3
=
r
VB2 − BM 2
SA
= x 2 + 2 xl
4

Vair= 1294.14 − 6  π × 3.253 
3

= 431.38..
= 431cm3 (3 s.f.)
431.38.. cm3 ×
sin90°
= 35.2°
5.20
so BVC =
2 × 35.2 =
70.5°
h
39
= = 6
6.5 6.5
d
sin90°
5.20
BVM
= sin−1 3
V =π r 2h =π × 3.252 × 39 =1294.14 cm3
c
62 + 62 = 8.49 cm
1
1
Then OC =AC =× 8.49 =
4.24 cm
2
2
b =
Vtr Vcone − Vcut
5 a
1
AC
2
AB2 + BC 2 =
AC =
160 π
= 168 cm3
3
=
1
× 6 × 6 × 3 = 36 cm3
3
b W = 12 × V = 12 × 36 = 432 grams
4r 3 = 32
4 a
1
( base area × height )
3
7 a V
=
1 + 1 5 + 0 3 + 3 
,
,
= =

2
2 
 2
4
π × 1.53 =
74.2 m3
3
SA =
h × 2π r + 4π r 2
= 8.5 × 2π × 1.5 + 4π × 1.52 = 108 cm2
c
(1,2.5,3)
d=
( x2 − x1 )
+ ( y2 − y1 ) + ( z2 − z1 )
d =
(7 − 1)
(
d=
62 + 15 + 72 = 10
© Oxford University Press 2019
2
2
+
2
15 − 0
)
2
2
+ (10 − 3)
2
10
Worked solutions
9
Area =
1
ab sin C
2
V = 3x 2
3
x 300 − 3x 2
4
9
x 100 − x 2
=
4
1
× 6 × 4 sin30°
2
12
= 12 sin30=
°
= 6 cm2
2
(
tan 45° =
4 3 4
πr =
π × 83 = 2144.66 cm3
3
3
Then
= 171573 cm3
Then
1 2
π r h 171573 cm3
=
3
tan30° =
=
OA
70
= 397 m
tan10°
65.4
m
1 km
60 min
×
×
=
3.92 km / h
min 1000 m
1h
15 We can form a triangle with the bottom of
the building, O
CD
10
° 58°
ACO= 90° − 32=
tan15° =
=
CD 10 =
tan60° 10 3 m
b
CD
OA
CO =
10 3
= 30 m
tan30°
and tan58° =
12 4 faces, equilateral triangles
=
°
tan58
62 − 32 = 3 3
A=
=
x
= 6 x 2 + 8 xh
82 + 72 − 62
= 0.6875
2×8×7
=
θ cos−1 0.6875
= 46.6°
6 x 2 + 8 xh =
600
b
8=
xh 600 − 6 x 2
600 − 6 x 2 300 − 3x 2
=
8x
4x
V= base area × height=
x
(80 + x ) tan15°
−80 tan58° tan15°
= 60.1m
tan58° tan15° − 1
16
a cos θ
=
as required.
c
x
°
tan15
=
80 + x
x ( tan58° tan15° − 1) =
−80 tan58° tan15°
13 a =
A 2 ( x ) (3x ) + 2 xh + 2 (3x ) h
h
=
CO
80 + x
tan58° tan15° ( 80 + x ) =
x
1
1
bh =
×6×3 3 = 9 3
2
2
AT =×
4 9 3 =
36 3 cm2
b
x
CO
x
tan15°
AB = OA − OB = 30 − 10 = 20 m
h=
70
OB
Δd 397 − 70
= = 65.4 m / min
Δt
5
1
π × 402 h =
171573
3
tan60° =
70
OA
distance at angle 10°
=
OB
171573
=
h 3= 102 cm
1600 π
11 a
)
70
= 70 m
tan 45°
=
OA
tan10° =
V=
80V=
80 × 2144.66
T
B
b=
Vcone
)
14 Distance at angle 45°
d 16
=
= 8 cm
2
2
VB =
(
=
Area =
10 a =
r
300 − 3x 2
4x
A=
1
1
ab sin C =
× 8 × 7 × sin 46.6°
2
2
= 20.3 cm2
(3x ) ( x ) h
17 a
2
CB=
AC 2 + AB2 − 2 × AC × AB × cos BAC
CB2= 152 + 342 − 2 × 15 × 34 × cos 25°
CB = 21.4 m
b
ACB= 180° − 25° − 85=
° 70°
© Oxford University Press 2019
11
Worked solutions
Then
sin ACB sin ABC
=
AB
AC
sin130° sin BCA
=
100
70
sin70° sin85°
=
34
AC
sin BCA = 70
sin85°
=
AC 34
= 36.0 m
sin70°
BCA
= sin−1 70
360° − 32.4° − (180° − 150=
° ) 298°
1
=
× 34 × 36.0 × sin25°= 259 m2
2
d
c
sin ACB sin ABC
=
AB
AC
=
BC 2 1002 + 702
− 2 × 100 × 70 × cos17.6°
BC = 39.4km
20 a
sin110° sin 45°
=
34
AC
b
sin QRS sin QSR
=
QS
QR
sin26.5°
=
PQ 119
= 57.1m
sin68.5°
sin 42°
=
= 20.2 m
QS 30
sin85°
c
° 53°
SQR= 180° − 85° − 42=
1
× 20.2 × 30 × sin53°= 242 m2
2
sin85°
=
PA 119
= 127.4 m
sin68.5°
sin PAG =
1
Area = × PQ × QS sin θ =141
2
sin θ =
=
PG 127.4
=
sin26.5° 56.8 m
242
= 86.7°
12 × 20.2
21 a
and the obtuse angle,
180° − 86.7
=
° 93.3°
b
S =4 ×
2
PS=
PQ2 + QS 2 − 2 × PQ × QS × cos θ
2
PS
=
122 + 20.22
− 2 × 12 × 20.2 × cos 93.3°
PS = 24.1m
ABC
= 360° − (180° − 100° ) − 150°
= 130°
b
sin ABC sin BCA
=
AC
AB
5×7
+ 52 =95 cm2
2
h = 72 − 2.52 =6.54 cm
V =
d We choose the obtuse angle,
=
θ 93.3°
19 a
PG
PA
PG
sin26.5° =
127.4
242
12 × 20.2
=
θ sin−1
sin QPA sin PQA
=
QA
PA
sin68.5° sin85°
=
119
PA
1
Area = × QS × QR × sin SQR 2
c
sin PAQ sin QPA
=
PQ
QA
sin26.5° sin68.5°
=
PQ
119
sin 42° sin85°
=
30
QS
Area=
° 85°
PQA= 180° − 95=
QPA
= 180° − 85° − 26.5
=
° 68.5°
sin 45°
=
AC 34
= 25.6 m
sin110°
b
CAB
= 180° − 130° − 32.4
=
° 17.6°
2
BC=
AC 2 + AB2 − 2 × AC × AB × cos CAB
ACB
= 180° − 70
=
° 110°
and so ABC= 180° − 110° − 25=
° 45°
18 a
sin130°
= 32.4°
100
Then the bearing is given by
1
A = × AB × AC × sin BAC
2
c
sin130°
100
M1A1
M1A1
1
× 52 × 6.538... = 54.5 cm3
3
M1A1
22 a
b
l=
2
M1A1
2
10 − 3 = 9.54 m
6 × 9.538...
S= 4 ×
2
= 114.47...
= 114 m2
c
d
2
M1A1
2
h=
42 + 9.538... − 3
= 51.1 m
 3 
arccos 
 = 72.5º
 10 
© Oxford University Press 2019
M1A1
M1A1
12
Worked solutions
e CP 6=
=
tan 60º 10.4 m
23 a
b
M1A1
M1A1
52 + 32 = 5.83 cm
l =
S = 2 × (π × 3 × 5.83...) = 110 cm2
M1A1
1
2 × × π × 32 × 5
3
31.9%
× 100% =
π × 3.052 × 10.1
M1A1
22 − 12
= 5
2
24
a x
=
A1
132 − 52= 12
22 + 12
b=
12 204 cm2
A
×=
2
 5 
c C =
90º + sin−1 
112.6º
=
 13 
M1A1
d
M1A1
h=
25 a
M1A1
M1A1
172 + 122 = 20.8 cm
AC =
^
A1
A B C = 135º
2
AC
=
^
d The angle C D B can either be acute or
obtuse
A1
and the two possible values add up to
180º.
A1
28 a
b
sin C
sin135º
=
20
41.61...
M1
26 a
b
c
d
𝐹𝐹𝐹𝐹 = √82 + 102 + 62 = 10√2
42 + 02 + 62 = 2 13
FM =
2
CM =
2
2
4 + 10 + 0 = 2 29
p = 2 13 + 2 29 + 10 2 cm
e
27 a
b
6
3
𝑡𝑡𝑡𝑡𝑡𝑡 𝑀𝑀 = 4 = 2
𝑐𝑐𝑐𝑐𝑐𝑐 𝑀𝑀 =
A=
1
9
� +1
4
=
M1A1
A1
M1A1
29 a
tan32º =
⇒
=
x
b
c
30 a
b
c
A1
30
x
30
= 48.0 metres
tan32º
𝐴𝐴𝐴𝐴 = 𝑦𝑦 = �(3 + 48.0)2 + 302
= 59.2 metres
M1A1AG
0 + 8 0 + 0 6 + 6
M
,
,
 = ( 4, 0, 6 ) M1A1
2
2 
 2
M1A1A1
1
× 2π × 32 + 2π × 3 × 7 + π × 32
2
= 60
=
π 188 cm (3 s.f.)
^
A1
C = 19.9º
Therefore the bearing of A from point C
M1A1
is 360 − 105 − 19.9 =
235.1o
S =
2
^
b
1 4π
×
× 33 + π × 32 × 7
2 3
= 81
=
π 254 cm3 (3 s.f.)
2
20 + 25 − 2 × 20 × 25 × cos135º
M1A1
A1
AC = 41.6 km
V =
30
M1A1A1
M1A1
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 �51.0� = 30.5º
M1A1
= 89.7 metres
M1A1A1
482 + 572 − 2 × 48 × 57 cos117º
BC
=
1
A = × 48 × 57 sin (117º )
2
= 1219 sq metres (3 s.f.)
𝑠𝑠𝑠𝑠𝑠𝑠 𝐵𝐵
48
=
𝑠𝑠𝑠𝑠𝑠𝑠 117º
89.7
⇒ 𝐵𝐵 = 28.5º
M1A1
M1A1A1
A1
M1A1
M1
2√13
13
1
25
× 5 × 10 sin30º =
2
2
M1A1AG
M1A1
BD2 = 52 + 102 − 2 × 5 × 10 cos 30º
M1A1
=
BD
125 − 50 3
A1
=
BD
25 5 − 2 3
M1
(
)
=
BD 5 5 − 2 3
AG
^
c
sin C D B
sin 45º
=
13
5 5−2 3
^
sin C D B =
13 2
10 5 − 2 3
M1A1
A1
© Oxford University Press 2019
13
Worked solutions
12
Periodic relationships: trigonometric functions
Skills check
1 a
2 a
b
3 a
b
2
2
b
3
c
3
2
( −0.618, 0) ,(1, 0) ,(1.62, 0)
(0.633, 0)
( −1.61, 0.199)
(2.21, 0.792)
i
j
b
c
d
b
c
g
h
i
j
2 a
π
18π
=
180 10
225π
5π
225
=
°
=
180
4
b
c
d
e
f
g
h
80π
4π
=
180
9
200π
10π
=
°
=
200
180
9
80
=
°
120π
2π
=
180
3
135π
3π
135
=
°
=
180
4
=
°
120
π
=×
6 6
π
π
180°
π
= ×
10 10
f
85π
≈ 1.48
180
12.8π
h 12.8
=
°
≈ 0.233
180
g
=
°
85
i
=
°
37.5
j =
1°
37.5π
≈ 0.654
180
π
180
≈ 0.0175
180°
4 a 1c =
1×
57.3°
=
π
180°
b 2c =
=
2×
115°
π
c
180°
0.63c =0.63 ×
d 1.41c =1.41 ×
=
18°
5π
5π 180°
150°
= ×
=
6
6
π
180°
3π =
3π ×
=
540°
180°
π
7π
7π 180°
= ×
=
63°
20 20
π
4π
4π 180°
= ×
=
144°
5
5
π
7π
7π 180°
315°
= ×
=
4
4
π
14π
14π 180°
=
×
=280°
9
9
π
e 1.55c =1.55 ×
180°
f
3c =
3×
g
0.36c =0.36 ×
π
π
=80.8°
180°
π
=88.8°
172°
=
h 1.28c =1.28 ×
i
0.01c = 0.01 ×
j
2.15c =2.15 ×
© Oxford University Press 2019
=36.1°
π
=°
30
180°
π
110π
≈ 1.92
180
75π
75
=
°
≈ 1.31
180
e 110
=
°
270π
3π
=
°
=
270
180
2
360π
360
=
°
= 2π
180
π
25π
≈ 0.436
180
300π
300
=
°
≈ 5.24
180
25
=
°
d
π
45π
=
°
=
45
180
4
π
60π
=
°
=
60
180 3
e 18
=
°
f
10π
≈ 0.175
180
40π
=
°
≈ 0.698
40
180
3 a 10
=
°
Exercise 12A
1 a
5π
5π 180°
= ×
=
300°
3
3
π
13π
13π 180°
=
×
=585°
π
4
4
180°
π
180°
π
180°
π
180°
π
=20.6°
=73.3°
= 0.573°
=
123°
Worked solutions
1
Worked solutions
Exercise 12C
Exercise 12B
1 a i
l = rθ = 14 ×
ii
π
1 a 130° is obtuse, hence we have a
negative cosine
= 7π cm
2
3π
= 9π cm
l = rθ =12 ×
4
b
5π
5π
= cm
6
2
14π
70π
iv l =
=
rθ =
15 ×
cm
9
3
c
iii l =rθ =
3×
b i
1 2
1
π
r θ = × 142 × = 49π cm2
2
2
2
1
1
3π
A = r 2θ = × 122 ×
=54π cm2
2
2
4
A=
ii
1 2
1
5π
15π
cm2
r θ = × 32 ×
=
2
2
6
4
1
1
14π
iv A = r 2θ = × 152 ×
=
175π cm2
2
2
9
iii A =
2
A=
r 2 = 3π × 2 ×
=
r
12
π
= 72
=
72 6 2 cm
3 a=
A
1 2
=
r θ 36π cm2
2
36π × 2 π
θ
=
=
144
2
b
l = rθ = 12 ×
π
2
sin36
=
° sin (180° −=
36° ) sin144°
b
sin50
=
° sin (180° −=
50° ) sin130°
c
sin85
=
° sin (180° − 85
=
° ) sin95°
d
sin=
460° sin ( 460° − 360
=
° ) sin100°
e
= 6π
Then P = 2r + l = 2 × 12 + 6π = 42.8 m
4 Area of sector:
1
1
As = r 2θ = × 102 × 1.5 =75
2
2
Area of triangle:
1 2
1 2
=
At
=
r sin θ
10=
sin1.5 49.9
2
2
note that the angle is in radians
5
l per second: l =rθ =4 ×
12
8π
2π
 8π

− 2π=
sin = sin 
 sin
3
3
3


cos110
=
° cos (360° −=
110° ) cos 250°
c
cos 300
=
° cos (360° − 300
=
° ) cos 60°
d
cos=
500° cos (500° − =
360° ) cos140°
e
15π
π
π

cos= cos  2π − =
 cos
8
8
8


f
π
π 
19π

cos = cos  2π − =
 cos
10
10 
10

3π
3π 
π

cos = cos  2π − =
 cos
2
2 
2

9π
π
 9π

− 2π=
h cos = cos 
 cos
4
4
 4

g
=
3
10800
1 π
=
2 6
angles
1
6 We have
° per minute, which in radians
60
1
π
π
is
=
°
=
60
60 × 180 10800
π
sin−1
π
π
5π

sin = sin  π − =
 sin
6
6
6

60 times in a minute:
Then l =
rθ =
6371 ×
cos
=
40° cos (360° −
=
40° ) cos 320°
b
π
60l = 20π m
1.85km
=
5π

 = sin
7

h
A = As − At = 75 − 49.9 = 25.1 units2
π
2π
2π

= sin  π −
7
7

sin
4 a
Area of shaded region:
π
π
2π

sin = sin  π − =
 sin
3
3
3


π
π
4π

sin = sin  π − =
 sin
5
5
5

g
3 a
1
× 122 × θ =
36π
2
sin225°
is negative sine
cos 225°
divided by negative cosine hence we
have a positive tangent
tan225° =
2 a
f
1 2
1
π
r θ = × r2 ×
= 3π
2
2
12
320° is obtuse, hence we have a
negative sine
b
cos−1
π 5π
6
,
6
2 π
=
2
4
π
π
7π

cos= cos  2π − =
 cos
4
4
4

angles
© Oxford University Press 2019
π 7π
4
,
4
2
Worked solutions
c
tan−1 3 =
2 sin θ − 1 =
0
π
3
sin θ =
π
4π
π

tan= tan  + π=
 tan
3
3
3

angles
5 a
π
5π
−1 1
=
=
θ sin
and
2 6
6
π 4π
3
,
4 a
3
2
2
 8 
sin2 θ= 1 − 

 17 
tan x =
2
 8 
sin θ =
1−

 17 
15
=
±
17
and 0.93 + π =
4.07
b
15
sin θ
15
17
tan
=
θ
= =
8
cos θ
8
17
sin x
1
= −
cos x
2
tan x = −
1 a=
θ cos−1 =
0.6 53.1°
b=
θ sin−1 0.15
= 8.63°
c
and 180° − 8.63
=
° 171.4 °
tan2 x − tan x − 2 =
0
0
( tan x − 2) ( tan x + 1) =
c=
θ tan−1 =
0.2 11.3°
tan x = 2 and tan x = −1
and 180° + 11.3
=
° 191.3°
−1
Then
=
x tan
=
2 1.107..
= 1.11 (3 s.f.)
d θ = tan−1 − 0.76 = 322.8°
+ π 4.249..
= 4.25 (3 s.f.)
and 1.107=
and 322.8° − 180
=
° 142.8°
θ = cos−1 − 0.43 = 115.5°
and=
x tan−1 ( −=
1) 5.50
and 360° − 115.5
=
° 244.5°
and 5.50 − π =
2.36
−1
2 a θ sin
=
=
0.82 0.96
d
and π − 0.96 =
2.18
2 cos2 x + sin x =
1
2(1 − sin2 x ) + sin x =
1
b θ = tan−1 − 0.94 = 5.53
−2 sin2 x + sin x + 1 =
0
and 5.53 − π =
2.39
−(sin x − 1)(2 sin x + 1) =
0
θ = cos−1 − 0.94 = 2.79
so sin x 1,
=
=
then x 1.57
and 2π − 2.79 =
3.49
1
and sin x =
− , then x =
5.76 and 3.67
2
−1
d θ cos
=
=
0.77 0.69
e
1
2
1
x= tan−1 − = 5.82 and 2.68
2
and 360° − 53.1
=
° 306.9°
c
2 sin x + cos x =
0
2 sin x = − cos x
Exercise 12D
e
4
3
−1 4
=
x tan
=
0.93
3
We take the positive value for θ acute
b
4 cos x = 3 sin x
sin x
4
=
cos x
3
sin θ + cos θ =
1
2
1
2
and 2π − 0.69 =
5.59
5 a
θ =sin − 0.23 =6.05
−1
=
θ cos
=
0.3 1.27
−1
and π − 6.05 =
−2.91 =
2π − 2.91 =
3.37
3 2 sin θ + 5 sin θ =
3
cos θ = 0.3
and 2π − 1.27 =
5.02
For 2π ≤ θ ≤ 3π
2
0
( sin θ + 3) (2 sin θ − 1) =
2π + 1.27 =
7.55
For −π ≤ θ ≤ 0
Then sin θ + 3 =
0
5.02 − 2π =
−1.27
sin θ = −3
=
θ sin−1 − 3
This is outside of the domain for the sine
function. Second equation gives us
© Oxford University Press 2019
3
Worked solutions
b
c
tan θ = 1.61
θ tan
1.61 1.01
=
=
−1
and 1.01 + π =
4.16
Because we have 3x , and
−180° ≤ x ≤ 180° , then we use
−540° ≤ 3x ≤ 540° .
4.16 + 2π =
10.4
sin θ = −2 cos θ
sin θ
= −2
cos θ
3x= 60°, −60°,300°, −300°, 420°, −420°
Then x= 20°, −20°,100°, −100°,
tan θ = −2
140°, −140°
θ=
tan−1 − 2 =−1.11
d
2.03
π − 1.11 =
2 tan2 θ + 5 tan θ =
3
x
+3 =
0
2
x
3
=
− =
−1
2
3
tan
1
−3 and tan θ =
tan θ =
2
Because we have
x
, and
2
−180° ≤ x ≤ 180° , then we use
x
−90° ≤
≤ 90° .
2
1
2
θ = −1.25
and θ =
0.46 =
0.46 − 2π =
−5.82
as well as −1.25 − π =
−4.39
and − 5.82 + π =
−2.68
2 a
6 3 cos x = 5 sin x
tan x =
3
5
sin−1
3
= 3θ
2
π 2π 7π 8π 13π 14π
3
,
3
then θ =
b
as well as 31° + 180
=
° 211°
,
3
,
3
,
,
3
3
π 2π 7π 8π 13π 14π
9
,
9
,
9
,
9
,
9
,
9
cos 3θ − 1 =
0
3θ = cos−1 1
Because we have 3θ , and 0 ≤ θ ≤ 2π ,
then we use 0 ≤ 3θ ≤ 6π .
Exercise 12E
3
= 2x
2
3θ = 0,2π , 4π ,6π then θ = 0,
Because we have 2x , and
−180° ≤ x ≤ 180° , then we use
−360° ≤ 2 x ≤ 360° .
2 x= 30°, −30°,330°, −330°
c
sin
θ
2
=
2π 4π
,
,2π
3 3
2
2
θ
2
= sin−1
2
2
Then x= 15°, −15°,165°, −165°
cos−1
x
−90°
=
−45° and so x =
2
3θ =
3
=
x tan−1 = 31°
5
cos−1
Then
Because we have 3θ , and 0 ≤ θ ≤ 2π ,
then we use 0 ≤ 3θ ≤ 6π .
sin x
3
=
cos x 5
b
3 tan
0
( tan θ + 3) (2 tan θ − 1) =
θ =
tan−1 − 3 andθ =
tan−1
1 a
1
2
3x = cos−1
2π + 1.01 =
7.30
d
1
2
cos 3x =
For 2π ≤ θ ≤ 4π
c
2 cos 3x − 1 =
0
3
= 3x
2
Because we have
Because we have 3x , and
−180° ≤ x ≤ 180° , then we use
−540° ≤ 3x ≤ 540° .
then we use 0 ≤
θ
=
4
,
3x= 30°, −30°,330°, −330°,390°, −390°
Then x= 10°, −10°,110°, −110°,
Then θ =
© Oxford University Press 2019
2
2
, and 0 ≤ θ ≤ 2π ,
≤π
π 3π
2
130°, −130°
θ
θ
4
π 3π
2
,
2
4
Worked solutions
d
sin2
sin
2θ
−1 =
0
3
sin2 θ =1 −
2θ
= ±1
3
sin θ =
2θ
, and 0 ≤ θ ≤ 2π ,
3
2θ
4π
then we use 0 ≤
.
≤
3
3
Because we have
π
2
×
3 3π
=
2
4
1 a
c
3
1
 1
−
cos 2θ =
cos2 θ − sin2 θ =−

 − =
4
2
 2
d
sin2θ
tan2
θ
=
=
cos 2θ
2
2 sin5 cos =
5 sin2 × =
5 sin10 by the
double angle formula
π
π
2
 1
2
1
 −  + cos θ =
 8
π
c
2 sin 4π cos 4π = sin2 × 4π = sin8π by the
double angle formula
cos2 θ =
1−
d
cos2 0.4 − sin2 0.4 = cos 2 × 0.4 = cos 0.8
cos θ = −
f
by the double angle formula
double angle formula
π
= 2×−
π
π
b
2
2
acute cos θ =
b
2 2
3
sin2θ = 2 sin θ cos θ = 2 ×
1 2 2 4 2
×
=
3
3
9
2
c
d
3 a
8 1
7
cos 2θ =
cos2 θ − sin2 θ =−   =
9 3
9
tan2
=
θ
sin2θ
=
cos 2θ
4 2
4 2
9
=
7
7
9
sin2 θ + cos2 θ =
1
63
32
cos
=
2θ cos2 θ − sin2 θ
=
1
2
1
  + cos θ =
3
We take only the positive value as θ is
1
63
×−
=
8
8
2
sin2 θ + cos2 θ =
1
1 8
1
cos θ =1 −   =1 − =
9 9
3
63
8
2

63 
 1
= −
− − 


8 
 8

2 a We use the Pythagorean identity
2
1
63
=
64 64
sin2θ = 2 sin θ cos θ
2 cos2 6 −=
1 cos 2 ×=
6 cos12 by the
1 − 2 sin2 = cos 2 × = cos by the
4
4
2
double angle formula
3
sin2 θ + cos2 θ =
1
cos = sin2 × = sin π by the
2
2
2
double angle formula
e
3
−
2
=
1
−
2
θ is obtuse, so we take the negative
value of the cosine
b
2 sin
3
1
3
×− = −
2
2
2
sin2θ = 2 sin θ cos θ = 2 ×
4 a
Exercise 12F
3
2
b
2θ
π
= sin−1 ±=
1
3
2
Then θ =
1 3
=
4 4
63
1
62 31
−
=
=
64 64 64 32
63
32
=
31
32
c
sin2θ
=
=
θ
tan2
cos 2θ
d
sin 4θ = 2 sin2θ cos 2θ
=
2×
5 a
63
31
63 31 31 63
×
=
32 32
512
sin2θ = sin θ
this is true for θ = 0,2π ,π
divide by sin θ
2 cos θ = 1
cos θ =
1
2
π 5π
−1 1
=
θ cos
=
,
2 3 3
2
 1
sin2 θ +  −  =
1
 2
© Oxford University Press 2019
5
Worked solutions
b
6 a
cos 2θ + sin θ =
0
32 sin x cos x
1 − 2 sin θ + sin θ =
0
=2 × 16 × sin x cos x =16 sin2 x
0
− ( sin θ − 1) (2 sin θ + 1) =
Then a=16, b=2
2
b 16 sin2 x = 8
sin θ = 1
2
and sin θ = −
θ = sin−1 −
π
and so x =
11π
6
=
6
7
sin2θ = 3 cos θ
1
Area =
× 15 x sin2θ =
10
2
15 x sin θ cos θ = 10
π 3π
2
,
2
15 x sin θ
divide by cos θ
2 sin θ = 3
sin θ =
d
π 2π
3
10
( 1 − sin θ ) =
with sin θ =
3
2
then θ =
,
3
cos θ = sin θ sin2θ
cos θ = sin θ 2 sin θ cos θ
true for θ =
,
2
1
we have that
4
15 x ×
2 
1 
1
× 1−   =
10
4 
4 


15 x ×
1
15
10
×
=
4
4
15
x = 10
16
π 3π
2
π 5π
,
12 12
1
15x × 2 × sin θ cos θ =
10
2
2 sin θ cos θ = 3 cos θ
this is true for θ =
8
1
=
16 2
Note that 0 ≤ 2 x ≤ 2π , so
1
π 5π
sin−1= 2=
,
x
2
6 6
1
2
1
π 7π
=π +
=
2
6
6
and θ = 2π −
c
so sin2=
x
π
−1
=
=
1
θ sin
2
=
x
divide by cos θ
160 32
=
15
3
2 sin2 θ = 1
sin2 θ =
Exercise 12G
1
2
sin θ = ±
1 a
1
2
π 3π 5π 7π
Then θ =sin−1 ± 1 / 2 = ,
,
,
4 4 4 4
e
cos 2θ = cos θ
2 cos2 θ − 1 =
cos θ
2 cos2 θ − cos θ − 1 =
0
(cos θ − 1)(2 cos θ + 1) =
0
cos θ = 1
cos−1 1= θ= 0,2π
2 cos θ + 1 =
0
cos θ = −
1
2
1
2
θ= cos−1 − =
2π 4π
,
3 3
© Oxford University Press 2019
6
Worked solutions
b
c period of 2π , amplitude of 2 , then
y = −2 cos x
d period of 2π , amplitude of 2 , then
=
y 2 sin ( − x )
4 a The amplitude is 6 = 6
b The period is
2π
=4
π /2
Exercise 12H
c
1 a amplitude 3, period
2π
= π . Option iv
2
b amplitude 2, period 2π and vertical
shift +1 . Option ii
1
π
π
, horizontal shift −
or
2
2
2
units to the right. Option i
c amplitude
d amplitude 1, period
2π
= 4π , vertical
1/2
shift 2. Option iii
d
2 a period 2π , amplitude
max − min 5 − 1
= = 2 , vertical shift +3
2
2
, =
y 2 sin x + 3
b period 2π , amplitude
max − min 2 − 0
= = 1 , vertical shift +1
2
2
horizontal shift π , =
y cos ( x − π ) + 1
2 a The amplitude is 1 = 1 and the period
is
2π
.
3
2π
=π .
2
c The amplitude is −4 =
4 and the period
is
2π
.
3
d The amplitude is −
period is
2π
, amplitude
3
max − min −0.5 + 1.5
= = 0.5 ,
2
2
vertical shift
−1 , y 0.5 sin3x − 1
=
d period
b The amplitude is 0.5 = 0.5 and the
period is
c period 2π , amplitude
max − min 0 + 4
= = 2 , vertical shift −2
2
2
=
y 2 cos x − 2
3 a vertical shift 2, amplitude 3, period π ,
plot given
1 1
= and the
2 2
2π
= 6π .
1
3
3 a period of π , amplitude of 1 , then
y = sin2 x
b period of π , amplitude of 3 , then
y = 3 cos 2 x
© Oxford University Press 2019
7
Worked solutions
b horizontal shift −
π
, amplitude 0.5 ,
3
period 2π , plot given
1
=
y , find zero at
2
π 5π 13π 17π
x = ,
,
,
6 6
6
6
6 Plot sin x −
7 Plot e x − cos x =
y between −2 and −1 .
Find zero at ( −1.29, 0 )
8 a c is the horizontal shift:
π
2
b The graph of y = cos x may be
c vertical shift −1 , amplitude 1 ,
period 2π , horizontal shift −π , plot
given
translated
π
horizontally to the right to
2
become the graph of y = sin x
9 a
d vertical shift 2, amplitude 2, period π ,
plot given
b
10 a
b
c
0.6075,1.571,2.534, 4.712
(2.36, −1)
2π
= 4π
1
2
x = 0,1.26,3.77, 4.19
Exercise 12I
1 a
e vertical shift 1 , amplitude 2 ,
2π
period
, horizontal shift −π ,
3
plot given
4
( −0.824, 0) ,(0.824, 0)
5 Plot 2 sin x − x − 1 =
y . Find zero at
( −2.38, 0)
© Oxford University Press 2019
8
Worked solutions
b
c
f
2 a 1.25, 4.39
b 1.13, 4.53
c
–π ,0,π
d
−0.903, 0.677,1.98,2.61
Exercise 12J
π

1 a Minimum when sin  ( t − 5)  =
−1 , i.e.
6


h ( t ) =−5 + 7 =2 m
π

Maximum when sin  ( t − 5)  =
1 , i.e.
6


h ( t ) = 5 + 7 = 12 m
d
b
π

sin  ( t − 5)  =
1
6


π
 π
⇒t = 8
 ( t − 5)  =
6

 2
High tide at 8 am
c
π

sin  ( t − 5)  =
−1
6


π
π

 ( t − 5)  =− ⇒ t =2
6
2


Low tide at 2 am
e
π

d =
h ( 9 ) 5 sin  ( 9 − 5)  + 7
6


2π
= 5 sin
+7
3
=
© Oxford University Press 2019
5 3
+7
= 11.3 m
2
9
Worked solutions
π
so −12 cos
4
4
+ sin−1   , − sin−1   ,
5
5
4
4
π + sin−1   , 2π − sin−1  
5
5
(t − 5) =−π
6
c
6 a
0:46 am, 3:14 am, 12:46 am, 3:14 pm
(multiply decimals by 60 to convert the
decimal number of hours into minutes)
2 a 13000 on February 1
7000 on August 9th
c
3000 cos 0.5 ( 4 − 1) + 10000 =
10212
)
b 10
cos
20
π
20
23 − 21
1
= −
10
−20
x =
cos−1 −
1
=x =10.64 s
10
60 − 40
= 10
2
period 2 × (1.8 − 0.3) =
3 , then b =
35 − 5
= 15m
2
vertical shift
d=
a 15, 20
=
c vertical shift
=
e horizontal distance between
maximums: 4 − 0 =
4 (period)
b is calculated as=
b
π
2
t + 20 . Find y = 30 at
60 + 40
= 50
2
 2π
50 + 10 cos 
y =
(t − 0.3) 
3


b
 2π
y =
50 + 10 cos 
43.3m
(17.2 − 0.3)  =
 3

maximum is at 2m + 24 m = 26 m
above the ground
minimum is at 2 m above the ground
26 − 2
= 12 , increases then
2
decreases, so a = −12
amplitude
max + min 26 + 2
= = 14 m
2
2
π
h (t ) =
−12 cos
x + 14
20
π
20
c Plotting the function and the line y = 59
gives t = 0.0847 s
Chapter review
1 a
b
30 ×
π
π
=
6
π
5π
150 ×
=
180
6
180
7π
=
4
π
2π
d 120 ×
=
180
3
2π
π
=
b =
40 20
b At t = p , −12 cos
2π
3
maximum at 0.3 instead of 0 , so
horizontal shift of 0.3
π
2π
=
4
2
5 a radius=12 m, so diameter=24 m
=
c
π
7 a amplitude
b 5m
t = 0.535
2π
π
=
40 20
π
b for h =
−20 cos
x + 21 ,
23 m =
20
d 16 fish
4 a 35 m
g=
plot y 15 cos
max = 20 × 2 + 1 = 41m , min = 1m .
Then amplitude is 20, a = −20
h (t ) =
−20 cos
x + 21
20
20 − 0
c amplitude
= 10 , vertical shift 10
2
. Then y 10 sin0.5x + 10
period ≈ 4π=
f
40
= 13.3 s
3
π
3 a 20
c
20
6 , i.e. cos
p=
2π
then=
p
3
p=
period
st
b
π
20
vertical shift =
max + min 41 + 1
= = 21m
2
2
t = 0.771, 3.229, 12.771, 15.229
(
π
1
π
p= − .
20
2
2π
Then the required angle is θ =
3
π
e=
h ( t ) 5 sin  ( t − 5
=
)  + 7 3
6


4
π
−1 
(t − 5=) sin  − 5 
6


p + 14 =
20
c
315 ×
e
−20 ×
f
π
180
π
=
−
π
9
4π
−240 ×
=
−
180
3
© Oxford University Press 2019
180
π
10
Worked solutions
3π
=
−
2
4π
π
h 144 ×
=
180
5
g
2 a
b
c
d
e
f
g
h
−270 ×
π
c
180
6 a
b
7π 180
×
=
−105°
12
π
π 180
×
=
20°
π
9
2x =
6
,
6
6
rad
5
A=
1 2
1
6
r θ = × 25 × = 15
2
2
5
6
,
6
π 5π 13π 17π
,
,
,
12 12 12 12
2
 12 
2
1
−
 + sin θ =
 13 
sin θ =
b
cos2 θ + 0.62 =
1
5
13
cos
=
2θ cos2 θ − sin2 θ
2
8 a
cos θ = ±0.8
2
5
sin (θ + π ) =
− sin θ =
−
13
2 sin2 θ + sin θ − 1 =
0
(sin θ + 1)(2 sin θ − 1)
We take the positive value for acute θ
Then sin θ = −1
cos θ = 0.8
and sin θ =
sin θ
0.6
= = 0.75
cos θ
0.8
b
2 sin x = tan x
144
25
=
169 169
119
 12 
 5 
=
−
 −
 =
169
 13 
 13 
c
cos2 θ =
1 − 0.36 =
0.64
holds for x = 0 . We divide by tan x to
get
1
2
3π
sin−1 − 1 =
2
and sin−1
2 cos x = 1
1 π 5π
= ,
2 6 6
2
1
2
holds for x =
,
7 a the angle is obtuse, so we need a
positive sine
cos2 θ + sin2 θ =
1
b
π 5π 13π 17π
sin2 θ =
1−
θ =
9 a
π
3
and x = −
π
7π 11π
,
6
6
3
2
1
  + cos θ =
4
cos2 θ =
1−
3
9
7
=
16 16
Obtuse angle, so we take the negative
1
π
π
=π +
and2π −
2
6
6
So θ =
1
= 2x
2
so x =
11π 180
×
=330°
6
π
34π 180
×
= 408°
15
π
sin−1 −
1
2
Note that 0 ≤ 2 x ≤ 4π , so
4 a We use the Pythagorean identity
5 a
4 sin2 x = 2
sin−1
7π 180
×
= 420°
3
π
11π 180
−
×
=
−66°
30
π
cos x =
5π
4
8 sin x cos x =
4 × 2 sin x cos x =
4 sin2 x
sin2 x =
6= 5 × θ
b
π
and π=
+
4
2 sin2 x = 1
−
and tan
=
θ
4
so a = 4 and b = 2
3π 180
×
=270°
π
2
7π 180
×
=210°
6
π
3 a We have that rθ = l so
b
π
−1
tan
1
=
cosine cos θ =
b
− 7
4
cos
2θ cos2 θ − sin2 θ
=
2 π
π 7π
−1
cos=
and 2=
π−
2
4
4
4
© Oxford University Press 2019
11
Worked solutions
2
15 a Plot
=
y cos x − x 2 and find zeros for
2

7
1
3
=
−
 −
 −   =
4
4
8
 


π
10 period:
, amplitude =
2
30 + 30
= 30 .
2
2π
= 4 and a = 30
b
Then=
π
2
11 a period is π − 0 =
π
b 3
12 a
b
2π
= 2
π
2
, then x = 0.824
b Plot =
y 4 sin π x − 4e − x + 3 for
0.5 ≤ x ≤ 1.5 and find zero at x = 1.14
1
1
× 10 × 8 × sin θ = 10
ab sin θ =
2
2
16 A =
sin θ =
1
4
Obtuse angle and positive sine,
π
π − 0.25 =
2.89
5
2π
π
−1 1
=
θ sin
=
0.25
4
c 3
d =
b
0≤ x ≤
17 Area of the sector is
1
1
Asector = r 2 θ = × 102 × 0.8 =40
2
2
=4
2
ˆ =
Then OAN
π − 0.8 =
0.77
c
Using the sine rule
sin
ˆ
ˆ
sin ANO
sin OAN
=
AO
ON
π
2 = sin 0.77
10
ON
=
=
ON 10
sin 0.77 6.96
1
Then Atriangle = × 10 × 6.96 × sin 0.8 =25
2
Then
Ashaded
= Asect − Atriangle
= 40 − 25 = 15 cm2
18 a Using the sine rule
Asector − Atriangle
13 A=
shaded
Ashaded
=
1 2
1
r θ − r 2 sin θ
2
2
Ashaded =
π 1
π
1
× 82 × − × 82 × sin
2
6 2
6
= 0.755
14 a =
A
1
2 × 4 × sin=
θ 2
2
2 2 sin θ = 2
sin
=
θ
1
3π
=
4
2
for obtuse θ
b Then CBD =π −
and so ABDC =
3π
π
=
4
4
1
π
× 42 × = 2π
2
4
sin ADO sin AOD
=
AO
AD
sin AOD
sin 0.8
=
AD AO = 8= 14.7 cm
sin ADO
sin 0.4
b
DAO =
π − ADO − AOD
=
π − 0.8 − 0.4 =
1.94
sin DAO sin ADO
=
OD
AO
=
OD
sin DAO
sin1.94..
AO
= 8= 19.1cm
sin ADO
sin 0.4
1 2
1
r θ = × 82 × 0.8 = 25.6 cm2
2
2
c
A=
d
1
Atriangle = × AD × OD × sin ADO
2
1
54.9 cm2
=× 14.7 × 19.1 × sin 0.4 =
2
=
AABCD Atriangle − Asector
= 54.9 − 25.6 = 29.3 cm2
© Oxford University Press 2019
12
Worked solutions
19 a
f ( x ) =2 × 10 × sin3x cos 3x =10 sin6 x
22 a
q
p = 3 ,=
b 10 sin 6 x = 0
sin 6 x = 0
−d =−
for 0 ≤ 6 x ≤ 2π . Then 6 x = 0,π ,2π
and x = 0,
20 a =
f (θ )
π π
b
,
6 3
(2 cos
2
c
d=
f ( x ) 3 sin
l = rθ = 15 ×
A=
π
1
2
Wheel then turns a further
π
6
2π
complete a total turn of
.
3
hA = 15 + 15 sin
π
6
= 15 ×
π
2
, A is 15
radians to
3
= 22.5 m
2
e
π 
h (0) =
15 − 15 cos 2   =
4.39 m
8
π

When cos 2  t +  is −1 , we have a
8

maximum. Then
t =
π
2
−
π
8
= 1.18 s
x +5
π
−(3 sin ( x − 3) + 5) =
fˆ ( x ) =
g (x)
4
A ( t ) = (2 cos t − 1) ( cos t − 1)
24 a
b
A1A1
0
(2 cos t − 1) ( cos t − 1) =
2 cos t − 1 = 0 ⇒ cos t =
1
2
π 5π
⇒ t =,
3 3
π 
π π 
h   = 15 − 15 cos 2  +  = 25.6 m
4
4 8
π

2 t +  =
π
8

4
reflected in the x axis
d
π

−1
cos 2  t +  =
8

π
π
fˆ ( x ) 3 sin ( x − 3) + 5
=
4
1 2
1
π
r θ = × 152 × = 58.9 m2
2
2
6
m above the ground.
8+2
=5
2
translated (3,0)
= 7.85m
6
2π
π
=
8
4
b 10 − 2 =
8 =period,=
b
cos θ = 0
⇒θ =
−90°, 90°
c When wheel turns through
f
23 a 3
c
⇒θ =
−120°,120°
b
4
3 cos
=
x sin2 x − 1
3
cos θ (2 cos θ + 1) =
0
2 cos θ + 1 =0 ⇒ cos θ =−
21 a
2
=−1 , so d = 1
2
The intersection points can be found as
x = 1.30,3.41,6.19
θ − 1) + cos θ + 1
= 2 cos2 θ + cos θ
b This is a quadratic equation in cos θ
with positive discriminant, so there are 2
distinct values of cos θ which satisfy it.
2π
4
2π
,=
=
b = 2
3π
3
π
2
M1A1
cos t − 1 = 0 ⇒ cos t = 1
M1A1
t = 0,2π
25 2 cos x = sin2 x
2
M1
⇒ 2 cos2 x − 2 sin x cos x =
0
2 cos x ( cos x − sin x ) =
0
=
cos x 0,
=
cos x sin x
x =
x =
π
2
π
4
M1
M1
, x =
3π
2
A1
, x =
5π
4
A1
26 a Correct attempt to at least one
parameter
=
a
=
b
=
c
© Oxford University Press 2019
M1
14 − 8
= 3
2
A1
2π
= 2
A1
14 + 8
= 11
2
A1
π
13
Worked solutions
b
ii θ =
29 a i
A1
2
A1
a = −2
A1
=
b
A1 for trigonometric scale and correct
domain,
A1 for correct max/min,
27 a
2
2
S ( x ) = sin
2x
2x + 2
x cos
144
4 2+ cos
4 4 43
1 sin2
4 44 2
4 42
43x
AG
b A1 for correct shape, A1 for 2 cycles,
A1 for correct max/min
M1A1
A1
−2 cos π x + 1 = 0 ⇒ cos π x =
 π π 5π 7π 
, ,
,
,
 3 3 3 3 
M1A1A1
= 1 + sin 4x
2π
= π
2
c =1
c
A1 for two complete cycles
A1
−1 ≤ y ≤ 3
ii
b
π
18
π x ∈ −
M1
 1 1 5 7 
x ∈  − , , , ,
 3 3 3 3 
M1
30 a i =
x 0,
=
x
ii
b
1
M1
2
π
=
,x π
2
π
A1
A1
2
iii ¡
A1
b=2
A1
d =1
A1
c The first point of inflexion occurs at
π
R1
x =
4
c i
ii
π
A1
2
0≤y ≤2
d <Please insert the graph of
=
y cos (2 x ) − 1 for 0 ≤ x ≤ 2π >
e i
ii
1
2
ii
b i
A3
A=
1
2π
4π
× 22 ×
=
2
3
3
2π
4π
l =
2×
=
3
3
rθ =
r
2
θ
2
π
3
=π
 π
−3
a tan  −  =
4
1 4 2 43 
A1
a=3
AG
−1
p = π (or any π + 2nπ , n ∈ ¢ )
A1
A1
π 
f   = −2
8
  π π 
⇒ f ( x ) =a tan  2  −   + 1 =−2
  8 4 
M1
A1
A1
q = −2
28 a i
d
e
π
A1
8
5π
9π
and
8
8
A1
31 a
A1
Solve simultaneously
M1
r = 6 cm
A1
−2 cos2 x + sin x + 3
(
)
=
−2 1 − sin2 x + sin x + 3
A1
A1
A1
= 2 sin2 x + sin x + 1
b
M1
A1
−2 cos2 x + sin x + 3 =
2
⇒ 2 sin2 x + sin x + 1 =
2
M1
2
2 sin x + sin x − 1 =
0
⇒ (2 sin x − 1) ( sin x + 1) =
0
© Oxford University Press 2019
M1
14
Worked solutions
sin x =
1
, sin x = −1
2
A1
π
7π π 5π 3π 
 11π
x ∈ −
,− ,−
, ,
,

6
2
6 6 6 2 

A2
Award A1 for two correct
solutions
© Oxford University Press 2019
15