Chapter 4 Trigonometry
Chapter 4 Prerequisite Skills
Chapter 4 Prerequisite Skills
Question 1 Page 200
a)
cos θ =
5
3
θ
4
3
, tan θ =
5
4
4
b)
Using the CAST rule; sine and tangent are negative:
sin θ = !
12
12
, tan θ = 5!
13
5
θ
12
13
c)
25
7
x
Using the CAST rule; cosine and tangent are negative:
7
24
sin x =
, cos x = !
25
25
24
d)
15
8
x
Using the CAST rule; cosine and sine are negative:
15
8
cos x = ! , tan x =
17
15
17
MHR • Advanced Functions 12 Solutions
400
Chapter 4 Prerequisite Skills
Question 2 Page 200
a) positive (sine in A quadrant); 0.2588
b) positive (cosine in A quadrant); 0.5592
c) positive (tangent in A quadrant); 3.7321
d) positive (sine in S quadrant); 0.9848
e) negative (cosine in S quadrant); –0.9205
f)
positive (tangent in T quadrant); 2.7475
g) negative (sine in C quadrant); –0.8480
h) positive (cosine in C quadrant); 0.9781
Chapter 4 Prerequisite Skills
Question 3 Page 200
a) 41°
c) 83°
b) 65°
Chapter 4 Prerequisite Skills
a)
1
2
d) 117°
Question 4 Page 200
b) 3
c)
Chapter 4 Prerequisite Skills
5
3
d)
2
3
Question 5 Page 200
a)
sec x =
5
4
5
3
, cot x =
3
4
x
3
b)
csc θ = !
θ
13
12
, cot θ = !
5
5
12
5
13
MHR • Advanced Functions 12 Solutions
401
c)
csc x =
25
7
25
25
, sec x = !
7
24
x
24
d)
sec θ = !
17
8
, cot θ =
8
15
8
θ
15
17
Chapter 4 Prerequisite Skills
Question 6 Page 200
a)
1
= 1.7434
sin 35°
b)
1
= –1.2361
cos 216°
c)
1
= 2.1445
tan 25°
d)
1
= 1.1792
sin122°
e)
1
= –1.2690
cos142°
f)
1
= 1.0724
tan 223°
g)
1
= –1.5890
sin 321°
h)
1
= 1.0038
cos 355°
Chapter 4 Prerequisite Skills
" 1 %
a) sin !1 $
= 53°
# 1.25 '&
c)
" 1 %
= 18°
tan !1 $
# 3.1416 '&
Question 7 Page 200
" 7%
b) cos !1 $ ' = 54°
# 12 &
" 1 %
d) cos !1 $
= 139°
# !1.32 '&
MHR • Advanced Functions 12 Solutions
402
Chapter 4 Prerequisite Skills
Question 8 Page 200
60°
2
2
45°
3
b)
1
θ
sin θ
cos θ
tan θ
30°
1
2
1
1
3
2
1
2
2
3
2
1
2
45°
c)
1
ds
1
30°
a)
45°
60°
Chapter 4 Prerequisite Skills
1
θ
1
3
Question 9 Page 201
& 1 1 #
P$
,
!
% 2 2"
a)
3
b) x =
1
2
,y=
1
2
1
c) csc 45° =
2
1
2 , sec 45° =
2 , cot 45° = 1
2
Chapter 4 Prerequisite Skills
! 1 1 "
Q$#
,
%
2 2'
&
1
2
& 1 1 #
P$
,
!
% 2 2"
1
1
2
2
sin 315° = –
2
1
1
1 "
! 1
R$#
,#
%
2
2'
&
sin 225° = –
1
1
1
sin 135° =
Question 10 Page 201
1
1
2
2
1 "
! 1
S$
,#
%
2'
& 2
1
2
1
, cos 135° = –
2
1
2
, cos 225° = –
, cos 315° =
1
2
1
, tan 135° = –1, csc 135° =
2
1
2
2 , sec 135° = – 2 , cot 135° = –1
, tan 225° = 1, csc 225° = – 2 , sec 225° = – 2 , cot 225° = 1
, tan 315° = –1, csc 315° = – 2 , sec 315° =
2 , cot 315° = –1
MHR • Advanced Functions 12 Solutions
403
Chapter 4 Prerequisite Skills
Question 11 Page 201
a) AB = (10 ! 6)2 + (8 ! 5)2
b) CD = (!5 ! 7)2 + (!3 ! 2)2
c)
= 16 + 9
= 144 + 25
= 25
=5
= 169
= 13
EF = (!8 ! 7)2 + (4 ! 12)2
d) GH = (!3 ! 3)2 + (!6 ! 2)2
= 225 + 64
= 36 + 64
= 289
= 17
= 100
= 10
Chapter 4 Prerequisite Skills
Question 12 Page 201
a) a2 + 2ab + b2
b) c2 – d2
c) 6x2 – 4xy + 3xy – 2y2 = 6x2 – xy – 2y2
d) sin2 x + 2 sin x cos y + cos2 y = 1 + 2 sin x cos y
Chapter 4 Prerequisite Skills
Question 13 Page 201
MHR • Advanced Functions 12 Solutions
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Chapter 4 Prerequisite Skills
Question 14 Page 201
MHR • Advanced Functions 12 Solutions
405
Section 1
Radian Measure
Chapter 4 Section 1
Question 1 Page 208
a)
!
3
b)
!
2
c)
Chapter 4 Section 1
a)
!
12
b)
!
2
b)
!
18
c)
!
8
b)
3!
4
!
12
c)
b) 10 !
"
7"
=
180 6
e)
d)
5!
4
!
20
d)
!
60
"
"
=
180 18
c)
315 !
"
5"
=
180 3
f)
75 !
300 !
"
7"
=
180 4
"
5"
=
180 12
Question 6 Page 208
!
=& 0.40
180
b) 51 ×
!
=& 2.23
180
e) 240 ×
d) 128 ×
!
36
Question 5 Page 208
Chapter 4 Section 1
a) 23 ×
d)
Question 4 Page 208
"
2"
=
180 9
d) 210 !
!
24
c) π
Chapter 4 Section 1
a) 40 !
5!
6
Question 3 Page 208
Chapter 4 Section 1
a)
d)
Question 2 Page 208
Chapter 4 Section 1
a)
2!
3
Chapter 4 Section 1
!
=& 0.89
180
c) 82 ×
!
=& 4.19
180
f)
!
=& 1.43
180
330 ×
!
=& 5.76
180
Question 7 Page 208
a)
! 180°
= 36°
"
5
!
b)
! 180°
= 20°
"
9
!
c)
5! 180°
= 75°
"
12
!
d)
5! 180°
= 50°
"
18
!
e)
3! 180°
= 135°
"
4
!
f)
3! 180°
= 270°
"
2
!
MHR • Advanced Functions 12 Solutions
406
Chapter 4 Section 1
Question 8 Page 208
a) 2.34 ×
180°
=& 134.1°
!
b) 3.14 ×
180°
=& 179.9°
!
c) 5.27 ×
180°
=& 301.9°
!
d) 7.53 ×
180°
=& 431.4°
!
e) 0.68 ×
180°
=& 39.0°
!
f)
1.72 ×
180°
=& 98.5°
!
Chapter 4 Section 1
!=
Question 9 Page 208
a
r
a = !r
a = (4.75)(25)
a = 118.75 cm
Chapter 4 Section 1
Question 10 Page 208
a) 360° × 2 = 720°/s
Chapter 4 Section 1
b) 2π × 2 = 4π rad/s
Question 11 Page 208
Let x represent the equal angles and y represent the other angle.
2x + y = π, the sum of the angles of a triangle.
Given, x = 2y:
( )
2 2y + y = !
5y = !
!
y=
5
2!
x=
5
The three angles are
Chapter 4 Section 1
! 2!
2!
,
, and
.
5 5
5
Question 12 Page 208
Answers may vary depending on speed of animation. A sample solution is shown.
The time for five complete revolutions was 20 s.
!
5 ! 360
5 ! 2"
a)
= 90°/s
b)
=
rad/s
20
20
2
MHR • Advanced Functions 12 Solutions
407
Chapter 4 Section 1
Question 13 Page 209
1
"
!
=& 0.000 291
60 180
a)
b) 0.000 291 × 6 400 000 =& 1862
A nautical mile is approximately 1862 m.
c) Answers may vary. A sample solution is shown.
The earth approximates a sphere. Its radius varies from place to place. It varies from
6356.750 km (polar radius) to 6378.135 km (equatorial radius).
Chapter 4 Section 1
a) ! =
Question 14 Page 209
a
r
=
(1)
(1000)
=
1
rad
1000
b) r =
=
a
!
(2)
(0.25)
= 8 km
Chapter 4 Section 1
Question 15 Page 209
3480
384 400 " 6400
=& 0.009 206 rad
180°
! =& 0.009 206 #
$
=& 0.5°
! =&
Chapter 4 Section 1
1.2
2.4
= 0.5 rad
!=
Chapter 4 Section 1
Question 16 Page 209
180°
#
=& 28.6°
! = 0.5 "
Question 17 Page 209
12 000 ! 2"
= 400π rad/s
60
The angular velocity of the engine is 400π rad/s or approximately 1256.6 rad/s.
MHR • Advanced Functions 12 Solutions
408
Chapter 4 Section 1
a)
Question 18 Page 209
a = !r
#"&
= % ( (80)
$ 3'
=
80"
3
The length of the on-ramp is
80!
m.
3
b) The approximate length of the on-ramp is 83.8 m.
Chapter 4 Section 1
Question 19 Page 209
a) It must follow the rotation of the earth.
b) 24 h, since it takes the earth 24 h (one day) to rotate.
c) Dividing 2π by 24 h, 60 min, and 60 s; the angular velocity is approximately
0.000 023π rad/s.
d) It is the same. For the satellite to stay above the same point, it must travel at the same angular
velocity.
Chapter 4 Section 1
a)
90°
x
=
360° 400
1
x
=
4 400
x = 100
100 grads
Chapter 4 Section 1
Question 20 Page 209
b)
x 150
=
2! 400
300!
x=
400
3!
x=
4
Question 21 Page 210
Solutions to Achievement Check questions are provided in the Teacher’s Resource.
Chapter 4 Section 1
Question 22 Page 210
Radius of orbit is about 35 900 + 6400 = 42 300 km or 42 300 000 m.
Divide the length of orbit in one day by 24 h, 60 min, and 60 s:
2! " 42 300 000
=& 3076 m/s
24 " 60 " 60
The orbital speed of a geostationary satellite is approximately 3076 m/s.
MHR • Advanced Functions 12 Solutions
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Chapter 4 Section 1
Question 23 Page 210
1852
× 30 =& 15.4
divide by 3600 to get metres per second, multiply by 30 to get 30 s.
3600
Using the modern definition of a nautical mile, 1852 m, the knots are approximately 15.4 m apart.
Chapter 4 Section 1
Question 24 Page 210
The angular velocities are the same because we all pass through a 360° rotation in a day.
Chapter 4 Section 1
a)
Question 25 Page 210
A
"
=
2
2!
!r
1
A = r 2"
2
b)
A=
"!%
1
(12)2 $ '
2
# 5&
=& 45.24 cm 2
Chapter 4 Section 1
Question 26 Page 210
" !%
" !%
" 7! %
" 3! %
a) A $ 1, ' , B $ 2, ' , C $ 2, ' , D $ 2, '
# 3&
# 4&
# 4 &
# 2&
b) i)
To find r, use the Pythagorean Theorem:
r = 12 + 12
r= 2
The angle is shown on the diagram.
"
!%
$# 2, 4 '&
ii) r = (!3)2 + (4)2
r = 25
r =5
Since tan θ =
(5, 2.21)
4
, θ is about 2.21 rad.
!3
iii) r = 5
(0, –5) is on the y-axis, so ! =
3"
.
2
" 3! %
$# 5, 2 '&
MHR • Advanced Functions 12 Solutions
410
Section 2
Trigonometric Ratios and Special Angles
Chapter 4 Section 2
Question 1 Page 216
a) Set mode to degrees. Screen shot for i):
i)
0.4226
ii) 0.3090
iii) –2.1445
iv) 0.2588
iii) –2.1452
iv) 0.2586
b) Set mode to radians. Screen shot for i):
i)
0.4223
ii) 0.3087
c) The degree measures are approximately the same as the radian measures.
For example, 25° =& 0.436.
Chapter 4 Section 2
Question 2 Page 216
a) Set the mode to radians.
i) 0.9356
ii) –0.8187
iii) –0.0918
iv) 0.0076
b) Set the mode to degrees.
i) 0.9336
ii) –0.8192
iii) –0.0875
iv) 0.0000
c) The degree measures are approximately the same as the radian measures.
For example, 1.21 =& 69°.
MHR • Advanced Functions 12 Solutions
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Chapter 4 Section 2
Question 3 Page 216
Set mode to radians. Screen shot for a):
a) 0.7071
b) 0.9010
c) –0.5774
Chapter 4 Section 2
d) 0.4142
Question 4 Page 216
Set mode to degrees. Screen shot for a):
a) 3.8637
b) 1.6243
c) –0.6745
Chapter 4 Section 2
d) –2.6695
Question 5 Page 216
Set mode to radians. Screen shots for b) and c):
a) 1.2123
b) –3.7599
c) 14.5955
d) 1.0582
MHR • Advanced Functions 12 Solutions
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Chapter 4 Section 2
Question 6 Page 216
Set mode to radians. Screen shot for a):
a) –2.0000
b) 1.5270
c) –0.3249
Chapter 4 Section 2
a)
d) –2.7475
Question 7 Page 216
1
P
2225
2!
3
!
3
–1
0
–1
1
The terminal arm of an angle of
!
intersects the unit circle at a point with coordinates
3
! 1 3$
2!
is in the second quadrant, the
# , & . Since the terminal arm of an angle of
2
2
3
"
%
" 1 3%
coordinates of the point of intersection are P $ ! , ' .
# 2 2 &
sin
2!
3
2!
1
2!
=
,cos
= " , tan
=" 3
3
2
3
2
3
MHR • Advanced Functions 12 Solutions
413
b)
1
P
5!
6
!
–1
6
0
1
–1
The terminal arm of an angle of
!
intersects the unit circle at a point with coordinates
6
! 3 1$
5!
is in the second quadrant, the
, & . Since the terminal arm of an angle of
#
6
" 2 2%
"
3 1%
coordinates of the point of intersection are P $ !
, '.
# 2 2&
sin
5! 1
5!
3
5!
1
= ,cos
="
, tan
="
6 2
6
2
6
3
c)
1
3!
2
–1
P
!
2
1
–1
!
intersects the unit circle at a point with coordinates (0, 1).
2
3!
Since the terminal arm of an angle of
is in the third quadrant, the coordinates of the point
2
of intersection are P(0, –1).
The terminal arm of an angle of
sin
3!
3!
3!
= –1, cos
= 0, tan
is undefined
2
2
2
MHR • Advanced Functions 12 Solutions
414
d)
1
7!
!
4
–1
1
4
P
–1
!
intersects the unit circle at a point with coordinates
4
! 1 1 $
7!
,
#"
&% . Since the terminal arm of an angle of 4 is in the fourth quadrant, the
2 2
The terminal arm of an angle of
" 1
1 %
coordinates of the point of intersection are P $
,!
'.
# 2
2&
sin
7!
1
7!
1
7!
="
,cos
=
, tan
= "1
4
4
4
2
2
Chapter 4 Section 2
Question 8 Page 216
1
a)
7!
6
1
!
–1
6
P
–1
The terminal arm of an angle of
!
intersects the unit circle at a point with coordinates
6
! 3 1$
7!
is in the third quadrant, the coordinates
, & . Since the terminal arm of an angle of
#
6
" 2 2%
"
3 1%
of the point of intersection are P $ !
,! ' .
2&
# 2
sin
7!
1
7!
3
7!
1
7!
7!
2
7!
= " ,cos
="
, tan
=
,csc
= "2,sec
="
,cot
= 3
6
2
6
2
6
6
6
6
3
3
MHR • Advanced Functions 12 Solutions
415
b)
1
4!
3
1
–1
P
–1
The terminal arm of an angle of
!
intersects the unit circle at a point with coordinates
3
! 1 3$
4!
is in the third quadrant, the coordinates
# , & . Since the terminal arm of an angle of
3
"2 2 %
" 1
3%
of the point of intersection are P $ ! ,!
'.
2 &
# 2
sin
4!
3
4!
1
4!
4!
2
4!
4!
1
="
, cos
= " , tan
= 3, csc
="
, sec
= "2, cot
=
3
2
3
2
3
3
3
3
3
3
c)
1
5!
4
1
–1
P
–1
!
intersects the unit circle at a point with coordinates
4
! 1 1 $
5!
,
#"
&% . Since the terminal arm of an angle of 4 is in the third quadrant, the
2 2
The terminal arm of an angle of
" 1
1 %
coordinates of the point of intersection are P $ !
,!
'.
#
2
2&
sin
5!
1
5!
1
5!
5!
5!
5!
="
,cos
="
, tan
= 1,csc
= " 2,sec
= " 2,cot
=1
4
4
4
4
4
4
2
2
MHR • Advanced Functions 12 Solutions
416
d)
1
π
P
–1
1
–1
The terminal arm of an angle of π intersects the unit circle at a point with coordinates (0, –1).
sin π = 0, cos π = –1, tan π = 0, csc π is undefined, sec π = –1, cot π is undefined
Chapter 4 Section 2
Question 9 Page 217
y1
40 m
!
3
y2
40 m
!
4 x1
x2
a)
cos
! x2
=
4 40
40
x2 =
2
x2 " x1 =
sin
! x1
=
3 40
x1 = 20
40 2
" 20
2
= 20
b)
cos
( 2 " 1) m
! y1
=
3 40
y1 = 20 3
y1 " y2 = 20
( 3 " 2) m
! y2
=
4 40
40 2
y2 =
2
sin
c) Approximately 8.3 m horizontally and 6.4 m vertically.
MHR • Advanced Functions 12 Solutions
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Chapter 4 Section 2
Question 10 Page 217
y2
60 m
40 m
y1
x1
x2
cos
a)
! x2
=
3 60
x2 = 30
(
cos
)
! x1
=
4 40
x1 = 20 2
x2 " x1 = 30 " 20 2 m
b) The kite moves farther from Lynda, since the horizontal distance of the kite at
!
has
3
increased.
sin
c)
! y2
=
3 60
y2 = 30 3
(
sin
)
! y1
=
4 40
y1 = 20 2
y2 " y1 = 30 3 " 20 2 m
The altitude increases since the vertical distance of the kite at
!
has increased.
3
d) Approximately 1.7 m horizontally and 23.7 m vertically.
Chapter 4 Section 2
a) i)
3 1
!
2
3 = 1 ! 2 or
1
2
Question 11 Page 217
2
ii) 1+
1
3
! 3 = 1+ 1
=2
2
b) i) Both sides from part a) equal about 1.4142.
ii) Both sides from part a) equal 2.
MHR • Advanced Functions 12 Solutions
418
Chapter 4 Section 2
Question 12 Page 217
1 # 1 &
! "%!
(
2 $
1
3'
=
" 2
a) i)
1
2 3
" 1 %
ii) 1+ $ !
' ( ! 3 = 1+ 1
#
3&
( )
=2
2
=
b) i)
1
6
Both sides from part a) equal about 0.167.
ii) Both sides from part a) equal 2.
Chapter 4 Section 2
Question 13 Page 217
" ! % AC
a) cos $ ' =
# 4 & 60
AC =
1
2
" ! % AB
b) cos $ ' =
# 6 & AC
( 60
AB =
AB = 15 6 m
AC = 30 2 m
Chapter 4 Section 2
a)
!
2
b)
3
( 30 2
2
Question 14 Page 217
2!
3
c) 9:00
d) 11:00
Chapter 4 Section 2
e)
5!
4
Question 15 Page 218
a), b)
0.500π radians
c), d)
The values are approximately the same.
MHR • Advanced Functions 12 Solutions
419
Chapter 4 Section 2
a) i)
!
Question 16 Page 218
3 1 #
3& # 1&
3
3
" +%!
+
( "%! ( = !
2 2 $ 2 ' $ 2'
4
4
=0
#
&
#
1
1
1
1 & 1 1
ii) !
"%!
!
"
!
(
%
(= +
2 $
2'
2 $
2' 2 2
=1
b) i)
ii)
Chapter 4 Section 2
a) i)
Question 17 Page 218
3
3 # 1& 1 3 1
!
" " ! = +
2
2 %$ 2 (' 2 4 4
=1
ii)
1
2
!
# 1 & 1
1 1
+%"
!
= "
(
2 $
2'
2 2 2
1
=0
b) i)
ii)
MHR • Advanced Functions 12 Solutions
420
Chapter 4 Section 2
a) i)
( ) =0
1! (1) ( !1)
1+ !1
Question 18 Page 218
! 3!
ii)
1
3 =
" 1 %
1+ ! 3 $
# 3 '&
( )
! 3!
1
3
1! 1
! 3!
=
0
1
3 is undefined
b) i)
Chapter 4 Section 2
cos
! OB
=
6 10
OB = 10cos
cos
Question 19 Page 218
!
=
4
!
6
OC
!
6
!
!
OC = 10cos cos
6
4
10cos
Chapter 4 Section 2
Question 20 Page 219
Solutions to Achievement Check questions are provided in the Teacher’s Resource.
MHR • Advanced Functions 12 Solutions
421
Chapter 4 Section 2
a)
Question 21 Page 219
x 150
=
2! 400
3
x = " 2!
8
3!
x=
4
sin (150 grads) =
csc (150 grads) =
1
2
, cos (150 grads) = –
1
2
, tan (150 grads) = –1,
2 , sec (150 grads) = – 2 , cot (150 grads) = –1
b) Answers may vary. A sample solution is shown.
!
!
The special angles 30° and 60°, or
and
in radians, would be in fraction form.
6
3
Chapter 4 Section 2
Question 22 Page 219
a)
b), c)
MHR • Advanced Functions 12 Solutions
422
d)
Domain: {x ∈ R, 0.00 ≤ x ≤ 0.31}
e)
0.00 ≤ x ≤ 0.144
Chapter 4 Section 2
Question 23 Page 219
a) Answers may vary. A sample solution is shown.
!
3
! cos 6
2
sin "
= 1"
!
2
3
tan
3
1
= 1"
2
1
=
2
MHR • Advanced Functions 12 Solutions
423
Chapter 4 Section 2
Question 24 Page 219
Method 1:
!5
, the angle can be drawn in a 5-12-13 right triangle.
13
Extend the adjacent side so it is as long as the hypotenuse and label all the vertices.
Since the sum of the angles in a triangle equals 180° or π, the other two angles in the isosceles
! "#
triangle each equal
.
2
B 1
12
A
Since sin θ =
D
θ
Since ∠ACB =
5
'
!
! "# $ !
!
– θ, ∠BCD =
– & " # ) or .
2
2
2
%2
(
!
!
From the diagram, sin
=
2
26
13
1
26
or
.
26
26
2
–θ
C
A
Method 2:
" 5%
sin !1 $ ! ' =& !0.395 Using the CAST rule, ( is in quadrant 3 or 4.
# 13 &
sin
0.395
=& 0.196
2
0.196 =&
Using the CAST rule,
(
is in the quadrant 2.
2
26
26
A
Chapter 4 Section 2
Since sin θ =
tan θ =
B
4
3
Question 25 Page 219
!4
, draw the right triangle with θ and look at the tangent ratio.
5
4
5
θ
3
MHR • Advanced Functions 12 Solutions
424
Section 3
Equivalent Trigonometric Expressions
Chapter 4 Section 3
Question 1 Page 225
!
!
!
lies in the first quadrant, it can be expressed as the difference
– .
3
2
6
Now apply the cofunction identity:
#! !&
!
cos
= cos % " (
3
$ 2 6'
Since an angle of
!
6
= sin
=
1
2
Chapter 4 Section 3
Question 2 Page 225
!
!
!
lies in the first quadrant, it can be expressed as the difference
– .
4
2
4
Now apply the cofunction identity:
#! !&
!
sin
= sin % " (
4
$ 2 4'
Since an angle of
= cos
=
!
4
1
2
Chapter 4 Section 3
Question 3 Page 225
2!
!
!
lies in the second quadrant, it can be expressed as the sum
+ .
3
2
6
Now apply the cofunction identity:
"! !%
2!
cos
= cos $ + '
3
# 2 6&
Since an angle of
= –sin
=–
!
6
1
2
MHR • Advanced Functions 12 Solutions
425
Chapter 4 Section 3
Question 4 Page 225
3!
!
!
lies in the second quadrant, it can be expressed as the sum
+ .
4
2
4
Now apply the cofunction identity:
"! !%
3!
sec
= sec $ + '
4
# 2 4&
Since an angle of
= –csc
!
4
=– 2
Chapter 4 Section 3
Question 5 Page 225
!
!
5!
lies in the first quadrant, it can be expressed as the difference
–
.
7
2
14
Now apply the cofunction identity:
# ! 5! &
!
cos
= cos % " (
7
$ 2 14 '
Since an angle of
= sin
So, y =
5!
14
5!
.
14
Chapter 4 Section 3
Question 6 Page 225
4!
!
!
lies in the first quadrant, it can be expressed as the difference
–
.
9
2 18
Now apply the cofunction identity:
#! ! &
4!
cot
= cot % " (
9
$ 2 18 '
Since an angle of
= tan
So, z =
!
18
!
.
18
MHR • Advanced Functions 12 Solutions
426
Chapter 4 Section 3
Question 7 Page 225
13!
!
2!
lies in the second quadrant, it can be expressed as the sum
+
.
18
2
9
Now apply the cofunction identity:
" ! 2! %
13!
cos
= cos $ +
18
# 2 9 '&
Since an angle of
= –sin
So, y =
2!
9
2!
.
9
Chapter 4 Section 3
Question 8 Page 225
13!
!
3!
lies in the second quadrant, it can be expressed as the sum
+
.
14
2
7
Now apply the cofunction identity:
" ! 3! %
13!
cot
= cot $ + '
14
#2 7 &
Since an angle of
= –tan
So, z =
3!
7
3!
.
7
Chapter 4 Section 3
Question 9 Page 225
5!
!
3!
lies in the first quadrant, it can be expressed as the difference
–
.
22
2
11
Now apply the cofunction identity:
# ! 3! &
5!
sin
= sin % " (
22
$ 2 11 '
a) Since an angle of
3!
11
=& 0.6549
= cos
17!
!
3!
lies in the second quadrant, it can be expressed as the sum
+
.
22
2
11
Now apply the cofunction identity.
" ! 3! %
17!
sin
= sin $ + '
22
# 2 11 &
b) Since an angle of
3!
11
=& 0.6549
= cos
MHR • Advanced Functions 12 Solutions
427
Chapter 4 Section 3
Question 10 Page 225
5!
!
2!
lies in the first quadrant, it can be expressed as the difference
–
.
18
2
9
Now apply the cofunction identity:
# ! 2! &
5!
cot
= cot % "
18
$ 2 9 ('
a) Since an angle of
2!
9
=& 0.8391
= tan
13!
!
2!
lies in the second quadrant, it can be expressed as the sum
+
.
18
2
9
Now apply the cofunction identity:
" ! 2! %
13!
cot
= cot $ +
18
# 2 9 '&
b) Since an angle of
2!
9
=& –0.8391
= –tan
Chapter 4 Section 3
Question 11 Page 226
#!
&
Since the angle a lies in the first quadrant, use the cofunction identity sec 1.45 = csc % " 1.45( .
$2
'
Find the measure of angle a.
!
a = – 1.45
2
a =& 0.12
Chapter 4 Section 3
Question 12 Page 226
#!
&
Since the angle b lies in the first quadrant, use the cofunction identity csc 0.64 = sec % " 0.64( .
$2
'
Find the measure of angle b.
!
b = " 0.64
2
b =& 0.93
MHR • Advanced Functions 12 Solutions
428
Chapter 4 Section 3
Question 13 Page 226
Since the angle a lies in the second quadrant, use the cofunction identity.
"!
%
sec 0.75 = –csc $ + 0.75' . Find the measure of angle a.
#2
&
!
+ 0.75
2
a =& 2.32
a=
Chapter 4 Section 3
Question 14 Page 226
Since the angle b lies in the second quadrant, use the cofunction identity.
"!
%
csc 1.34 = –sec $ + 1.34' . Find the measure of angle b.
#2
&
!
+ 1.34
2
b =& 2.91
b=
Chapter 4 Section 3
Question 15 Page 226
sin(π – x) = sin x, cos(π – x) = –cos x, tan(π – x) = –tan x,
csc(π – x) = csc x, sec(π – x) = –sec x, cot(π – x) = –cot x
Chapter 4 Section 3
Question 16 Page 226
sin(π + x) = –sin x, cos(π + x) = –cos x, tan(π + x) = tan x,
csc(π + x) = –csc x, sec(π + x) = –sec x, cot(π + x) = cot x
Chapter 4 Section 3
Question 17 Page 226
# 3!
&
# 3!
&
# 3!
&
sin %
" x ( = –cos x, cos %
" x ( = –sin x, tan %
" x ( = cot x,
$ 2
'
$ 2
'
$ 2
'
# 3!
&
# 3!
&
# 3!
&
csc %
" x ( = –sec x, sec %
" x ( = –csc x, cot %
" x ( = tan x
$ 2
'
$ 2
'
$ 2
'
Chapter 4 Section 3
Question 18 Page 226
" 3!
%
" 3!
%
" 3!
%
sin $
+ x ' = –cos x, cos $
+ x ' = sin x, tan $
+ x ' = –cot x,
# 2
&
# 2
&
# 2
&
" 3!
%
" 3!
%
" 3!
%
csc $
+ x ' = –sec x, sec $
+ x ' = csc x, cot $
+ x ' = –tan x
# 2
&
# 2
&
# 2
&
MHR • Advanced Functions 12 Solutions
429
Chapter 4 Section 3
Question 19 Page 226
sin(2π – x) = –sin x, cos(2π – x) = cos x, tan(2π – x) = –tan x,
csc(2π – x) = –csc x, sec(2π – x) = sec x, cot(2π – x) = –cot x
Chapter 4 Section 3
Question 20 Page 226
Answers may vary. A sample solution is shown.
A. sin(π – x) = sin x
B. cos(π + x) = –cos x
# 3!
&
C. tan %
" x ( = cot x
$ 2
'
MHR • Advanced Functions 12 Solutions
430
" 3!
%
D. csc $
+ x ' = –sec x
# 2
&
E. sec(2π – x) = sec x
Chapter 4 Section 3
Question 21 Page 226
Answers may vary. A sample solution is shown.
" ! 5! %
9!
9!
5!
Since sin
= sin $ + ' , by the cofunction identity, sin
= cos
.
13
13
26
# 2 26 &
Also, since sin
# 3! 21! &
9!
9!
21!
= sin %
, by the cofunction identity, sin
= –cos
.
"
(
13
13
26
26 '
$ 2
Chapter 4 Section 3
a)
r=
$!
'
v2
tan & " # )
g
%2
(
=
v2
cot #
g
=
v2
g tan #
b) r =
Question 22 Page 226
502
9.8 tan
!
4
2500
9.8
=& 255 m
=
MHR • Advanced Functions 12 Solutions
431
Chapter 4 Section 3
Question 23 Page 226
Solutions to Achievement Check questions are provided in the Teacher’s Resource.
Chapter 4 Section 3
Question 24 Page 227
a) Answers may vary. A sample solution is shown.
#!
&
csc x = sec % " x ( cofunction identity
$2
'
)! #
#
!&
!&,
csc % 6b + ( = sec + " % 6b + ( .
8'
8'$
*2 $
)# !
& !,
= sec +% " 6b( " .
' 8*$ 2
"
"
!%
!%
Since csc $ 6b + ' = sec $ 2b ( '
8&
8&
#
#
!
" 6b
2
!
8b =
2
!
b=
16
2b =
b) Check that x =
!
is a point of intersection.
16
MHR • Advanced Functions 12 Solutions
432
Chapter 4 Section 3
Question 25 Page 227
a) Answers may vary. A sample solution is shown.
#
#
"&
"&
cot % 4c ! ( = ! tan % 2c + (
4'
4'
$
$
)#
#
"&
"& ",
cot % 4c ! ( = cot +% 2c + ( + . cofunction identity
4'
4' 2$
*$
)
",
= cot + 2c + " ! .
4*
!
So 4c = 2c + π, or c = .
2
(
)
b) Answers may vary. A sample solution is shown.
#
#
#
#
"&
"&
"&
"&
cot % 4c ! ( + tan % 2c + ( = cot % 2" ! ( + tan % " + (
4'
4'
4'
4'
$
$
$
$
# 7" &
# 5" &
= cot % ( + tan % (
$ 4 '
$ 4'
= !1+ 1
=0
Chapter 4 Section 3
Question 26 Page 227
Answers may vary. A sample solution is shown.
"
!%
a) Determine a value of m such that sin (6m + π) + sin $ 2m + ' = 0.
3&
#
b)
sin (! + x ) = "sin x
cofunction identity
sin ( 6m + ! ) = " sin ( 6m )
!&
#
sin ( 6m + ! ) = " sin % 2m + (
$
3'
6m = 2m +
!
3
!
m=
12
!
3
4m =
MHR • Advanced Functions 12 Solutions
433
Chapter 4 Section 3
A=
Question 27 Page 227
1
base ! height
2
h
h
c
h = c sin B
1
A = ac sin B
2
1 ! sin C $
A = a# a
sin B
2 " sin A &%
!
c
a
sin C $
#" Using the sine law sin C = sin A ,c = a sin A &%
1 a 2 sin C sin B
2
sin A
2
a sin Bsin C
A=
2sin(B + C)
! Using the cofunction identity and sum of angles in a triangle $
#" sin A = sin(' ( A) = sin(B + C)
&%
sin B =
A=
Chapter 4 Section 3
Question 28 Page 227
a) i)
" !%
$# 1, 3 '&
#
"&
ii) % 5,! (
6'
$
b) i)
! 3 3 3$
, &
#
" 2 2%
ii) (–4, 0)
c) i)
!
+ 2! k, k "Z
6
ii) !
"
+ 2" k, k #Z
3
MHR • Advanced Functions 12 Solutions
434
Chapter 4 Section 3
Question 29 Page 227
Since ∠PQR is subtended by central angle ∠POQ, ∠PQR =
PS
1
PS = sin !
sin ! =
1
!.
2
(the radius of a unit circle is 1)
OS
1
OS = cos !
cos ! =
QS = OS + QO
= cos ! + 1
1
PS
tan ! =
2
QS
sin !
=
cos ! + 1
MHR • Advanced Functions 12 Solutions
435
Section 4
Compound Angle Formula
Chapter 4 Section 4
Question 1 Page 232
"! ! %
" 3! ! %
a) sin $ + ' = sin $
+
# 4 12 &
# 12 12 '&
= sin
= sin
3
2
=
c)
!
3
=
"! ! %
" 3! ! %
cos $ + ' = cos $
+
# 4 12 &
# 12 12 '&
= cos
!
3
!
6
3
2
Question 2 Page 232
" 3! ! %
" 9! ! %
a) sin $
+ ' = sin $
+
# 5 15 &
# 15 15 '&
= sin
c)
1
2
#! ! &
# 3! ! &
d) cos % " ( = cos %
"
$ 4 12 '
$ 12 12 ('
=
Chapter 4 Section 4
=
!
6
= cos
1
2
=
#! ! &
# 3! ! &
b) sin % " ( = sin %
"
$ 4 12 '
$ 12 12 ('
2!
3
3
2
=0
= sin
="
" 2! 5! %
" 4! 5! %
cos $
+ ' = cos $
+
# 9 18 &
# 18 18 '&
= cos
# 7! ! &
# 21! ! &
b) sin %
" ( = sin %
"
$ 5 15 '
$ 15 15 ('
!
2
4!
3
3
2
# 10! 5! &
# 20! 5! &
d) cos %
" ( = cos %
"
18 '
$ 9
$ 18 18 ('
= cos
="
5!
6
3
2
MHR • Advanced Functions 12 Solutions
436
Chapter 4 Section 4
Question 3 Page 233
"! !%
!
!
!
!
a) sin $ + ' = sin cos + cos sin
3
4
3
4
# 3 4&
3 1 1 1
(
+ (
2
2 2
2
=
3 +1
=
2 2
"! !%
!
!
!
!
b) cos $ + ' = cos cos ( sin sin
3
4
3
4
# 3 4&
=
=
c)
1 1
3 1
)
(
)
2
2 2
2
1( 3
2 2
# 2! ! &
2!
!
2!
!
cos %
" ( = cos
cos + sin
sin
3
4
3
4
$ 3 4'
1 1
3 1
=" )
+
)
2
2 2
2
=
"1+ 3
2 2
# 2! ! &
2!
!
2!
!
" ( = sin
cos " cos
sin
d) sin %
3
4
3
4
$ 3 4'
=
=
3 1 1 1
)
+ )
2
2 2
2
3 +1
2 2
MHR • Advanced Functions 12 Solutions
437
Chapter 4 Section 4
" 4! 3! %
7!
= sin $
+
12
# 12 12 '&
a) sin
Question 4 Page 233
b) sin
# 8! 3! &
5!
= sin %
"
12
$ 12 12 ('
"! !%
= sin $ + '
# 3 4&
# 2! ! &
= sin %
"
$ 3 4 ('
!
!
!
!
cos + cos sin
3
4
3
4
3 1 1 1
=
(
+ (
2
2 2
2
= sin
=
" 8! 3! %
11!
= cos $
+
12
# 12 12 '&
b) cos
2!
!
2!
!
cos ( sin
sin
3
4
3
4
= cos
a) sin
" 3! ! %
= sin $
+
# 4 3 '&
3!
!
3!
!
cos + cos sin
4
3
4
3
1 1 " 1 %
3
=
( +$)
(
'
2 2 #
2& 2
= sin
=
=
2 2
" 9! 4! %
13!
= sin $
+
12
# 12 12 '&
1) 3
2 2
2!
!
2!
!
cos + sin
sin
3
4
3
4
1 1
3 1
=" )
+
)
2
2 2
2
(1( 3
Chapter 4 Section 4
# 8! 3! &
5!
= cos %
"
12
$ 12 12 ('
# 2! ! &
= cos %
"
$ 3 4 ('
1 1
3 1
=( )
(
)
2
2 2
2
=
2 2
Question 5 Page 233
" 2! ! %
= cos $
+
# 3 4 '&
= cos
3 +1
=
2 2
2!
!
2!
!
cos " cos
sin
3
4
3
4
3 1 1 1
)
+ )
2
2 2
2
=
3 +1
Chapter 4 Section 4
a) cos
= sin
"1+ 3
2 2
Question 6 Page 233
b) cos
" 8! 9! %
17!
= cos $
+
12
# 12 12 '&
" 2! 3! %
= cos $
+ '
4&
# 3
2!
3!
2!
3!
cos
( sin
sin
3
4
3
4
1 " 1 %
3 1
= ( )$(
(
)
'
2 #
2& 2
2
= cos
=
1( 3
2 2
MHR • Advanced Functions 12 Solutions
438
Chapter 4 Section 4
a) sin
" 9! 10! %
19!
= sin $
+
12
# 12 12 '&
Question 7 Page 233
b) cos
" 8! 15! %
23!
= cos $
+
12
# 12 12 '&
" 2! 5! %
= cos $
+ '
4&
# 3
" 3! 5! %
= sin $
+ '
6&
# 4
2!
5!
2!
5!
cos
( sin
sin
3
4
3
4
1 " 1 %
3 " 1 %
= ( )$(
(
)$(
'
'
2 #
2& 2 #
2&
3!
5!
3!
5!
cos
+ cos sin
4
6
4
6
1 "
3% " 1 % 1
=
($)
' +$)
'(
2& 2
2 # 2 & #
= cos
= sin
=
) 3 )1
=
2 2
Chapter 4 Section 4
1+ 3
2 2
Question 8 Page 233
Use the Pythagorean Theorem to get the remaining side.
5
13
3
x
y
4
5
a =5 !3
= 25 ! 9
= 16
a=4
4
a) cos x =
5
2
12
12
2
2
b = 132 ! 52
= 169 ! 25
= 144
b = 12
12
b) sin y =
13
2
MHR • Advanced Functions 12 Solutions
439
Chapter 4 Section 4
a) sin(x + y) = sin x cos y + cos x sin y
Question 9 Page 233
b) sin(x ! y) = sin x cos y ! cos x sin y
3 5 4 12
= ! + !
5 13 5 13
15 48
=
+
65 65
63
=
65
c)
cos(x + y) = cos x cos y ! sin x sin y
3 5 4 12
= " ! "
5 13 5 13
15 48
=
!
65 65
33
=!
65
d) cos(x ! y) = cos x cos y + sin x sin y
4 5 3 12
" ! "
5 13 5 13
20 36
=
!
65 65
16
=!
65
4 5 3 12
" + "
5 13 5 13
20 36
=
+
65 65
56
=
65
=
Chapter 4 Section 4
=
Question 10 Page 233
Use the Pythagorean Theorem to find the other side as in question 8.
Using the CAST rule we know that cos x will be negative and sin y will be positive.
a 2 = 132 ! 52
b2 = 52 ! 32
= 169 ! 25
= 144
a = 12
a) cos x = !
= 25 ! 9
= 16
b=4
12
13
b)
sin y =
4
5
MHR • Advanced Functions 12 Solutions
440
Chapter 4 Section 4
a) sin(x + y) = sin x cos y + cos x sin y
=
Question 11 Page 233
b)
5 3 # 12 & 4
! + "
!
13 5 %$ 13 (' 5
15 48
"
65 65
33
="
65
=
c)
cos(x + y) = cos x cos y ! sin x sin y d)
12 3 5 4
" ! "
13 5 13 5
36 20
=! !
65 65
56
=!
65
=!
Chapter 4 Section 4
sin(x ! y) = sin x cos y ! cos x sin y
=
5 3 # 12 & 4
" ! !
"
13 5 %$ 13 (' 5
15 48
+
65 65
63
=
65
=
cos(x ! y) = cos x cos y + sin x sin y
12 3 5 4
" + "
13 5 13 5
36 20
=! +
65 65
16
=!
65
=!
Question 12 Page 233
sin 2θ = sin(θ + θ)
= sin θ cos θ + cos θ sin θ
= 2sin θ cos θ
Chapter 4 Section 4
Question 13 Page 233
cos 2x = cos(x + x)
= cos x cos x – sin x sin x
= cos2 x – sin2 x
Chapter 4 Section 4
Question 14 Page 233
a) From question 13, cos 2x = cos2 x – sin2 x.
Rearranging the Pythagorean identity: cos2 x = 1 – sin2 x
Substitute into the above equation.
cos 2x = (1! sin 2 x) ! sin 2 x
= 1! 2sin 2 x
b) From question 13, cos 2x = cos2 x – sin2 x.
Rearranging the Pythagorean identity: cos2 x = 1 – sin2 x
Substitute into the above equation.
cos 2x = cos 2 x ! (1! cos 2 x)
= 2cos 2 ! 1
MHR • Advanced Functions 12 Solutions
441
Chapter 4 Section 4
Question 15 Page 233
Using the Pythagorean Theorem:
a 2 = 252 ! 7 2
= 625 ! 49
= 576
a = 24
24
25
b) sin 2! = 2sin ! cos !
Since θ is in the second quadrant, cosine is negative: cos ! = "
a) cos 2! = 1" 2sin 2 !
# 7&
= 1" 2 % (
$ 25 '
" 7 % " 24 %
= 2$ ' $ ( '
# 25 & # 25 &
2
98
625
527
=
625
=(
= 1"
336
625
c)
The approximate measure for θ is 2.86 radians.
d) 2θ is about 2(2.8578) = 5.7156
527
= 0.8432.
625
336
For part b), sin 5.7156 =& –0.5376 and –
= 0.5376.
625
For part a), cos 5.7156 =& 0.8432 and
MHR • Advanced Functions 12 Solutions
442
Chapter 4 Section 4
Question 16 Page 233
For question 12:
For question 13:
For question 14:
a)
b)
MHR • Advanced Functions 12 Solutions
443
Chapter 4 Section 4
a) sin x =
Question 17 Page 234
h1
12
h1 = 12sin x
b)
6
a
-1
2x
6
h2
b
x
b
6
b = 6sin x
h2 = 6sin x + 6sin 2x
sin x =
a
6
a = 6sin 2x
sin 2x =
h2 = 6sin x + 6(2sin x cos x)
h2 = 6sin x(1+ 2cos x)
Chapter 4 Section 4
Question 18 Page 234
a) Since sin 180° = 0, P = 0 when x = 180° – 113.5° or 66.5°.
The angle of latitude at which the power drops to 0 is approximately 66.5°.
The Sun is not seen at all at this latitude.
b) The sine function has its maximum (1) at 90°. So P has its maximum when 90° = x – 113.5°,
or when x = –23.5°.
The negative sign represents a latitude in the southern hemisphere.
The Sun appears directly overhead at noon.
Chapter 4 Section 4
Question 19 Page 234
Solutions to Achievement Check questions are provided in the Teacher’s Resource.
MHR • Advanced Functions 12 Solutions
444
Chapter 4 Section 4
a)
tan(x + y) =
=
Question 20 Page 234
sin(x + y)
cos(x + y)
sin x cos y + cos x sin y
cos x cos y ! sin x sin y
sin x cos y + cos x sin y
cos x cos y ! sin x sin y
sin x cos y cos x sin y
+
cos x cos y cos x cos y
=
cos x cos y sin x sin y
!
cos x cos y cos x cos y
tan x + tan y
=
1! tan x tan y
b) tan(x + y) =
c)
L.S. = tan(x + y)
" 2! ! %
= tan $
+
# 3 6 '&
" 5! %
= tan $ '
# 6&
=(
1
3
R.S. =
tan x + tan y
1! tan x tan y
2"
"
+ tan
3
6
=
2"
"
1! tan
tan
3
6
1
! 3+
3
=
# 1 &
1! ! 3 %
$ 3 ('
tan
( )
!3 + 1
=
=
3
2
!2
2 3
1
=!
3
2!
!
Since L.S. = R.S., the formula is valid for x =
and y = .
3
6
MHR • Advanced Functions 12 Solutions
445
Chapter 4 Section 4
a)
tan(x + (! y)) =
=
tan x + tan(! y)
1! tan x tan(! y)
tan x ! tan y
1+ tan x tan y
#! !&
b) L.S. = tan % " (
$ 3 6'
#!&
= tan % (
$ 6'
=
Question 21 Page 234
1
3
Since tan (– θ) = –tan θ.
!
!
" tan
3
6
R.S. =
!
!
1+ tan tan
3
6
1
3"
3
=
# 1 &
1+ 3 %
$ 3 ('
2
tan
( )
=
=
3
2
1
3
!
!
Since the L.S. = R.S., the formula is valid for x = and y = .
3
6
Chapter 4 Section 4
a)
Question 22 Page 235
tan 2x = tan(x + x)
tan x + tan x
1! tan x tan x
2 tan x
=
1! tan 2 x
=
b)
MHR • Advanced Functions 12 Solutions
446
c)
Both sides of the formula equal approximately 1.7036.
Chapter 4 Section 4
Question 23 Page 235
a) L.S. = sin x + sin y
! x + y$
! x ' y$
R.S. = 2sin #
cos #
&
" 2 %
" 2 &%
= sin
=
2!
!
+ sin
3
3
! 2( ( $
! 2( ( $
# 3 + 3&
# 3 ' 3&
= 2sin #
cos
&
#
&
# 2 &
# 2 &
"
%
"
%
3
3
+
2
2
= 3
!($
!($
= 2sin # & cos # &
" 2%
" 6%
! 3$
= 2(1) #
&
" 2 %
= 3
Since the L.S. = R.S., the formula is valid for x =
2!
!
and y = .
3
3
b) sin x + (! sin y) = sin x + sin(! y)
" x + (! y) %
" x ! (! y) %
= 2sin $
cos $
'
'&
2
2
#
&
#
" x ! y%
" x + y%
= 2sin $
cos $
'
# 2 &
# 2 '&
Chapter 4 Section 4
a) From question 14 part a):
! x$
! x$
cos 2 # & = 1' 2sin 2 # &
" 2%
" 2%
Question 24 Page 235
b)
From question 14 part b):
! x$
! x$
cos 2 # & = 2cos 2 # & ' 1
" 2%
" 2%
! x$
2sin 2 # & = 1' cos x
" 2%
! x$
2cos 2 # & = cos x + 1
" 2%
! x $ 1' cos x
sin 2 # & =
2
" 2%
! x $ cos x + 1
cos 2 # & =
2
" 2%
! x$
1' cos x
sin # & = ±
2
" 2%
! x$
cos x + 1
cos # & = ±
2
" 2%
MHR • Advanced Functions 12 Solutions
447
Chapter 4 Section 4
Question 25 Page 235
The point of intersection is (1, 3) from the graph.
From the graph, the point where the altitude meets the base is (1, 0), so the height is 3.
For the x-intercepts:
let y = 0
0 = 4x ! 1
0 = !2x + 5
x=
1
4
5
2
x=
(1, 3)
x
y
3
!1 $
#" , 0&%
4
!5 $
#" , 0&%
2
(1, 0)
3
3
tan x = 4
tan y = 2
3
3
1
1
tan x =
tan y =
4
2
x =& 0.2450
y =& 0.4636
x + y =& 0.71
! =& 0.71 opposite angle theorem
)
Chapter 4 Section 4
Question 26 Page 235
Using the fact that the sum of the angles in a triangle is 180° and tan 180° = 0:
tan A + tan(B + C)
tan( A + B + C) =
1! tan A tan(B + C)
0=
tan A + tan(B + C)
1! tan A tan(B + C)
0 = tan A + tan(B + C)
tan B + tan C
1! tan B tan C
tan A(1! tan B tan C) + tan B + tan C
0=
1! tan B tan C
0 = tan A ! tan A tan B tan C + tan B + tan C
0 = tan A + tan B + tan C ! tan A tan B tan C
0 = tan A +
tan A tan B tan C = tan A + tan B + tan C
MHR • Advanced Functions 12 Solutions
448
Chapter 4 Section 4
Question 27 Page 235
a)
θ
!3
!–
6
sin θ
0.01
0.05
0.10
0.15
0.25
0.35
0.010 00
0.049 98
0.099 83
0.149 44
0.247 40
0.342 85
0.010 00
0.049 98
0.099 83
0.149 44
0.247 40
0.342 90
0.01
0.05
0.10
0.15
0.25
0.35
0.999 95
0.998 75
0.995 00
0.988 75
0.968 75
0.938 75
0.999 95
0.998 75
0.995 00
0.988 77
0.968 91
0.939 37
b)
θ
!2
1–
2
cos θ
Chapter 4 Section 4
a) sin θ =& ! "
!3
6
b) cos θ =& 1!
"2
2
c) tan θ =& θ +
!3
3
Question 28 Page 235
MHR • Advanced Functions 12 Solutions
449
Section 5
Prove Trigonometric Identities
Chapter 4 Section 5
Question 1 Page 240
L.S. = cos 2x
= cos(x + x)
= cos x cos x ! sin x sin x
R.S. = 2cos 2 x ! 1
= 2cos 2 x ! (sin 2 x + cos 2 x)
= cos 2 x ! sin 2 x
= cos 2 x ! sin 2 x
Since L.S. = R.S., cos 2x = 2 cos2 x – 1 is an identity.
Chapter 4 Section 5
L.S. = cos 2x
= cos(x + x)
= cos x cos x ! sin x sin x
Question 2 Page 240
R.S. = 1! 2sin 2 x
= (sin 2 x + cos 2 x) ! 2sin 2 x
= cos 2 x ! sin 2 x
= cos 2 x ! sin 2 x
Since L.S. = R.S., cos 2x = 1 – 2 sin2 x is an identity.
Chapter 4 Section 5
L.S. = sin(x + ! )
Question 3 Page 240
R.S. = –sin x
= sin x cos ! + cos x sin !
= sin x("1) + cos x(0)
= " sin x
Since L.S. = R.S., sin (x + π) = –sin x is an identity.
Chapter 4 Section 5
# 3!
&
L.S. = sin %
" x(
$ 2
'
Question 4 Page 240
R.S. = –cos x
3!
3!
cos x " cos sin x
2
2
= ("1)cos x " (0)sin x
= sin
= " cos x
# 3!
&
Since L.S. = R.S., sin %
" x ( = " cos x is an identity.
$ 2
'
MHR • Advanced Functions 12 Solutions
450
Chapter 4 Section 5
L.S. = cos(! " x)
Question 5 Page 240
R.S. = –cos x
= cos ! cos x + sin ! sin x
= ("1)cos x + (0)sin x
= " cos x
Since L.S. = R.S., cos (π – x) = –cos x is an identity.
Chapter 4 Section 5
" 3!
%
L.S. = cos $
+ x'
# 2
&
Question 6 Page 240
R.S. = sin x
3!
3!
cos x ( sin sin x
2
2
= (0)cos x ( ((1)sin x
= cos
= sin x
" 3!
%
Since L.S. = R.S., cos $
+ x ' = sin x is an identity.
# 2
&
Chapter 4 Section 5
L.S. = cos x
Question 7 Page 240
R.S. = sin x cot x
! cos x $
= sin x #
" sin x &%
quotient identity
= cos x
Since L.S. = R.S., cos x = sin x cot x is an identity.
Chapter 4 Section 5
L.S. = 1 + sin x
Question 8 Page 240
R.S. = sin x(1+ csc x)
!
1 $
= sin x # 1+
" sin x &%
reciprocal identity
= sin x + 1
Since L.S. = R.S., 1 + sin x = sin x (1 + csc x) is an identity.
MHR • Advanced Functions 12 Solutions
451
Chapter 4 Section 5
a)
Question 9 Page 241
L.S. = 1! 2cos 2 x
= sin 2 x + cos 2 x ! 2cos 2 x Pythagorean identity
= sin 2 x ! cos 2 x
R.S. = sin x cos x(tan x ! cot x)
" sin x cos x %
= sin x cos x $
!
quotient identity and reciprocal identity
# cos x sin x '&
= sin 2 x ! cos 2 x
Since L.S. = R.S., 1! 2cos 2 x = sin x cos x(tan x ! cot x) is an identity.
b)
Chapter 4 Section 5
a) L.S. = csc 2 x
Question 10 Page 241
R.S. = 1+ cot 2 x
cos 2 x
reciprocal and quotient identity
sin 2 x
sin 2 x cos 2 x
= 2 +
sin x sin 2 x
sin 2 x + cos 2 x
=
sin 2 x
1
= 2
Pythagorean identity
sin x
Since L.S. = R.S., csc2 x = 1 + cot2 x is an identity.
=
1
sin 2 x
b) L.S. = sec 2 x
reciprocal identity
= 1+
R.S. = 1+ tan 2 x
1
sin 2 x
=
1+
cos 2 x
cos 2 x
sin 2 x + cos 2 x
cos 2 x sin 2 x
=
=
+
cos 2 x
cos 2 x cos 2 x
sin 2 x cos 2 x
=
+
cos 2 x cos 2 x
Since L.S. = R.S., sec2 x = 1 + tan2 x is an identity.
=
MHR • Advanced Functions 12 Solutions
452
Chapter 4 Section 5
1! sin 2 x
cos x
cos 2 x
=
Pythagorean identity
cos x
= cos x
L.S. =
Since L.S. = R.S.,
Question 11 Page 241
sin 2x
2sin x
sin x cos x + cos x sin x
=
Compound angle formula
2sin x
2cos x sin x
=
2sin x
= cos x
R.S. =
1! sin 2 x sin 2x
is an identity.
=
cos x
2sin x
Chapter 4 Section 5
Question 12 Page 241
csc 2 x ! 1
R.S. = 1 – sin2 x
2
csc x
csc 2 x
1
=
!
2
csc x csc 2 x
= 1! sin 2 x
reciprocal identity
L.S. =
Since L.S. = R.S.,
csc 2 x ! 1
= 1! sin 2 x is an identity.
csc 2 x
Chapter 4 Section 5
csc x
cos x
1
1
=
sin x cos x
L.S. =
Since L.S. = R.S.,
reciprocal identity
Question 13 Page 241
R.S. = tan x + cot x
sin x cos x
=
+
quotient identity
cos x sin x
sin x sin x cos x cos x
=
!
+
!
cos x sin x sin x cos x
sin 2 x + cos 2 x
=
sin x cos x
1
=
Pythagorean identity
sin x cos x
csc x
= tan x + cot x is an identity.
cos x
Chapter 4 Section 5
Question 14 Page 241
Answers may vary. A sample solution is shown.
a) 5 s
b) 20 s
c) 75% time saved.
d) 10 s
MHR • Advanced Functions 12 Solutions
453
Chapter 4 Section 5
L.S. = 2sin x sin y
Question 15 Page 241
R.S. = cos(x ! y) ! cos(x + y)
= cos x cos y + sin x sin y ! cos x cos y + sin x sin y
= 2sin x sin y
Since L.S. = R.S., 2sin x sin y = cos(x ! y) ! cos(x ! y) is an identity.
Chapter 4 Section 5
Question 16 Page 241
L.S. = sin 2x + sin 2 y
= 2sin x cos x + 2sin y cos y double angle formula
= 2(sin x cos x + sin y cos y)
R.S. = 2sin(x + y)cos(x ! y)
= 2 "#sin x cos y + cos x sin y $% "#cos x cos y + sin x sin y $% compound angle formula
= 2 "#sin x cos x cos 2 y + sin 2 x sin y cos y + cos 2 x sin y cos y + sin 2 y cos x sin x $%
= 2 "#sin x cos x(cos 2 y + sin 2 y) + sin y cos y(sin 2 x + cos 2 x) $%
= 2(sin x cos x + sin y cos y) Pythagorean identity
Since L.S. = R.S., sin 2x + sin 2 y = 2sin(x + y)cos(x ! y) is an identity.
Chapter 4 Section 5
Question 17 Page 241
a)
Yes, the graphs appear to be the same.
MHR • Advanced Functions 12 Solutions
454
b) L.S. = cos 2x
= cos x cos x ! sin x sin x
= cos 2 x ! sin 2 x
1! tan 2 x
R.S. =
1+ tan 2 x
sin 2 x
1!
cos 2 x
=
quotient identity
sin 2 x
1+
cos 2 x
cos 2 x ! sin 2 x cos 2 x + sin 2 x
=
÷
cos 2 x
cos 2 x
cos 2 x ! sin 2 x cos 2 x
=
"
Pythagorean identity
1
cos 2 x
= cos 2 x ! sin 2 x
1! tan 2 x
Since L.S. = R.S., cos 2x =
is an identity.
1+ tan 2 x
Chapter 4 Section 5
Question 18 Page 241
a) Answers may vary. Graphs are different.
L.S. ≠ R.S.
Therefore, sin x = 1! cos 2 x is not an identity.
b) While the left results in both positive and negative values, the right side is restricted to
positive values only.
Chapter 4 Section 5
Question 19 Page 241
Solutions to Achievement Check questions are provided in the Teacher’s Resource.
MHR • Advanced Functions 12 Solutions
455
Chapter 4 Section 5
Question 20 Page 241
a)
Yes, the graphs are the same.
L.S. = sin 6 x + cos6 x Factor using sum of cubes
b)
(x 6 + y 6 ) = (x 2 + y 2 )(x 4 ! x 2 y 2 + y 4 )
L.S. = (sin 2 x + cos 2 x)(sin 4 x ! sin 2 x cos 2 x + cos 4 x)
= 1 "#sin 2 x(1! cos 2 x) ! sin 2 x cos 2 x + cos 2 x(1! sin 2 x) $% Pythagorean identity
= sin 2 x ! sin 2 x cos 2 x ! sin 2 x cos 2 x + cos 2 x ! sin 2 x cos 2 x
= (sin 2 x + cos 2 x) ! 3sin 2 x cos 2 x Pythagorean identity
= 1! 3sin 2 x cos 2 x
R.S. = 1! 3sin 2 x cos 2 x
Since L.S. = R.S., sin 6 x + cos6 x = 1! 3sin 2 x cos 2 x is an identity.
Chapter 4 Section 5
Question 21 Page 241
L.S. = cos 4 x ! sin 4 x
Factor a difference of squares
R.S. = cos 2x
= cos 2 x ! sin 2 x Double angle formula
x 4 ! y 4 = (x 2 ! y 2 )(x 2 + y 2 )
L.S. = (cos 2 x ! sin 2 x)(cos 2 x + sin 2 x)
= cos 2 x ! sin 2 x
Pythagorean identity
Since L.S. = R.S., cos x ! sin 4 x = cos 2x is an identity.
4
Chapter 4 Section 5
Question 22 Page 241
tan x + tan x
1! tan x tan x
2 tan x
=
1! tan 2 x
tan 2x =
MHR • Advanced Functions 12 Solutions
456
Chapter 4 Section 5
Question 23 Page 241
a) Draw a triangle with sin C = x (hypotenuse of 1).
C
1
a
B
A
x
Use the Pythagorean Theorem (x2 + a2 = 1) to express the missing side, a =
cos C =
1! x 2 .
1! x 2
1
cos(sin !1 x) = 1! x 2
b) Draw triangles with cos A = a and cos B = b (hypotenuse of 1).
A
B
1
a
C
1
b
D
1! a 2
E
1! b2
F
L.S. = cos–1 a + cos–1 b
=A+B
cos(A + B) = cos A cos B ! sin A sin B
=
a b
1! a 2
1! b2
" !
"
1 1
1
1
= ab ! 1! a 2 1! b2
(
A + B = cos !1 ab ! 1! a 2 1! b2
So the L.S. = R.S..
)
MHR • Advanced Functions 12 Solutions
457
Chapter 4 Review
Chapter 4 Review
a) 33 ×
!
=& 0.58
180
Question 1 Page 244
b) 138 ×
!
!
!
=& 2.41 c) 252 ×
=& 4.40 d) 347 ×
=& 6.06
180
180
180
Chapter 4 Review
Question 2 Page 244
a) 1.24 ×
180
=& 71.0°
!
b) 2.82 ×
180
=& 161.6°
!
c) 4.78 ×
180
=& 273.9°
!
d) 6.91 ×
180
=& 395.9°
!
Chapter 4 Review
a) 75 !
"
5"
=
180 12
Question 3 Page 244
b) 20 !
"
"
=
180 9
Chapter 4 Review
a)
2! 180
"
= 72°
5
!
b)
4! 180
"
= 80°
9
!
d) 9 !
"
"
=
180 20
c)
7! 180
"
= 105°
12
!
d)
11! 180
"
= 110°
18
!
Question 5 Page 244
12 ! 360°
= 72°/s
60 s
b)
Chapter 4 Review
a), b)
Revolutions per
minute
Degrees per
second
Radians per
second
"
"
=
180 15
Question 4 Page 244
Chapter 4 Review
a)
c) 12 !
12 ! 2" 2"
rad/s
=
60 s
5
Question 6 Page 244
16
33
1
3
45
78
96
200
270
468
8!
15
10!
9
3!
2
13!
5
MHR • Advanced Functions 12 Solutions
458
Chapter 4 Review
Question 7 Page 244
sin
4!
=& 0.9096
11
cos
4!
=& 0.4154
11
tan
4!
=& 2.1897
11
csc
4!
=& 1.0993
11
sec
4!
=& 2.4072
11
cot
4!
=& 0.4567
11
Chapter 4 Review
Question 8 Page 244
1
tan
a)
!
cos "
3
!
4
=
1
sin
!
2
#
&
! % 1 (
!
3 2
1
"
+
b) cos " %
( + sin =
!
6 %
4
2
3
2
sin (
$
'
3
1
1
1 1
"
2 1
= 1+
1
1
2
=2
=
=
Chapter 4 Review
1
2
2 +1
2
Question 9 Page 244
Let a be the length of the common side.
! a
cos =
3 15
" 1%
a = 15 $ '
# 2&
15
2
!
b
cos =
6 15
2
15
!
b = cos
2
6
a=
b=
15
3
"
2
2
b=
15 3
4
The base of the lodge is
15 3
m.
4
MHR • Advanced Functions 12 Solutions
459
Chapter 4 Review
Question 10 Page 244
2!
!
lies in the first quadrant, it can be expressed as a difference between
7
2
and an angle z. Find the measure of angle a.
2! !
=
–z
7
2
!
2!
z=
–
2
7
7! 4!
z=
"
14 14
3!
z=
14
Since an angle of
Chapter 4 Review
Question 11 Page 244
5!
!
lies in the second quadrant, it can be expressed as a sum between
and an
9
2
angle y. Find the measure of angle y.
5! !
=
+y
9
2
5!
!
y=
–
9
2
10! 9!
y=
"
18 18
!
y=
18
Since an angle of
MHR • Advanced Functions 12 Solutions
460
Chapter 4 Review
Question 12 Page 244
a) Apply the cofunction identity.
# ! 4! &
!
cot
= cot % "
18
$ 2 9 ('
4!
9
= 5.6713
= tan
b) Method 1:
Apply the cofunction identity.
13!
26!
tan
= tan
9
18
# 3! ! &
= tan %
"
$ 2 18 ('
!
18
= 5.6713
= cot
Method 2:
Apply the cofunction identity.
"
13!
4! %
tan
= tan $ ! +
9
9 '&
#
4!
9
= 5.6713
= tan
Chapter 4 Review
Question 13 Page 244
Since an angle of x lies in the second quadrant, it can be expressed as the sum of
!
and an
2
3!
.
11
! 3!
x= +
2 11
11! 6!
=
+
22 22
17!
=
22
angle
MHR • Advanced Functions 12 Solutions
461
Chapter 4 Review
Question 14 Page 244
" 5! ! %
" 5! 3! %
a) sin $
+ ' = sin $
+
# 12 4 &
# 12 12 '&
= sin
=
c)
# 5! ! &
# 5! 3! &
b) sin %
" ( = sin %
"
$ 12 4 '
$ 12 12 ('
2!
3
= sin
3
2
=
" 5! ! %
" 5! 3! %
cos $
+ ' = cos $
+
# 12 4 &
# 12 12 '&
= cos
=(
1
2
# 5! ! &
# 5! 3! &
d) cos %
" ( = cos %
"
$ 12 4 '
$ 12 12 ('
2!
3
= cos
1
2
Chapter 4 Review
!
6
=
!
6
3
2
Question 15 Page 245
Use the Pythagorean Theorem to get the remaining side.
5
4
25
13
y
7
x
24
3
a 2 = 52 ! 42
= 25 ! 16
=9
a=3
a) cos x =
3
5
b) sin y =
24
25
c)
b2 = 252 ! 7 2
= 625 ! 49
= 576
b = 24
sin(x + y) = sin x cos y + cos x sin y
4 7 3 24
! + !
5 25 5 25
28 + 72
=
125
100
=
125
4
=
5
=
MHR • Advanced Functions 12 Solutions
462
Chapter 4 Review
x
25
25
Question 16 Page 245
7
24
a) cos 2x = cos 2 x ! sin 2 x
2
" 24 % " 7 %
=$ ' !$ '
# 25 & # 25 &
b) sin 2x = 2sin x cos x
! 7 $ ! 24 $
= 2# & # &
" 25 % " 25 %
2
576 49
!
625 625
527
=
625
=
=
Chapter 4 Review
cos
336
625
Question 17 Page 245
" 9! 4! %
13!
= cos $
+
12
# 12 12 '&
" 3! ! %
= cos $
+
# 4 3 '&
3!
!
3!
!
cos ( sin sin
4
3
4
3
" 1 % " 1% " 1 % " 3%
= $(
'
'$ ' ($
'$
#
2 & # 2& # 2 & # 2 &
= cos
=
(1( 3
2 2
Chapter 4 Review
Question 18 Page 245
L.S. = sin(2! " x)
R.S. = –sin x
= sin 2! cos x " sin x cos 2! Compound angle formula
= 0 # cos x " sin x(1)
= " sin x
Since L.S. = R.S., sin(2! " x) = " sin x is an identity.
MHR • Advanced Functions 12 Solutions
463
Chapter 4 Review
a) L.S. = sec x
Since L.S. = R.S., sec x =
Question 19 Page 245
2(cos x sin 2x ! sin x cos 2x)
sin 2x
2cos x sin 2x 2sin x cos 2x
=
!
sin 2x
sin 2x
2sin x(2cos 2 x ! 1)
= 2cos x !
double angle formula
2sin x cos x
2cos 2 x ! 1
= 2cos x !
cos x
1
= 2cos x ! 2cos x +
cos x
= 0 + sec x reciprocal identity
R.S. =
2(cos x sin 2x ! sin x cos 2x)
is an identity.
sin 2x
b)
Chapter 4 Review
L.S. = 2sin x cos y
Question 20 Page 245
R.S. = sin(x + y) + sin(x ! y) compound angle formula
= sin x cos y + sin y cos x + sin x cos y ! sin y cos x
= 2sin x cos y
Since L.S. = R.S., 2sin x cos y = sin(x + y) + sin(x ! y) is an identity.
MHR • Advanced Functions 12 Solutions
464
Chapter 4 Review
Question 21 Page 245
a) No, the graphs are not the same for all values.
b) cos 3x = cos x cos 2x ! sin x sin 2x
= cos x(cos 2 x ! sin 2 x) ! sin x(2sin x cos x)
= cos3 x ! cos x sin 2 x ! 2sin 2 x cos x
= cos3 x ! 3cos x sin 2 x
Chapter 4 Review
Let x = 0:
Question 22 Page 245
L.S. = cos 2(0)
=1
R.S. = 2sin(0)sec(0)
=0
Since L.S. ≠ R.S., cos 2x = 2sin x sec x is not an identity.
Chapter 4 Review
Question 23 Page 245
! sin x cos x $
(sin 2x)(tan x + cot x) = (2sin x cos x) #
+
quotient and reciprocal identities
" cos x sin x &%
2sin 2 x cos x 2sin x cos 2 x
+
cos x
sin x
2
2
= 2sin x + 2cos x
=
= 2(sin 2 x + cos 2 x)
= 2(1)
Pythagorean identity
=2
Chapter 4 Problem Wrap-Up
Solution to the Chapter Problem Wrap-Up is provided in the Teacher’s Resource.
MHR • Advanced Functions 12 Solutions
465
Chapter 4 Practise Test
Chapter 4 Practise Test
Question 1 Page 246
The correct solution is B.
105 !
"
7"
=
180 12
Chapter 4 Practise Test
Question 2 Page 246
The correct solution is C.
13! 180
"
= 97.5°
24
!
Chapter 4 Practise Test
Question 3 Page 246
The correct solution is C.
Chapter 4 Practise Test
Question 4 Page 246
The correct solution is D.
Cofunction identity
"
!%
cos $ x + ' = ( sin x
2&
#
"
!%
sin x = ( cos $ x + '
2&
#
MHR • Advanced Functions 12 Solutions
466
Chapter 4 Practise Test
Question 5 Page 246
The correct solution is B.
"
2! %
2!
2!
cos $ ! +
= cos ! cos
( sin ! sin
'
3&
3
3
#
" 3%
" 1%
= ((1) $ ( ' ( (0) $
'
# 2&
# 2 &
=
1
2
Chapter 4 Practise Test
Question 6 Page 246
The correct solution is C.
From the CAST rule;
secant is negative sec ! = " 2
2
45
1
45
1
Chapter 4 Practise Test
Question 7 Page 246
The correct solution is A.
"! !%
11!
cos $ + ' = cos
30
# 5 6&
Chapter 4 Practise Test
360°
=& 13°/day
27.3
b) a = ! r
a)
Question 8 Page 246
2!
=& 0.23 rad/day
27.3
# 2" &
=%
) 384 400
$ 27.3 ('
=& 88 470.93 km/day
The moon moves along an arc of its orbit approximately 88 471 km/day.
Chapter 4 Practise Test
3 "
3%
!$!
'
2 # 2 &
1! (!1)(1)
=
Question 9 Page 246
3
2
MHR • Advanced Functions 12 Solutions
467
Chapter 4 Practise Test
a) Let h be the shared side.
! 40
sin =
3 h
40
h=
!
sin
3
= 40 ÷
=
3
2
80
3
Question 10 Page 246
cos
! x
=
4 h
80
!
x=
cos
4
3
80
1
=
"
3
2
80
=
6
b) x =& 32.7 m
Chapter 4 Practise Test
a) Since an angle of
Question 11 Page 247
!
!
is in the first quadrant, it can be expressed as a difference between
9
2
and an angle.
!
2!
sin = sin
9
18
# 9! 7! &
= sin %
"
$ 18 18 ('
# ! 7! &
= sin % "
$ 2 18 ('
7!
18
=& 0.3420
= cos
b) Since an angle of
8!
!
lies in the second quadrant, it can be expressed as a sum between
9
2
and an angle.
8!
16!
sin
= sin
9
18
" 9! 7! %
= sin $
+
# 18 18 '&
" ! 7! %
= sin $ +
# 2 18 '&
7!
18
=& 0.3420
= cos
MHR • Advanced Functions 12 Solutions
468
Chapter 4 Practise Test
Question 12 Page 247
a) Answers may vary. A sample solution is shown.
" 9! 8! %
17!
sin
= sin $
+
12
# 12 12 '&
" 3! 2! %
= sin $
+
3 '&
# 4
3!
2!
2!
3!
cos
+ sin
cos
4
3
3
4
1 " 1%
3 " 1 %
=
($) ' +
($)
'
2 # 2& 2 #
2&
= sin
)1) 3
=
2 2
b) Answers may vary. A sample solution is shown.
" 15! 2! %
" 5! ! %
sin $
+
= sin $
+
'
# 12 12 &
# 4 6 '&
5!
!
!
5!
cos + sin cos
4
6
6
4
" 1 % " 3% 1 " 1 %
= $(
' + $(
'$
'
#
2&# 2 & 2#
2&
= sin
=
( 3 (1
2 2
Chapter 4 Practise Test
25
13
x
24
Question 13 Page 247
7
121
2
y
13
13
13
5
12
Using the CAST rule we know that cos x will be negative and sin y will be positive.
a 2 = 252 ! 7 2
= 625 ! 49
b2 = 132 ! 52
= 169 ! 25
= 576
a = 24
= 144
b = 12
a) cos x = !
24
25
b) sin y =
12
13
MHR • Advanced Functions 12 Solutions
469
c)
cos(x ! y) = cos x cos y + sin x sin y
24 5 7 12
" + "
25 13 25 13
!120 + 84
=
325
!36
=
325
=!
Chapter 4 Practise Test
Question 14 Page 247
2800
! 2" =& 293.2 rad/s
60
Yes, the engine’s maximum velocity (293.2 rad/s) is slower than the maximum velocity of the
propeller (300 rad/s).
Chapter 4 Practise Test
Question 15 Page 247
In the diagram, the total distance north is 50 + 50 cos
In the diagram, the total distance east is 50 sin
!
.
6
!
.
6
50
Use the Pythagorean theorem:
2
"
!% "
!%
a 2 = $ 50 + 50cos ' + $ 50sin '
6& #
6&
#
!
6
2
a
2
2
"
3% "
1%
= $ 50 + 50 (
' + 50 ( '
2 & $#
2&
#
50
2
"
3%
2
= 2500 $ 1+
' + (25)
2
#
&
"
3%
= 2500 $ 1+ 3 + ' + 625
4&
#
= 2500 + 2500 3 +
7500
+ 625
4
= 5000 + 2500 3
a = 5000 + 2500 3
MHR • Advanced Functions 12 Solutions
470
Chapter 4 Practise Test
Question 16 Page 247
L.S. = (cos x ! sin x)2
R.S. = 1! sin 2x
= 1! 2sin x cos x
= cos x ! 2sin x cos x + sin x
2
2
Double angle formula
= (cos 2 x + sin 2 x) ! 2sin x cos x
= 1! 2sin x cos x
Pytagorean identity
Since L.S. = R.S., (cos x ! sin x)2 = 1! sin 2x is an identity.
Chapter 4 Practise Test
Question 17 Page 247
R.S. = cos(x + y) + cos(x ! y)
L.S. = 2 cos x cos y
= cos x cos y ! sin x sin y + (cos x cos y + sin x sin y)
= 2cos x cos y
Since L.S. = R.S., 2cos x cos y = cos(x + y) + cos(x ! y) is an identity.
Chapter 4 Practise Test
Question 18 Page 247
Answers may vary. A sample solution is shown.
!
Let x = 0, y = .
2
"
"
"!%
!%
!%
R.S. = 2sin $ 0 + ' cos $ 0 ( '
L.S. = cos 2(0) + sin 2 $ '
2&
2&
#
#
# 2&
= 2(1)(0)
= 1+ 0
=1
=0
Since L.S. ≠ R.S., cos 2x + sin 2 y = 2sin(x + y)cos(x ! y) is not an identity.
Chapter 4 Practise Test
L.S. = (csc x ! cot x)2
" 1
cos x %
=$
!
# sin x sin x '&
" 1! cos x %
=$
# sin x '&
=
2
2
(1! cos x)2
sin 2 x
Since L.S. = R.S., (csc x ! cot x)2 =
Question 19 Page 247
1! cos x
1+ cos x
1! cos x 1! cos x
=
"
1+ cos x 1! cos x
(1! cos x)2
=
1! cos 2 x
(1! cos x)2
=
sin 2 x
R.S. =
1! cos x
is an identity.
1+ cos x
MHR • Advanced Functions 12 Solutions
471
Chapter 4 Practise Test
! 2.4
=
3 CB
2.4
CB =
!
tan
3
2.4
=
3
Question 20 Page 247
! 2.4
=
4 DB
2.4
DB =
!
tan
4
= 2.4
tan
tan
CD = DB ! CB
CD = 2.4 !
2.4
3
" 3 ! 1%
= 2.4 $
'
3 &
#
Chapter 4 Practise Test
Question 21 Page 247
a)
!4.712389 = !
3"
2
The ticks on the horizontal axis are !
The distance between each tick is
3"
"
"
3"
.
, !" , ! , 0, , " ,
2
2
2
2
!
.
2
MHR • Advanced Functions 12 Solutions
472
b)
"!
%
$# 3 ,0.5'&
cos
" 5!
%
$# 3 ,0.5'&
!
5!
= cos
3
3
#
!&
= cos % 2! " (
3'
$
cos x = cos(2! " x)
c)
"! 1 %
$# 4 ,
'
2&
cos
# 5!
1 &
%$ 4 ,"
(
2'
!
5!
= " sin
4
4
# 3! ! &
= " sin %
"
$ 2 4 ('
# 3!
&
cos x = " sin %
" x(
$ 2
'
d) No. An identity must be proven algebraically.
MHR • Advanced Functions 12 Solutions
473