Module 2: Trigonometry
Section 1 Trigonometric Functions
Lesson 1 The Trigonometric Values for θ, 0°≤ θ ≤ 360° 186
Lesson 2 Solving Trigonometric Equations for 0° ≤ θ ≤ 360° 183
Lesson 3 Special Trigonometric Values 185
Lesson 4 The Unit Circle 187
Lesson 5 The Circular Functions 189
Lesson 6 Radian Measures 193
Review 201
Section 2: Graphs of Trigonometric Functions
Lesson 1 Sine and Cosine Graphs 205
Lesson 2 Transformations of the Sine and Cosine Functions 207
Lesson 3 Graphs of the Remaining Circular Functions 215
Review 219
Section 3: Trigonometric Identities
Lesson 1 Elementary Identities 225
Lesson 2 Using Elementary Identities 231
Lesson 3 Sum and Difference Identities 237
Lesson 4 Double Angle Identities 243
Review 255
Section 4: Problem Solving
Lesson 1 Modelling with Trigonometric Functions 261
Lesson 2 Solving Trigonometric Equations UsingYour Graphing Calculator 265
Lesson 3 Solving Trigonometric Equations WithGeneral Solution 265
Review 266
Principles of Mathematics 12 Answer Key, Contents 185
Module 2
Module 2
Lesson 1Answer Key
1. 2.
8 4sin10 5
6 3cos10 58 4tan6 3
θ
θ
θ
= =
= − = −
= − = −
11sin1221cos122
tan 11
θ
θ
θ
=
=
=
y
x
8�
–6
Q(–6,8)
θ
r = 10
y
x
10
5
� P(1,11)
θ
r = √122
Principles of Mathematics 12 Section 1, Answer Key, Lesson 1 187
3. 4.
3 1sin3 5 5
6 2cos3 5 53 1tan6 2
θ
θ
θ
= − = −
= − = −
= − =
12sin13
5cos13
12tan5
θ
θ
θ
= −
=
= −
y
x
3
–3
–5
S(–6,–3)�
θ
r = √45or 3√5
y
x
–10
R(5,–12)�
r = 13
θ5
188 Section 1, Answer Key, Lesson 1 Principles of Mathematics 12
Module 2
5. 6.
7.
8.
2sin 120cos 022tan which is undefined0
θ
θ
θ
= − = −
= − =
= −
0sin 03
3cos 130tan 03
θ
θ
θ
= =
= − = −
= − =
y
x
�K(0,–2)
θ
r = 2}
y
x�
L(–3,0)
θr = 3
4sin 140cos 044tan which is undefined0
θ
θ
θ
= =
= =
=
0sin 055cos 150tan 05
θ
θ
θ
= =
= =
= =
y
x
� N(0,4)
θ
r = 4{ 4
y
x
4
�
M(5,0)r = 5 }
Principles of Mathematics 12 Section 1, Answer Key, Lesson 1 189
Module 2
Lesson 2Answer Key
1. a) Find the reference angles for 116°, 265°, 289°, 323°.
64°, 85°, 71°, 37°b) Evaluate the three trigonometric ratios for the angles
above.
c) Evaluate the three trigonometric ratios for the angles above.
i) 476° or –244°ii) 625° or –95°iii) 649° or –71°iv) 683° or –37°
2.
3.
1
2 4
5 tan 4 2 tan5 tan 4
4tan7
4tan 30º7
150º , 330º
θ θθ
θ
θ
θ θ
−
+ = −= −
= −
= ≈ = =
R
13 4
3sin 22sin3
2sin 42 , 222º , 318º3
θ
θ
θ θ θ−
= −
= −
= ≅ ° = = R
Principles of Mathematics 12 Section 1, Answer Key, Lesson 2 191
Module 2
b) 116° 265° 289° 323° sin 0.899 –0.996 –0.946 –0.602 cos –0.438 –0.087 0.326 0.799 tan –2.05 11.43 –2.90 –0.754
4.
5.
6.
7.
0°; 180° or π. (360° is not valid because of the allowed interval for θ.)
8. 4 tan2x – 13 tan x = 1276°, 143°, 256°, 323°
22 sin 3 sinθ θ=
[ ]
1
7 cos 8 412cos7
12cos No Solution7
Remember that 1 cos 1
θ
θ
θ
θ
−
− =
=
= − ≤ ≤
R
( )( )( )
( )( )
2
1
1
1 4
3cos 14cos 5 0cos 5 3cos 1 0
Case 1 cos 5 0cos 5
cos 5 No Solution
Case 2 3cos 1 01cos3
1cos 71º3
71º , 289º
θ θθ θ
θθ
θ
θ
θ
θ
θ θ
−
−
+ − =+ − =
+ == −
=
− =
=
= ≈ = =
R
R
1
3 4
3sin 1 01sin3
1sin 19º3
199º , 341º
θ
θ
θ
θ θ
−
+ =
= −
= ≈ = =
R
192 Section 1, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
Lesson 3Answer Key
1.
2.
2 2
2
32
2
2 2
1 1 3 1 3a)
2 22 21
b) 3 1 1 1 23
3 1 3 1c) 1
2 2 4 4
3 1 1 3d) 2
2 22
1 3 1 3e)
2 2 123
3 1 1 3 1 1f) 1
2 2 2 4 4 2
1 1 3 3 1 1 3 1g) 0
2 2 2 2 2 4 4 2
−⋅ − =
⋅ + = + =
+ − = + =
− − = − =
− − =
− − + = − + = ⋅ + − − − = − + =
210° 225° 240° 270° 300° 315° 330° 360°
sin θ –1 0
cos θ 0 1
tan θ 1 undefined –1 0
12
− 12
− 12
− 12
− 12
− 12
− 12− 3
2
− 32
− 32
13 3 − 3 − 1
3
12
32
0° 30° 45° 60° 90° 120° 135° 150° 180°
sin θ 0 1 012
12
32
32
12
12
cos θ 1 0 –1
tan θ 0 1 undefined –1 0
32
12
12
− 12
− 12
− 32
13 3 − 3 − 1
3
Principles of Mathematics 12 Section 1, Answer Key, Lesson 3 193
Module 2
3.
(Note: cos θ = 1 produces no solutions since 0° < θ < 360°.)
( )( )( )
( )
2
o o
o o o
o o
i) 2 tan 3 0 j) 2sin sin 1 0
tan 3 2sin 1 sin 1 0
120 , 300 2sin 1 0 or sin 1 01sin or sin 12
210 , 330 , 90
k) cos cos 1 0
cos 0 or cos 1
90 , 270
θ θ θ
θ θ θ
θ θ θ
θ θ
θ
θ θ
θ θ
θ
+ = − − =
=− + − =
= + = − =
=− =
=− =
= ==
2
o o o o
1h) cos41cos2
60 , 120 , 240 , 300
θ
θ
θ
=
=±
=
( )
o o o
g) sin 2sin 1 0
1sin 0 or sin
2180 , 210 , 330
θ θ
θ θ
θ
+ =
= =−
=
o o o o
o o o
o
o o
a) 45 , 135 b) 60 , 240
c) 30 , 330 d) 240
e) 120 f) cos 1
, since 0 and 360are not allowed.
=∅θ
194 Section 1, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
Lesson 4Answer Key
1. a) E b) Ic) H d) Je) Q f) Sg) L h) Hi) A j) Mk) D l) Hm) K n) Lo) G p) Nq) M r) S
2. a) Sb) Hc) k is a multiple of 4.
k is an even, non-multiple of 4.k is an odd integer.
3. a) B, F, J, or N.b) S, D, H, or L.
4. a) b)45 3c) d)3 2
e) f)6 3
3g) 2 h) 4
7i) j)6 2
π π
π π
π π
ππ
π π
SH
D or L
Principles of Mathematics 12 Section 1, Answer Key, Lesson 4 195
Module 2
Careful: 0 is not positive.It is non-negative.
5.
Note: This exercise was pretty straight forward. However, justbecause it is simple does not mean it is not important. In fact, it is avery important assignment as it forms the basis for the rest of themodule. You must know where you have stopped for every specialvalue of θ. Conversely, if you know where you are on the unit circle,you must know how to get there!
a) b)
c) d)
e) f)
g) h)
i) j)
− −
− −
− −
− −
− −
74
3 211
623
254
56 4
π π
π π
π π
π π
π π
196 Section 1, Answer Key, Lesson 4 Principles of Mathematics 12
Module 2
Lesson 5Answer Key
1. Question
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
(r)
θθ P(θθ) cos θθ sin θθ tan θθ
23π
−
1 3,
2 2− 1
23
2− 3
27π (–1, 0) –1 0 0
76π
− −
3 1,
2 2− 3
2− 1
2
13
154π − −
1 1,2 2
− 12
− 12
− π6
−
3 1,
2 23
2− 1
2− 1
3
14π (1, 0) 1 0 0
(0, –1) 0 –1 undefined− 52π
53π (–1, 0) –1 0 0
− 116π
3 1,
2 23
212
13
173π
−
1 3,
2 212
− 32
− 3
(0, 1) 0 1 undefined
–π (–1, 0) –1 0 0
(0, –1) 0 –1 undefined
π2
32π
43π
− −
1 3,
2 2− 1
2 − 32
3
56π
−
3 1,
2 2− 3
212
− 13
–1154π −
1 1,2 2
12
− 12
4π (1, 0) 1 0 0
− 73π
−
1 3,
2 212
− 32
− 3
Principles of Mathematics 12 Section 1, Answer Key, Lesson 5 197
Module 2
2.
3.
( )( )
2h) 4 cos 1 0
2 cos 1 2 cos 1 0
1cos
22 4 5
, , ,3 3 3 3
θ
θ θ
θ
π π π πθ
− =− + =
=±
=
( )g) sin 2sin 1 0
1sin 0 or sin2
7 110, , ,6 6
θ θ
θ θ
π πθ π
+ =
= =−
=
f) cos 10
θθ
==
e)23π
4d)
3π11
c) ,6 6π π
4b) ,
3 3π π3
a) ,4 4π π
( )
( )
( ) ( )
( )
( )( )
2
1a) 2 1 1 1 22
1 1b) 1 13 3
1 1c)33
d) cos 0 1
1 1 3 1 3 2e) 3 12 3 2 3 6
1f) cos3 2
3 3 3g) 12 2 4
h) 1 1 1
π
+ = + = =
= =
+ − − + − − = + =
=
− =− − =−
198 Section 1, Answer Key, Lesson 5 Principles of Mathematics 12
Module 2
4.
5.
( )d) cot 1 sin 0
cot 0 or sin 1
cot 0 or 2
3,2 2
θ θ
θ θπ
θ θ
π πθ
− == =
= =
=
c) tan 1 and sin 074
θ θπθ
= − <
=
1b) sin2
3,4 4
θ
π πθ
=
=
1a) cos
25
,3 3
θ
π πθ
=
=
22
2
22
1 2 4 1a) 2 1 1 1 0 b) (2) 4 5
2 3 33
1c) ( 3) 3 d) ( 3) 1
3
1e) ( 2) 1 f) 4(2) 8
2
− = − = + = + = − = − − =
= =
( )k) cos cos 1 0
cos 0 or cos 13
, , 02 2
θ θ
θ θπ π
θ
− =
= =
=
( )( )
2j) 2 sin sin 1 0
2 sin 1 sin 1 0
1sin , or sin 1
27 11
, ,6 6 2
θ θ
θ θ
θ θ
π π πθ
− − =+ − =
=− =
=
( )i) 2 tan 3 0
tan 32 5
,3 3
θ
θπ π
θ
+ =
=−
=
Principles of Mathematics 12 Section 1, Answer Key, Lesson 5 199
Module 2
( )( )
2
1f) sin 0
sin
1 sin 0
1 sin 1 sin 0
sin 13
,2 2
θθ
θ
θ θ
θπ π
θ
− =
− =− + =
=±
=
( )( )e) tan 1 tan 1 0
tan 13 5 7
, , ,4 4 4 4
θ θ
θπ π π π
θ
− + =
=±
=
200 Section 1, Answer Key, Lesson 5 Principles of Mathematics 12
Module 2
Lesson 6Answer Key
1.
2.
3.
4.
5.
6.
7. Use the CAST rule.
a) IV b) I or IIIc) II d) II or III
S AT C
Cos is positive in Quadrant IV.All are positive in Quadrant I.Sine is positive in Quadrant II.Tan is positive in Quadrant III.
2 2 1524 24 . Therefore, or .
180 15 15 2x
xπ π π π ° = = = =
The other base angle is also
The third angle =
27
227
37
π
π π π
.
.− FHGIKJ =
180 130 50 50180
518
°− ° = ° = FHGIKJ =π π
o oSince 90 and 180 :2
5 5 7;
2 12 12 12 12
π π
π π π π ππ
= =
− = − =
a) b)
c) d)
− FHGIKJ = − ° F
HGIKJ = °
FHGIKJ = ° − F
HGIKJ = − °
76
180210
1112
180165
2 634180
150 9 0 9825180
56 3
ππ
ππ
π π. . . .
a) b)
c) d)
25180
536
125180
2536
460180
239
330180
116
π π π π
π π π π
FHGIKJ = − F
HGIKJ = −
FHGIKJ = F
HGIKJ =
Principles of Mathematics 12 Section 1, Answer Key, Lesson 6 201
Module 2
8. Since s = θr, it follows that 12.3 = θ(8.2).
9.
10.
11. a) 1 rev. = 2π; therefore, setting up a ratio where x = measure inrevolutions
b) 1 rev. = 360°; therefore, setting up a ratio where x = measure in revolutions
o o
1 rev360 75
75 5revolutions
360 24
x
x
=
= =
1 rev2 3
3revolutions
2
x
x
π
π
=
=
30Since , it follows that 5 ; therefore, or 60 .
2 3πθ θ θ
π = = = °
s r
Since 230 2302
C rr
r
ππ
π
==
=
2 2 2
2 2 2
Using the Pythagorean Theorem:
AB AC BC
AB 6 8AB = 10Circumference = 2 = 2 (5) 10 cm.π π π
= +
= +
=r
Hence, orθπ
= FHGIKJ = °15 15
18085 9. . . .
202 Section 1, Answer Key, Lesson 6 Principles of Mathematics 12
Module 2
12.
13. a) Reference arc = sin–1(0.123)= 0.123
Therefore, x = 0.12 to two decimal places.sin is positive in Quadrants I and II.x = 0.12 is the measure of the angle in Quadrant I.In Quadrant II, x = (π – 0.12)
x = 3.02
∴ x = 0.12 or 3.02
b) Reference arc = cos–1(0.123)= 1.4475
Therefore, x = 1.45 to two decimal places.cos is negative in Quadrants II and III.In Quadrant II, x = (π – 1.45)
x = 1.69In Quadrant III, x = (π + 1.45)
x = 4.59∴ x = 1.69 or 4.59
a b
c d
e f
g h
i j
k l
m n
o
o
) . ) .
) . )cos .
.
)tan
. ) .
) . ) .
)sin
. ) .
)tan .
. ) .
) . )sin
.
) .
2 23716 0 93969
0 67546115
1413683
13
7 01525 0 42262
0 61104 0 05489
15
1 04284 0 59448
10 5
183049 014112
0 540301
414 33559
218504
b g
b g
b g
b g
d i
=
= −
−
= −
= −
=
Principles of Mathematics 12 Section 1, Answer Key, Lesson 6 203
Module 2
c) Reference arc = tan–1 2= 1.1072
Therefore, x = 1.11 to two decimal places.tan is positive in Quadrants I and III.x = 1.11 in Quadrant I.In Quadrant III, x = (π + 1.11)
x = 4.25∴ x = 1.11 or 4.25
d) Reference arc = sin–1 0.25= 0.2527= 0.25 (rounded to two decimal places)
sin is negative in Quadrants III and IV.In Quadrant III, x = (π + 0.25)
x = 3.39In Quadrant IV, x = (2π – 0.25)
x = 6.03∴ x = 3.39 or 6.03
e) Reference arc = cos–1 (0.675)= 0.82983
x = 0.83cos is positive in Quadrants I and IV.In Quadrant I, x = 0.83In Quadrant IV, x = (2π – 0.83)
x = 5.45∴ x = 0.83 or 5.45
204 Section 1, Answer Key, Lesson 6 Principles of Mathematics 12
Module 2
f) sec x = 2.5
Reference arc = cos–1(0.4)= 1.1593= 1.16
cos is positive in Quadrants I and IV.In Quadrant I, x = 1.16In Quadrant IV, x = (2π – 1.16)
x = 5.12
∴ x = 1.16 or 5.12
g) cot x = –4
Reference arc = tan–1(0.25)= 0.245
tan is negative in Quadrants II and IV.In Quadrant II, x = (π – 0.245)
x = 2.90In Quadrant IV, x = (2π – 0.245)
x = 6.04∴ x = 2.90 or 6.04
1 4tan
1tan 0.254
x
x
= −
= − = −
1 2.5cos
1cos 0.42.5
x
x
=
= =
Principles of Mathematics 12 Section 1, Answer Key, Lesson 6 205
Module 2
h) csc x = 6
Reference arc = sin–1(0.16667)= 0.16745= 0.17
sin is positive in Quadrants I and II.In Quadrant I, x = 0.17In Quadrant II, x = (π – 0.17)
= 2.97∴ x = 0.17 or 2.97
14.
Note: Since x < 360° in solving the equation, you use 2x < 720° (twice around the unit circle).
o oc) sin 1 when 90 or (360 90) 4502 90 or 450
45 or 225
x
x
θ θ= = + == ° °
∴ = ° °
o o
sinb) 1, provided cos 0cos
tan 1
45 , or 225
x xx
x
x
= ≠
=
=
( )
2a) 2 sin sin 0
sin 2 sin 1 0
1sin 0 or sin2
0 , 30 , 150 , 180
x x
x x
x x
x
− =
− =
= =
= ° ° ° °
1 6sin
1sin 0.16676
x
x
=
= =
206 Section 1, Answer Key, Lesson 6 Principles of Mathematics 12
Module 2
15.
b) (twice around)
5 9 132 , , ,4 4 4 4
5 9 13, , ,8 8 8 8
x
x
π π π π
π π π π
=
=
( )a) cos cos 2 0
cos 0 or cos 23
, cos 22 2
x x
x x
x xπ π
− =
= =
= ≠
( )( )
2
1 o
o o o o
o o o o
f) 2 tan tan 1 0
2 tan 1 tan 1 0
1tan or tan 12
1Related tan 26.572
tan is negative in Quadrants II and IV.
(180 26.57) , (360 26.57) , or 45 , 225
153.43 , 333.43 , 45 , 225
x x
x x
x x
x
x
−
− − =
+ − =
−= =
∠= =
= − −
=
4e) csc 2
2
1sin
2
30 , 150
x
x
x
= =
=
= ° °
o o o o
o o o o
o o o o
1d) cos 22
1cos when 60 , 300 , 420 , 6602
(twice around again)
2 60 , 300 , 420 , 660
30 , 150 , 210 , 330
x
xx
θ
=
= =
==
Principles of Mathematics 12 Section 1, Answer Key, Lesson 6 207
Module 2
( )f) sin tan 1 0
sin 0 or tan 1
3 70, , ,4 4
x x
x x
x π ππ
+ =
= =−
=
3e) ,
2 2 2
, 2 (not allowed)
Use .2 2
0,
x
x
x
x
π π π
π π
π π
π
− =
=
− =−
=
( ) 1d) sin
2
5,
6 67 11
, (both within range)6 6
x
x
x
π
π ππ
π π
− =
− =
=
1 1c)
sin cos
sin1
cos
1 tan
5,
4 4
x x
xx
x
x π π
=
=
=
=
208 Section 1, Answer Key, Lesson 6 Principles of Mathematics 12
Module 2
ReviewAnswer Key
1.
2.
( )k) tan tan 1 0
tan 0, 15
0, , ,4 4
θ θ
θπ π
θ π
− =
=
=
( )( )
2j) 2 cos cos 1 0
2 cos 1 cos 1 0
1cos , 1
25
, ,3 3
θ θ
θ θ
θ
π πθ π
+ − =− + =
= −
=
i) cot 35 11
,6 6
θπ πθ
= −
=
2 1h) cos41cos2
2 4 5, , ,3 3 3 3
θ
θ
π π π πθ
=
= ±
=
( )g) cos 2 cos 1 0
1cos 0, cos2
3 2 4, , ,2 2 3 3
θ θ
θ θ
π π π πθ
+ =
= =−
=
5 7 3 7a) , b) ,4 4 4 4
5 7c) , d)3 3 6
11 3 11e) f) cos , ,6 2 6 6
π π π π
π π π
π π πθ θ= ∴ =
a) b)
c) d)
e) f)
1212
12
3112
2
2 2 1 4 1 1 0
2 2 3 4 3 1 1 0 1
2 2
2 2
FHGIKJFHGIKJ =
−= −
− − = − − − − =
− − = − = − + = −
b g b g b ge j e j
Principles of Mathematics 12 Section 1, Answer Key, Review 209
Module 2
3.
4.
5.
a)
b)
c) d)
e) f)
( )( )
2
o o o o
2 tan tan 1 0
2 tan 1 tan 1 0
1tan , 12
26.565 , 206.565 , 135 , 315
x xx x
x
x
+ − =− + =
= −
=
csc 21sin2
7 11,6 6
θ
θ
π πθ
= −
= −
=
o o o o
o o o o
1sin 2
22 30 , 150 , 390 , 510
15 , 75 , 195 , 255
=
=
=
x
xx
2 3
232
θ π π
θ π π=
=
,
,
sin1
costan 1
45 , 225
=
== ° °
xxxx
( )
22 sin sin 0
sin 2 sin 1 0
1sin 0,
25
0, , ,6 6
θ θ
θ θ
θ
π πθ π
− =− =
=
=
5 180 11 180a) 150 b) 1656 12
180 180c) 3.8 217.7 d) 0.5432 31.1
π ππ π
π π
= ° − = − °
= ° − = − °
a) b)
c) d)
45180 4
150180
56
500180
259
600180
103
π π π π
π π π π
FHGIKJ = − F
HGIKJ = −
FHGIKJ = F
HGIKJ =
210 Section 1, Answer Key, Review Principles of Mathematics 12
Module 2
g) h)
i) ( )
o o o o
cos tan 1 0
cos 0, tan 1
90 , 270 , 135 , 315
x x
x xx
+ =
= =−=
3 5, , ,. . .2 2 2
3, , ,. . .2 2 23,
2 2
π π πθ π
π π πθ
π πθ
+ =
= −
=
1cos2 2
5. . ., , ,2 3 3 3
5 13. . ., , ,6 6 6
5,6 6
πθ
π π π πθ
π π πθ
π πθ
− =
− = −
=
=
Principles of Mathematics 12 Section 1, Answer Key, Review 211
Module 2
Module 2
Lesson 1Answer Key
1. Domain is ℜ
2.
3. y-intercept = 1
4. Zeros are the odd integral multiples of
This is the same as the sequence . . .
5. Period = 2π
6. Cos (x) = cos (−x); the transformation of cos (x) to cos (−x) is areflection across the y-axis, and yields the same graph. Thiskind of symmetry is called "even."
Cos (x) ≠ −cos (−x); when lines are drawn through the origin tothe cosine graph, the origin is rarely the midpoint of those lines.Note that this is the opposite case from the sine curve. Thus, thecosine curve is not symmetric about the origin, but the sinecurve is.
We’ll ask you to distinguish "even" from "odd"symmetry several timesin the course. How canyou remember thatcos(x) has "even" andsin(x) has "odd"
3 3 5– , – , , , . . .
2 2 2 2 2π π π π π
( )or 2 1 | .2 2
k kπ π + ∈ Ι
[ ]{ }Range is 1,1 | 1 1y y− ∈ ℜ − ≤ ≤
θ
y
2π−π −π 2
π2
1
π 3π 2
−1
Principles of Mathematics 12 Section 2, Answer Key, Lesson 1 213
The graph of f(x) = x2 The graph of f(x) = x3
symmetry? One trick is to know where the even-odd termscome from. The parabola f(x) = x2 is familiar to you, and youcan see right away that f(x) = f(−x). Like the cosine graph,the parabola graph has even symmetry. And it just sohappens that exponent 2 in f(x) = x2 is an even number.
Now consider the graph of f(x) = x3. It has symmetry like thesine curve-through the origin, which we call odd symmetry-just as odd as the 3 in its equation!
In fact, the graphs of x1, x3, and x5 all display "odd"symmetry like the sine curve, while the graphs of x0, x2, andx4 all have "even" symmetry like the cosine curve. You mayuse your graphing calculator to see this.
By recalling the shapes of the x2 and x3 curves, you can usethe symmetry terms "even" and "odd" with confidence.
7. Amplitude = 1.
8. In one revolution, cos θ > 0 in Quadrants I and IV; cos θ < 0in Quadrants II and III.
9. In one revolution, the cosine curve is increasing inQuadrants III and IV and decreasing in Quadrants I and II.
214 Section 2, Answer Key, Lesson 1 Principles of Mathematics 12
Module 2
Lesson 2Answer Key
1.
a)
b) y
− 1
1
xπ2
π
x
y
2π
2
π
− 2
Question Domain Range Amplitude y-intercept Period
(a) ℜ [–2, 2] 2 0 2π
(b) ℜ [–1, 1] 1 0 π
(c) ℜ [–1, 1] 1 0 2π
(d) ℜ [–2, 2] 2 2 2π
(e) ℜ [–1, 1] 1 –1 π
(f) ℜ [–2, 0] 1 2π
(g) ℜ [0, 1] not a wave 0 π
(h) ℜ [–3, 3] 3 0 π
(i) ℜ [–1, 1] 1
(j) ℜ [0, 3] not a wave 1
(k) ℜ [–1, 1] 1 1
(l) ℜ [–2, 2] 2 0
2π/3
2π
2
4π
3 3
2
−
1
Principles of Mathematics 12 Section 2, Answer Key, Lesson 2 215
Module 2
c)
d)
e)
f)
x
y
−π
2
π
−1
−
344
x
y
−π π 2π1
−1
y
2
−2
xπ 2π
y
1
−1
xπ 2π
216 Section 2, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
g)
x makes all the values of x < 0 behave as values of x > 0.
h)
i)
j)
x
y
3
π3
π π2
2
1
x
y
−1
1
π3
π
π2
y
−3
3
π3
π π2
x
y
− 2π
1
−1
ππ
Principles of Mathematics 12 Section 2, Answer Key, Lesson 2 217
Module 2
k)
Cosine sine with period =
l)
Stretch by a factor of 2 in x-direction and y-direction; followed bya reflection in the x-axis.
2. Some possible answers. There are others. a) a = –2 b = 1 c = 0
b) a = 3 b = 1 c =
3. Some possible answers. There are others.
a) a = 2 b = 1 c = π
b) a = –3 b = 1 c = 0
a = 3 b = 1 c = π
–2π
x
y
−2
−2π
2
2π
22.
ππ
=
x
y
21
−1
−1
1
218 Section 2, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
4.
5. Note: Be sure you know how to denote various sets of integers.• All integers are denoted by k ∈ I.• Even integers are denoted by 2k.• Odd integers are denoted by 2k + 1 or 2k – 1, since odd integers
are one greater or one less than any even integer denoted by2k.
This question is equivalent to solving equations that you arefamiliar with except that now the answers are not restricted to aninterval.
( )
b) 0 cos23 52 . . ., , , , , . . .
2 2 2 23 5. . ., , , , , . . .
4 4 4 42 1
I4
x
x
x
kx k
π π π π
π π π π
π∈
=
= −
= −
+ =
a) 0 sin33 . . ., , 0, , 2 , 3 , . . .
2. . ., , 0, , , . . .3 3 3
I3
xx
x
kx k
π π π ππ π π
π∈
== −
= −
=
a) b)
c) d)
e) f)
22
22
2 23
24 2
212
4
π π ππ
π π
π π π π
= =
= =
Principles of Mathematics 12 Section 2, Answer Key, Lesson 2 219
Module 2
( )f) 0 3sin 2
2 . . ., , 0, , 2 , 3 , . . .1 1 3
. . ., ,0, , 1, , . . .2 2 2
I2
x
x
x
kx k
π
π π π π π
∈
=− −
− = −
= − − −
=
( )
2 2 2
2
e) 0 cos
3 5. . ., , , , , . . .
2 2 2 23
. . ., , , , . . .2 2 2
2 1I
2
x
x
x
kx k
ππ π π π
ππ π π
π∈
=
= −
= −
+ =
I
d) 0 sin. . ., , 0, , 2 , 3 , . . .
. . ., 1, 0, 1, 2, 3, . . .
xxxx
ππ π π π π
∈
== −
= −
( )
c) 0 sin2
. . ., , 0, , 2 , 3 , . . .2
. . ., , , , 2 , . . .2 2 2 23 3
. . ., , , , , . . .2 2 2 22 1
I2
x
x
x
x
kx k
π
ππ π π π
π π π ππ π π
π π π π
π∈
= −
− = −
+ − − =
− − =
+ =
220 Section 2, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
Extra for Experts6.
⇓
7.
⇓
x
y
−π π 2π
y
x−π 2ππ
sin x
|sin x|
x
y
−2
−π 2
cos x −
π2
|cos x|
3π 2
x
y
π
−1
−π
1 cos x
π2
|cos x|
Principles of Mathematics 12 Section 2, Answer Key, Lesson 2 221
Module 2
Lesson 3Answer Key
1.
y
−2π −π π2
2πx
1
−1
y = sec x
π
3π 2
π2−
3π 2−
y
−2π −π π 2π x
1
−1
y = csc x
Principles of Mathematics 12 Section 2, Answer Key, Lesson 3 223
Module 2
Note: cot x = 0 where tan x is undefined and tan x = 0 where cot xis undefined.
2.
Function Asymptotes Period SymmetryQuadrant
Sign
Increasing or
Decreasing
cot x {x ≠ kπ|k ∈ I} π odd+ in I & III
– in II & IValways decreasing
csc x {x = kπ|k ∈ I} 2π odd+ in I & II
– in III & IV
inreasing: II & III
decreasing: I & IV
sec x 2π even+ in I & IV
– in II & III
inreasing: I & II
decreasing: III & IV( )
π ≠ + ∈
| 2 1 , I2
x x k k
Function Domain Range y-intercept Zeros
cot x {x ≠ kπ|k ∈ I} ℜ none
csc x {x ≠ kπ|k ∈ I} (–∞, –1]∪[1,∞) none none
sec x (–∞, –1]∪[1,∞) 1 none
( )π
= + ∈
2 1 | I2
x k k
( )π
≠ + ∈
| 2 1 , I2
x x k k
y
2π
−3π 2
x
y = cot x
−π 2
π2
3π 2
π
224 Section 2, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
3. a)
b)
c)
d)coty x=
x
y
−π −π 2
π2
3π 2
csc2
y x π = −
x
y
−π 2
1
−1
π2
3π 2
sec 1y x= +
x
y
−π π
1
2
tany x=
x
y
π2
−π 2
−π π
Principles of Mathematics 12 Section 2, Answer Key, Lesson 3 225
Module 2
e)
f)2secy x=
x
y
−π π
1
2
−1
−2
tan 1= −y x
x
y
−π π
−1
−π 2
π2
3π 2
226 Section 2, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
ReviewAnswer Key
1.
2. a)
b)
c)
1
−1
2π
y
xπ
1
−1
ππ2
x
y
1
−1
ππ 2π x
y
a)
b)
c)
22
22
21
2
π π
πππ π
=
=
=
Principles of Mathematics 12 Section 2, Answer Key, Review 227
Module 2
d)
e)
f)
g)
1
−1
2πx
y
π−π
1
−1
ππ2
x
y
1
−1
ππ2
x
y
2
−2
π 2π x
y
228 Section 2, Answer Key, Review Principles of Mathematics 12
Module 2
h)
i)
j)
k)
−2
x
y
π2
3π 2
π2−
3
π2
x
y
3π 2
π2−
1
y
−1 x2π
4π
1
−1
2πx
y
π−π
Principles of Mathematics 12 Section 2, Answer Key, Review 229
Module 2
l)
The sketches can be used to fill in many parts of the followingtable.
_____________In question (h): the period is mostly 2π but has a “break” atthe y-intercept. So strictly speaking, there is not a patternthat repeats across the whole function. Consider either “none”or “2π” as correct.
1
x
y
π 2π−π
230 Section 2, Answer Key, Review Principles of Mathematics 12
Module 2
Question Domain Range Amplitude y-intercept Period(a) ℜ [–1, 1] 1 0 2π
(b) ℜ [–1, 1] 1 1 π
(c) ℜ [–1, 1] 1 0 2π
(d) ℜ [–2, 2] 2 0 2π
(e) ℜ [–1, 1] 1 0 π
(f) ℜ [–1, 1] 1 1 π
(g) ℜ [0, 1] none 0 π
(h) ℜ [–1, 1] 1 0 none or 2π
(i) ℜ [–1, 1] 1 1 4π
(j) ℜ none 0 π
(k) (–∞, –2]∪[0, ∞) none 0 2π
(l) [1, ∞) none none π
( )x k kπ ≠ + ∈
2 1 I2
( )x k kπ ≠ + ∈
2 1 I2
x k kπ ≠ ∈
I2
3.
Range ℜ ℜ
Domain [–2, 4] [–2, 8]period π 4π
amplitude 3 5phase shift π/4 –π
y-intercept –2 3
y
π
3
4π
2−
x−2π
8 ∗
∗∗
∗ ∗
∗ ∗
2π 3ππ
( )12b) 5cos 3y x π=− + +
y
π4
1
5π
4−1
2−
x−3π
9π44
4∗ ∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗∗
( )4a) 3sin 2 1y x π= − +
Principles of Mathematics 12 Section 2, Answer Key, Review 231
Module 2
( )43sin 2 1y x π= − + ( )125cos 3y x π=− + +
4.
Compare your answers to question 5 with the graphs of 2 = sin 2 and 2 = cos 2, you should notice that:
sin cos2
cos sin2
x x
x x
π
π
= − = −
y
π
x
−1
5π
1
2 2∗
∗
∗
∗
∗
∗
∗
∗
b) cos cos2 2
y x xπ π = − − = −
y
π
x
−1
5π
1
2 2∗
∗
∗ ∗ ∗ ∗
a) sin sin2 2
y x xπ π = − − =− −
232 Section 2, Answer Key, Review Principles of Mathematics 12
Module 2
Module 2
Lesson 1Answer Key
1. Using the CAST rule:a) II b) IV c) III d) IV
2. a) The third side is a2 + 52 = 132 or a = 12. Furthermore, since sinθ > 0 and tan θ < 0, it follows that θ ∈ II. Therefore,
b) The third side is a2 + 32 = 52 or a = 4. Furthermore, since sin x< 0 and cos x < 0, it follows that x ∈ III. Therefore,
5 a
3 5
−4 5
,
3x −
sin x tan x csc x sec x cot x
− 45
43
− 54
− 53
34
3
5
θ
a
cos θ tan θ csc θ sec θ cot θ
−1213
− 512
135
−1312
−125
Principles of Mathematics 12 Section 3, Answer Key, Lesson 1 233
c) The third side is 52 + 12 = a2 or a = . Furthermore,since tan θ > 0 and cos θ < 0, it follows that θ ∈ III. Therefore,
d) Since csc θ = 2, it follows that sin θ = The third side is
a2 + 12
= 22 or a = Furthermore, since sin θ > 0, it follows that θ ∈ I or II. Therefore,
2
a
,− 3 122
1
θ
,3 122
sin θ cos θ tan θ sec θ cot θ
12 ± 3
2± 1
3± 2
3± 3
3.
12
.
5a
1 −5,
1
−
θ 26 26
sin θ cos θ csc θ sec θ cot θ
− 526
− 126
− 265
− 2615
26
234 Section 3, Answer Key, Lesson 1 Principles of Mathematics 12
Module 2
e)
f) Since cot x = it follows that tan x = The third
side is 152 + 82 = a2 or a = 17. Furtherfore, since tan x < 0, itfollows that θ ∈ II or IV. Therefore,
15a
8
, 817
1517
−
x
, 817
1517
−
sin x cos x tan x csc x sec x
±1517
± 817
−158
±1715
±178
−158
.− 815
,
13 a
5
,
x
513
1213
sin x cos x tan x csc x cot x
1213
513
125
1312
512
2 2 2
13 5Since sec , it follows that cos . The third side 5 13
is 5 13 or 12. Furthermore, since cos 0 and tan 0 then I . Therefore,
x x
a a xx x ∈
= =
+ = = >>
Principles of Mathematics 12 Section 3, Answer Key, Lesson 1 235
Module 2
3. One solution is provided but there are other solutions.
2 2
2 2
2 2
j) LHS
(tan 1) (cot 1)
tan 1 cot 1
tan cot RHS
θ θθ θθ θ
+ − += + − −= − =
2
22
i) LHS(1 sin ) (1 sin )(1 sin )(1 sin )
21 sin
2 2sec RHScos
θ θθ θ
θ
θθ
+ + −+ −
=−
= = =
2 2 2 2
2 2
2
h) LHS
(cos sin )(cos sin )
1((1 sin ) sin )
1 2 sin RHS
x x x xx xx
+ −= − −= − =
2 2
2
g) LHS
(1 sin ) sin
1 2 sin RHS
x x
x
− −
= − =
f) LHS
1cos
1sin
1 sincos 1
sincos
tan RHS
x
x
xx
xx
x
= ⋅
=
= =
a) LHS RHS
b) LHS RHS
c) LHS RHS
d) LHS RHS
e) LHS RHS
= FHGIKJ = =
= FHGIKJ = =
= FHGIKJ = =
= = =
= = =
coscos
sinsin
tantan
cossin
sin cos
sincos
cos sin
xx
xx
xx
xx
x x
xx
x x
11
11
11
236 Section 3, Answer Key, Lesson 1 Principles of Mathematics 12
Module 2
2
2
2
2
2
2
2
2
n) RHS
12 1
sin
21
sin
2 sinsin
2 (1 cos )sin
1 cosLHS
sin
x
x
xx
xx
xx
−
= −
−=
− −=
+= =
m) RHS
LHS
1
1
1
1
2 2
2 2
sincos
cossin
sin coscos sin
( )cos sin
sin cos
cos sincos sin
xx
xx
x xx xx xx x
x x x x
+
=+
=+
= = =
2
2
3
l) LHS
sec (1 tan )
sec (sec )
sec RHS
θ θθ θθ
+== =
k) LHS
RHS
1 1
1
1
1
11
1
2
2
2
2
2
2
2
2
2 2
sin
sin
sinsin
sin
sinsin
sin
sin cos
θ
θ
θθ
θ
θθ
θ
θ θ
−
=
−
= − ⋅
= − = =
Principles of Mathematics 12 Section 3, Answer Key, Lesson 1 237
Module 2
4.
2 2
2 2
2
2
coscos sinsinb) 1sin cos1cos
sin
1c) sin cos cossin
cos sin 1d) cscsin sin
e) cos (sin cos ) cos (1) cos
1 cotf) 1
1 cot
xx xxx xx
x
x x xx
x x xx x
x x x x x
xx
= ⋅ =
=
+ = =
+ = =
+ =+
22
2 2 2
sin cos csc sina) cos1csc csc
sin
sinsin cos sin cos 11
x x x x xx x
x
xx x x x
+ = +
= + = + =
238 Section 3, Answer Key, Lesson 1 Principles of Mathematics 12
Module 2
Lesson 2Answer Key
1. a)
b)
c)
2. a)
b) 3cos25 7 11, , ,
6 6 6 6
x
x π π π π
= ±
=
2
2
2(1 sin ) 1 sin
0 2 sin sin 10 (2 sin 1)(sin 1)
1sin or sin 1
27 11
, ,6 6 2
x xx xx x
x x
x π π π
− − = −
= − −= + −
= − =
=
o o o
tan sin tan 0tan (1 sin ) 0
tan 0 or sin 1
0 , 180 , 90
x x xx x
x xx
− =− =
= ==
o o o o
o o o o
sin 2 1 provided cos 2 0cos 2tan 2 1
2 45 , 225 , 405 , 585
22.5 , 112.5 , 202.5 , 292.5
x xxxxx
= ≠
===
2 2
2
o o o o
tan (1 tan ) 3
2 tan 2tan 1
45 , 135 , 225 , 315
x xxxx
+ + =
== ±=
Principles of Mathematics 12 Section 3, Answer Key, Lesson 2 239
Module 2
c)
3. a)
Check:
b) Be careful! Since the LHS is 0, this equation can be done quiteeasily.
0 : sin 0 0 and 1 cos 0 1 1 0
: sin 1 and 1 cos 1 0 12 2 23 3 3
: sin 1 and 1 cos 1 0 12 2 2
3Therefore, is not a root.
2
Solution is: 0, .2
x
π π π
π π π
π
π
= − = − =
= − = − =
= − − = − =
=
3Possible values of 0, , .
2 2x π π=
(square both sides)2 2
2 2
2
sin 1 cos
sin 1 2 cos cos
(1 cos ) 1 2 cos cos
0 2 cos 2 cos0 2 cos (cos 1)
cos 0 or cos 1
x xx x xx x x
x xx x
x x
= −= − +
− = − += −= −= =
2 1sin
2 4
1sin
2 25 7 11
. . ., , , , , , . . .2 6 6 6 6 6
2 4 5, , ,
3 3 3 3
x
x
x
x
π
π
π π π π π π
π π π π
− = − = ±
− = −
=
240 Section 3, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
(Note: is out of range.)73π
c)
The discriminant (b2 – 4ac) of the quadratic is (–4)2 – 4(2)3 < 0. Hence, no solution! Observe the originalequation as sin x – cos x = 2. It should be apparent thatno solution exists since sin x ≤ 1 and cos x ≤ 1, but theyare never equal to 1 at the same time.
4. a)
b)
c)
The identity seems to be true since it was valid in all the abovecases.
cos cos cos sin sin4 4 4
1 1 11 02 2 2
3 1But, cos cos . They are equal!4 4 2
π π ππ π π
π ππ
− = + = − + = −
− = = −
o o o o o o
o o o
cos(150 30 ) cos 150 cos 30 sin 150 sin 30
3 3 1 1 12 2 2 2 2
1But, cos(150 30 ) cos 120 . They are equal!
2
− = +
= − + = −
− = = −
o o o o o o
o o o
cos(90 30 ) cos 90 cos 30 sin 90 sin 30
3 1 10 1 .
2 2 2
1But, cos(90 30 ) cos 60 . They are equal!
2
− = +
= + =
− = =
(square both sides)2 2
2 2
2
sin 4 sin 4 cos
sin 4 sin 4 (1 sin )
2 sin 4 sin 3 0
x x xx x xx x
− + =
− + = −− + =
sin cossin 1costan 1
3 7Therefore, , .4 4
x xxxx
x π π
= −
= −
= −
=
Principles of Mathematics 12 Section 3, Answer Key, Lesson 2 241
Module 2
5.
Note:
6. The form of the identities is where A and B
have names which differ by the prefix “co.” For example, sine andcosine are two such functions.
( )2
A x B xπ − =
oo
oo
oo
1 1d) sec(90 ) csc
cos(90 ) sin
1 1e) csc(90 ) sec
sin(90 ) cos
1 1f) cot(90 ) tan
tan(90 ) cot
x xx x
x xx x
x xx x
− = = =−
− = = =−
− = = =−
90 . Therefore, the identities are valid with 2either measure 90 .
2or
π
π= °
°
a) cos cos cos sin sin2 2 2
0(cos ) 1 sinsin
b) Use part(a), replacing with .2
sin cos cos2 2 2
sincos2c) tan cot
2 sincos2
x x x
x xx
x x
x x x
xxx xxx
π π π
π
π π π
ππ
π
− = + = +=
−
− = − − =
− − = = = −
242 Section 3, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
7.
8.
{ }I
b) sin 12cos 1 (co-function)
related arc: 0
( )2
x
x
x k k
π
π ∈
− = =
=
( )
I
a) cos 12
sin 1 co-function
related arc: 2
( )22
π
π
π π ∈
− = =
= +
x
x
x k k
2 2
a) sin ( ) cos( ) cos2
b) tan ( ) cot( ) cot2
c) cos cos ( ) sin( ) sin2 2
cosd) sin( )( cot ) sin cot sin cos
sin
1e) (sin ) ( csc ) (sin ) sin
sin
1f)
csc
x x x
x x x
x x x x
xx x x x x xx
x x x xx
π
π
π π
− − = − =
− − = − = −
+ = − − = − = −
− − = = =
− = − = −
sin xx=
Principles of Mathematics 12 Section 3, Answer Key, Lesson 2 243
Module 2
There are other forms of these answers. This answer may also bestated in degrees since there were no instructions on the values ofx. In degrees, the answers matching the above form are:
{ }{ }{ }
o o
o
o o
I
I
I
a) 90 (360 )
b) (360 )
c) 90 (180 )
x k k
x k k
x k k
∈
∈
∈
= +
=
= +
I
c) sin 02
sin 02cos 0
3related arc: ,2 2
2
x
x
x
x k k
π
π
π π
π π ∈
− = − − = − =
= +
244 Section 3, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
Lesson 3Answer Key
1.
2.
3.7cos cos cos cos sin sin12 4 3 4 3 4 3
1 1 1 3 1 3 2 6 or 2 2 42 2 2 2
7 2 6 2 6Therefore, the coordinates are: P , .12 4 4
π π π π π π π
π
= + = −
− −= ⋅ − ⋅ =
− + =
7sin sin sin cos cos sin
12 4 3 4 3 4 3
1 1 1 3 1 3 2 6 or
2 2 42 2 2 2
π π π π π π π = + = +
+ += ⋅ + ⋅ =
5 1a) sin sin9 36 4 2
2 3b) sin sin 05 5
7 1c) cos cos12 4 3 2
5 2 1d) cos cos9 9 3 2
π π π
π π π
π π π
π π π
+ = =
+ = =
− = =
+ = = −
Principles of Mathematics 12 Section 3, Answer Key, Lesson 3 245
Module 2
4.
Note: The answers appear to
be different. However, they are not; hence, you can use either form.
5.
22 2 2
7Since sin and P( ) QI, it follows that
25
P( ) QII, so cos 0.
7Since sin cos 1, cos 1.
2524
Therefore, cos .25
9Similarly, since cos and P( ) QI, it foll ows that
41P( ) QIV, so sin 0
α α
α α
α α α
α
β β
β β
= ∉
∈ <
+ = + =
= −
= ∉
∈ <2
2 2 2
.
9Since sin cos 1, sin 1.
4140
Therefore, sin .41
β β β
β
+ = + =
= −
o o o o o o o
o
o oo o o
o o
b) sin 105 sin(60 45 ) sin 60 cos 45 cos 60 sin 45
3 2 1 2 6 22 2 2 2 4
6 2sin 105 6 24c) tan 105cos 105 2 6 2 6
4
(This is the easy way after (a) and (b). )
tan 60 tan 45 3 1tan 105 tan(60 45 )
1 tan 60 tan 45 1 3
T
= + = +
+= + =
+° += = =° − −
+ += + = =− −
he answers are equal!
22
is the rationalized form of 12
.
o o o o o o oa) cos 105 cos(60 45 ) cos 60 cos 45 sin 60 sin 45
1 2 3 2 2 62 2 2 2 4
= + = −
−= − =
246 Section 3, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
6.
7.
8.
b) Extra for Experts
2 2 2 2
2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2
2 2
cos( )cos( )(cos cos sin sin )(cos cos sin sin )
cos cos sin sin
cos cos 0 sin sin
cos cos (cos sin cos sin ) sin sin
cos (cos sin ) sin (cos sin )
cos (1) sin
x y x yx y x y x y x yx y x yx y x yx y x y x y x yx y y y x xx y
+ −= − += −
= + −
= + − −
= + − += −
2 2
(1)
cos sinx y= −
a) sin( ) sin( )sin cos cos sin sin cos cos sin2sin cos
α β α βα β α β α β α βα β
+ + −= + + −=
1tantan tan 3 tan 136tan6 1 3 tan1 tan tan 1 tan
6 3
π θθπ θθ π θθ θ
−− − − = = = + + +
1 3cos cos cos sin sin cos sin3 3 3 2 2π π πθ θ θ θ θ + = − = −
a) sin( ) sin cos cos sin
7 9 24 40 63 960 102325 41 25 41 1025 1025
b) cos( ) cos cos sin sin
24 9 7 40 216 280 6425 41 25 41 1025 1025
1023sin( )
c) tan( )cos( )
α β α β α β
α β α β α β
α βα βα β
+ = ++ = + − − = =
+ = −− + = − − − = =
++ = =+
1023102564 64
10251 1 1025
d) sec( )64cos( ) 64
1025e) In Quadrant I, since both sin( ) 0 and
cos( ) 0.
α βα β
α βα β
=
+ = = =+
+ >+ >
Principles of Mathematics 12 Section 3, Answer Key, Lesson 3 247
Module 2
9. a) sin(2 3 ) sin 5b) tan( 3 ) tan 4c) sin( 6 ) sin 7d) cos 3 cos 3 sin 3 sin 3 cos(3 3 ) cos 6e) sin 5 cos 5 sin 5 cos 5 sin(5 5 ) sin10f) cos(2 3 ) cos( ) cos , since cos is an even f unction
x x xx x x
x x x x x x x
a a a a
α α α
α α α α α α α
+ =+ =+ =
− = + =+ = + =
− = − =
248 Section 3, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
Lesson 4Answer Key
1. sin 4A = 2 sin 2A cos 2A= 2(2 sin A cos A)(cos2A – sin2A)
The above is one possibility.
2. cos 4θ = cos2(2θ) – sin2(2θ) = (–0.309)2 – (0.951)2 = –0.809sin 4θ = 2 sin 2θ cos 2θ = 2(0.951)(–0.309) = –0.588. Therefore,P(4θ) = (–0.809, –0.588).
3. Use the following two forms of cos 2θ, and replace 2θ by θ:
(Note:
4.2 2
a) cos 8
(Using the identity cos2 cos sin .)b) sin 4
(Using the identity sin2 2sin cos .)c) 2 sin 2d) cos 10e) cos 20
2sin cos sin 2f) sin cos2 2
α α α
α α α
αα
α α αα α
= −
=
= =
x
x
x
I, then I.)Both positive since 2θθ ∈ ∈
22
22
2 2
2 2
2 2
2 cos 12 cos 1 2cos 2 becomes cos 1 2sin 1 2sin
2
and now solve for cos sin .2 2
cos 2 cos 1 cos 1 2 sin2 2
0.588 2 cos 1 0.588 1 2 sin2 2
1.588 1 .588cos sin2 2 2 2
cos 0.892
θθθ θ
θθ
θ θ
θ θθ θ
θ θ
θ θ
θ
− − = = − −
= − = −
= − = −
−= =
=
and
1 sin 0.4542θ =
Principles of Mathematics 12 Section 3, Answer Key, Lesson 4 249
Module 2
5.
6.
7.
8. Since sin α > 0 and not in Quadrant I, it follows P(α) is inQuadrant II. Similarly, since cos β > 0 and not in Quadrant I, itfollows that P(β) is in Quadrant IV.
2 22 2
4 3 24a) sin 2 2 sin cos 2
5 5 25
3 5 4 12b) cos( ) cos cos sin sin
5 13 5 1315 48 63
65 65
5 12 25 144 119c) cos 2 cos sin
13 13 169 169
α α α
α β α β α β
β β β
= = − = − − = + = − + −
− −= = −
− = − = − − = = −
22
22
4 3cos 1. Hence, in II cos and
5 5
5 12sin 1. Hence, in IV sin .
13 13
α α
β β
+ = = −
+ = = −
2
2 2
2 2
4 2
4 2
cos 4 2 cos 2 1
2(2cos 1) 1
2(2 1) 1
2(4 4 1) 1
8 8 1
pp p
p p
θ θθ
= −= − −= − −= − + −= − +
22
22
I.
3Since 0 and sin , it follows that 2 5
3 4cos 1, so cos with 5 5
3 4 24sin 2 2 sin cos 2 and 5 5 25
3 18 7cos 2 1 2 sin 1 2 1 .5 25 25
πθ θ
θ θ θ
θ θ θ
θ θ
∈
< < =
+ = = = = =
= − = − = − =
2 2cos 2 cos 1 2(0.95) 1 0.8055 10π π = − = − =
250 Section 3, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
9.
10.
2
2 3
2 3
3 3
3
e) sin(2 ) sin 2 cos cos 2 sin
(2 sin cos )cos (1 2 sin )sin
2 sin cos sin 2 sin
2 sin (1 sin ) sin 2 sin
2 sin 2 sin sin 2 sin
3 sin 4 sin
x x x x x xx x x x xx x x xx x x xx x x xx x
+ = += + −= + −= − + −= − + −= −
2 2
2 2
2 2
1 11cos cosd) LHS
1 2 cos 1 2 cos 12cos cos
1sec 2
cos 2
x xx x
x x
xx
= = =− −−
= =
2 2
2 2 2 2
2 2
sin cos sin cos 1a) LHScos sin cos sin cos sin
2 2 2 csc 22 cos sin sin 2
b) LHS (cos sin )(cos sin )
(1)(cos sin ) cos 2
2 sin 1c) LHS sec2 sin cos cos
x x x xx x x x x x
xx x x
x x x x
x x x
x xx x x
+= + = =
= = =
= + −
= − =
= = =
sin 2b)
cos
2 sin coscos
2 sin
xx
x xx
x
=
=
2
2 2
a) (sin cos )
sin 2 sin cos cos
1 2sin cos
1 sin 2
x x
x x x x
x x
x
+
= + +
= +
= +
d) sin( ) sin cos cos sin
4 5 3 12 20 36 565 13 5 13 65 65
α β α β α β+ = ++ = + − − = =
Principles of Mathematics 12 Section 3, Answer Key, Lesson 4 251
Module 2
2
3 2
3 2
3 3
3
f) cos(2 ) cos 2 cos sin 2 sin
(2 cos 1) cos 2 sin cos sin
2 cos cos 2 cos sin
2 cos cos 2 cos (1 cos )
2 cos cos 2 cos 2 cos
4 cos 3 cos
x x x x x xx x x x xx x xx x x xx x x xx x
+ = −
= − −= − −= − − −= − − +
= −
252 Section 3, Answer Key, Lesson 4 Principles of Mathematics 12
Module 2
ReviewAnswer Key
1.
2.
2 2
2 2
d) LHS
(tan 1) (csc 1)
tan csc RHS
θ θθ θ
= + + −= + =
2
2
2
2
2
c) LHS1 sin 1 sin(1 sin )(1 sin )
21 sin
2cos2 sec
2(tan 1)
2 tan 2 RHS
θ θθ θ
θ
θθθθ
+ + −=+ −
=−
=
== += + =
2 2 2 2
2
b) LHS
(cos sin )(cos sin )1(cos 2 )
2cos 1 RHS
x x x xxx
+ −=
= − =
2 2 2
2
a) LHS
cos sin sin
1 sin RHS
x x xx
+ +
= + =
2 2
2 2
1 sina) sin cos sinsin cos
b) sin (sin cos ) sin (1) sin
1c) 2 sin cos 2 sincos
d) cos cos ( sin ) sin cos sin 1
e) 2(2 sin 4 cos 4 ) 2 sin 8
f) sin(5 2 ) sin 3
xx x xx x
x x x x x
x x xx
x x x x x x
x x x
x x x
=
+ = =
=
− − = + =
=
− =
Principles of Mathematics 12 Section 3, Answer Key, Review 253
Module 2
3. a) The third side is a2 + 52 = 132 or a = 12. Furthermore, since sinθ < 0 and tan θ < 0, it follows that θ ∈ QIV. Therefore,
b) The third side is a2 + 32 = 52 or a = 4. Furthermore, since cos x> 0 and csc x < 0, it follows that x ∈ QIV. Therefore,
c) The third side is 42 + 32 = a2 or a = 5. Furthermore, since tan θ< 0 and cos θ < 0, it follows that θ ∈ QII. Therefore,
d) The third side is 82 + 152 = a2 or a = 17. Furthermore, since cotx > 0, it follows that x ∈ QI or QIII. Therefore,
sin x cos x csc x sec x tan x
±1517
± 817
±1715
±178
158
cos θ sin θ csc θ sec θ cot θ
− 45
35
53
− 54
43
−
sin x tan x csc x sec x cot x
− 45
− 43
− 54
53
− 34
cos θ tan θ csc θ sec θ cot θ
1213
− 512
−135
1312
−125
2 2
2
2
f) RHS
sin 2 (1 2 sin 2 )
1 sin 2
cos 2 LHS
x xx
x
= + −= −= =
2
2 2 2
2 2
2
e) LHS
(2 sin cos )sin cos sin
2 sin cos sin
sin (2 cos 1)
sin cos 2 RHS
x x x x xx x x
x xx x
= −
= −= −
= =
254 Section 3, Answer Key, Lesson 4 Principles of Mathematics 12
Module 2
4.
5. a)
o o o o
2 sin cos sin 0sin (2 cos 1) 0
1sin 0, cos2
0 , 180 , 60 , 300
x x xx x
x x
x
− =− =
= =
=
positiveQuadrant I
d) Since both sin( ) and cos( ) are , it follows that ( ) terminates in .
α β α βα β+ ++
3 4 24c) sin 2 2 sin cos 2
5 5 25α α α = = − = −
4 5 3 12b) cos( ) cos cos sin sin
5 13 5 1320 36 16
65 65
α β α β α β + = − = − − − − += =
3 5 4 12a) sin( ) sin cos cos sin
5 13 5 1315 48 63
65 65
α β α β α β + = + = + − − += =
it follows that
22 2 2
3Since sin and P( ) QI, P( ) QII, 5
so cos 0.
3Since sin cos 1, cos 1. 5
4Therefore, cos .5
5Similarly, since cos and P( ) QI, it foll ows that13
P( ) QIV, so sin 0.
α α α
α
α α α
α
β β
β β
= ∉ ∈
<
+ = + =
= −
= ∉
∈ <2
2 2 2 5Since sin cos 1, sin 1. 13
12Therefore, sin .13
β β β
β
+ = + =
= −
Principles of Mathematics 12 Section 3, Answer Key, Lesson 4 255
Module 2
b)
6.
2
2
2
2
2
2
b) Use the identity cos 2 2 cos 1 where .8
cos 2 2 cos 1 8 8
cos 2 cos 14 82 2 cos 1
2 82 22 cos
8 22 2cos
8 4
2 2cos 8 4
(positive, since is in Quadrant I)8
x x x π
π π
π π
π
π
π
π
π
= − =
= −
= −
= −
+=
+=
+=
a) sin sin sin cos cos sin12 4 6 4 6 4 6
2 3 2 1 6 22 2 2 2 4
π π π π π π π = − = −
− = − =
o o o o
2 sin cos cos 0cos (2 sin 1) 0
1cos 0, sin
290 , 270 , 210 , 330
x x xx x
x x
x
+ =+ =
= = −
=
256 Section 3, Answer Key, Lesson 4 Principles of Mathematics 12
Module 2
7. Extra for Experts
The sketch can be used to fill in the following:
2 2
1 cos 1 sin
1 1 2
1 12 cos sin2 2
2 cos4
y x x
r
y x x
y x π
= +
= + =
= + = −
Domain Range Amplitude y-intercept Period
ℜ 1 2π− 2 2, 2
1
y
π4
x
− 2
2
5π 4
9π 4
Principles of Mathematics 12 Section 3, Answer Key, Review 257
Module 2
Lesson 1Answer Key
1.
2. a)
b)
−5
5
π2
π4
π−
x
y
π2
π3
π6
π6−
1
−1
−3
x
y
Question Period Frequency Phase Shift
(a)
(b) none
(c) none
(d) 2π π
23π 3
2π
12π
1π
14
− π3
22π π=
22
4π π÷ =
Principles of Mathematics 12 Section 4, Answer Key, Lesson 1 259
c)
d)
3. a)
b) 2Period is: 12.5 hours.
0.16ππ=
9.2
6.25 12.5−6.25
18.2
t
d
−12.5
π 2ππ
1
23
4
x
y
x
y
4
6
1 3 5−1−3
260 Section 4, Answer Key, Lesson 1 Principles of Mathematics 12
Module 2
c)
−0.17778
t
9.2
3.48 9.02
18.2
t
d
14.5
1
4.5 cos(0.16 ) 13.7 14.54.5 cos(0.16 ) 0.8
0.8cos(0.16 ) 0.177784.5
Related arc is: cos (0.17778) 1.392.Therefore, ( 1.392) 0.16 ( 1.392)
3.48 9.02.Hence, the carrier can dock safely for (9.
tt
t
tt
ππ
π
π π π
−
− + ≥− ≥
≤ = −−
=− ≤ ≤ +≤ ≤
02 3.48) 5.54 hours.− =
Principles of Mathematics 12 Section 4, Answer Key, Lesson 1 261
Module 2
262 Section 4, Answer Key, Lesson 2 Principles of Mathematics 12
Module 2
Lesson 2Answer Key
1. 5 cos2x – 6 cos x = 8, 0 ≤ x < 2π
Y1 = 5[cos(x)]2 – 6 cos(x) – 8
Y2 =
Y3 =
Y4 =
Solutions: 2.50, 3.79[–0.1, 7] [–12, 3]Xmin Xmax Ymin Ymax
2. 3 tan2x = tan x + 4, 0 ≤ x < 2π
Y1 = 3[tan(x)]2 – tan(x) – 4
Y2 =
Y3 =
Y4 =
Solutions: 0.93, 2.36, 4.07, 5.50[–0.1, 7] [–10, 20]Xmin Xmax Ymin Ymax
3. The system: y = 2 sin2x – 11 tan x and y = –4, 0 ≤ x < 2π
Y1 = 2[sin(x)]2 – 11 tan(x)
Y2 = –4
Y3 =
Y4 =
Solutions: (0.37, –4) (3.51, –4)[–0.1, 7] [–10, 20] The next solution (6.65, –4)
Xmin Xmax Ymin Ymax is out of the specified domain.
Lesson 3Answer Key
1. Solving over the domain
and
Then add multiples of the period 2π until you reach 6π
and
2. Solve over the domain
then add multiples of the period.
where n ∈ integers.These two answers can be combined to one answer , where n ∈ integers.
3x n
π π= +
343
x n
x n
π π
π π
= +
= +
2tan 2 3
tan 34
,3 3
x
x
xπ π
=
=
=
0 2θ π≤ <2tan 2 3 0x − =
723 35 1123 3
132 23 35 172 23 3
x
x
x
x
π ππ
π ππ
π ππ π
π ππ π
= + =
= + =
= + + =
= + + =
522 23 722 2
92 22 23 112 22 2
x
x
x
x
π ππ
π ππ
π ππ π
π ππ π
= + =
= + =
= + + =
= + + =
1cos
25
,3 3
x
xπ π
=
=
cos 03,
2 2
x
xπ π=
=
0 2θ π≤ <22cos cos 0
cos (2cos 1) 0x x
x x− =
− =
Principles of Mathematics 12 Section 4, Answer Key, Lesson 3 263
Module 2
3. Solve over the domain
and
Add multiples of the period
4. Solve over the domain
and which has no solution
Add multiples of the period
where n ∈ integers.
nx
nx
ππ
ππ
26
11
26
7
+=
+=
2sin02sin
==−
xx
611,
67
21sin
1sin201sin2
ππ=
−=
−==+
x
x
xx
( )
( )( ) 02sin1sin202sin3sin2
0sin32sin20sin31sin2
2
2
2
=−+=−−
=−−=−−
xxxxxxxx
0 2θ π≤ <0sin3cos2 2 =− xx
nxnx
nx
nnx
ππ
ππ
ππππ
+=
+=
+=
=+=
43
474
30
47,
43
1tanππ=
−=
x
x0
0tan==
xx
tan (tan 1) 0x x + =
0 2θ π≤ <2tan tan 0x x+ =
264 Section 4, Answer Key, Lesson 3 Principles of Mathematics 12
Module 2
5. Solve a simpler version of the problem by solving
over the domain
Now replace so the solutions become
and
Lastly, add multiples of the period. Since the period of
is 4π, the final solutions are
where n ∈ integers.
6.
bnbn
π
π
243.2
276.0
+
+
nx
nx
ππ
ππ
42
5
42
3
+=
+=
x21cos
25
452
45
21
ππ
π
=
=
=
x
x
23
432
43
21
ππ
π
=
=
=
x
x
x21→θ
45,
43
21cos
1cos2
01cos2
ππθ
θ
θ
θ
=
−=
−=
=+
0 2θ π≤ <01cos2 =+θ
Principles of Mathematics 12 Section 4, Answer Key, Lesson 3 265
Module 2
ReviewAnswer Key
1. a)
b)
c)
2. a)
b) 285°
c) 42.4°
3. a)
b)
4. θ = 360 – 58.9 = 301.1°
5. a) –1.2187
b)
( )( )
512512
6 4
6 4
6 4 6 4
6 4 6 4
3 1 1 12 22 2
31 1 12 22 2
cos5cot12 sin
cos
sin
cos cos – sin sin
sin cos cos sin
–
6 – 22 6
π
π
π π
π π
π π π π
π π π π
π =
+=
+
=+
• •=
• + •
=+
902
orπ °
( )23 32 2
5 53 25 5
π ππ
π ππ
− =
− =
2374.136
180π =
tan 4A = −
17csc
4A =
1cos17
A = −
266 Section 4, Answer Key, Review Principles of Mathematics 12
Module 2
c)
6. a)
b)
c) 3 tan 2sin 0sin
3 2sin 0cos
3sin 2 0
cos
3sin 0 or 2 0
cos3
sin 0 or cos2
110, , , or
6 6
x xx
xx
xx
xx
x x
xπ ππ
− =
− =
− =
= − =
= =
=
( )( )23sin sin 2 0
3sin 2 sin 1 0
2sin or 13
221.8 or 318.2 or 90221.8 360 or 318.2 360 or 90 360
where integern n n
n
θ θθ θ
θ
θθ
− − =+ − =
= −
= ° ° °= ° + ° ° + ° ° + °
∈
( )2cos 2cos 0
cos cos 2 0
cos 0 or cos 23,
2 2cos 2 is undefined.
x x
x x
x x
x
x
π π
− =− =
= =
=
=
14 23 3
14 1 1 2csc3 sin sin 3π ππ = = =
Principles of Mathematics 12 Section 4, Answer Key, Review 267
Module 2
7.
Note: It is possible to simplify further using a double angleidentity.
( )12
1 1 2csc2sin cos sin2
θθ θ θ
= =
( ) ( )( )( )
angle
2 2 2
22 2
2
2 2
2
1sin
cos
a) tan cos sin
sincos sin
cossin sin
2sinb) 4sin2 cos2 is a version of the double identity
2sin cos sin2 so4sin2 cos2
2 2sin 2 cos 2
2sin2 2
2sin4
sincsc – sinc)
cot
A A A
θθ
θ θ θ
θ θ θθθ θ
θθ θ
θ θ
θ θ
θθ
θθ θθ
+
• +
+
=
=
=
=
=
−=
sin
2
2
2 2
Multiply numerator and denominator by si n
1 – sin=
coscoscos
cos1 1 cos sin
d)tan cot sin cos
cos sinsin cos
1sin cos
θ
θ
θθθθθ
θ θθ θ θ θθ θθ θ
θ θ
=
=
+ = +
+=
=
268 Section 4, Answer Key, Review Principles of Mathematics 12
Module 2
8.
9.
10.
( )a) sin75 cos25 cos75 sin25
sin 75 25
sin50
° ° − ° °= ° − °
= °
( )
2period = 4 2
phase shift 2
3 53cos4 has zeros at , , , etc.2 8 8 8
3 53sec4 has asymtotes at , , , etc.2 8 8 8
2 1domain = | , 1
8
range = ( , 3] [3, )
π π
π
π π π π
π π π π
π
=
=
− ∴ − + ≠ ∈ −∞ − ∪ ∞
x
x
kx x k
y
x−2
∗
∗
∗
∗
∗
2
5
8
4 8
∗ ∗∗∗∗
∗
[ ]
amplitude 3 3
2period = 4
2phase shift 4vertical displacement 5domain =
range = 2,8
ππ
= − =
=
==
ℜ
Principles of Mathematics 12 Section 4, Answer Key, Review 269
Module 2
11.2
1cos
2
2
sec sec 1a) Prove 1 cos sin
1RHS
1 cos1 cos 1
cos 1 cos1 cos
x
x xx x
xx
x xx
+=−
+=−+= •
−+=
( )1
cos 1 cosx x•
+ ( )
( )( )
( )
2
2
2
2
2
2
1 cos
1 1=cos 1 cos
sec1 cos
2b) Prove 1 cot1 cos2
2LHS Double angle identity1 1 2sin
21 2sin
1sin
=csc
1 cot Pythagorean Identity
RHS
x
x xx
x
θθ
θ
θ
θθ
θ
−
•−
=−
= +−
=− +
=+
=
= +
=
2b) 1 2sin12
cos212
cos6
32
π
π
π
−
= =
=
270 Section 4, Answer Key, Review Principles of Mathematics 12
Module 2
Factoring 1– cos2 x as adifference of squares.
12.a) max = 7.4 mmin = 3.8 m
axis occurs at
amplitude =
period = 2(11:45 – 5:30)= 2(6:15)= 12:30= 12.5 hours
b =
Since the max occurs at 5:30 a.m., we can use a cosinefunction to model the tide flow.phase shift = 5.5 hours
b) 10:30 p.m. is 22.5 hours from midnight
h(22.5) = 1.8 cos 0.16 π (22.5 – 5.5) + 5.6
= 4.45 m
c)
h
t
∗∗
5:30
5.6
3.8
7.4
1:00 a.m.
∗∗
∗
11:45 7:00
( )
( )
1
1.8 cos 0.16 5.5 5.6 6.0
0.4cos 0.16 5.5
1.81 0.4
5.5 cos0.16 1.8
8.18 hours or 8:11 a.m.
t
t
t
π
π
π−
− + =
− =
= + =
( )( ) 1.8cos 0.16 5.5 5.6π= − +h t t
2 2 4 0.16period 12.5 25
π π π π= = =
7.4 3.81.8 m
2+ =
7.4 3.85.6 m
2+ =
Principles of Mathematics 12 Section 4, Answer Key, Review 271
Module 2
Using the graphing calculator
Set Y1 = 1.8 cos (0.16π(x – 5.5)) + 5.6
Set Y2 = 6.0
Solving we get 8.18 or 15.32 or 20.68
The next time after 11.45 a.m. is 3.19 p.m.
13.Set Y1 = 4 tan2 θ – sin θ
Set Y2 = 1 + 3 tan θ
Solutions: 0.849, 2.845,
3.845, 6.078[–0.1, 7] [ –2, 20]Xmin Xmax Ymin Ymax
Another solution,using one graph:
Set Y1 = 4 tan2 θ – sin θ – 1 – 3 tan θ
Solutions: 0.849, 2.845,
3.845, 6.078[–0.1, 7] [ –2, 20]Xmin Xmax Ymin Ymax
(15 hours 12 3.320.32 hours 19 min)
− =≈
272 Section 4, Answer Key, Review Principles of Mathematics 12
Module 2
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