Principles of Mathematics 12 Answer Key, Contents...

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

186 Answer Key, Contents Principles of Mathematics 12

Module 2

Notes

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 }


Module 2

Notes


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


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


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


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


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


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


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


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

+ = + = =

= =

+ − − + − − = + =

=

− =− − =−


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

+ =

=−

=


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

− + =

=±

=


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 =


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. . . .


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

=

= −

−

= −

= −

=


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


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

=

= =


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

=

= =


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

=

= =

==


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

=

=

=

=


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

− =

− = −

=

=


Module 2

Notes


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


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.


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


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


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


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


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

= =

= =


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∈

= −

− = −

+ − − =

− − =

+ =


Module 2

Notes


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


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


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


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


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

=

=

=


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


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


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


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= − +


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 = − − =− −


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


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


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 ∈

= =

+ = = >>


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


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

−

=

−

= − ⋅

= − = =


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

+ = +

= + = + =


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

+ + =

== ±=


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

− = − = ±

− = −

=


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

= −

= −

= −

=


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

− = + = +=

−

− = − − =

− − = = = −


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=


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 ∈

− = − − = − =

= +


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

+ = =

+ = =

− = =

+ = = −


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

= + = −

−= − =


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

=

+ = = =+

+ >+ >


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

+ =+ =+ =

− = + =+ = + =

− = − =


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 =


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 = − = − =


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+ = ++ = + − − = =


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

+ = −

= − −= − −= − − −= − − +

= −


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

=

+ = =

=

− − = + =

=

− =


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

= −

= −= −

= =


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

+ = + =

= −


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

+ =+ =

= = −

=


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


Module 2

Notes


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÷ =


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


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.− =


Module 2


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− =

− =


Module 2


and

Add multiples of the period


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+ =


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


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 = −


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 = = =


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

=

=

+ = +

+=

=


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


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

−

= =

=


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+ =


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)

− =≈


Module 2

Principles of Mathematics 12 Answer Key, Contents...

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