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www.math30.ca Answer Key Trigonometry Lesson One: Degrees and Radians Note: For illustrative purposes, all diagram angles will be in degrees. 120° -120° ť a) The rotation angle between the initial arm and the terminal arm is called the standard position angle. Example 1: b) An angle is positive if we rotate the terminal arm counter- clockwise, and negative if rotated clockwise. c) The angle formed between the terminal arm and the x-axis is called the reference angle. d) If the terminal arm is rotated by a multiple of 360° in either direction, it will return to its original position. These angles are called co-terminal angles. e) A principal angle is an angle that exists between 0°and 360°. f) The general form of co-terminal angles is ť c ť p + n(360°) using GHJUHHV RU ť c ť p Q using radians. 420° 60° 45°, 405°, 765°, 1125°, 1485° 45°, -315°, -675°, -1035°, -1395° 57.3° Example 2: 150° 30° a) i. One degree is defined as 1/360 th of a full rotation. ii. One radian is the angle formed when the terminal arm swipes out an arc that has the same length as the terminal arm. One radian is approximately 57.3°. iii. One revolution is defined as 360º, or 2pi. It is one complete rotation around a circle. b) i. 0.40 rad ii. 0.06 rev iii. 148.97° iv. 0.41 rev v. 270° vi. 4.71 rad c) L UDG LL UDG Conversion Multiplier Reference Chart degree radian revolution degree radian revolution 180° 180° 1 rev 1 rev 1 rev 360° 1 rev 360° Example 3: a) 3.05 rad b) UDG c) 1/3 rev d) 143.24° e) 270° f) 4.71 rad g) 1/4 rev h) 180° i) UDG 30° = 45° = 60° = 0° = 330° = 315° = 300° = 90° = = 270° = 120° = 135° = 150° = 180° = 210° = 225° = 240° 360° = Example 4: 210° 30° ť -260° 80° ť 304° ť 56° 135° ť 45° 309° 51° ť Example 5: a) ť r = 30° b) ť r = 80° c) ť r = 56° (or 0.98 rad) d) ť r = 45° RU UDG e) ť r = 51° RU UDG 120° ť 60° 156° ť 24° 225° ť 45° 210° ť 30° Example 6: a) ť p = 210°, ť r = 30° b) ť p = 225°, ť r = 45° c) ť p = 156°, ť r = 24° d) ť p = 120°, ť r = 60° RU Ŧ p = , Ŧ r = RU Ŧ p = , Ŧ r = Example 7: a) ť = 60°, ť p = 60° b) ť = -495°, ť p = 225° c) ť = 675°, ť p = 315° d) ť = 480°, ť p = 120° ť c = -855°, -135°, 225°, 585° ť c = -300°, 420°, 780° Example 8: a) ť p = 93° b) ť p = 148° c) ť p = 144° d) ť p = 330° (or 2.58 rad) (or rad) (or rad) Example 9: a) ť c = 1380° b) ť c c) ť c = 20° d) ť c 60° 225° ť c = -45°, 315° ť c = -960°, -600°, -240°, 120°, 840°, 1200° RU Ŧ c = , 5.50) (RU Ŧ c = , -10 , 2 20 120° 315° 93° 148° 144° 330°

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Answer KeyTrigonometry Lesson One: Degrees and Radians Note: For illustrative purposes,

all diagram angles will be in degrees.

120°

-120°

ť

a) The rotation angle between the initial arm and the terminal arm is called the standard position angle.

Example 1:b) An angle is positive if we rotate the terminal arm counter-clockwise, and negative if rotated clockwise.

c) The angle formed between the terminal arm and the x-axis is called the reference angle.

d) If the terminal arm is rotated by a multiple of 360° in either direction, it will return to its original position. These angles are called co-terminal angles.

e) A principal angle is an angle that exists between 0°and 360°.

f) The general form of co-terminal angles is ťc� �ťp + n(360°) using GHJUHHV��RU�ťc� �ťp���Q�����using radians.

420° 60°

45°,405°, 765°, 1125°, 1485°

45°, -315°, -675°, -1035°, -1395°

57.3°

Example 2:

150°

30°

a) i. One degree is defined as 1/360th of a full rotation.

ii. One radian is the angle formed when the terminal arm swipes out an arc that has the same length as the terminal arm. One radian is approximately 57.3°.

iii. One revolution is defined as 360º, or 2pi. It is one complete rotation around a circle.

b) i. 0.40 rad ii. 0.06 rev iii. 148.97° iv. 0.41 rev v. 270° vi. 4.71 rad c) L�������UDG����LL������UDG

Conversion Multiplier Reference Chart

degree radian revolution

degree

radian

revolution

�180°

�180°

��1 rev

��1 rev

1 rev

360°

1 rev

360°

Example 3: a) 3.05 rad b)������UDG���c) 1/3 rev d) 143.24° e) 270° f) 4.71 rad g) 1/4 rev h) 180° i)����UDG

30° =

45° =

60° =

0° =

330° =

315° =

300° =

90° =

= 270°

= 120°

= 135°

= 150°

= 180°

= 210°

= 225°

= 240°

360° =

Example 4:

210°

30°ť

-260°

80°

ť

304°

ť

56°

135°

ť45°

309°

51°

ť

Example 5:a) ťr = 30° b) ťr = 80° c) ťr = 56°

(or 0.98 rad)d) ťr = 45°�RU�����UDG�

e) ťr = 51°�RU������UDG�

120°

ť

60°156°

ť24°

225°

ť45°

210°

ť30°

Example 6:a) ťp = 210°, ťr = 30° b) ťp = 225°, ťr = 45° c) ťp = 156°, ťr = 24° d) ťp = 120°, ťr = 60°

�RU�Ŧp = ����, Ŧr = ����� �RU�Ŧp = ����, Ŧr = ����

Example 7:

a) ť = 60°, ťp = 60° b) ť = -495°, ťp = 225°

c) ť = 675°, ťp = 315° d) ť = 480°, ťp = 120°

ťc = -855°, -135°, 225°, 585°ťc = -300°, 420°, 780° Example 8:a) ťp = 93° b) ťp = 148° c) ťp = 144° d) ťp = 330°

(or 2.58 rad) (or ���� rad) (or ����� rad)

Example 9:a) ťc = 1380° b) ťc� �������� c) ťc = 20° d) ťc� �����

60°

225°

ťc = -45°, 315° ťc = -960°, -600°, -240°, 120°,840°, 1200°�RU�Ŧc = ������, 5.50)

(RU�Ŧc = ������,-10����������, 2������������20����

120°

315°

93°

148° 144°

330°

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Example 10:a) ťp = 112.62°, ťr = 67.38° b) ťp = 303.69°, ťr = 56.31°

Example 11:a) sinť��4,��+��4,,��+��4,,,��-��4,9��-b) cosť� QI +��4,,��-��4,,,��-��4,9��+c) tanť� QI +��4,,��-��4,,,��+��4,9��-d) cscť��4,��+��4,,��+��4,,,��-��4,9��-e) secť� QI +��4,,��-��4,,,��-��4,9��+f) cotť� QI +��4,,��-��4,,,��+��4,9��-g) VLQť��FVFť�VKDUH�WKH�VDPH�TXDGUDQW�VLJQV�FRVť��VHFť�VKDUH�WKH�VDPH�TXDGUDQW�VLJQV�WDQť��FRWť�VKDUH�WKH�VDPH�TXDGUDQW�VLJQV

Example 12: a) i. QIII or QIV ii. QI or QIV iii. QI or QIII b) i. QI ii. QIV iii. QIII c) i. none ii. QIII iii. QI

-12

-513

22.62°202.62°

ť

73

154.62°

25.38°

ť

Example 13:a) ťp = 202.62°, ťr = 22.62° b) ťp = 154.62°, ťr = 25.38°

�RU�Ŧp� ������UDG��Ŧr� ������UDG� �RU�Ŧp� ������UDG��Ŧr� ������UDG�

Example 14:

-2

3

33.69°

326.31°

ť

4

-35

323.13°

36.87°

ť

a) ťp = 323.13°, ťr = 36.87° b) ťp = 326.31°, ťr = 33.69°

Example 15:

If the angle ť�could exist in either quadrant ___ or ___ ...

The calculator alwayspicks quadrant

I or III or IIII or IVII or IIIII or IVIII or IV

IIIIIIVIV

a)

b) Each answer is different because the calculator is unawareRI�ZKLFK�TXDGUDQW�WKH�WULDQJOH�LV�LQ��7KH�FDOFXODWRU�DVVXPHVMark’s triangle is in QI, Jordan’s triangle is in QII, and Dylan’striangle is in QIV.

Example 16:a) The arc length can be found by multiplying the circumference by the sector percentage. 7KLV�JLYHV�XV�D� ���U�ù�ť���� �Uť�b) 13.35 cmc) 114.59°d) 2.46 cme) Q� �����

Example 17:a) The area of a sector can be found by multiplying the area of the full circle by the sector percentage to get the area of the sector.7KLV�JLYHV�XV�D� ��U2�ù�ť���� �U2ť���b) ������FP2

c) ���FP2

d) ������FP2

e) ����FP2

Example 18:a) 600°/sb) 0.07 rad/sc) 1.04 kmd) 70 rev/se) 2.60 rev/s

Example 19:a) �������UDG�Vb) 468.45 km

ť-5

1312

112.62°

67.38°2

-356.31°

ť

303.69°

Answer Key

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Answer KeyTrigonometry Lesson Two: The Unit Circle

Example 1:

-10

10

10 -10

10

10

(0.6, 0.8)

(0.5, 0.5)

a) i. ii. b) i. Yes ii. No

Example 2: See Video.

a)Example 3: b) -1 c)

d) e) 0 f) 0 g) h)

a)Example 4: b) 1 c)

d) e) -1 f) g) h)

, , , ,

, , , ,

Example 5:

a)

b)

, , , ,

, , , ,

Example 6:

a)

b)

Example 7: See Video.

Example 8:

a) -2 b) undefined c) d) e) f) -1 g) 0 h)

Example 9:

a) b) 1 c) d)

Example 10:

a) 1 b) c) d)

Example 11:

a) -1 b) c) undefined d) undefined

Example 12: See Video.

Example 13:a)�3������PHDQV��SRLQW�FRRUGLQDWHV�DW�����.

b) c)

d) e) 3���� ������������������

Example 14:a)�&� ��� b)�7KH�FHQWUDO�DQJOH�DQG�DUF�OHQJWK�RI�

WKH�XQLW�FLUFOH�DUH�HTXDO�WR�HDFK�RWKHU�c)�D� ����� d)�D� �����

Example 15:a)�7KH�XQLW�FLUFOH�DQG�WKH�OLQH�\� ��GR�QRW�LQWHUVHFW��VR�LWV�LPSRVVLEOH�IRU�VLQť�WR�HTXDO���

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b)

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Example 17:a) �������������b)������������

Example 18:a) See Video b)�����P

c) i. ii.

LLL���\� �� LY�

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Answer KeyTrigonometry Lesson Three: Trigonometric Functions I

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Example 2: a) y = sinť��b) D� �����c)�3� ���d)�F� �����e)�G� �����f)�ť� �Q���QŢ,���g)�������h)�'RPDLQ��ť�Ţ�5��5DQJH�������\������

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Answer KeyExample 10:

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Example 16:

a) b)

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Answer Key

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Example 18: a) y = cscť��b) 3� ��� c) 'RPDLQ��ť�Ţ�5��ť���Q���QŢ,��5DQJH��\�������\����d) ť� �Q���QŢ,

Example 19: a) y = cotť b) 3� �� c) 'RPDLQ��ť�Ţ�5��ť���Q���QŢ,��5DQJH��\Ţ5d) ť� �Q���QŢ,

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Answer KeyTrigonometry Lesson Four: Trigonometric Functions II

Example 1:

-2

2

� �� �� Ŧ

y

1

-1

-2

2

180º Ŧ

y

1

-1

360º 540º

a) b)

-2

2

t

h

1

-1

15 30 45 60

-2

2

x

y

1

-1

88 16 24

Example 2:

a) b)

-26

4 8

y

x

-24-22-20-18-16-14-12-10-8-6-4-2

1 2 3 5 6 7

x-225-200 -150

-175 -125 -75 -25-100 -50

025

50 10075 125 175

150 200 250225 275

25

50

75y

Example 3:

a) b)

-8

y

x

-6-4-202468

10121416

120110100908070605040302010

1820

y

x

2

54321-1-2-3

468

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Example 4:

a) b)

Example 5:

a)

b)

c)

d)

Example 6:

a)

b)

c)

d)

5e)

Example 7:

a)

b)

c)

d)

Example 8:

a) b) The b-parameter is doubled when the period is halved. The a, c, and d parameters remain the same.

c) The d-parameter decreases by 2 units, giving us d = 4. All other parameters remain unchanged.

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Answer Key

t

h(t)

5 10

45

75

105

Example 9:

a)

b) c) If the wind turbine rotates counterclockwise, we still get the same graph.

Example 10:

a)

Ŧ

h(Ŧ)

90°

c) The angle of elevation increases quickly at first, but slows down as the helicopter reaches greater heights. The angle never actually reaches 90°.

b) ,

Example 11:

a)

b) c) 2.86 m

Example 12:

a)

c) 28.14 mb)

t

h(t)

5.2

4.0

2.8

1 2

t

h(t)

0 25 50 75 100 125 150 175 200

1

16

31

d) 0.26 s

Example 13:

b)

d) 15.86 h

n

d(n)

-50 0 50 100 150 200 250 300 350 400

4

8

12

16

20

24

� � � �2

5.525cos 11 12.295365

d n nSª º � � �« »¬ ¼

c)

a) Decimal daylight hours: 6.77 h, 12.28 h, 17.82 h, 12.28 h, 6.77 h

e) 64 days

Example 14:

b)

d) 10.75 m

a) Decimal hours past midnight: 2.20 h, 8.20 h, 14.20 h, 20.20 h

e) 32.3%

t

h(t)

4

8

12

16

0 4 8 12 16 20 24

c)

Population

Time(years)

200

250

300

8000

12000

16000

1 2 3 4 5 60

Mic

eO

wls

7 8

M(t)

O(t)

Example 15:

a) b) See Video.

Example 16: 2.5 m

Example 17: 15.6 s and 18.3 s

t

h(t)

1.5

15

3.0

30 45 60

(8, 2.5)

t

h(t)

10

15

19

30 45 60

(18.3, 13.1)

1

(15.6, 10.5)

d) 26.78 s

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Answer KeyTrigonometry Lesson Five: Trigonometric Equations

Example 1:

b) , c) d)

Note:�Q��,�IRU�DOO�JHQHUDO�VROXWLRQV�

Example 2:

-1

1

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Example 3:

360°

90°

180°

270°

30°30°

30°150°

360°

90°

180°

270°

30°

150°

30°

210°

360°

90°

180°

270°

45°

45°

45°

225°

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Answer KeyExample 5:

60°

60°

120°

240°

-3

3

-2

-1

2

1

���

-3

3

-2

-1

2

1

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Example 6:

a)���������DQG�������� b) ��������DQG�������� c) ��������DQG�������� d) ��������DQG��������

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Answer Key

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Answer KeyExample 14:

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Answer KeyTrigonometry Lesson Six: Trigonometric Identities I

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Answer KeyExample 6:

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Answer KeyExample 9:�6HH�9LGHR Example 10:�6HH�9LGHR Example 11:�6HH�9LGHR

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Answer KeyTrigonometry Lesson Seven: Trigonometric Identities II

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