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Form 4 Chapter 9: Trigonometry II

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Quadrants and angles in the unit circle

0°

90°

180°

270°

360°

Quadrant I

Quadrant II

Quadrant III

Quadrant IV

Cartesian plane can be divided into four parts called quadrants.

Quadrants are named in the anticlockwise direction.

y

x

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Quadrants and angles in the unit circle

O

Angle is measured by rotating the line OP in the anticlockwise direction from the positive x-axis at the origin, O.

y

x

P

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Verify sin = y-coordinate in quadrant I of the unit circle

O

sin =

=

= y

y

x

P (x, y)

1

y

x Q

sin = y-coordinate

OP

PQ

1

y

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Verify cos = x-coordinate in quadrant I of the unit circle

O

cos =

=

= x

P

y

x

(x, y)

1

y

x Q

cos = x-coordinate

OP

OQ

1

x

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Verify tan = in quadrant I of the unit circle

O

tan =

=

P

y

x

(x, y)

1

y

x Q

OQ

PQ

x

y

coordinate-

coordinate-

x

y

tan =

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y

xIII

III IV

Allsin

tan cos

Determine whether the value is positive or negative

Quadrant I = All positive

Quadrant II = sin positive

Quadrant III = tan positive

Quadrant IV = cos positive

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Example 1: sin 213°

y

x

Determine whether the value is positive or negative

The angle 213° lies in quadrant III.

Therefore, the value of sin 213° is negative.

213°

O

Sin is positive in quadrant II.

Notquadrant II

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y

x

Determine whether the value is positive or negative

The angle 321° lies in quadrant IV.

Therefore, the value of cos 321° is positive.

321°

O

Cos is positive in quadrant IV.

It isquadrant IV.

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y

x

Determine whether the value is positive or negative

The angle 123° lies in quadrant II.

Therefore, the value of tan 123° is negative.

123°

O

Tan is positive in quadrant III.

Notquadrant III

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y

x

Determine whether the value is positive or negative

The angle 32° lies in quadrant I.

Therefore, the value of sin 32° is positive.

32°O

It isquadrant I.

All positive in quadrant I.

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Determine the values of sine, cosine and tangent for special angles

45°

45°

1

1

sin 45° =

cos 45° =

tan 45° = 1

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Determine the values of sine, cosine and tangent for special angles

30°

60°

1

2

sin 30° =

cos 30° =

tan 30° =

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Determine the values of sine, cosine and tangent for special angles

30°

60°

1

2cos 60° =

sin 60° =

tan 60° =

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Determine the values of sine, cosine and tangent for special angles

y

xO (1, 0)

(0, 1)

(–1, 0)

(0, –1)

0° 90° 180° 270° 360°

sin 0 1 0 –1 0

cos 1 0 –1 0 1

tan 0 0 0

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Summary:

Determine the values of sine, cosine and tangent for special angles

0° 30° 45° 60° 90° 180° 270° 360°

sin 0 1 0 –1 0

cos 1 0 –1 0 1

tan 0 1 0 0

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Determine the values of sine, cosine and tangent for special angles

Question 1: Calculate the values of the following:

7 sin 90° + 4 cos 180 °

Solution:

7 sin 90° + 4 cos 180 ° = 7 × (1) + 4 × (–1)

= 7 – 4

= 3

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Values of angles in quadrant II

y

xO

between x-axis and line OP

= corresponding angle in quadrant I

P

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Values of angles in quadrant II

y

x

O

P

where = 180° – sin = + sin

cos = – cos

tan = – tan

= – 90°X

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Values of angles in quadrant III

y

x

O

P

where = – 180°

sin = – sin

cos = – cos

tan = + tan

= 270° – X

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Values of angles in quadrant IV

y

x

O

P

where = 360° – sin = – sin

cos = + cos

tan = – tan

= – 270°X

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Solution:

= – sin 51°

Finding the value of an angle

Question 1: Find the value of sin 231°.

231°

y

xO

P

231° quadrant III

sin 231° negative

sin 231° = – sin (231° – 180°)

= – 0.7771

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Solution:

= cos 56° 43'

Finding the value of an angle

Question 2: Find the value of cos 303° 17‘.

303° 17' quadrant IV

cos 303° 17' positive

cos 303° 17' = cos (360° – 303° 17')

= 0.5488

303° 17'

x

y

O

P

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Solution:

= – tan 62° 47'

Finding the value of an angle

117° 13' quadrant II

tan 117° 13' negative

tan 117° 13' = – tan (180° – 117° 13')

= – 1.945

Question 3: Find the value of tan 117° 13'.

117° 13'

x

y

O

P

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Solution:

Finding angles between 0° and 360°

0.9511 positive

Therefore, the acute angle is in quadrant I or II.

Question 1: For sin x = 0.9511 where 0° ≤ x ≤ 360°, find the value of x.

x

O72°

P

Quadrant I: x =

Quadrant II: x =

72°x

P

x

y

and x

y Corresponding acute angle, x = 72°

72°180° – 72° = 108°

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Solution:

Finding angles between 0° and 360°

– 1.746 negative

Therefore, the acute angle is in quadrant II or IV.

Question 2: For tan x = – 1.746 where 0° ≤ x ≤ 360°, find the value of x.

60° 12'

O

P

xx

y

Quadrant IV: x =

Quadrant II: x =60° 12'

P

xand x

y Corresponding acute angle, x = 60° 12'

180° – 60° 12' = 119° 48'

360° – 60° 12' = 299° 48'

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Solution:

Finding angles between 0° and 360°

0.5 positive

Therefore, the acute angle is in quadrant I or IV.

Question 3: For cos x = 0.5 where 0° ≤ x ≤ 360°, find the value of x.

60°

O

P

x

x

y

Quadrant IV: x =

Quadrant I: x =60°

P

xand x

y Corresponding acute angle, x = 60°

60°

360° – 60° = 300°

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Solve problems involving sine, cosine and tangent

Question: In the diagram below, HMS and JHN are straight lines. H is the midpoint of JN. Given that HM = 12 cm, MN = 13 cm and FJ = 4 cm, calculate:(a) the length of HN,

(b) the value of cos x°,

(c) the value of tan y°.

N

H MS

JF

y°x°

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Solution:

Solve problems involving sine, cosine and tangent

N

H MS

JF

y°x°

HN2 =

(a)

= 169 – 144

= 25

HN = 5 cm

Pythagoras’ theorem

132 – 122

12 cm12 cm

13 cm13 cm

4 cm4 cm

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Solve problems involving sine, cosine and tangent

N

H MS

JF

y°x°

Solution:

x° = 180° – HMN

(b)

cos x° =

HMS is a straight line

=

– cos HMN

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Solve problems involving sine, cosine and tangent

N

H MS

JF

y°x°

Solution:

y° = 180° – FHJ

(c)

tan y° =

JHN is a straight line

=

– tan FHJ

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