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The Values of sin  , cos  , tan . Quadrants and angles in the unit circle. y. 90 °. Quadrant II. Quadrant I. 0 °. 180 °. x. 360 °. Quadrant IV. Quadrant III. 270 °. The Values of sin  , cos  , tan . Cartesian plane can be divided into four parts called quadrants .

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The Values of sin , cos, tan 

y

90°

180°

x

360°

270°

The Values of sin , cos, tan 

• Cartesian plane can be divided into four parts called quadrants.

• Quadrants are named in the anticlockwise direction.

The Values of sin , cos, tan 

Quadrants and angles in the unit circle

y

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

x

O

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

P

(x, y)

y

x

The Values of sin , cos, tan 

y

sin  =

=

= y

1

x

Q

O

 sin  = y-coordinate

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

The Values of sin , cos, tan 

y

cos  =

=

= x

P

(x, y)

1

y

x

x

Q

O

 cos = x-coordinate

 tan =

The Values of sin , cos, tan 

Verify tan  = in quadrant I of the unit circle

y

tan  =

=

P

(x, y)

1

y

x

x

Q

O

Determine whether the value is positive or negative

y

x

The Values of sin , cos, tan 

• Quadrant I = All positive

sin

All

• Quadrant II = sin  positive

II

I

IV

III

• Quadrant III = tan  positive

tan

cos

• Quadrant IV = cos  positive

• Sin  is positive in quadrant II.

Not

The Values of sin , cos, tan 

Determine whether the value is positive or negative

y

• The angle 213° lies in quadrant III.

213°

x

O

• Therefore, the value of sin 213° is negative.

Example 2: cos 321°

• Cos  is positive in quadrant IV.

It is

The Values of sin , cos, tan 

Determine whether the value is positive or negative

y

• The angle 321° lies in quadrant IV.

321°

x

O

• Therefore, the value of cos 321° is positive.

• Tan  is positive in quadrant III.

Not

The Values of sin , cos, tan 

Determine whether the value is positive or negative

y

• The angle 123° lies in quadrant II.

123°

x

O

• Therefore, the value of tan 123° is negative.

• All positive in quadrant I.

It is

The Values of sin , cos, tan 

Determine whether the value is positive or negative

y

• The angle 32° lies in quadrant I.

32°

x

O

• Therefore, the value of sin 32° is positive.

• sin 45° =

45°

• cos 45° =

1

45°

1

The Values of sin , cos, tan 

• tan 45° = 1

• sin 30° =

30°

• cos 30° =

2

• tan 30° =

60°

1

The Values of sin , cos, tan 

• sin 60° =

30°

• cos 60° =

2

• tan 60° =

60°

1

The Values of sin , cos, tan 

y

(0, 1)

x

O

(–1, 0)

(1, 0)

(0, –1)

The Values of sin , cos, tan 

Summary: angles

The Values of sin , cos, tan 

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

7 sin 90° + 4 cos 180 °

Solution:

The Values of sin , cos, tan 

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

7 sin 90° + 4 cos 180 ° =

7 × (1) + 4 × (–1)

= 7 – 4

= 3

y

P

x

O

The Values of sin , cos, tan 

 = corresponding angle in quadrant I

between x-axis and line OP

angles   where  = 180° – 

sin  = + sin 

cos  = – cos 

tan  = – tan 

P

 = – 90°

The Values of sin , cos, tan 

Values of angles in quadrant II

y

x

O

X

   where  =  – 180°

sin  = – sin 

cos  = – cos 

tan  = + tan 

P

 = 270°–

The Values of sin , cos, tan 

y

x

O

X

   where  = 360° – 

P

sin  = – sin 

cos  = + cos 

tan  = – tan 

 = – 270°

The Values of sin , cos, tan 

y

x

O

X

Solution:

y

x

O

P

The Values of sin , cos, tan 

Question 1: Find the value of sin 231°.

• sin 231°  negative

• sin 231° = – sin (231° – 180°)

231°

= – sin 51°

= – 0.7771

Solution: angles

y

x

O

P

The Values of sin , cos, tan 

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')

303° 17'

= cos 56° 43'

= 0.5488

Solution: angles

y

P

x

O

The Values of sin , cos, tan 

Finding the value of an angle

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

• 117° 13'  quadrant II

• tan 117° 13'  negative

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

117° 13'

= – tan 62° 47'

= – 1.945

Finding angles between angles0° and 360°

Solution:

P

y

y

P

x

x

72°

72°

x

x

O

The Values of sin , cos, tan 

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

• Corresponding acute angle, x = 72°

• 0.9511  positive

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

and

72°

180° – 72° = 108°

Finding angles between angles0° and 360°

Solution:

y

y

P

x

60° 12'

x

x

x

O

60° 12'

P

The Values of sin , cos, tan 

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

• Corresponding acute angle, x = 60° 12'

• – 1.746  negative

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

and

180° – 60° 12' = 119° 48'

360° – 60° 12' = 299° 48'

Solution: angles

y

y

P

x

60°

x

x

x

O

60°

P

The Values of sin , cos, tan 

Finding angles between 0° and 360°

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

• Corresponding acute angle, x = 60°

• 0.5  positive

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

and

60°

360° – 60° = 300°

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:

N

• the length of HN,

• the value of cos x°,

• the value of tan y°.

S

H

M

F

J

The Values of sin , cos, tan 

Solution: angles

Pythagoras’ theorem

The Values of sin , cos, tan 

Solve problems involving sine, cosine and tangent

(a)

N

132 – 122

13 cm

HN2 =

= 169 – 144

= 25

S

H

12 cm

M

HN = 5 cm

F

J

4 cm

HMS angles is a straight line

The Values of sin , cos, tan 

Solve problems involving sine, cosine and tangent

Solution:

(b)

N

x° = 180° – HMN

cos x° =

– cos HMN

S

H

M

=

F

J

JHN angles is a straight line

The Values of sin , cos, tan 

Solve problems involving sine, cosine and tangent

Solution:

(c)

N

y° = 180° – FHJ

tan y° =

– tan FHJ

S

H

M

=

F

J

The End angles