the values of sin , cos , tan
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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 . - PowerPoint PPT PresentationTRANSCRIPT
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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
OPPQ
1y
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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
OPOQ
1x
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Verify tan = in quadrant I of the unit circle
O
tan =
=P
y
x
(x, y)1
y
x Q
OQPQ
xy
coordinate-coordinate-
xy
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 0cos 1 0 –1 0 1tan 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 IIIsin 231° negativesin 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 IVcos 303° 17' positivecos 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 IItan 117° 13' negativetan 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 positiveTherefore, 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 negativeTherefore, 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
Px
x
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 positiveTherefore, 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
Px
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 cm
13 cm
4 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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