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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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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. • Quadrants are named in the anticlockwise direction.
P 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
Example 1: sin 213° • Sin is positive in quadrant II. Not quadrant II 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 quadrant IV. 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.
Example 3: tan 123° • Tan is positive in quadrant III. Not quadrant III 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.
Example 4: sin 32° • All positive in quadrant I. It is quadrant I. 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.
Determine the values of sine, cosine and tangent for special angles • sin 45° = 45° • cos 45° = 1 45° 1 The Values of sin , cos, tan • tan 45° = 1
Determine the values of sine, cosine and tangent for special angles • sin 30° = 30° • cos 30° = 2 • tan 30° = 60° 1 The Values of sin , cos, tan
Determine the values of sine, cosine and tangent for special angles • sin 60° = 30° • cos 60° = 2 • tan 60° = 60° 1 The Values of sin , cos, tan
Determine the values of sine, cosine and tangent for special angles y (0, 1) x O (–1, 0) (1, 0) (0, –1) The Values of sin , cos, tan
Summary: The Values of sin , cos, tan 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: 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
Values of angles in quadrant II y P x O The Values of sin , cos, tan = corresponding angle in quadrant I between x-axis and line OP
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
Values of angles in quadrant III where = – 180° sin = – sin cos = – cos tan = + tan P = 270°– The Values of sin , cos, tan y x O X
Values of angles in quadrant IV where = 360° – P sin = – sin cos = + cos tan = – tan = – 270° The Values of sin , cos, tan y x O X
Finding the value of an angle Solution: y x O P The Values of sin , cos, tan Question 1: Find the value of sin 231°. • 231° quadrant III • sin 231° negative • sin 231° = – sin (231° – 180°) 231° = – sin 51° = – 0.7771
Solution: 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: 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 0° 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 • Quadrant I: x = 72° 180° – 72° = 108° • Quadrant II: x =
Finding angles between 0° 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 • Quadrant II: x = 180° – 60° 12' = 119° 48' • Quadrant IV: x = 360° – 60° 12' = 299° 48'
Solution: 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 • Quadrant I: x = 60° 360° – 60° = 300° • Quadrant IV: x =
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: N • the length of HN, • the value of cos x°, • the value of tan y°. x° y° S H M F J The Values of sin , cos, tan
Solution: Pythagoras’ theorem The Values of sin , cos, tan Solve problems involving sine, cosine and tangent (a) N 132 – 122 13 cm HN2 = x° = 169 – 144 = 25 y° S H 12 cm M HN = 5 cm F J 4 cm
HMS is a straight line The Values of sin , cos, tan Solve problems involving sine, cosine and tangent Solution: (b) N x° = 180° – HMN x° cos x° = – cos HMN y° S H M = F J
JHN is a straight line The Values of sin , cos, tan Solve problems involving sine, cosine and tangent Solution: (c) N y° = 180° – FHJ x° tan y° = – tan FHJ y° S H M = F J