# The Values of sin , cos , tan 

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The Values of sin , cos , tan 

The Values of sin , cos , tan 

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. P x O

The Values of sin , cos , tan 
Verify sin  = y-coordinate in quadrant I of the unit circle y sin  = = = y P (x, y) 1 y x x O Q  sin  = y-coordinate

The Values of sin , cos , tan 
Verify cos  = x-coordinate in quadrant I of the unit circle y cos  = = = x P (x, y) 1 y x O x Q  cos  = x-coordinate

The Values of sin , cos , tan 
Verify tan  = in quadrant I of the unit circle y tan  = = P (x, y) 1 y x O x Q  tan  =

The Values of sin , cos , tan 
Determine whether the value is positive or negative y x Quadrant I = All positive sin All Quadrant II = sin  positive II I III IV Quadrant III = tan  positive tan cos Quadrant IV = cos  positive

The Values of sin , cos , tan 
Determine whether the value is positive or negative Example 1: sin 213° Sin  is positive in quadrant II. y The angle 213° lies in quadrant III. 213° x O Therefore, the value of sin 213° is negative. Not quadrant II

The Values of sin , cos , tan 
Determine whether the value is positive or negative Example 2: cos 321° Cos  is positive in quadrant IV. y The angle 321° lies in quadrant IV. 321° x O Therefore, the value of cos 321° is positive. It is quadrant IV.

The Values of sin , cos , tan 
Determine whether the value is positive or negative Example 3: tan 123° Tan  is positive in quadrant III. y The angle 123° lies in quadrant II. 123° x O Therefore, the value of tan 123° is negative. Not quadrant III

The Values of sin , cos , tan 
Determine whether the value is positive or negative Example 4: sin 32° All positive in quadrant I. y The angle 32° lies in quadrant I. 32° x O Therefore, the value of sin 32° is positive. It is quadrant I.

The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles sin 45° = 45° 1 cos 45° = tan 45° = 1

The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles sin 30° = 30° 60° 1 2 cos 30° = tan 30° =

The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles sin 60° = 30° 60° 1 2 cos 60° = tan 60° =

The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles 90° 180° 270° 360° sin  1 –1 cos  tan  y x O (0, 1) (–1, 0) (1, 0) (0, –1)

The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles Summary: 30° 45° 60° 90° 180° 270° 360° sin  1 –1 cos  tan 

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: 7 sin 90° + 4 cos 180 ° = 7 × (1) + 4 × (–1) = 7 – 4 = 3

The Values of sin , cos , tan 
Values of angles in quadrant II y x O P  = corresponding angle in quadrant I between x-axis and line OP

The Values of sin , cos , tan 
Values of angles in quadrant II    where  = 180° –  sin  = + sin  cos  = – cos  tan  = – tan  y P x O X  =  – 90°

The Values of sin , cos , tan 
Values of angles in quadrant III    where  =  – 180° sin  = – sin  cos  = – cos  tan  = + tan  y x P O X  = 270° – 

The Values of sin , cos , tan 
Values of angles in quadrant IV    where  = 360° –  sin  = – sin  cos  = + cos  tan  = – tan  y x O X P  =  – 270°

The Values of sin , cos , tan 
Finding the value of an angle Question 1: Find the value of sin 231°. Solution: y x O 231°  quadrant III sin 231°  negative sin 231° = – sin (231° – 180°) 231° = – sin 51° = – P

The Values of sin , cos , tan 
Finding the value of an angle Question 2: Find the value of cos 303° 17‘. Solution: x y O 303° 17'  quadrant IV cos 303° 17'  positive cos 303° 17' = cos (360° – 303° 17') 303° 17' = cos 56° 43' = P

The Values of sin , cos , tan 
Finding the value of an angle Question 3: Find the value of tan 117° 13'. Solution: x y O 117° 13'  quadrant II P tan 117° 13'  negative tan 117° 13' = – tan (180° – 117° 13') 117° 13' = – tan 62° 47' = – 1.945

The Values of sin , cos , tan 
Finding angles between 0° and 360° Question 1: For sin x = where 0° ≤ x ≤ 360°, find the value of x. Solution: 72° x P y x y Corresponding acute angle, x = 72° x O 72° P  positive Therefore, the acute angle is in quadrant I or II. and Quadrant I: x = 72° Quadrant II: x = 180° – 72° = 108°

The Values of sin , cos , tan 
Finding angles between 0° and 360° Question 2: For tan x = – where 0° ≤ x ≤ 360°, find the value of x. Solution: Corresponding acute angle, x = 60° 12' x y x y –  negative 60° 12' O P x 60° 12' P x 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'

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. Solution: x y x y Corresponding acute angle, x = 60° 60° P x 60° O P x 0.5  positive Therefore, the acute angle is in quadrant I or IV. and Quadrant I: x = 60° Quadrant IV: x = 360° – 60° = 300°

The Values of sin , cos , tan 
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: the length of HN, the value of cos x°, the value of tan y°. N H M S J F

The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent Solution: (a) N Pythagoras’ theorem HN2 = 132 – 122 13 cm = 169 – 144 = 25 S H 12 cm M  HN = 5 cm F J 4 cm

The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent Solution: HMS is a straight line (b) N x° = 180° – HMN cos x° = – cos HMN S H M = F J

The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent Solution: JHN is a straight line (c) N y° = 180° – FHJ tan y° = – tan FHJ S H M = F J

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