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The Values of sin  , cos  , tan 

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 

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  1. The Values of sin , cos, tan 

  2. 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.

  3. 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

  4. 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

  5. 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

  6.  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

  7. 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

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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

  13. 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 

  14. 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 

  15. 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 

  16. Summary: The Values of sin , cos, tan  Determine the values of sine, cosine and tangent for special angles

  17. 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

  18. 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

  19.    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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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 =

  26. 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'

  27. 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 =

  28. 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 

  29. 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

  30. 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

  31. 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

  32. The End

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