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Trigonometry on the Unit Circle

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- 1. Trig on the Unit Circle Monday 23rd May 2011
- 2. y P(x, y) 1 x
- 3. y P(x, y) 1 x
- 4. y P(x, y) 1 xAngles are always taken in an anticlockwise direction from the positive x axis
- 5. y P(x, y) 1 θ xAngles are always taken in an anticlockwise direction from the positive x axis
- 6. y P(x, y) 1 θ x
- 7. y P(x, y) 1 θ xConsider the coordinates of P
- 8. y P(x, y) 1 θ x xConsider the coordinates of P
- 9. y y P(x, y) 1 θ x xConsider the coordinates of P
- 10. y P(x, y) 1 θ x
- 11. y P(x, y) 1 θ xThis creates a right angled triangle with an angle θ
- 12. y P(x, y) 1 θ xThis creates a right angled triangle with an angle θ
- 13. y P(x, y) 1 θ x xThis creates a right angled triangle with an angle θ
- 14. y P(x, y) 1 y θ x xThis creates a right angled triangle with an angle θ
- 15. y P(x, y) 1 y θ x x
- 16. y P(x, y) 1 y θ x xUsing our standard trigonometric ratios
- 17. y y tan θ = x P(x, y) 1 y θ x xUsing our standard trigonometric ratios
- 18. y y tan θ = x P(x, y) y sin θ = = y 1 1 y θ x xUsing our standard trigonometric ratios
- 19. y y tan θ = x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x xUsing our standard trigonometric ratios
- 20. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x
- 21. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x xIn the ﬁrst quadrant all the trig ratios are positive
- 22. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x
- 23. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x1st Identity:
- 24. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x y sin θ1st Identity: tan θ = = x cosθ
- 25. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x
- 26. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x2nd Identity (By Pythagoras):
- 27. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x 2 2 x + y =12nd Identity (By Pythagoras):
- 28. y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x 2 2 x + y =12nd Identity (By Pythagoras): ∴sin 2 θ + cos 2 θ = 1
- 29. Two important Trig Identities
- 30. Two important Trig Identities sin θ tan θ = cosθ
- 31. Two important Trig Identities sin θ tan θ = cosθ 2 2 sin θ + cos θ = 1
- 32. y2nd Quadrant 1st Quadrant tan θ = y xP(−x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x
- 33. y 2nd Quadrant 1st Quadrant tan θ = y x P(−x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x2nd Quadrant (90 - 180 degrees) - θ is always from the x axis
- 34. y y 2nd Quadrant 1st Quadrant tan θ = ytan θ = x −x P(−x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x 2nd Quadrant (90 - 180 degrees) - θ is always from the x axis
- 35. y y 2nd Quadrant 1st Quadrant tan θ = ytan θ = x −x y P(−x, y) ysin θ = = y sin θ = = y 1 1 1 y x cosθ = = x θ 1 x x 2nd Quadrant (90 - 180 degrees) - θ is always from the x axis
- 36. y y 2nd Quadrant 1st Quadrant tan θ = ytan θ = x −x y P(−x, y) ysin θ = = y sin θ = = y 1 1 1 −x y xcosθ = = −x cosθ = = x 1 θ 1 x x 2nd Quadrant (90 - 180 degrees) - θ is always from the x axis
- 37. y y 2nd Quadrant 1st Quadrant tan θ = ytan θ = x −x y ysin θ = = y sin θ = = y 1 1 −x xcosθ = = −x cosθ = = x 1 x 1 θ x y 1 P(−x, −y) 3rd Quadrant
- 38. y y 2nd Quadrant 1st Quadrant tan θ = ytan θ = x −x y ysin θ = = y sin θ = = y 1 1 −x xcosθ = = −x cosθ = = x 1 x 1 θ x y 1 P(−x, −y) 3rd Quadrant 3rd Quadrant (180 - 270 degrees)
- 39. y y 2nd Quadrant 1st Quadrant tan θ = ytan θ = x −x y ysin θ = = y sin θ = = y 1 1 −x xcosθ = = −x cosθ = = x 1 x 1 θ x −y y ytan θ = = 1 −x x P(−x, −y) 3rd Quadrant 3rd Quadrant (180 - 270 degrees)
- 40. y y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x −x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y y ytan θ = = 1 −x x −ysin θ = = −y P(−x, −y) 1 3rd Quadrant 3rd Quadrant (180 - 270 degrees)
- 41. y y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x −x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y y y tan θ = = 1 −x x −ysin θ = = −y P(−x, −y) 1 −x 3rd Quadrantcosθ = = −x 1 3rd Quadrant (180 - 270 degrees)
- 42. y −y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y y y tan θ = = 1 −x x −ysin θ = = −y P(x, −y) 1 −x 3rd Quadrant 4th Quadrantcosθ = = −x 1
- 43. y −y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y y y tan θ = = 1 −x x −ysin θ = = −y P(x, −y) 1 −x 3rd Quadrant 4th Quadrantcosθ = = −x 1 4th Quadrant (270 - 360 degrees)
- 44. y −y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y −y y y tan θ = tan θ = = 1 x −x x −ysin θ = = −y P(x, −y) 1 −x 3rd Quadrant 4th Quadrantcosθ = = −x 1 4th Quadrant (270 - 360 degrees)
- 45. y −y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y −y y y tan θ = tan θ = = 1 x −x x −y −y sin θ = = −ysin θ = = −y P(x, −y) 1 1 −x 3rd Quadrant 4th Quadrantcosθ = = −x 1 4th Quadrant (270 - 360 degrees)
- 46. y −y 2nd Quadrant 1st Quadrant tan θ = y tan θ = x x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = −x cosθ = = x 1 x 1 θ x −y −y y y tan θ = tan θ = = 1 x −x x −y −y sin θ = = −ysin θ = = −y P(x, −y) 1 1 −x 3rd Quadrant 4th Quadrant cosθ = x = xcosθ = = −x 1 1 4th Quadrant (270 - 360 degrees)
- 47. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = x cosθ = = −x 1 1 x −y y −y tan θ = = tan θ = −x x x −y −ysin θ = = −y sin θ = = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360°
- 48. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = x cosθ = = −x 1 1 x −y y −y tan θ = = tan θ = −x x x −y −ysin θ = = −y sin θ = = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360° which ratio is positive in each of the quadrants?
- 49. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 cosθ = −x 1 = −x All x cosθ = = x 1 x −y y −y tan θ = = tan θ = −x x x −y −ysin θ = = −y sin θ = = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360° which ratio is positive in each of the quadrants?
- 50. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 cosθ = −x 1 = −x sin All x cosθ = = x 1 x −y y −y tan θ = = tan θ = −x x x −y −ysin θ = = −y sin θ = = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360° which ratio is positive in each of the quadrants?
- 51. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 cosθ = −x 1 = −x sin All x cosθ = = x 1 x −y y −y tan θ = = tan θ =sin θ = −x x −y = −y tan sin θ = −y x = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360° which ratio is positive in each of the quadrants?
- 52. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 cosθ = −x 1 = −x sin All x cosθ = = x 1 x −y y −y tan θ = = tan θ =sin θ = −x x −y = −y tan cos sin θ = −y x = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360° which ratio is positive in each of the quadrants?
- 53. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 −x x cosθ = = x cosθ = = −x 1 1 x −y y −y tan θ = = tan θ = −x x x −y −ysin θ = = −y sin θ = = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360°
- 54. 2nd Quadrant y 1st Quadrant 90 - 180° 0 - 90° y y tan θ = tan θ = x −x y y sin θ = = y sin θ = = y 1 1 cosθ = −x 1 = −x S A x cosθ = = x 1 x −y y −y tan θ = = tan θ =sin θ = −x x −y = −y T C sin θ = −y x = −y 1 1 −x xcosθ = = −x cosθ = = x 1 3rd Quadrant 4th Quadrant 1 180 - 270° 270 - 360° All Stations To Central
- 55. Example 1Determine whether sin 243° is positive or negative
- 56. Example 1 Determine whether sin 243° is positive or negativeStep 1 : Determine which Quadrant the angle is in.
- 57. Example 1 Determine whether sin 243° is positive or negativeStep 1 : Determine which Quadrant the angle is in. y S A x T C
- 58. Example 1 Determine whether sin 243° is positive or negativeStep 1 : Determine which Quadrant the angle is in. y S A x T C Remember the angle is always taken anticlockwise from the positive x - axis
- 59. Example 1 Determine whether sin 243° is positive or negativeStep 1 : Determine which Quadrant the angle is in. y S A 243° x T C Remember the angle is always taken anticlockwise from the positive x - axis
- 60. Example 1Determine whether sin 243° is positive or negative y S A 243° x T C
- 61. Example 1Determine whether sin 243° is positive or negative So the angle is in the 3rd Quadrant y S A 243° x T C
- 62. Example 1Determine whether sin 243° is positive or negative So the angle is in the 3rd Quadrant y S A 243° x T C Thus sin 243°is negative
- 63. Example 2For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6
- 64. Example 2 For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6Step 1: Find the corresponding acute angle (i.e in the 1st Q)
- 65. Example 2 For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6Step 1: Find the corresponding acute angle (i.e in the 1st Q) ∴θ = 37° (to nearest degree from Calculator)
- 66. Example 2 For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6Step 1: Find the corresponding acute angle (i.e in the 1st Q) ∴θ = 37° (to nearest degree from Calculator) Step 2: Find other quadrants where the ratio is positive
- 67. Example 2 For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6Step 1: Find the corresponding acute angle (i.e in the 1st Q) ∴θ = 37° (to nearest degree from Calculator) Step 2: Find other quadrants where the ratio is positive y S A x T C
- 68. Example 2 For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6Step 1: Find the corresponding acute angle (i.e in the 1st Q) ∴θ = 37° (to nearest degree from Calculator) Step 2: Find other quadrants where the ratio is positive y S A As sin is positive it must be an angle in the 1st or 2nd Quadrant x T C
- 69. Example 2 For 0 ≤θ≤360° ﬁnd all possible values of θ such that sin θ = 0.6Step 1: Find the corresponding acute angle (i.e in the 1st Q) ∴θ = 37° (to nearest degree from Calculator) Step 2: Find other quadrants where the ratio is positive y S A As sin is positive it must be an angle in the 1st or 2nd Quadrant x T C Step 3: Find the angle in the other quadrant(s)
- 70. Example 2 y S A37° 37° x T C
- 71. Example 2Step 3: Find the angle in the other quadrant(s) y S A 37° 37° x T C
- 72. Example 2Step 3: Find the angle in the other quadrant(s) y So the two angles are: 37° and S A 37° 37° x T C
- 73. Example 2Step 3: Find the angle in the other quadrant(s) y So the two angles are: 37° and 180°-37°=143° S A 37° 37° x T C

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