Unit Circle - Trigonometry

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Trigonometry on the Unit Circle

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  • Unit Circle - Trigonometry

    1. 1. Trig on the Unit Circle Monday 23rd May 2011
    2. 2. y P(x, y) 1 x
    3. 3. y P(x, y) 1 x
    4. 4. y P(x, y) 1 xAngles are always taken in an anticlockwise direction from the positive x axis
    5. 5. y P(x, y) 1 θ xAngles are always taken in an anticlockwise direction from the positive x axis
    6. 6. y P(x, y) 1 θ x
    7. 7. y P(x, y) 1 θ xConsider the coordinates of P
    8. 8. y P(x, y) 1 θ x xConsider the coordinates of P
    9. 9. y y P(x, y) 1 θ x xConsider the coordinates of P
    10. 10. y P(x, y) 1 θ x
    11. 11. y P(x, y) 1 θ xThis creates a right angled triangle with an angle θ
    12. 12. y P(x, y) 1 θ xThis creates a right angled triangle with an angle θ
    13. 13. y P(x, y) 1 θ x xThis creates a right angled triangle with an angle θ
    14. 14. y P(x, y) 1 y θ x xThis creates a right angled triangle with an angle θ
    15. 15. y P(x, y) 1 y θ x x
    16. 16. y P(x, y) 1 y θ x xUsing our standard trigonometric ratios
    17. 17. y y tan θ = x P(x, y) 1 y θ x xUsing our standard trigonometric ratios
    18. 18. y y tan θ = x P(x, y) y sin θ = = y 1 1 y θ x xUsing our standard trigonometric ratios
    19. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 29. Two important Trig Identities
    30. 30. Two important Trig Identities sin θ tan θ = cosθ
    31. 31. Two important Trig Identities sin θ tan θ = cosθ 2 2 sin θ + cos θ = 1
    32. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 55. Example 1Determine whether sin 243° is positive or negative
    56. 56. Example 1 Determine whether sin 243° is positive or negativeStep 1 : Determine which Quadrant the angle is in.
    57. 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. 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. 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. 60. Example 1Determine whether sin 243° is positive or negative y S A 243° x T C
    61. 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. 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. 63. Example 2For 0 ≤θ≤360° find all possible values of θ such that sin θ = 0.6
    64. 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. 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. 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. 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. 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. 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. 70. Example 2 y S A37° 37° x T C
    71. 71. Example 2Step 3: Find the angle in the other quadrant(s) y S A 37° 37° x T C
    72. 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. 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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