Trig identities

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Basic Trig Identities

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  • Trig identities

    1. 1. Trig Identities Tuesday 14th June 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 x(By Pythagoras):
    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 2 2 x + y =1(By Pythagoras):
    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 x 2 2 x + y =1(By Pythagoras): 2 2 ∴sin θ + cos θ = 1
    24. 24. VERY important Trig Identity
    25. 25. VERY important Trig Identity 2 2 sin θ + cos θ = 1
    26. 26. or rearranging
    27. 27. or rearranging 2 2sin θ + cos θ = 1
    28. 28. or rearranging 2 2sin θ + cos θ = 1 2 2sin θ = 1 − cos θ
    29. 29. or rearranging 2 2sin θ + cos θ = 1 2 2sin θ = 1 − cos θ 2 2cos θ = 1 − sin θ
    30. 30. or dividing both sides
    31. 31. or dividing both sides 2 2 sin θ + cos θ = 1
    32. 32. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2sin θ sin θ sin θ
    33. 33. or dividing both sides
    34. 34. or dividing both sides 2 2 sin θ + cos θ = 1
    35. 35. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2sin θ sin θ sin θ
    36. 36. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 sin θ sin θ sin θ 2 2∴1 + cot θ = cosec θ
    37. 37. or dividing both sides
    38. 38. or dividing both sides 2 2 sin θ + cos θ = 1
    39. 39. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2sin θ sin θ sin θ
    40. 40. or dividing both sides
    41. 41. or dividing both sides 2 2 sin θ + cos θ = 1
    42. 42. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
    43. 43. or dividing both sides
    44. 44. or dividing both sides 2 2 sin θ + cos θ = 1
    45. 45. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
    46. 46. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
    47. 47. or dividing both sides
    48. 48. or dividing both sides 2 2 sin θ + cos θ = 1
    49. 49. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
    50. 50. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
    51. 51. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ 2 2 ∴ tan θ = sec θ − 1
    52. 52. or dividing both sides
    53. 53. or dividing both sides 2 2 sin θ + cos θ = 1
    54. 54. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
    55. 55. or dividing both sides
    56. 56. or dividing both sides 2 2 sin θ + cos θ = 1
    57. 57. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
    58. 58. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
    59. 59. or dividing both sides
    60. 60. or dividing both sides 2 2 sin θ + cos θ = 1
    61. 61. or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
    62. 62. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
    63. 63. or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ 2 2 ∴ tan θ = sec θ − 1
    64. 64. But easier to remember how to get them!
    65. 65. 2 2 sin θ + cos θ = 1But easier to remember how to get them!
    66. 66. 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θBut easier to remember how to get them!
    67. 67. 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1But easier to remember how to get them!
    68. 68. 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1But easier to remember how to get them!
    69. 69. 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1 2 2∴ tan θ = sec θ − 1But easier to remember how to get them!
    70. 70. 2 2 sin θ + cos θ = 1 2 2 sin θ = 1 − cos θ 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1 2 2∴ tan θ = sec θ − 1But easier to remember how to get them!
    71. 71. 2 2 sin θ + cos θ = 1 2 2 sin θ = 1 − cos θ 2 2 cos θ = 1 − sin θ 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1 2 2∴ tan θ = sec θ − 1But easier to remember how to get them!
    72. 72. ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ
    73. 73. ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ Only work on one side at a time!
    74. 74. ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ 2 1 − sin θ LHS = 2 2 sin θ + cos θ
    75. 75. ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ 2 1 − sin θ LHS = 2 2 sin θ + cos θ 2 cos θ = 1
    76. 76. ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ 2 1 − sin θ LHS = 2 2 sin θ + cos θ 2 cos θ = 1 2 = cos θ
    77. 77. Example cosθProve − tan θ = sec θ 1 − sin θ
    78. 78. Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ
    79. 79. Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) Common = − (1 − sin θ )cosθ cosθ (1 − sin θ ) Denominator
    80. 80. Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) = − (1 − sin θ )cosθ cosθ (1 − sin θ ) 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ
    81. 81. Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) = − (1 − sin θ )cosθ cosθ (1 − sin θ ) 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ 1 − sin θ = (1 − sin θ )cosθ
    82. 82. Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) = − (1 − sin θ )cosθ cosθ (1 − sin θ ) 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ 1 − sin θ 1 = = (1 − sin θ )cosθ cosθ
    83. 83. Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) Common = − (1 − sin θ )cosθ cosθ (1 − sin θ ) Denominator 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ 1 − sin θ 1 = = = sec θ (1 − sin θ )cosθ cosθ
    84. 84. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5
    85. 85. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant
    86. 86. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!
    87. 87. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved 3 O since sin θ = = 5 H θ
    88. 88. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved 3 O since sin θ = = 5 5 H 3 θ
    89. 89. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved 5 By Pythagoras! 3 θ 4
    90. 90. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved −3 tan θ = 5 4 3 θ 4
    91. 91. Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved −3 5 tan θ = 4 3 θ −4 cosθ = 4 5
    92. 92. Now do Ex 8.13

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