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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