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Week 12 - Trigonometry

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Carlos Vázquez
Carlos Vázquez

Keynote de la semana 12 de mi curso de trignometria.

EducationTechnology
1 of 81
Day 56 1. Opener
2. Finding an Equation from a Graph
2. Finding an Equation from a Graph Amplitude: A = 3
2. Finding an Equation from a Graph Amplitude: A = 3 Period:
2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: ω
2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω
2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1
2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1 Cosine reflected
2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1 Cosine reflected y = A cos (ω x − φ )
2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1 Cosine reflected y = A cos (ω x − φ ) y = −3cos x
2. Finding an Equation from a Graph
2. Finding an Equation from a Graph Amplitude: A = 1
2. Finding an Equation from a Graph Amplitude: A = 1 Period:
2. Finding an Equation from a Graph Amplitude: A = 1 2π Period: ω
2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3
2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3
2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3 Sine
2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3 Sine y = Asin (ω x − φ )
2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3 Sine y = Asin (ω x − φ ) y = sin 3x
2. Finding an Equation from a Graph
2. Finding an Equation from a Graph Amplitude: A = 4
2. Finding an Equation from a Graph Amplitude: A = 4 Period:
2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: ω
2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω
2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π
2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π Sine
2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π Sine y = Asin (ω x − φ )
2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π Sine y = Asin (ω x − φ ) y = 4 sin π x
2. Finding an Equation from a Graph
2. Finding an Equation from a Graph Amplitude: A = 1/2
2. Finding an Equation from a Graph Amplitude: A = 1/2 Period:
2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: ω
2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω
2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2
2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2 Cosine reflected
2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2 Cosine reflected y = Asin (ω x − φ )
2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2 Cosine reflected y = Asin (ω x − φ ) 1 y = − cos 2x 2
Day 57 1. Opener
1. Opener
1. Opener
1. Opener
1. Opener
2. Example
2. Example Amplitude: A = 2
2. Example Amplitude: A = 2
2. Example Amplitude: A = 2
2. Example Amplitude: A = 2 Period:
2. Example Amplitude: A = 2 2π Period: ω
2. Example Amplitude: A = 2 2π Period: = 2π ω
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift φ = ω
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift φ = −π 4 ω
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift φ = −π 4 ω φ = −π 4
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift Cosine reflected φ = −π 4 ω φ = −π 4
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift Cosine reflected φ = −π 4 y = A cos (ω x − φ ) ω φ = −π 4
2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift Cosine reflected φ = −π 4 y = A cos (ω x − φ ) ω φ = −π 4 ( y = −2 cos x + π 4 )
3. Classwork Solve problems 32 - 40 in your copies. Even numbered problems
4. Homework Solve problems 31 - 39 in your copies. Odd numbered problems
Day 58 1. Opener La siguiente figura corresponde a un trozo de cartulina y en ella se realiza un doblez tomando como eje una recta que pase por los puntos D y B, de tal manera que el triángulo DBC quede sobre el triángulo ABD. ¿Qué figura se observará posteriormente? A A) B) D B C C) D)
Day 58 1. Opener La siguiente figura corresponde a un trozo de cartulina y en ella se realiza un doblez tomando como eje una recta que pase por los puntos D y B, de tal manera que el triángulo DBC quede sobre el triángulo ABD. ¿Qué figura se observará posteriormente? A A) B) D B C C) D)
1. Opener This is a graph of my elevation over sixteen seconds. What am I doing? Where am I? How do you know? 12 10 Elevation (feet) 8 6 4 2 2 4 6 8 10 12 14 16 18 Time (seconds)
1. Opener Solve the equations: 1. cos x = 1 2. cos x = −1 3. tan x = −1 4. sin x = − 3 2
1. Opener Solve the equations: 1. cos x = 1 1. x = 0, 2π 2. cos x = −1 2. x = π 3. tan x = −1 3. x = 3 4 π , 7 4 π 4. sin x = − 3 2 4. x = 4π 3 , 5π 3
1. Opener Find f-1(x): x 1. f (x) = e x 2. f (x) = 10
2. Inverse Trigonometric Functions Inverse Sine Function
2. Inverse Trigonometric Functions Inverse Sine Function
2. Inverse Trigonometric Functions Inverse Sine Function
2. Inverse Trigonometric Functions Inverse Sine Function
2. Inverse Trigonometric Functions Inverse Sine Function −π 2 ≤ x ≤ π 2
2. Inverse Sine Function −1 y = sin x means x = sin y where −π 2 ≤ y ≤ π 2 and −1 ≤ x ≤ 1
3. Examples Find the exact value of sin-1 1
3. Examples Find the exact value of sin-1 1
3. Examples Find the exact value of sin-1 1
3. Examples Find the exact value of sin-1 1 sin 1 = π 2 −1
3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠
3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠
3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠
3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠ −1 sin ⎛ 3 ⎞ = π ⎝ 2 ⎠ 3
4. Inverse Cosine Function −1 y = cos x means x = cos y where 0≤ y≤π and −1 ≤ x ≤ 1
5. Classwork Solve problems 1 - 11 in your copies. Odd numbered problems.

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Week 12 - Trigonometry

  • 2. 2. Finding an Equation from a Graph
  • 3. 2. Finding an Equation from a Graph Amplitude: A = 3
  • 4. 2. Finding an Equation from a Graph Amplitude: A = 3 Period:
  • 5. 2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: ω
  • 6. 2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω
  • 7. 2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1
  • 8. 2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1 Cosine reflected
  • 9. 2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1 Cosine reflected y = A cos (ω x − φ )
  • 10. 2. Finding an Equation from a Graph Amplitude: A = 3 2π Period: = 2π ω ω =1 Cosine reflected y = A cos (ω x − φ ) y = −3cos x
  • 11. 2. Finding an Equation from a Graph
  • 12. 2. Finding an Equation from a Graph Amplitude: A = 1
  • 13. 2. Finding an Equation from a Graph Amplitude: A = 1 Period:
  • 14. 2. Finding an Equation from a Graph Amplitude: A = 1 2π Period: ω
  • 15. 2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3
  • 16. 2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3
  • 17. 2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3 Sine
  • 18. 2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3 Sine y = Asin (ω x − φ )
  • 19. 2. Finding an Equation from a Graph Amplitude: A = 1 2π 2π Period: = ω 3 ω=3 Sine y = Asin (ω x − φ ) y = sin 3x
  • 20. 2. Finding an Equation from a Graph
  • 21. 2. Finding an Equation from a Graph Amplitude: A = 4
  • 22. 2. Finding an Equation from a Graph Amplitude: A = 4 Period:
  • 23. 2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: ω
  • 24. 2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω
  • 25. 2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π
  • 26. 2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π Sine
  • 27. 2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π Sine y = Asin (ω x − φ )
  • 28. 2. Finding an Equation from a Graph Amplitude: A = 4 2π Period: =2 ω ω =π Sine y = Asin (ω x − φ ) y = 4 sin π x
  • 29. 2. Finding an Equation from a Graph
  • 30. 2. Finding an Equation from a Graph Amplitude: A = 1/2
  • 31. 2. Finding an Equation from a Graph Amplitude: A = 1/2 Period:
  • 32. 2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: ω
  • 33. 2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω
  • 34. 2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2
  • 35. 2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2 Cosine reflected
  • 36. 2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2 Cosine reflected y = Asin (ω x − φ )
  • 37. 2. Finding an Equation from a Graph Amplitude: A = 1/2 2π Period: =π ω ω =2 Cosine reflected y = Asin (ω x − φ ) 1 y = − cos 2x 2
  • 44. 2. Example Amplitude: A = 2
  • 45. 2. Example Amplitude: A = 2
  • 46. 2. Example Amplitude: A = 2
  • 47. 2. Example Amplitude: A = 2 Period:
  • 48. 2. Example Amplitude: A = 2 2π Period: ω
  • 49. 2. Example Amplitude: A = 2 2π Period: = 2π ω
  • 50. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1
  • 51. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift
  • 52. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift φ = ω
  • 53. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift φ = −π 4 ω
  • 54. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift φ = −π 4 ω φ = −π 4
  • 55. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift Cosine reflected φ = −π 4 ω φ = −π 4
  • 56. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift Cosine reflected φ = −π 4 y = A cos (ω x − φ ) ω φ = −π 4
  • 57. 2. Example Amplitude: A = 2 2π Period: = 2π ω ω =1 Phase shift Cosine reflected φ = −π 4 y = A cos (ω x − φ ) ω φ = −π 4 ( y = −2 cos x + π 4 )
  • 58. 3. Classwork Solve problems 32 - 40 in your copies. Even numbered problems
  • 59. 4. Homework Solve problems 31 - 39 in your copies. Odd numbered problems
  • 60. Day 58 1. Opener La siguiente figura corresponde a un trozo de cartulina y en ella se realiza un doblez tomando como eje una recta que pase por los puntos D y B, de tal manera que el triángulo DBC quede sobre el triángulo ABD. ¿Qué figura se observará posteriormente? A A) B) D B C C) D)
  • 61. Day 58 1. Opener La siguiente figura corresponde a un trozo de cartulina y en ella se realiza un doblez tomando como eje una recta que pase por los puntos D y B, de tal manera que el triángulo DBC quede sobre el triángulo ABD. ¿Qué figura se observará posteriormente? A A) B) D B C C) D)
  • 62. 1. Opener This is a graph of my elevation over sixteen seconds. What am I doing? Where am I? How do you know? 12 10 Elevation (feet) 8 6 4 2 2 4 6 8 10 12 14 16 18 Time (seconds)
  • 63. 1. Opener Solve the equations: 1. cos x = 1 2. cos x = −1 3. tan x = −1 4. sin x = − 3 2
  • 64. 1. Opener Solve the equations: 1. cos x = 1 1. x = 0, 2π 2. cos x = −1 2. x = π 3. tan x = −1 3. x = 3 4 π , 7 4 π 4. sin x = − 3 2 4. x = 4π 3 , 5π 3
  • 65. 1. Opener Find f-1(x): x 1. f (x) = e x 2. f (x) = 10
  • 66. 2. Inverse Trigonometric Functions Inverse Sine Function
  • 67. 2. Inverse Trigonometric Functions Inverse Sine Function
  • 68. 2. Inverse Trigonometric Functions Inverse Sine Function
  • 69. 2. Inverse Trigonometric Functions Inverse Sine Function
  • 70. 2. Inverse Trigonometric Functions Inverse Sine Function −π 2 ≤ x ≤ π 2
  • 71. 2. Inverse Sine Function −1 y = sin x means x = sin y where −π 2 ≤ y ≤ π 2 and −1 ≤ x ≤ 1
  • 72. 3. Examples Find the exact value of sin-1 1
  • 73. 3. Examples Find the exact value of sin-1 1
  • 74. 3. Examples Find the exact value of sin-1 1
  • 75. 3. Examples Find the exact value of sin-1 1 sin 1 = π 2 −1
  • 76. 3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠
  • 77. 3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠
  • 78. 3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠
  • 79. 3. Examples Find the exact value of sin −1 ⎛ 3 2 ⎞ ⎝ ⎠ −1 sin ⎛ 3 ⎞ = π ⎝ 2 ⎠ 3
  • 80. 4. Inverse Cosine Function −1 y = cos x means x = cos y where 0≤ y≤π and −1 ≤ x ≤ 1
  • 81. 5. Classwork Solve problems 1 - 11 in your copies. Odd numbered problems.

Editor's Notes

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  9. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  10. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  11. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  12. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  13. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  14. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
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  24. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
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  44. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
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  68. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  69. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  70. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n
  71. Louisianans, Mainers, Nutmeggers, New Jersey-ites, Bay Staters (Massholes, more locally).\n