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Trigonometric Equations Mastery

Learn to solve, verify, and find extraneous solutions in trigonometric equations, with practical examples and in-depth explanations to enhance your understanding.

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Trigonometric Equations Mastery

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 13–4) CCSS Then/Now New Vocabulary Example 1: Solve Equations for a Given Interval Example 2: Infinitely Many Solutions Example 3: Real-World Example: Solve Trigonometric Equations Example 4: Determine Whether a Solution Exists Example 5: Solve Trigonometric Equations by Using Identities Lesson Menu

  3. Find the exact value of sin 2 when cos  = – and 180° <  < 270°. A. B. C. D. 7 __ 8 5-Minute Check 1

  4. Find the exact value of sin 2 when cos  = – and 180° <  < 270°. A. B. C. D. 7 __ 8 5-Minute Check 1

  5. A. B. C. D. 5-Minute Check 2

  6. A. B. C. D. 5-Minute Check 2

  7. A. B. C. D. Find the exact value of sin 67.5° by using half-angle formulas. 5-Minute Check 3

  8. A. B. C. D. Find the exact value of sin 67.5° by using half-angle formulas. 5-Minute Check 3

  9. A. B. C. D. Find the exact value of cos 22.5° by using double-angle formulas. 5-Minute Check 4

  10. A. B. C. D. Find the exact value of cos 22.5° by using double-angle formulas. 5-Minute Check 4

  11. Find the exact value of tan by using double-angle formulas. A. B. C. D. 5-Minute Check 5

  12. Find the exact value of tan by using double-angle formulas. A. B. C. D. 5-Minute Check 5

  13. Simplify the expression tan x (cot x + tan x). A. sin2x B. cos2x C. sec2x D. csc2x 5-Minute Check 6

  14. Simplify the expression tan x (cot x + tan x). A. sin2x B. cos2x C. sec2x D. csc2x 5-Minute Check 6

  15. Content Standards F.TF.8 Prove the Pythagorean identity sin2 (θ) + cos2 (θ) = 1 and use it to find sin (θ), cos (θ), or tan (θ) given sin (θ), cos (θ), or tan (θ) and the quadrant of the angle. Mathematical Practices 4 Model with mathematics. 6 Attend to precision. CCSS

  16. You verified trigonometric identities. • Solve trigonometric equations. • Find extraneous solutions from trigonometric equations. Then/Now

  17. trigonometric equations Vocabulary

  18. Solve Equations for a Given Interval Solve 2cos2 – 1 = sin  if 0≤  180. 2cos2 – 1 = sin  Original equation 2(1 – sin2 ) – 1 – sin  = 0 Subtract sin  from each side. 2 – 2 sin2 – 1 – sin  = 0 Distributive Property –2 sin2 – sin  + 1 = 0 Simplify. 2 sin2 + sin  – 1 = 0 Divide each side by –1. (2 sin  – 1)(sin  + 1) = 0 Factor. Example 1

  19. Solve Equations for a Given Interval Now use the Zero Product Property. 2 sin  – 1 = 0 sin  + 1 = 0 2 sin  = 1 sin  = –1  = 270°  = 30° or 150° Answer: Example 1

  20. Solve Equations for a Given Interval Now use the Zero Product Property. 2 sin  – 1 = 0 sin  + 1 = 0 2 sin  = 1 sin  = –1  = 270°  = 30° or 150° Answer: Since 0°≤ ≤ 180°, the solutions are 30°, and 150°. Example 1

  21. Find all solutions of sin2 + cos 2 – cos  = 0 for the interval 0≤ ≤ 360. A. 0°, 90°, 180° B. 0°, 180°, 270° C. 90°, 180°, 270° D. 0°, 90°, 270° Example 1

  22. Find all solutions of sin2 + cos 2 – cos  = 0 for the interval 0≤ ≤ 360. A. 0°, 90°, 180° B. 0°, 180°, 270° C. 90°, 180°, 270° D. 0°, 90°, 270° Example 1

  23. A. Solve cos  + = sin2for all values of  if  is measured in degrees. Look at the graph of y = cos  – sin2 to find solutions of cos  – sin2 = – Infinitely Many Solutions Example 2

  24. Infinitely Many Solutions The solutions are 60°, 300°, and so on, and –60°, –300°, and so on. The period of the function is 360°. So the solutions can be written as 60° + k ● 360° and 300° + k ● 360°, where k is measured in degrees. Answer: Example 2

  25. Infinitely Many Solutions The solutions are 60°, 300°, and so on, and –60°, –300°, and so on. The period of the function is 360°. So the solutions can be written as 60° + k ● 360° and 300° + k ● 360°, where k is measured in degrees. Answer: 60° + k ● 360° and 300° + k ● 360°, where k is measured in degrees. Example 2

  26. Infinitely Many Solutions B. Solve 2cos  = –1 for all values of  if  is measured in radians. 2cos  = –1 Example 2

  27. The solutions are , and so on, and , and so on. The period of the cosine function is 2 radians. So the solutions can be written as , where k is any integer. Infinitely Many Solutions Answer: Example 2

  28. The solutions are , and so on, and , and so on. The period of the cosine function is 2 radians. So the solutions can be written as , where k is any integer. Answer: , where k is any integer. Infinitely Many Solutions Example 2

  29. A. Solve cos 2 sin  + 1 = 0 for all values of  if  is measured in degrees. A.0° + k● 360° and 45° + k ● 360° where k is any integer B.45° + k● 360° where k is any integer C.0° + k● 360° and 90° + k ● 360° where k is any integer D.90° + k● 360° where k is any integer Example 2

  30. A. Solve cos 2 sin  + 1 = 0 for all values of  if  is measured in degrees. A.0° + k● 360° and 45° + k ● 360° where k is any integer B.45° + k● 360° where k is any integer C.0° + k● 360° and 90° + k ● 360° where k is any integer D.90° + k● 360° where k is any integer Example 2

  31. A. B. C. D. B. Solve 2 sin  = –2 for all values of  if  is measured in radians. Example 2

  32. A. B. C. D. B. Solve 2 sin  = –2 for all values of  if  is measured in radians. Example 2

  33. AMUSEMENT PARKSWhen you ride a Ferris wheel that has a diameter of 40 meters and turns at a rate of 1.5 revolutions per minute, the height above the ground, in meters, of your seat after t minutes can be modeled by the equation h = 21 – 20 cos 3t. How long after the Ferris wheel starts will your seat first be meters above the ground? Solve Trigonometric Equations Example 3

  34. Replace h with Solve Trigonometric Equations Original equation Subtract 21 from each side. Divide each side by –20. Take the Arccosine. Example 3

  35. The Arccosine of Solve Trigonometric Equations Divide each side by 3. Answer: Example 3

  36. The Arccosine of Answer: Solve Trigonometric Equations Divide each side by 3. Example 3

  37. AMUSEMENT PARKS When you ride a Ferris wheel thathas a diameter of 40 meters and turns at a rate of 1.5 revolutions per minute, the height above the ground,in meters, of your seat after t minutes can be modeled by the equation h = 21 – 20 cos 3t. How long after theFerris wheel starts will your seat first be 11 meters above the ground? A. about 7 seconds B. about 10 seconds C. about 13 seconds D. about 16 seconds Example 3

  38. AMUSEMENT PARKS When you ride a Ferris wheel thathas a diameter of 40 meters and turns at a rate of 1.5 revolutions per minute, the height above the ground,in meters, of your seat after t minutes can be modeled by the equation h = 21 – 20 cos 3t. How long after theFerris wheel starts will your seat first be 11 meters above the ground? A. about 7 seconds B. about 10 seconds C. about 13 seconds D. about 16 seconds Example 3

  39. sin  = cos  Divide each side by cos  Determine Whether a Solution Exists A. Solve the equation sin cos  = cos2 if 0 ≤  ≤ 2. sin cos = cos2 Original equation sin cos  – cos2 = 0 Subtract cos2 from each side. cos  (sin  – cos )= 0 Factor. cos  = 0 or sin  – cos  = 0 Zero Product Property Example 4

  40. Determine Whether a Solution Exists Divide each side by cos . tan  = 1 Example 4

  41. ? ? ? ? Determine Whether a Solution Exists Check   Example 4

  42. ? ? ? ? Determine Whether a Solution Exists   Answer: Example 4

  43. Answer: ? ? ? ? Determine Whether a Solution Exists   Example 4

  44. Determine Whether a Solution Exists B. Solve the equation cos  = 1 – sin if 0° ≤ < 360°. cos  = 1 – sin Original equation cos2 = (1 – sin)2 Square each side. 1 – sin2= (1 – 2 sin + sin2 ) cos2 = 1 – sin2 0 = 2 sin2 – 2 sin Simplify. 0 = sin (2 sin – 2) Factor. sin  = 0 or 2 sin  – 2 = 0 Zero Product Property  = 0 or 180 sin  = 1 Solve for sin   = 90 Example 4

  45. ? ? cos (0°) = 1 – sin (0°) cos (180°) = 1 – sin (180°) ? ? 1 = 1 – 0 –1 = 1 – 0 ? cos (90°) = 1 – sin (90°) ? 0 = 1 – 1 Determine Whether a Solution Exists Check cos  = 1 – sin  cos  = 1 – sin  1 = 1 –1 = 1   cos  = 1 – sin  0 = 0  Answer: Example 4

  46. ? ? cos (0°) = 1 – sin (0°) cos (180°) = 1 – sin (180°) ? ? 1 = 1 – 0 –1 = 1 – 0 ? cos (90°) = 1 – sin (90°) ? 0 = 1 – 1 Determine Whether a Solution Exists Check cos  = 1 – sin  cos  = 1 – sin  1 = 1 –1 = 1   cos  = 1 – sin  0 = 0  Answer: 0° and 90° Example 4

  47. A. B. C. D. A. Solve the equation cos  = (1 – sin2 ) if 0 ≤  < 2. Example 4

  48. A. B. C. D. A. Solve the equation cos  = (1 – sin2 ) if 0 ≤  < 2. Example 4

  49. B. Solve the equation sin  cos  = sin2 if 0 ≤  < 360. A. 0°, 45°, 180°, 225° B. 0°, 90°, 180°, 270° C. 30°, 45°, 225°, 330° D. 30°, 90°, 180°, 330° Example 4

  50. B. Solve the equation sin  cos  = sin2 if 0 ≤  < 360. A. 0°, 45°, 180°, 225° B. 0°, 90°, 180°, 270° C. 30°, 45°, 225°, 330° D. 30°, 90°, 180°, 330° Example 4

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