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Double-Angle and Half-Angle Identities

1 2. = 2 – Replace with exact values. = – Simplify. 3 2. 3 2. Double-Angle and Half-Angle Identities. ALGEBRA 2 LESSON 14-7. Use a double-angle identity to find the exact value of sin 600°. sin 600° = sin 2(300°) Rewrite 600 as 2(300).

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Double-Angle and Half-Angle Identities

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  1. 1 2 = 2 – Replace with exact values. = – Simplify. 3 2 3 2 Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 Use a double-angle identity to find the exact value of sin 600°. sin 600° = sin 2(300°) Rewrite 600 as 2(300). = 2 sin 300° cos 300° Use a sine double-angle identity. 14-7

  2. sin 2 (1 – sin2 ) sin 2 (1 – sin2 ) = 2 tan . 2 sin • cos cos2 = double angle and Pythagorean identities 2 sin cos = Simplify. = 2 tan Tangent Identity Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 Verify the identity = 2 tan . 14-7

  3. 150° 2 150° 2 sin 75° = sin Rewrite 75° as . = Use the principal square root, since sin 75° is positive. (1 – cos150°) 2 = Substitute the exact value for cos 150°. 2 + 3 2 2 + 3 4 – 1 – 2 = Simplify. 3 2 = Simplify. Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 Use the half-angle identities to find each exact value. a. sin 75° 14-7

  4. 135° 2 135° 2 cos 67.5° = cos Rewrite 65.7° as . = Use the principal square root, since sin 75° is positive. (1 + cos 135°) 2 = Substitute the exact value for cos 135°. 2 – 2 2 2 – 2 4 – 1 + 2 = Simplify. 2 2 = Simplify. Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 (continued) b. cos 67.5° 14-7

  5. 12 13 2 2 2 Since 90° < < 180°, 45° < < 90° and is in Quadrant I. 2 1 – cos 2 sin = ± half-angle identity 12 13 – 1 – 2 Substitute. Choose the positive square root since is in Quadrant I. = 2 25 26 = Simplify. 5 26 26 = Simplify. Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 Given cos = – and 90° < < 180°, find sin . 14-7

  6. pages 810–811  Exercises 1. – 2. – 3. – 3 4. 1 5. – 6. 3 7. – 8. – 9. sin 2 = sin ( + ) = sin cos + cos • sin = 2 sin cos 10. tan 2 = tan ( + ) = = 11. 12. 7 – 4 3 13. 14. 15. 16. 3 – 2 2 17. 0 18. 19. 20. 21. 3 22. 23. 24. – 25. – 4 26. – 17 2 tan 1 – tan2 tan + tan 1 – tan tan 17 17 10 10 1 2 1 3 1 2 1 2 2 + 2 2 2 + 3 2 2 – 3 2 2 – 2 2 2 – 2 + 3 2 3 2 3 2 4 17 17 3 10 10 Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 14-7

  7. 27. sin 2R = 2 sin R cos R = 2 • = 28. cos 2R = cos2R – sin2R = – = – = 29. sin 2S = 2 sin S cos S = 2 • • = = 2 sin R cos R = sin 2R 30. sin2 = sin = ± = = = – = 31. tan = = = • = = = S 2 s t S 2 1 – cos S 1 + cos S r t 2rs t2 2 32. tan2 = (tan )2 = ± = = 33. No; since the sine function is periodic, A and B can have many different values. 34. – 35. – 36. 37. – 1 – cos S 1 + cos S t – r t + s t2 – s2 (t+ s)2 r2 (t+ s)2 s2 – r2 t2 = s t s2 t2 r2 t2 2 2 r t s t r t 2sr t2 S 2 2 2 S 2 1 – cos S 2 24 25 r t 1 – 2 1 – cos S 2 t – r 2t 1 2 r 2t 7 25 s t r t 24 7 1 – 1 – 1 – cos R 1 + cos R R 2 r t s t 25 24 1 + 1 + t – s t + s t + s t + s r t + s Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 14-7

  8. 45. 4 cos2 – 1 = 0 , , , 46. 1 47. –cos 48. cos – sin 49. Answers may vary. Sample: a. sin 60° = cos 60° = b. sin 120° = c. cos 30° = 50. Answers may vary. Sample: No; the graphs of y = and y = tan are only equal at certain finite values. 38. 39. 40. – 41. –2 42. cos (8 sin – 3) = 0 , , 0.384, 2.757 43. sin (4 cos – 3) = 0 0, , 0.723, 5.560 44. cos (2 sin2 – 1) = 0 , , , , , 1 2 2 3 2 4 3 2 3 2 5 4 3 4 2 3 4 3 5 3 7 4 1 2 2 5 5 3 2 5 5 3 2 3 2 tan 4 2 Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 14-7

  9. 51–56.Answers may vary. 51. 4 sin cos (cos2 – sin2 ) 52. 8 cos4 – 8 cos2 + 1 53. 54. ± + 55. ± ± + cos 56. ± A 2 1 – cos A 1 + cos A 1 – cos A 1 + cos A 57.a. tan = ± = ± • = ± = ± = Since tan and sin A always have the same sign, only the positive sign occurs. b. tan = ± = ± • = ± = ± = Since tan and sin A always have the same sign, only the positive sign occurs. 1 + cos A 1 + cos A sin2 A (1 + cos A)2 sin A 1 + cos A (1 – cosA)2 1 + cos2A (1 – cosA)2 sin2 A 1– cos2 A (1 + cos A)2 4 tan (1 – tan2 ) tan4 – 6 tan2 + 1 A 2 cos 2 1 2 1 2 1 2 ± 1 2 1 2 1 2 A 2 1 – cos A 1 + cos A 1 – cos A 1 + cos A 1 2 1 2 1 – cos A 1 – cos A 1± + cos 1 2 1 2 1± + cos 1 – cos A sin A A 2 Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 14-7

  10. 58. D 59. G 60.[2] sin = = = = [1] appropriate methods with minor error 61.[4] tan = = 0.5 = tan 2 = = = = = [3] appropriate method with minor error [2] correct answer without work shown [1] recognized using double-angle identities but did not apply them properly 62. 63. 64. 3 65. 12, about 4.1 66. 3, 2 1 2 0.5 1 2 tan 1 – tan2 135° 2 1 – cos 135° 2 1 3 4 4 3 1 – 2 2 2 2 – 2 2 1 2 2 2 2 + 2 2 1 2 2 1 – 1 3 2 2 2 1 4 1 – Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 14-7

  11. 3 2 – sin 3 = sin (2 + ) = sin 2 cos + cos 2 sin = (2 sin cos )cos + (cos2 – sin2 )sin = 3 sin cos2 – sin3 = 3 sin (1 – sin2 ) – sin3 = 3 sin – 4 sin3 2 + 2 2 2 20 29 3 58 2 – Double-Angle and Half-Angle Identities ALGEBRA 2 LESSON 14-7 1. Use a double-angle identity to find the exact value of sin 660°. 2. Use an angle sum identity and a double-angle identity to verify the identity sin 3 = 3 sin – 4 sin3 . 3. Use a half-angle identity to find the exact value of sin 67.5°. 4. Given cos = – and 180° < < 270°, find the exact value of cos . 14-7

  12. 11. 60° + 360° • n, 240° + 360° • n 12. , 13. , 14. , 15. , 0.63 16. , 1.28 17. , 1.25 18. , 1.28 19. , 1.60 20. , 1.25 21. 8, 36.9°, 53.1° Page 816 1. csc 2. 1 3. csc2 4. csc cos tan = • • = = 1 5. csc2 – cot2 = 1 + cot2 – cot2 = 1 6. sec cot = • = = csc 7. sec2 – 1= 1 + tan2 – 1 = tan2 8. 60° + 360° • n, 120° + 360° • n 9. 30° + 360° • n, 330° + 360° • n 10. 180° + 360° • n sin cos cos 1 1 sin 4 6.4 cos sin sin cos 6.4 5 5 4 1 cos cos sin 1 sin 6.4 5 4 3 5 3 5 3 3 4 3 4 6.4 4 5 4 Trigonometric Identities and Equations ALGEBRA 2 CHAPTER 14 14-A

  13. 22. 15.6, 50.2°, 39.8° 23. 46.5, 24.3°, 65.7° 24. 6.1, 25.3°, 64.7° 25. 91.8 m2 26. 13.8 in. 27. 17.9 ft 28. 25.8° 29. 17.6° 30. No; the Law of Sines requires at least one angle in order to set up a ratio of side-to-angle. 31. 30.4 cm 32. 48.8° 33. 47.7° 34. –sin ( – ) = –sin (–( – )) = sin ( – ) = cos 35. csc ( + ) = = = = sec 36. csc ( – ) = = = = –sec 1 sin cos + cos sin 1 sin (0) + cos (1) 1 cos 1 sin (– ( – )) 1 – sin ( – ) 2 2 2 2 2 2 2 2 2 1 – cos Trigonometric Identities and Equations ALGEBRA 2 CHAPTER 14 14-A

  14. 37. cos (– – ) = cos (– ) cos + sin (– ) sin = cos ( )(0) + sin (– )(1) = sin (– ) 38. 0, 39. , 40. 0; 41. 42. 43. 3 44. 45. 1 46. –1 47. Check students’ work. 3 2 1 2 2 2 2 2 3 2 3 3 Trigonometric Identities and Equations ALGEBRA 2 CHAPTER 14 14-A

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