Chapter 4
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  • 1. 2 )] + 1 +θ 6 n(θ log 21 θ [si = 4 sin Chapter Four 2 θ) 1 cos 2 θ− − tan 2θ (sec 1 SOLUTIONθ) 2 ot +c 1 ( θ )] 2 − sin log 1−(1 [ By «Craig»
  • 2. 1 2 1 (sec θ −tan θ −cos θ ) [sin(θ + θ)] 2 2 2 log[1−(1−sin 2 θ )] = +1 2 log 2 16 (1+ cot θ) 4 sin θ At first glance, this problem can look a little bit complicated and intimidating. A good way to make it easier to solve is to break it down: Step 1- Solve the logarithm. Step 2- Prove the resulting identity.
  • 3. 1 1 (sec 2 θ −tan 2 θ −cos2 θ ) log[1−(1−sin 2 θ )] 2 (1+ cot θ) Step 1- Solve the logarithm.
  • 4. 1 1 1 (sec 2 θ −tan 2 θ −cos2 θ ) 2 2 2 (sec θ − tan θ − cos θ) log[1−(1−sin 2 θθ)] 2 2 [1− (1− sin )] (1+ cot θ) (1+ € € € Step 1- Solve the logartithm. € There are 3 parts to this logarithm: € Part A. Base Part B. Base of Argument Part C. Exponent of Argument
  • 5. 2 [1−(1−sin θ )] Step 1- Solve the logartithm. Part A. Simplify the Base.
  • 6. 2 [1−(1−sin θ )] Method 2 Method 1 Step 1- Solve the logartithm. Part A. Simplify the Base There are two methods that can be used to get the simplified version of this expression: Method 1 Method 2
  • 7. 2 [1−(1−sin θ )] Method 1 2 1− (1− sin θ) Step 1- Solve the logartithm. Part A. Simplify the Base €
  • 8. 2 [1−(1−sin θ )] Method 1 2 1−1+ sin θ Step 1- Solve the logartithm. Part A. Simplify the Base Method 1: Expand the original expression. €
  • 9. 2 [1−(1−sin θ )] Method 1 2 sin θ Step 1- Solve the logartithm. Part A. Simplify the Base Method 1: Simplify the resulting expression
  • 10. 2 [1−(1−sin θ )] Method 2 2 1− (1− sin θ) Step 1- Solve the logartithm. Part A. Simplify the Base €
  • 11. 2 [1−(1−sin θ )] Method 2 2 1− (cos θ) Step 1- Solve the logartithm. Part A. Simplify the Base sin2 θ + cos2 θ =1 so that Method 2: Rearrange the identity cos 2 θ = 1− sin2 θ and use it to simplify the expression. € €
  • 12. 2 [1−(1−sin θ )] Method 2 2 sin θ Step 1- Solve the logartithm. Part A. Simplify the Base Method 2: Simplify the resulting expression
  • 13. 2 [1−(1−sin θ )] 2 2 [1− (1− sin θ)] = sin θ Step 1- Solve the logartithm. Part A. Simplify the Base 2 sin θ Therefore, the Base is equal to € €
  • 14. 1 2 (1+ cot θ) Step 1- Solve the logartithm. Part B. Simplify the Base of Argument.
  • 15. 1 2 (1+ cot θ) Method 2 Method 1 € Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement There are two methods that can be used to get the simplified version of this expression: Method 1 Method 2
  • 16. 1 2 (1+ cot θ) Method 1 1 2 (1+ cot θ) € Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement
  • 17. 1 2 (1+ cot θ) Method 1 1 2 2 sin θ cos θ + € 2 2 sin θ sin θ Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement cos2 θ 2 Method 1: Recognize that cot θ is the same thing as and sin2 θ expand it as such. Also, rewrite “1” so it has the same LCD. € € €
  • 18. 1 2 (1+ cot θ) Method 1 1 2 2 sin θ + cos θ € 2 sin θ Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement Method 1: Add the two fractions together to get one fraction. €
  • 19. 1 2 (1+ cot θ) Method 1 1 1 € 2 sin θ Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement sin2 θ + cos2 θ =1) Method 1: Refer to the Pythagorean identity ( and use it to simplify the numerator of the fraction in the denominator. € €
  • 20. 1 2 (1+ cot θ) Method 1 2 sin θ € Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement Method 1: Multiply “1” by the reciprocal of the fraction in the 2 denominator ( sin θ ). €
  • 21. 1 2 (1+ cot θ) Method 2 1 2 (1+ cot θ) € Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement
  • 22. 1 2 (1+ cot θ) Method 2 1 2 csc θ € Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement Method 2: Recognize that the Pythagorean Identity csc2 θ − cot 2 θ = 1 applies to the denominator and simplify it. €
  • 23. 1 2 (1+ cot θ) Method 2 1 1 € 2 sin θ Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement csc2 θ is eqivalent to the reciprocal Method 2: Recognize that 2 of sin θ ( sin θ ) and rewrite it as such. 1 2 € € €€
  • 24. 1 2 (1+ cot θ) Method 2 2 sin θ € Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement Method 2: Multiply “1” by the reciprocal of the fraction in the 2 denominator ( sin θ ). €
  • 25. 1 2 (1+ cot θ) 1 2 sin θ = 2 €(1+ cot θ) Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement Therefore, the Base of the Argument 2 sin θ . is equal to €
  • 26. 1 1 1 (sec 2 θ −tan 2 θ −cos2 θ ) 2 2 2 (sec θ − tan θ − cos θ) log[1−(1−sin 2 θθ)] 2 2 [1− (1− sin )] (1+ cot θ) (1+ € € Step 1- Solve the logartithm. € So far the logarithm goes from this...
  • 27. 1 1 12 (sec 2 θθ−tan 2 θθ−cos2 θθ)) 2 2 2 (sec − tan − cos log[1−(1−sinθ2 θ )] sin θ sin2 2 (1+ cot θ) € € Step 1- Solve the logartithm. € ...to this.
  • 28. 1 1 12 (sec 2 θθ−tan 2 θθ−cos2 θθ)) 2 2 2 (sec − tan − cos log[1−(1−sinθ2 θ )] sin θ sin2 2 (1+ cot θ) € € Step 1- Solve the logartithm. € We can recognize this about the logarithm: The Base is the same as the Base of the Argument. If this is the case, no matter what the Exponent of the Argument is, the entire logarithm is equal to the Exponent of the Argument.
  • 29. 1 1 (sec2 θ −tan 2 θ −cos2 θ ) 1 log[1−(1−sin 2 θ )] =2 2 2 2 (1+ cot θ) (sec θ −tan θ − cos θ) Step 1- Solve the logartithm. 1 Therefore, the logarithm is equal to (sec 2 θ −tan 2 θ − cos2 θ) . €
  • 30. 1 2 1 (sec θ −tan θ −cos θ ) [sin(θ + θ)] 2 2 2 log[1−(1−sin 2 θ )] = +1 2 log 2 16 (1+ cot θ) 4 sin θ Step 1- Solve the logartithm. Now, we can take a look at the original identity. Since we have solved the logarithm the identity no longer looks like this...
  • 31. 2 1 [sin(θ + θ)] +1 = 2 2 2 log 2 16 (sec θ − tan θ − cos θ) 4 sin θ € Step 1- Solve the logartithm. ...it looks like this
  • 32. 2 1 [sin(θ + θ)] = +1 2 2 2 log 2 16 (sec θ − tan θ − cos θ) 4sin θ € Step 2- Prove the identity.
  • 33. 2 1 [sin(θ + θ)] +1 = 2 2 2 log 2 16 (sec θ − tan θ − cos θ) 4 sin θ € Step 2- Prove the identity. To prove the identity, solve the two sides separately: Side 1 Side 2
  • 34. 1 2 2 2 (sec θ − tan θ − cos θ) Step 2- Prove the identity. Side 1.
  • 35. 1 2 2 2 (sec θ − tan θ − cos θ) Method 2 Method 1 € Step 2- Prove the identity. Side 1 There are two methods that can be used to get the simplified version of this expression: Method 1 Method 2
  • 36. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 1 € 2 2 2 (sec θ − tan θ − cos θ) Step 2- Prove the identity. Side 1 €
  • 37. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 1 2 1 sin θ € 2 [( 2 ) − ( 2 ) − cos θ] cos θ cos θ Step 2- Prove the identity. Side 1 Method 1: Simplify the first two terms so that they are expressed in terms of sine and/or cosine. €
  • 38. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 1 2 € 1− sin θ 2 [( ) − cos θ] 2 cos θ Step 2- Prove the identity. Side 1 Method 1: Subtract the two fractions in the denominator to get one fraction. €
  • 39. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 1 2 € cos θ 2 [( 2 ) − cos θ] cos θ Step 2- Prove the identity. Side 1 2 2 Method 1: Due to the Pythagorean identity ( sin θ + cos θ =1), 2 cos θ the fraction in the denominator is simplified to cos θ 2 € . € €
  • 40. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 1 € 2 (1− cos θ) Step 2- Prove the identity. Side 1 Method 1: Simplify the fraction in the denominator to “1”. €
  • 41. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 1 € 2 sin θ Step 2- Prove the identity. Side 1 2 2 Method 1: Due to the Pythagorean identity ( sin θ + cos θ =1), 2 the denominator is simplified to sin θ . € € €
  • 42. 1 2 2 2 (sec θ − tan θ − cos θ) Method 1 2 csc θ € Step 2- Prove the identity. Side 1 csc 2 θ 1 Method 1: can also be rewritten as sin2 θ € €
  • 43. 1 2 2 2 (sec θ − tan θ − cos θ) Method 2 1 € 2 2 2 (sec θ − tan θ − cos θ) Step 2- Prove the identity. Side 1 €
  • 44. 1 2 2 2 (sec θ − tan θ − cos θ) Method 2 1 € 2 (1− cos θ) Step 2- Prove the identity. Side 1 Method 2: Recognize that the the Pythagorean Identity sec 2 θ −tan 2 θ =1 applies to the denominator of the expression and simplify it as such. €
  • 45. 1 2 2 2 (sec θ − tan θ − cos θ) Method 2 1 € 2 sin θ Step 2- Prove the identity. Side 1 2 θ + cos2 θ =1), Method 2: Using the Pythagorean identity ( sin 2 simplify the denominator to sin θ . € €
  • 46. 1 2 2 2 (sec θ − tan θ − cos θ) 2 csc θ € Step 2- Prove the identity. Side 1 csc 2 θ 1 Method 2: can also be rewritten as sin2 θ € €
  • 47. 2 1 [sin(θ + θ)] +1 = 2 2 2 log 2 16 (sec θ − tan θ − cos θ) 4 sin θ € Step 2- Prove the identity. So far the we have solved Side 1 of the identity. Now, it no longer looks like this...
  • 48. 2 [sin(θ + θ)] 2 csc θ = +1 log 2 16 4 sin θ € Step 2- Prove the identity. ...it looks like this
  • 49. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Step 2- Prove the identity. Side 2.
  • 50. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 Method 1 € Step 2- Prove the identity. Side 2 There are two methods that can be used to get the simplified version of this expression: Method 1 Method 2
  • 51. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 [sin(θ + θ)] +1 € log 2 16 4 sin θ Step 2- Prove the identity. Side 2
  • 52. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 [sin(θ + θ)] +1 € 4 4 sin θ Step 2- Prove the identity. Side 2 Method 1: Simplify the logarithm that is an exponent for one of the terms in the denominator. This can be done by converting it into a power (ie. 2 x =16 ). €
  • 53. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 (2sinθ cosθ) +1 € 4 4 sin θ Step 2- Prove the identity. Side 2 Method 1: The expression inside the brackets can be recognized as one of the “Double Angle Identity” and therefore can be simplified to 2sinθ cosθ .(see last slide)
  • 54. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 (2)(sinθ)(cosθ)(2)(sinθ)(cosθ) +1 € (2)(2)(sinθ)(sinθ)(sinθ)(sinθ) Step 2- Prove the identity. Side 2 Method 1: Now both the numerator and the denominator can be expanded. (Expand as much as possible).
  • 55. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 (2)(sinθ)(cosθ)(2)(sinθ)(cosθ) +1 € (2)(2)(sinθ)(sinθ)(sinθ)(sinθ) Step 2- Prove the identity. Side 2 Method 1: Now we can reduce many parts of the expression and simplify the remaining terms.
  • 56. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 (cosθ) +1 2 € (sinθ) Step 2- Prove the identity. Side 2 Method 1: This is the resulting expression. €
  • 57. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 2 cos θ sin θ +2 2 sin θ sin θ € Step 2- Prove the identity. Side 2 Method 1: From there we can rewrite “1” as a fraction with the same denominator as the resulting fraction. €
  • 58. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 2 cos θ + sin θ 2 sin θ € Step 2- Prove the identity. Side 2 Method 1: After that, it simply becomes a matter of adding the two fractions together...
  • 59. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 1 2 € sin θ Step 2- Prove the identity. Side 2 Method 1: ...and applying the Pythagorean Identity.
  • 60. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 1 2 csc θ € Step 2- Prove the identity. Side 2 csc 2 θ 1 Method 1: Fianlly, can also be rewritten as sin2 θ € €
  • 61. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 [sin(θ + θ)] +1 € log 2 16 4 sin θ Step 2- Prove the identity. Side 2
  • 62. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 [sin(θ + θ)] +1 € 4 4 sin θ Step 2- Prove the identity. Side 2 Method 2: The second method starts out the same as the first, solving the logarithm in the exponent of the denominator.
  • 63. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 (2sinθ cosθ) +1 € 4 4sin θ Step 2- Prove the identity. Side 2 Method 2: Also the same as the first method, we recognize the expression inside the brackets as a “Double Angle Identity”...
  • 64. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 (2)(sinθ)(cosθ)(2)(sinθ)(cosθ) +1 € (2)(2)(sinθ)(sinθ)(sinθ)(sinθ) Step 2- Prove the identity. Side 2 Method 2: The expression, once again, is then expanded; like terms are reduced; and remaining terms are simplified to get...
  • 65. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 (cosθ) +1 2 € (sinθ) Step 2- Prove the identity. Side 2 ...THIS! Method 2: €
  • 66. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 (cosθ) +1 2 € (sinθ) Step 2- Prove the identity. Side 2 Method 2: However, instead of rewriting “1” as a fraction, due to the fact that (cosθ) is an identity itself, 2 2 (sinθ) rewrite the expression as... € €
  • 67. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 cot θ +1 € Step 2- Prove the identity. Side 2
  • 68. 2 [sin(θ + θ)] +1 log 2 16 4 sin θ Method 2 2 csc θ € Step 2- Prove the identity. Side 2 2 2 Method 2: Now, apply the Pythagorean Identity csc θ − cot θ =1 csc 2 θ . and rewrite the expression as € €
  • 69. 2 2 [sin(θ + θ)] csc θ = +1 log 2 16 4 sin θ Step 2- Prove the identity. Now that we have solved Side 2, we can safely say... €
  • 70. Step 2- Prove the identity. ...SINCE... 1 2 = csc θ 2 2 2 (sec θ − tan θ − cos θ)
  • 71. Step 2- Prove the identity. ...AND... 2 [sin(θ + θ)] 2 = csc θ +1 log 2 16 4sin θ
  • 72. Step 2- Prove the identity. ...THEN... 2 1 [sin(θ + θ)] = +1 2 2 2 (sec θ − tan θ − cos θ) log 2 16 4sin θ €
  • 73. Step 2- Prove the identity. Q
  • 74. Step 2- Prove the identity. E
  • 75. Step 2- Prove the identity. D
  • 76. 1 2 1 (sec θ −tan θ −cos θ ) [sin(θ + θ)] 2 2 2 log[1−(1−sin 2 θ )] = +1 2 log 2 16 (1+ cot θ) 4 sin θ 2 2 csc θ = csc θ Q.E.D.
  • 77. http://www.slideshare.net/dkuropatwa/precal-40s-trigonometric-identities-math-dictionary/1