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數學題:三角應用
1. Solve the equatio4sinA=3tanAfor 0° ≤A ≤ 360°.The answer is=0°,180°,360°, 41.41°,318°,59° 2.Solve cos(3θ-45°)=1/2 where 0° ≤ x ≤360°.The answer is= 35°,115°,155°,235° ,275°,355° 我要求有列式，詳細步驟。
最佳解答:
4sinθ=3tanθ4sinθ=3(sinθ/cosθ) sinθ=0or cosθ=3/4[we cannot cancel out sinθ if we do not consider sinθ=0 case]θ= 0? or 180? or 360? [from sinθ=0] θ=41.41? or 318.59? [from cosθ=3/4, as cosθ>0, θ=41.41? in 1stQ while θ=360?-41.41?=318.59? in 4thQ] cos(3θ-45?)=1/2Firstly, let φ=3θ-45?0?≤θ≤360?0?≤3θ-45?≤1035?That means we shouldsolve φ in the interval of [0?, 1035?]cosφ=1/2 [As cosφ >0, there will be at least 2 answers lie on 1stQand 4thQ]φ= 60?, 300?Also, cosα = cos(360?+α)[property]Therefore,φ=60?, 300?, 420?, 660?, 780?, 1020? in the range of φsolve for θ.3θ-45?=φ θ=35?, 115?, 155?, 235?, 275?, 355?If you have anyproblem, feel free to e-mail me. 2013-05-21 00:12:22 補充： Also, cosα = cos(360?n+α) where n is an integer [property]
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數學題:三角應用
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發問:1. Solve the equatio4sinA=3tanAfor 0° ≤A ≤ 360°.The answer is=0°,180°,360°, 41.41°,318°,59° 2.Solve cos(3θ-45°)=1/2 where 0° ≤ x ≤360°.The answer is= 35°,115°,155°,235° ,275°,355° 我要求有列式，詳細步驟。
最佳解答:
4sinθ=3tanθ4sinθ=3(sinθ/cosθ) sinθ=0or cosθ=3/4[we cannot cancel out sinθ if we do not consider sinθ=0 case]θ= 0? or 180? or 360? [from sinθ=0] θ=41.41? or 318.59? [from cosθ=3/4, as cosθ>0, θ=41.41? in 1stQ while θ=360?-41.41?=318.59? in 4thQ] cos(3θ-45?)=1/2Firstly, let φ=3θ-45?0?≤θ≤360?0?≤3θ-45?≤1035?That means we shouldsolve φ in the interval of [0?, 1035?]cosφ=1/2 [As cosφ >0, there will be at least 2 answers lie on 1stQand 4thQ]φ= 60?, 300?Also, cosα = cos(360?+α)[property]Therefore,φ=60?, 300?, 420?, 660?, 780?, 1020? in the range of φsolve for θ.3θ-45?=φ θ=35?, 115?, 155?, 235?, 275?, 355?If you have anyproblem, feel free to e-mail me. 2013-05-21 00:12:22 補充： Also, cosα = cos(360?n+α) where n is an integer [property]
其他解答:
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