已知Sinα=15/17,α∈(π/2,π),求cos(Π/3+α),cos(π/3-α)

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已知Sinα=15/17,α∈(π/2,π),求cos(Π/3+α),cos(π/3-α)

郭敦顒回答:
∵Sinα=15/17,α∈(π/2,π),
∴α=118.072°,
又Π/3=60°,
∴cos(Π/3+α)= cos(60°+118.072°)= cos178.072°=-0.9994.
cos(π/3-α)= cos(118.072°-60°)= cos58.072°=0.5288.

已知sinα=15/17,α∈(π/2,π),求cos(Π/3+α),cos(π/3-α)
α∈(π/2,π),即α是钝角, 所以α的余弦为负值
sinα = 15/17 所以 cosα=-√(1 - sin²α) =-8/17
  *********上式计算用到 17²-15²=(17+15)(17-15)=64

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已知sinα=15/17,α∈(π/2,π),求cos(Π/3+α),cos(π/3-α)
α∈(π/2,π),即α是钝角, 所以α的余弦为负值
sinα = 15/17 所以 cosα=-√(1 - sin²α) =-8/17
  *********上式计算用到 17²-15²=(17+15)(17-15)=64

  cos(π/3 + α) =cos(π/3)cosα - sin(π/3) sinα
   =1/2 * cosα -√3/2 * sinα
   =1/2 * (-8/17) -√3/2 * (15/17)
   =-(8 + 15√3) / 34
  cos(π/3 - α) =cos(π/3)cosα + sin(π/3) sinα
   =1/2 * cosα +√3/2 * sinα
   =1/2 * (-8/17) +√3/2 * (15/17)
   =(15√3-8) / 34

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