已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,看补充.已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,求证α+β=-2π/3
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![已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,看补充.已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,求证α+β=-2π/3](/uploads/image/z/4498467-51-7.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E2%88%88%EF%BC%88-%CF%80%2F2%2C%CF%80%2F2%EF%BC%89%2C%CE%B2%E2%88%88%EF%BC%88-%CF%80%2F2%2C%CF%80%2F2%EF%BC%89%2Ctan%CE%B1%E4%BA%8Etan%CE%B2%E6%98%AF%E6%96%B9%E7%A8%8Bx%5E2%2B3%E2%88%9A3x%2B4%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9%2C%E7%9C%8B%E8%A1%A5%E5%85%85.%E5%B7%B2%E7%9F%A5%CE%B1%E2%88%88%EF%BC%88-%CF%80%2F2%2C%CF%80%2F2%EF%BC%89%2C%CE%B2%E2%88%88%EF%BC%88-%CF%80%2F2%2C%CF%80%2F2%EF%BC%89%2Ctan%CE%B1%E4%BA%8Etan%CE%B2%E6%98%AF%E6%96%B9%E7%A8%8Bx%5E2%2B3%E2%88%9A3x%2B4%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9%2C%E6%B1%82%E8%AF%81%CE%B1%2B%CE%B2%3D-2%CF%80%2F3)
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已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,看补充.已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,求证α+β=-2π/3
已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,看补充.
已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,求证α+β=-2π/3
已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,看补充.已知α∈(-π/2,π/2),β∈(-π/2,π/2),tanα于tanβ是方程x^2+3√3x+4=0的两个实根,求证α+β=-2π/3
∵tanα.tanβ是方程x^2+3√3x+4=0的两个实根,
∴tanα+tanβ=-3√3,tanα·tanβ=4.
∵tan(α+β)
=(tanα+tanβ)/(1-tanα·tanβ)
=-3√3/3=-√3
∵α∈(-π/2,π/2),β∈(-π/2,π/2)
∴α+β∈(-π,π)
∴α+β=2π/3