已知sinα是方程5x^2-7x-6=0的根求〔sin(-α-3/2π)sin(3/2π-α)*tan ^2(2π-α)〕/〔cos(π/2-α)cos(π/2 +α)*cos(π-α)〕
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![已知sinα是方程5x^2-7x-6=0的根求〔sin(-α-3/2π)sin(3/2π-α)*tan ^2(2π-α)〕/〔cos(π/2-α)cos(π/2 +α)*cos(π-α)〕](/uploads/image/z/2520054-54-4.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B1%E6%98%AF%E6%96%B9%E7%A8%8B5x%5E2-7x-6%3D0%E7%9A%84%E6%A0%B9%E6%B1%82%E3%80%94sin%EF%BC%88-%CE%B1-3%2F2%CF%80%EF%BC%89sin%EF%BC%883%2F2%CF%80-%CE%B1%EF%BC%89%2Atan+%5E2%EF%BC%882%CF%80-%CE%B1%EF%BC%89%E3%80%95%2F%E3%80%94cos%EF%BC%88%CF%80%2F2-%CE%B1%EF%BC%89cos%EF%BC%88%CF%80%2F2+%2B%CE%B1%EF%BC%89%2Acos%EF%BC%88%CF%80-%CE%B1%EF%BC%89%E3%80%95)
已知sinα是方程5x^2-7x-6=0的根求〔sin(-α-3/2π)sin(3/2π-α)*tan ^2(2π-α)〕/〔cos(π/2-α)cos(π/2 +α)*cos(π-α)〕
已知sinα是方程5x^2-7x-6=0的根
求〔sin(-α-3/2π)sin(3/2π-α)*tan ^2(2π-α)〕/〔cos(π/2-α)cos(π/2 +α)*cos(π-α)〕
已知sinα是方程5x^2-7x-6=0的根求〔sin(-α-3/2π)sin(3/2π-α)*tan ^2(2π-α)〕/〔cos(π/2-α)cos(π/2 +α)*cos(π-α)〕
〔sin(-α-3/2π)sin(3/2π-α)*tan(2π-α)〕/〔cos(π/2-α)cos(π/2+α)*cos(π-α)〕
=cosαcosαtanα/sinαsinαcosα
=sinα/sin²α
=1/sinα
5x²-7x-6=0
(5x+3)(x-2)=0
x=-3/5或x=2(舍去,|sinα|必须
晕了
sinα=-3/5;那么可以求的cosα=4/5或是-4/5,上式先用三角公式简化之后可算。