已知sinα是方程5x²-7x-6=0的根,求cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)cot(π-α)的值
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![已知sinα是方程5x²-7x-6=0的根,求cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)cot(π-α)的值](/uploads/image/z/963422-62-2.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B1%E6%98%AF%E6%96%B9%E7%A8%8B5x%26%23178%3B-7x-6%3D0%E7%9A%84%E6%A0%B9%2C%E6%B1%82cos%EF%BC%882%CF%80-%CE%B1%EF%BC%89cos%EF%BC%88%CF%80%2B%CE%B1%EF%BC%89tan%26%23178%3B%EF%BC%882%CF%80-%CE%B1%EF%BC%89%2Fsin%EF%BC%88%CF%80-%CE%B1%EF%BC%89sin%EF%BC%882%CF%80-%CE%B1%EF%BC%89cot%EF%BC%88%CF%80-%CE%B1%EF%BC%89%E7%9A%84%E5%80%BC)
已知sinα是方程5x²-7x-6=0的根,求cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)cot(π-α)的值
已知sinα是方程5x²-7x-6=0的根,求cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)cot(π-α)的值
已知sinα是方程5x²-7x-6=0的根,求cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)cot(π-α)的值
解方程5x²-7x-6=0
(5x+3)(x-2)=0
解得x=-3/5,或x=2
∵sinα是方程5x²-7x-6=0的根,sinα∈[-1,1]
∴sinα=-3/5
∴cosα=±4/5,tanα=±3/4
∴[cos(2π-α)cos(π+α)tan²(2π-α)]/[sin(π-α)sin(2π-α)cot(π-α)]
=[cosα(-cosα)tan²α]/[sinα(-sinα)(-cotα)]
=(-cos²α*sin²α/cos²α)/(sin²α*cosα/sinα)
=(-sin²α)/(sinα*cosα)
=-sinα/cosα
=-tanα
=3/4 或-3/4
〔sin(-α-3/2π)sin(3/2π-α)*tan(2π-α)〕/〔cos(π/2-α)cos(π/2+α)*cos(π-α)〕
=(-sin²α)/(sinα*cosα)
=-sinα/cosα
=-tanα
5x²-7x-6=0
(5x+3)(x-2)=0
x=-3/5或x=2(舍去,...
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〔sin(-α-3/2π)sin(3/2π-α)*tan(2π-α)〕/〔cos(π/2-α)cos(π/2+α)*cos(π-α)〕
=(-sin²α)/(sinα*cosα)
=-sinα/cosα
=-tanα
5x²-7x-6=0
(5x+3)(x-2)=0
x=-3/5或x=2(舍去,|sinα|必须<=1)
所以tanα=±3/4
原式=-tanα=±3/4
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