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)的值已知sinα是方程5x&

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
5x²-7x-6=0
(5x+3)(x-2)=0
x=-3/5 x=2>1
取sinα=-3/5
cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)cot(π-α)
=-cosαcosαtan²α/sinαsinαcotα
=-sinα/cosα=-sinα/[√(1-cosα²)]
=±(3/5)/(4/5)
=±3/5

〔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<=1)
所以
原式=1/(-3/5)=-5/3望楼主采纳!