对k∈Z,设sin(2kπ+α)与cos(2kπ+α)是方程2x²+(根号2+1)x+5m=0的两根求m与sin/(1-/tanα)+cosα/(1-tanα)的值

问题描述:

对k∈Z,设sin(2kπ+α)与cos(2kπ+α)是方程2x²+(根号2+1)x+5m=0的两根
求m与sin/(1-/tanα)+cosα/(1-tanα)的值

sin(2kπ+α)与cos(2kπ+α)是方程2x²+(根号2+1)x+5m=0的两根
sin(2kπ+α)+cos(2kπ+α)=-√2-1,
sin(2kπ+α)cos(2kπ+α)=5m
sinα+cosα=-√2-1……(1)
sinαcosα=5m……(2)
(1)平方得:1+2sinαcosα=3+2√2
10m=2+2√2
m=0.2+0.2√2
sinα/(1-1/tanα)+cosα/(1-tanα)
=sinαsinα/(sinα-cosα)+cosαcosα/(cosα-sinα)
=(sinα+cosα)(sinα-cosα)/(sinα-cosα)
=sinα+cosα
=-√2-1

sin(2kπ+α)=sinα cos(2kπ+α)=cosα
由韦达定理得
sinα+cosα=-(√2+1)/2
sinαcosα=5m/2
(sinα+cosα)²=[-(√2+1)/2]²
sin²α+cos²α+2sinαcosα=(3+2√2)/4
5m+1=(3+2√2)/4
m=(2√2-1)/20
sinα/(1- 1/tanα) +cosα/(1-tanα)
=sinα/(1- cosα/sinα) +cosα/(1- sinα/cosα)
=sin²α/(sinα-cosα)+cos²α/(cosα-sinα)
=sin²α/(sinα-cosα)-cos²α/(sinα-cosα)
=(sin²α-cos²α)/(sinα-cosα)
=(sinα+cosα)(sinα-cosα)/(sinα-cosα)
=sinα+cosα
=-(1+√2)/2