Question

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

Answer 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