Question

已知sinθ,cosθ是方程x2-(√(3)-1)+m两根求1)m的值 2)[sinθ/(1-cotθ)]+[cosθ/(1-tanθ)]值

Answer 1

1.根据韦达定理：sinθ+cosθ=√3 - 1 ， sinθcosθ=m
∵sin²θ+cos²θ=1
∴sin²θ+cos²θ=(sinθ+cosθ)² - 2sinθcosθ=(√3 - 1)² - 2m =4 - 2√3 - 2m=1
m=(3/2) - √3

2.[sinθ/(1-cotθ)]+[cosθ/(1-tanθ)]={sinθ/[1-(cosθ/sinθ)]} + {cosθ/[1-(sinθ/cosθ)]}
={sinθ/[(sinθ/sinθ)-(cosθ/sinθ)]} + {cosθ/[(cosθ/cosθ)-(sinθ/cosθ)]}
=[sin²θ/(sinθ-cosθ)] + [cos²θ/(cosθ-sinθ)]
=[sin²θ/(sinθ-cosθ)] - [cos²θ/(sinθ-cosθ)]
=(sin²θ-cos²θ)/(sinθ-cosθ)
=sinθ+cosθ
=√3 - 1

Answer 2

令x1=sinθ，x2=cosθ
x1^2+x2^2=1
x1^2+x2^2=(x1+x2)^2-2x1*x2=[√(3)-1]^2-2m=1
m=3/2-2√(3)
2）求出x1和x2，代入就可以得到结果