方程2x^2-〔(根号3)+1〕x+m=0的两根为 sin θ,cos θ ,则有:
sinθ+cosθ=(1+√3)/2
sinθ*cosθ=m/2
1.sinθ/(1-cotθ)+cosθ/(1-tanθ)化简为
sin^2 θ/(sinθ-cosθ)-cos^2 θ/(sinθ-cosθ)=(sin^2 θ-cos^2 θ)/(sinθ-cosθ)=sinθ+cosθ=(1+√3)/2
2.因为sinθ+cosθ=(1+√3)/2,两边平方得:
sin^2 θ+cos^2 θ+2sinθ*cosθ=1+2sinθ*cosθ=1+2*m/2=(1+√3)^2/4
解得m=√3/2
3.由两式sinθ+cosθ=(1+√3)/2
sinθ*cosθ=√3/4
解得sin θ=1/2 cos θ=√3/2 又因为θ∈(0,2π)此时θ=π/6
或者是sin θ=√3/2 cos =θ1/2此时,θ=π/3