秋风2000
幼苗
共回答了23个问题采纳率:95.7% 举报
关于x的方程2x^2-(根号3+1)x+m=0的两根为 sin θ,cos θ
所以 sinθ+cosθ=(√3+1)/2…………(1)
sinθcosθ=m/2………………(2)
(1)sin^2 θ/(sin θ-cos θ) + cos θ/(1-tan θ)
= sin^2 θ/(sin θ-cos θ) + cos θ/[1-(sinθ/cosθ)]
=sin^2 θ/(sin θ-cos θ) + cos θ/[(cosθ-sinθ)/cosθ]
=sin^2 θ/(sin θ-cos θ) + cos^2 θ/(cosθ-sinθ)
=sin^2 θ/(sin θ-cos θ) - cos^2 θ/(sinθ-cosθ)
=(sin^2 θ-cos^2 θ)/(sin θ-cos θ)
=(sin θ-cos θ)(sin θ+cos θ)/(sin θ-cos θ)
=sin θ+cos θ= (√3+1)/2
(2)将(1)式两边平方得到:
sin^2 θ+cos^2 θ +2sinθcosθ=(1/4)(3+2√3+1)
1+2sinθcosθ=1 + √3/2,
求得 sinθcosθ=√3/4,根据(2)式 sinθcosθ=m/2
解得 m=√3/2
1年前
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