作业帮已知方程2x²-(根号3+1)x+m=0的两个根分别为sinθ,cosθ,求[sin(π-θ)×tan(π+θ)/tanθ-1]+[cos(2π-θ)/1-tanθ]
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![已知方程2x²-(根号3+1)x+m=0的两个根分别为sinθ,cosθ,求[sin(π-θ)×tan(π+θ)/tanθ-1]+[cos(2π-θ)/1-tanθ]](/uploads/image/z/3024323-35-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8B2x%26%23178%3B-%28%E6%A0%B9%E5%8F%B73%2B1%29x%2Bm%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%E5%88%86%E5%88%AB%E4%B8%BAsin%CE%B8%2Ccos%CE%B8%2C%E6%B1%82%5Bsin%28%CF%80-%CE%B8%29%C3%97tan%28%CF%80%2B%CE%B8%29%2Ftan%CE%B8-1%5D%2B%5Bcos%282%CF%80-%CE%B8%29%2F1-tan%CE%B8%5D)
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已知方程2x²-(根号3+1)x+m=0的两个根分别为sinθ,cosθ,求[sin(π-θ)×tan(π+θ)/tanθ-1]+[cos(2π-θ)/1-tanθ]
已知方程2x²-(根号3+1)x+m=0的两个根分别为sinθ,cosθ,求[sin(π-θ)×tan(π+θ)/tanθ-1]+[cos(2π-θ)/1-tanθ]
已知方程2x²-(根号3+1)x+m=0的两个根分别为sinθ,cosθ,求[sin(π-θ)×tan(π+θ)/tanθ-1]+[cos(2π-θ)/1-tanθ]
x1+x2=-b/a,x1*x2=c/a
所以sinθ+cosθ=（√3+1）/2
sinθ*cosθ=m/2
(sinθ+cosθ)²=1+√3/2
即：sin²θ+cos²θ+2sinθ*cosθ=1+√3/2
1+m=1+√3/2
m=√3/2
又因为
sin(π-α) = sinα
tan（π+α）=tanα
cos(2π-α) = cosα
所以原式=[sinθ*tanθ/(tanθ-1)]+[cosθ/(1-tanθ)]
=sin²θ/(sinθ-cosθ)+cos²θ/(cosθ-sinθ)
=(sin²θ-cos²θ)/(sinθ-cosθ)
=sinθ+cosθ=1+√3/2