设X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的两根,求arctanx1+arctanx2的值 设arctanx1=a,arctanx2=b设X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的两根,求arctanx1+arctanx2的值 设arctanx1=a,arctanx2=b,则tana=x1,tanb=x2又因为x1+x2=sin(π/5),x1*x2=cos(4π/5)所以tan(a+b)=(tana+tanb)/(1-tanatanb)=(x1+x2)/(1-x1x2)=sin(π/5)/[1-cos(4π/5)]=tan(π/10)又因为x1+x2=sin(π/5)>0,x1*x2=cos(4π/5)0,x1*x2=cos(4π/5)
问题描述：
设X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的两根,求arctanx1+arctanx2的值 设arctanx1=a,arctanx2=b
设X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的两根,求arctanx1+arctanx2的值
设arctanx1=a,arctanx2=b,则tana=x1,tanb=x2
又因为x1+x2=sin(π/5),x1*x2=cos(4π/5)
所以tan(a+b)=(tana+tanb)/(1-tanatanb)=(x1+x2)/(1-x1x2)=sin(π/5)/[1-cos(4π/5)]=tan(π/10)
又因为x1+x2=sin(π/5)>0,x1*x2=cos(4π/5)0,x1*x2=cos(4π/5)
答
因为反正切函数的值域：arctanx属于（-π/2,π/2)
若X>0,则arctanx属于（0,π/2)
若X