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)

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)
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)

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(arctanx1+arctanx2)
=(x1+x2)/(1-x1x2)
=sin(π/5)/[1-cos(4π/5)]
=sin(π/5)/[2sin^2(2π/5)]
=sin(4π/5)/[2sin^2(2π/5)]
=2sin(2π/5)cos(2π/5)/[2sin^2(2π/5)]
=cot(2π/5)
故
arctanx1+arctanx2=π/2-2π/5=π/10

x1,x2是方程的两个根，由韦达定理得
x1+x2=sin(π/5)
x1x2=cos(4π/5)
设A=arctanx1，B=arctanx2
则tanA=x1，tanB=x2
tan(A+B)
=(tanA+tanB)/(1-tanAtanB)
=(x1+x2)/(1-x1x2)
=sin(π/5)/cos(4π/5)
=sin(π-π/5)/cos(4π/5)
=sin(4π/5)/cos(4π/5)
=tan(4π/5)
所以A+B=4π/5