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謝謝你的回答，能再幫忙看一下我的補(bǔ)充問題嗎？注意,x1,x2是數(shù)值，不是角度對啊。。所以呢？？所以你又問的問題不是問題
設(shè)X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的兩根,求arctanx1+arctanx2的值
設(shè)X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的兩根,求arctanx1+arctanx2的值
設(shè)arctanx1=a,arctanx2=b,則tana=x1，tanb=x2
又因?yàn)閤1+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?yàn)閤1+x2=sin(π/5)>0,x1*x2=cos(4π/5)0,x1*x2=cos(4π/5)
設(shè)arctanx1=a,arctanx2=b,則tana=x1，tanb=x2
又因?yàn)閤1+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?yàn)閤1+x2=sin(π/5)>0,x1*x2=cos(4π/5)0,x1*x2=cos(4π/5)
數(shù)學(xué)人氣：182 ℃時(shí)間：2020-06-15 01:30:42
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