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
设X1、X2是方程X^2-Xsin(π/5)+cos(4π/5)=0的两根,求arctanx1+arctanx2的值
2个回答
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