关于高一数学的小问题

来源：百度知道 编辑：UC知道 时间：2024/07/05 02:22:02
①设x1,x2是方程x^2-sin(π/5)x+cos(4π/5)x＝0的两根,则arctanx1+arctanx2的值是多少?
②已知sin(α/2)+cos(α/2)＝-3/(根号5),且2.5π＜α＜3π,则cot(α/4)＝___

①根据题意有x1+x2=sin(π/5),x1x2=cos(4π/5)
tan(arctanx1+arctanx2)=(x1+x2)/(1-x1x2)=sin(π/5)/[1-cos(4π/5)]
=sin(π/5)/[1+cos(π/5)]=tan(π/10)
∵-π/2<arctanx1<π/2,-π/2<arctanx2<π/2
∴-π<arctanx1+arctanx2<π
∴arctanx1+arctanx2=π/10
②[sin(α/2)+cos(α/2)]²=1+sinα=9/5
∴sinα=4/5
∵2.5π＜α＜3π
∴cosα=-3/5
∵1.25π＜α/2＜1.5π
∴sin(α/2)=-√[(1-cosα)/2]=-2/√5
cos(α/2)=-√[(1+cosα)/2]=-1/√5
∴cot(α/4)=1/tan(α/4)=[1+cos(α/2)]/sin(α/2)
=[1-(1/√5)]/(-2/√5)=(1-√5)/2