Question

证明 tan(π/7)tan(2π/7)tan(3π/7)=根号7 直接计算器打出来的不算。

RT 要过程

Answer 1

证明：
cos(π/7)cos(2π/7)cos(3π/7)
=2³sin(π/7)cos(π/7)cos(2π/7)cos(3π/7)/[2³sin(π/7)]
=2²sin(2π/7)cos(2π/7)cos(3π/7)/[8sin(π/7)]
=2sin(4π/7)cos(3π/7)/[8sin(π/7)]
=2sin(3π/7)cos(3π/7)/[8sin(π/7)]
=sin(6π/7)/[8sin(π/7)]
=1/8……(1)

cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7)
=cos(4π/7)2cos(4π/7)cos(2π/7)+cos(6π/7)cos(2π/7)
=cos(2π/7)×2cos²(4π/7)+cos(6π/7)cos(2π/7)
=cos(2π/7)(1+cos(8π/7))+cos(6π/7)cos(2π/7)
=cos(2π/7)+cos(2π/7)[cos(8π/7))+cos(6π/7)]
=cos(2π/7)+cos(2π/7)×2cosπcos(π/7)
=cos(2π/7)-2cos(2π/7)cos(π/7)
=cos(2π/7)-(cos(3π/7)+cos(π/7))
=cos(2π/7)+cos(4π/7)+cos(6π/7)
即cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7)-cos(2π/7)-cos(4π/7)-cos(6π/7)=0……(2)

[8sin(π/7)sin(2π/7)sin(3π/7)]²
=8[1-cos(2π/7)][1-cos(4π/7)][1-cos(6π/7)]
=8[1-(cos(2π/7)+cos(4π/7)+cos(6π/7))+cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7)-cos(2π/7)cos(4π/7)cos(6π/7)]
=8[1-(cos(2π/7)+cos(4π/7)+cos(6π/7))+cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7)-cos(2π/7)cos(4π/7)cos(6π/7)]
=8[1-(cos(2π/7)+cos(4π/7)+cos(6π/7))+cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7)-cos(2π/7)cos(3π/7)cos(π/7)]
将(1)(2)代入
[8sin(π/7)sin(2π/7)sin(3π/7)]²=8[1-0-1/8]=7
即sin(π/7)sin(2π/7)sin(3π/7))=√7/8
与(1)相除可得
tan(π/7)tan(2π/7)tan(3π/7)=√7