證明：cos(π/7)cos(2π/7)cos(3π/7) =2sin(π/7)cos(π/7)cos(2π/7)cos(3π/7)/[2sin(π/7)] =2sin(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 欲證tan(π/7)tan(2π/7)tan(3π/7)=√7 只需8sin(π/7)sin(2π/7)sin(3π/7)=√7 右端是無(wú)理數(shù),直接證明有困難,考慮平方,64sin(π/7)sin(2π/7)sin(3π/7)=7 即8[1-cos(2π/7)][1-cos(4π/7)][1-cos(6π/7)]=7 亦即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)]=7 而cos(2π/7)cos(4π/7)cos(6π/7)=cos(2π/7)cos(3π/7)cos(π/7)=1/8 故只需證 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) 左邊=cos(4π/7)[cos(2π/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)×2cos(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)=右邊