证明: 8sin(π/7)sin(2π/7)sin(3π/7)=√7 => 64sin²(π/7)sin²(2π/7)sin²(3π/7)=7 => 8[1-cos(2π/7)][1-cos(4π/7)][1-cos(6π/7)]=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)]=7 而cos(2π/7)cos(4π/7)cos(6π/7)=cos(2π/7)cos(3π/7)cos(π/7)=1/8 => 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) 左边=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)×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)[1+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)=右边 证毕 。
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