三角恒等式求证 sin(π/7)*sin (2π/7)*sin (4π/7)=(√7)/8
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)...全部
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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