证明三角恒等式证明三角恒等式证明
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
欲证tan(π/7)tan(2π/7)tan(3π/7)=√7
只需8sin(π/7)sin(2π/7)sin(3π/7)=√7
...全部
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
欲证tan(π/7)tan(2π/7)tan(3π/7)=√7
只需8sin(π/7)sin(2π/7)sin(3π/7)=√7
右端是无理数，直接证明有困难，考虑平方，
64sin³(π/7)sin³(2π/7)sin³(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
故只需证
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)+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)=右边。
收起
