19028：sin(2π/5) sin(4π/5) sin(6π/5) sin(8π/5)=？

問題19028(アナロジー問)

sin(2π/5) sin(4π/5) sin(6π/5) sin(8π/5)=？

解答

・わたしの...

５倍角の公式・・・チェビシェフの方程式がsinとcosとで同じに表されるのが不思議...♪
　sin5θ＝16sin⁵θ−20sin³θ＋5sinθ
　cos5θ＝16cos⁵θ−20cos³θ＋5cosθ

から...

sinθ...θがπ/5,2π/5,3π/5,4π/5 のとき、sin5θ=0をいずれも満たす...

cosθ...θがπ/5,-2π/5,3π/5,-4π/5のとき、cos5θ=-1をいずれも満たす...

so...16cos⁵θ−20cos³θ＋5cosθ+1=0

cosπ=-1

so...

sin(π/5) sin(2π/5) sin(3π/5) sin(4π/5)=5/16

cos(π/5) cos(-2π/5) cos(3π/5) cos(-4π/5)*(-1)

=cos(π/5) cos(2π/5) cos(3π/5) cos(4π/5)*(-1)

=-1/16

so...

cos(π/5) cos(2π/5) cos(3π/5) cos(4π/5)=1/16

{sin(π/5) sin(2π/5) sin(3π/5) sin(4π/5)}{cos(π/5) cos(2π/5) cos(3π/5) cos(4π/5)}

=(1/2)^4*{sin(2π/5)*sin(4π/5)*sin(6π/5)*sin(8π/5)}

=(1/2)^4*(5/16)*(1/16)

so...

sin(2π/5)*sin(4π/5)*sin(6π/5)*sin(8π/5)=5/16

そっか...

sin(6π/5)=-sin(π/5)

sin(8π/5)=-sin(3π/5)

だから...

sin(2π/5)*sin(4π/5)*sin(6π/5)*sin(8π/5)

=sin(-π/5) sin(2π/5) sin(-3π/5) sin(4π/5)

=sin(π/5) sin(2π/5) sin(3π/5) sin(4π/5)

=5/16 なのでしたわ... ^^;

cos(2π/5) cos(4π/5) cos(6π/5) cos(8π/5)

=cos(2π/5) cos(4π/5) cos(-π/5) cos(-3π/5)

=cos(π/5) cos(2π/5) cos(3π/5) cos(4π/5)

=1/16

とこちらも同じでしたわ...

複素平面で考えたら...

x1〜x4(5の倍数でなければ)は同じでしたか...

大した問題じゃじぇんじぇんなかったあるね Orz...