■正五角形と正十七角形(その24)
[1]
cosπ/7+cos3π/7+cos5π/7=1/2
cosπ/7・cos3π/7・cos5π/7=−1/8
[2]
−sin(π/7)+sin(3π/7)+sin(5π/7)=(√7)/2
sinπ/7・sin3π/7・sin5π/7=√7/8
であるが,
sinπ/7+sin3π/7+sin5π/7=?
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正弦・余弦の和公式において,α=π/(2n+1)とおくと,
Σsin(2k−1)π/(2n+1)=sin^2nπ/(2n+1)/sinπ/(2n+1)
sinπ/7+sin3π/7+sin5π/7=sin^23π/7/sinπ/7
としか書き様がない.
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しかし,
64x^6−112x^4+56x^2−7=0
はx^2に関する実質的3次方程式であるから,根と係数の関係より,
(sinπ/7)^2+(sin3π/7)^2+(sin5π/7)^2=112/64
(sinπ/7)^2(sin3π/7)^2+(sin3π/7)^2(sin5π/7)^2+(sin5π/7)^2(sinπ/7)^2=56/64
(sinπ/7)^2(sin3π/7)^2(sin5π/7)^2=7/64
となることがわかる.
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