正八角形の場合

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x6=1/(X^5-3X^4+X^3+3X^2+2)

x5=X/(X^5-3X^4+X^3+3X^2+2)

x4={X^2-X}/(X^5-3X^4+X^3+3X^2+2)

x3={X^3-2X^2+1}/(X^5-3X^4+X^3+3X^2+2)

x2={X^4-3X^3+X^2+2X}/(X^5-3X^4+X^3+3X^2+2)

x1=(X^5-4X^4+3X^3+3X^2-2X)/(X^5-3X^4+X^3+3X^2+2)

この中のいくつかは等しいと思われるが・・・

(X-1)/2=cos(360/8)=cosθ=√2/2

cos(2θ)=2{(X-1)/2}^2-1=(X^2-2X+1)/2-1=(X^2-2X-1)/2=0

cos(3θ)=4{(X-1)/2}^3-3{(X-1)/2}=(X-1)^3/2-(3X-3)/2=(X^3-3X^2+3X-1-3X+3)/2=(X^3-3X^2+2)/2=-cosθ=-(X-1)/2

cos(4θ)=8{(X-1)/2}^4-8{(X-1)/2}^2+1=(X-1)^4/2-4(X-1)^2/2+1=(X^4-4X^3+6X^2-4X+1)/2-4(X^2-2X+1)/2+2/2=(X^4-4X^3+2X^2+4X-1)/2＝-1

cos(5θ)=16{(X-1)/2}^5-20{(X-1)/2}^3+5{(X-1)/2}=(X-1)^5/2-5(X-1)^3/2+5{(X-1)/2}=(X^5-5X^4+10X^3-10X^2+5X-1)/2-5(X^3-3X^2+3X-1)/2+5(X-1)/2=(X^5-5X^4+5X^3+5X^2-5X-1)/2)＝cos3θ

X^2=2X+1

(X^3-3X^2+2)=-(X-1)

X^3=3X^2-X-1=5X+2

(X^4-4X^3+2X^2+4X+1)＝0

X^4=4X^3-2X^2-4X-1=12X+5

(X^5-5X^+5X^3+5X^2-5X-1)=-(X-1)

X^5=5X^4-5X^3-5X^2+4X+2=29X+12

x6=1/(4X+4)

x5=X/(4X+4)

x4={X+1}/(4X+4)

x3={X+1}/(4X+4)

x2={X}/(4X+4)

x1=(1)/(4X+4)

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x1+x3+x5+x7=(2x+2)/(4X+4)=1/2

x2+x4+x6+x8=(2x+2)/(4X+4)=1/2

やはり偶数角形では問題はないようである。

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