■正単体の正多角形断面(その33)
4sinxsin2xTncos2(n-1)α={-cos3x+cos(2n-3)x+cos(2n-1)x-cos(4n-1)x}
4sinxsin2xTnsin2(n-1)α={sin3x+sin(2n-3)x+sin(2n-1)x-sin(4n-1)x}
4sinxsin2xΣTncos2(n-1)α={-Σcos3x+Σcos(2n-3)x+Σcos(2n-1)x-Σcos(4n-1)x}
4sinxsin2xΣTnsin2(n-1)α={Σsin3x+Σsin(2n-3)x+Σsin(2n-1)x-Σsin(4n-1)x}
{-Σcos3x+Σcos(2n-3)x+Σcos(2n-1)x-Σcos(4n-1)x}^2+{Σsin3x+Σsin(2n-3)x+Σsin(2n-1)x-Σsin(4n-1)x}^2
d^2={(ΣTncos2(n-1)α)^2+(ΣTnsin2(n-1)α)}^2/(ΣTn)^2
d^2={(4sinxsin2xΣTncos2(n-1)α)^2+(4sinxsin2xΣTnsin2(n-1)α)}^2/(4sinxsin2xΣTn)^2
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4sinαsin2αΣTn・sinα=Σsinrxsin(r+1)x={(N)sin2x-sin2Nx}=(n+1)sin2α
4sinαsin2αΣTncos2(n-1)α・sinα=-(n+1)cos3xsinx
4sinαsin2αΣTnsin2(n-1)α・sinα=(n+1)sins3xsinx
XX=-cos3xsinx/sin2x
YY=sin3xsinx/sin2x
XX^2+YY^2=(sinx/sin2x)^2=1/(2cosx)^2
d^2→1/2
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検算してみたい。
n=2のとき、cosx=1/2→ d^2=1/4・4=1
n=3のとき、cosx=√2/2→ d^2=1/4・2=1/2
n=4のとき、cosx=τ/2→ d^2=1/4・4/τ^2=1/τ^2
n=5のとき、cosx=√3/2→ d^2=1/4・4/3=1/3
すべて一致した。
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