ΣTn

ΣTncos2(n-1)α

ΣTnsin2(n-1)α

と合致するか、確認しておきたい

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Tn=sin(n+1)π/(N+1))sin(nπ/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))

α=π/(N+1)

Tn=sin(n+1)αsinnα/sinαsin2α

n=0のときTn=0

n=1のとき、Tn=sin(2π/(N+1))sin(π/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))＝1

n=2のとき、Tn=sin(3π/(N+1))sin(2π/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))＝-4{sin(π/(N+1))}^2+3

X=1+2cos(2π/(n+1))＝1+2（1-2{sin(π/(N+1))}^2）より

n=2のとき、Tn=sin(3π/(N+1))sin(2π/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))＝-4{sin(π/(N+1))}^2+3=X

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X=1+2cos2π/(N+1)

Tn=sin(n+1)π/(N+1))sin(nπ/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))

もう一度問題を整理しておきたい

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Σsinrxsin(r+1)x={(n+1)sin2x-sin2(n+1)x}/4sinx,r=1〜N-1を用いると

r=1〜N-1

Σsinrxsin(r+1)x={(n+1)sin2x-sin2(n+1)x}/4sinxに訂正

n=0とおくと0

n=1とおくと

{2sin2x-sin4x}/4sinxsinxsin2x={2sin2x-2sin2xcos2x}/4sinxsinxsin2x

={2-2cos(2x)}/4sinxsinx=1

n=2とおくと

{3sin2x-sin6x}/4sinxsinxsin2x={3sin2x+4(sin2x)^3-3sin2x}/4sinxsinxsin2x

=(sin2x)^2/(sinx)^2=4(cosx)^2

X=1+2cos2α=4(cosx)^2-1

{3sin2x-sin6x}/4sinxsinxsin2x={3sin2x+4(sin2x)^3-3sin2x}/4sinxsinxsin2x

=(sin2x)^2/(sinx)^2=4(cosx)^2=1+X

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α=π/(N+1)とおくと

X=1+2cos2α

Tn=sinnαsin(n+1)α/sinαsin2α

Σsinrxsin(r+1)x={(n+1)sin2x-sin2(n+1)x}/4sinx

r=1〜N-1

求めたいのはr=N-1とおいて、

ΣTn=Σsinrxsin(r+1)x={(N)sin2x-sin2Nx}/4sinxsinαsin2α

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