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

α=π/(N+1)とおくと

X=1+2cos2α

Tn=sinnαsin(n+1)α/sinαsin2α

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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={Nsin2x-sin2Nx}/4sinx

x=αとおくと

ΣTn={Nsin2α-sin2Nα}/4sinαsinαsin2α

Sn=Tn/ΣTn=4sinαsinnαsin(n+1)α/{Nsin2α-sin2Nα}

Sncos(n-1)θ=Sncos2(n-1)α

Snsin(n-1)θ=Snsin2(n-1)α

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