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))

Tn=-1/2{cos(2n+1)α-cosα｝/sinαsin2α

次の課題は

Tncos2(n-1)α

Tnsin2(n-1)α

を求めることである。

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Tncos2(n-1)α=sin(n+1)αsinnαcos(2n-2)α/sinαsin2α

=1/4{-cos3x+cos(2n-3)x+cos(2n-1)x-cos(4n-1)x}/sinxsin2x

=1/4{-cos3x+cos2nxcos3x+sin2nxsin3x+cos2nxcosx+sin2nxsinx-cos4nxcosx-sin4nxsinx}/sinxsin2x

Tnsin2(n-1)α=sin(n+1)αsinnαsin(2n-2)α/sinαsin2α

=1/4{sin3x+sin(2n-3)x+sin(2n-1)x-sin(4n-1)x}/sinxsin2x

=1/4{sin3x+sin2nxcos3x-cos2nxsin3x+sin2nxcosx-cos2nxsinx-sin4nxcosx+cos4nxsinx}/sinxsin2x

Σsinrx=sinx+・・・+sinnx=sin((n+1)x/2)sin(nx/2)/sin(x/2)

Σcosrx=cosx+・・・+cosnx=cos((n+1)x/2)sin(nx/2)/sin(x/2)

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