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

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=1+2{2(cosx)^2-1}=4(cosx)^2-1

X=1+2cos2α=1+2{1-2(sinx)^2-1}=3-4(sinx)^2

Tn=sinnαsin(n+1)α/sinαsin2α

Tn=-1/2{cos(2n+1)α-cosα｝/sinαsin2α

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一方、n=1からｎ項までの和

　　ｃｏｓα＋ｃｏｓ３α＋ｃｏｓ５α＋・・・＋ｃｏｓ（２ｎ−１）α＝ｓｉｎ２ｎα／２ｓｉｎα

を用いる場合、n項の和

Σcos(2n+1)α=cos3α+・・・+cos(2n+1)α=ｓｉｎ２ｎα／２ｓｉｎα+cos(2n+1)α-ｃｏｓα

ΣTn=-1/2{Σcos(2n+1)α-Σcosα｝/sinαsin2α

n=0のとき、0

n=1のとき

Σcos(2n+1)α=cos3α+・・・+cos(2n+1)α=ｓｉｎ２α／２ｓｉｎα+cos3α-ｃｏｓα

=cos3x

Σcosα=cosx

ΣTn=-1/2{Σcos(2n+1)α-Σcosα｝/sinαsin2α=cos3x/sinxsin2x

=-1/2{4(cosx)^3-4cosx}/2(sinx)^2cosx

=-1/4{4(cosx)^2-4}/(sinx)^2=1

n=2のとき

Σcos(2n+1)α=cos3α+・・・+cos(2n+1)α=ｓｉｎ4α／２ｓｉｎα+cos5α-ｃｏｓα

=4(cosx)~3-2cosx+16(cosx)^5-20(cosx)^3+5cosx-cosx

Σcosα=2cosx

ΣTn=-1/2{-16(cosx)~3+16(cosx)^5}/2(sinx)^2cosx

-4(cosx)^2(-1+(cosx)^2)/(sinx)^2

=4(cosx)^2=X+1

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ΣTn=-1/2{Σcos(2n+1)α-Σcosα｝/sinαsin2α

=-1/2{sin2nx/2sinx+cos(2n+1)x-(n+1)cosx}/sinxsin2x

求めたいのはn=N-1での

ΣTn=-1/2{sin2(N-1)x/2sinx+cos(2N-1)x-(N)cosx}/sinxsin2x

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