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))

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)

Σsin2rx=sin((n+1)x)sin(nx)/sin(x)

Σcos2rx=cos((n+1)x)sin(nx)/sin(x)

Σsin4rx=sin(2(n+1)x)sin(2nx)/sin(2x)

Σcos4rx=cos(2(n+1)x)sin(2nx)/sin(2x)

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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

=

(Σcos3x)^2+(Σsin3x)^2

(Σcos(2n-3)x)^2+(Σsin(2n-3)x)^2

(Σcos(2n-1)x)^2+(Σsin(2n-1)x)^2

(Σcos(4n-1)x)^2+(Σsin(4n-1)x)^2

-2(Σcos3x)(Σcos(2n-3)x)+2(Σsin3x)(Σsin(2n-3)x)

-2(Σcos3x)(Σcos(2n-1)x)+2(Σsin3x)(Σsin(2n-1)x)

+2(Σcos3x)(Σcos(2n-1)x)-2(Σsin3x)(Σsin(2n-1)x)

+2(Σcos(2n-3)x)(Σcos(2n-1)x)+2(Σsin(2n-3)x)(Σsin(2n-1)x)

-2(Σcos(2n-3)x)(Σcos(4n-1)x)-2(Σsin(2n-3)x)(Σsin(2n-1)x)

-2(Σcos(2n-1)x)(Σcos(4n-1)x)-2(Σsin(2n-1)x)(Σsin(2n-1)x)

(Σcos(2n-3)x)=(Σcos2nxcos3x+Σsin2nxsin3x)=cos((n+1)x)sin(nx)cos3x/sin(x)+sin((n+1)x)sin(nx)sin3x/sin(x)

(Σsin(2n-3)x)=(Σsin2nxcos3x-Σcos2nxsin3x)=sin((n+1)x)sin(nx)cos3x/sin(x)-cos((n+1)x)sin(nx)sin3x/sin(x)

(Σcos(2n-1)x)=(Σcos2nxcosx+Σsin2nxsinx)=cos((n+1)x)sin(nx)cosx/sin(x)+sin((n+1)x)sin(nx)sinx/sin(x)

(Σsin(2n-1)x)=(Σsin2nxcosx-Σcos2nxsinx)=sin((n+1)x)sin(nx)cosx/sin(x)-cos((n+1)x)sin(nx)sinx/sin(x)

(Σcos(4n-1)x)=(Σcos4nxcosx+Σsin4nxsinx)=cos(2(n+1)x)sin(2nx)cosx/sin(2x)+sin(2(n+1)x)sin(2nx)sinx/sin(2x)

(Σsin(4n-1)x)=(Σsin4nxcosx-Σcos4nxsinx)=sin(2(n+1)x)sin(2nx)cosx/sin(2x)-cos(2(n+1)x)sin(2nx)sinx/sin(2x)

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