ΣTn

ΣTncos2(n-1)α

ΣTnsin2(n-1)α

と合致するか、確認しておきたい

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

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

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

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。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

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)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

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)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

n項まで

4sinxsin2xΣTncos2(n-1)α={-Σcos3x+Σcos(2n-3)x+Σcos(2n-1)x-Σcos(4n-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)

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

4sinxsin2xΣTncos2(n-1)α=-ncos3x+cos((n+1)x)sin(nx)(cos3x+cosx)/sin(x)+sin((n+1)x)sin(nx)(sin3x+sinx)/sin(x)

-cos(2(n+1)x)sin(2nx)cosx/sin(2x)-sin(2(n+1)x)sin(2nx)sinx/sin(2x)

4sinxsin2xΣTncos2(n-1)α=-ncos3x+cos((n+1)x)sin(nx)2(cos2xcosx)/sin(x)+sin((n+1)x)sin(nx)(2sin2xcosx)/sin(x)

-cos(2(n+1)x)sin(2nx)cosx/sin(2x)-sin(2(n+1)x)sin(2nx)sinx/sin(2x)

4sinxsin2xΣTnsin2(n-1)α={Σsin3x+Σsin(2n-3)x+Σsin(2n-1)x-Σsin(4n-1)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)

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

(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)

4sinxsin2xΣTnsin2(n-1)α=nsin3x+sin((n+1)x)sin(nx)(cos3x+cosx)/sin(x)-cos((n+1)x)sin(nx)(sin3x+sinx)/sin(x)

-sin(2(n+1)x)sin(2nx)cosx/sin(2x)+cos(2(n+1)x)sin(2nx)sinx/sin(2x)

4sinxsin2xΣTnsin2(n-1)α=nsin3x+sin((n+1)x)sin(nx)2(cos2xcosx)/sin(x)-cos((n+1)x)sin(nx)2(sin2xcosx)/sin(x)

-sin(2(n+1)x)sin(2nx)cosx/sin(2x)+cos(2(n+1)x)sin(2nx)sinx/sin(2x)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

α=π/(N+1)とおくと

求めたいのはr=N-1とおいて、

4sinxsin2xΣTn=Σsinrxsin(r+1)x={(N)sin2x-sin2Nx}/sinx

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

N=n-1

4sinxsin2xΣTncos2(n-1)α・sinx

=-(n-1)cos3xsinx+cos(nx)sin((n-1)x)2(cos2xcosx)+sin(nx)sin((n-1)x)(2sin2xcosx)

-cos(2nx)sin(2(n-1)x)sinxcosx/sin(2x)-sin(2nx)sin(2(n-1)x)(sinx)^2/sin(2x)

=-(n-1)cos3xsinx+2sin((n-1)x)cosx(cos(nx)cos2x+sin(nx)sin2x)

-sin(2(n-1)x)sinx(cos(2nx)cosx+sin(2nx)cosx(sinx))/sin(2x)

=-(n-1)cos3xsinx+2sin((n-1)x)(cos((n-2)x)cosx

-sin(2(n-1)x)sinx(cos((2n-1)x)/sin(2x)

4sinxsin2xΣTnsin2(n-1)α・sinx

=(n-1)sin3xsinx+sin(nx)sin((n-1)x)2(cos2xcosx)-cos(nx)sin((n-1)x)2(sin2xcosx)

-sin(2nx)sin(2(n-1)x)cosxsinx/sin(2x)+cos(2n)x)sin(2(n-1)x)(sinx)^2/sin(2x)

=(n-1)sin3xsinx+2sin((n-1)x)cosx(sin(nx)cos2x)-cos(nx)sin2x)

-sin(2(n-1)x)sinx{sin(2nx)cosx-cos(2nx)sinx}/sin(2x)

=(n-1)sin3xsinx+2sin((n-1)x)(sin((n-2)x)cosx

-sin(2(n-1)x)sinx{sin(2n-1)x}/sin(2x)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

ここで、x=π/(N+1)であるから

sin2Nx=sin(2(N+1)-2)x=-sin2x

{(N)sin2x-sin2Nx}=(n+1)sin2x

4sinxsin2xΣTncos2(n-1)α・sinx

=-(n-1)cos3xsinx+2sin((n+1-2)x)(cos((n+1-3)x)cosx-sin(2(n+1)-4)x)sinx(cos((2n+2)-3)x)/sin(2x)

=-(n-1)cos3xsinx-2sin(2x)cos(3x)cosx+sin(4x)sinx(cos(3x)/sin(2x)

=-(n-1)cos3xsinx-2sin(2x)cos(3x)cosx+2cos(2x)sinx(cos(3x)

4sinxsin2xΣTnsin2(n-1)α・sinx

=(n-1)sin3xsinx+2sin((n+1)-2)x)(sin((n+1-3)x)cosx-sin(2n+2-4)x)sinx{sin(2n+2-3)x}/sin(2x)

=(n-1)sin3xsinx+2sin(2x)(sin(3x)cosx-sin(4x)sinx{sin(3x)}/sin(2x)

=(n-1)sin3xsinx+2sin(2x)(sin(3x)cosx-2cos(2x)sinx{sin(3x)}

両者はだいぶ接近してきたようだが・・・

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝