Σ[k=1,n]sinkxsin(k+1)xなどの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^n \sin kx \sin (k+1)x=\frac{1}{4}\{(n+1) \sin 2x -\sin 2(n+1)x\} \csc x\\
&(2)  \displaystyle\sum_{k=1}^n \sin kx \sin (k+2)x=\frac{n}{2}\cos 2x -\frac{1}{2} \cos (n+3)x \sin nx \csc x\\
&(3)  \displaystyle\sum_{k=1}^n \sin kx \cos (2k-1)y\\
&=\frac{1}{2}\left\{\sin \left(ny+\frac{n+1}{2}x\right) \sin \frac{n(x+2y)}{2} \csc \frac{x+2y}{2}-\sin \left(ny-\frac{n+1}{2}x\right) \sin \frac{n(2y-x)}{2} \csc \frac{2y-x}{2}\right\}\\
\end{alignat}











<証明>

次の級数における等式を用います。[詳細はこちらです。(A)(B)(C)(D)]
\begin{alignat}{2}
&(A)  \displaystyle\sum_{k=1}^n \cos (2k-1)x=\frac{1}{2}\sin 2nx \csc x\\
&(B)  \displaystyle\sum_{k=1}^n \sin kx=\sin \frac{n+1}{2}x\sin \frac{nx}{2} \csc \frac{x}{2}\\
&(C)  \displaystyle\sum_{k=1}^n \cos kx=\cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}\\
&(D)  \displaystyle\sum_{k=0}^{n-1} \sin (x+ky)=\sin \left(x+\frac{n-1}{2}y\right)\sin \frac{ny}{2} \csc \frac{y}{2}
\end{alignat}






\begin{alignat}{2}
(1)  \displaystyle\sum_{k=1}^n \sin kx \sin (k+1)x&=-\frac{1}{2}\displaystyle\sum_{k=1}^n \{\cos (2k+1)x-\cos x\}\\
&=-\frac{1}{2}\displaystyle\sum_{k=1}^n\cos (2k+1)x+\frac{1}{2}\displaystyle\sum_{k=1}^n \cos x\\
&=-\frac{1}{2}\displaystyle\sum_{k=2}^{n+1}\cos (2k-1)x+\frac{n}{2}\cos x\\
&=\frac{1}{2}\cos x-\frac{1}{2}\displaystyle\sum_{k=1}^{n+1}\cos (2k-1)x+\frac{n}{2}\cos x\\
&=\frac{n+1}{2} \cos x-\frac{1}{4} \sin 2(n+1)x \csc x\\
&=\frac{n+1}{2} \cdot \frac{\sin x \cos x}{\sin x}-\frac{1}{4} \sin 2(n+1)x \csc x\\
&=\frac{n+1}{4} \sin 2x \csc x-\frac{1}{4} \sin 2(n+1)x \csc x\\
&=\frac{1}{4}\{(n+1) \sin 2x -\sin 2(n+1)x\} \csc x
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \sin kx \sin (k+1)x=\frac{1}{4}\{(n+1) \sin 2x -\sin 2(n+1)x\} \csc x$$








\begin{alignat}{2}
(2)  \displaystyle\sum_{k=1}^n \sin kx \sin (k+2)x&=-\frac{1}{2}\displaystyle\sum_{k=1}^n \{\cos (2k+2)x -\cos 2x\}\\
&=-\frac{1}{2}\displaystyle\sum_{k=1}^n \cos (2k+2)x+\frac{1}{2}\displaystyle\sum_{k=1}^n \cos 2x\\
&=\frac{n}{2}-\frac{1}{2}\displaystyle\sum_{k=1}^n \cos (2k+2)x\\
\end{alignat}右の級数を計算します。
\begin{alignat}{2}
\frac{1}{2}\displaystyle\sum_{k=1}^n \cos (2k+2)x&=-\frac{1}{2}\displaystyle\sum_{k=1}^n (\cos 2kx \cos 2x-\sin 2kx \sin 2x)\\
&=\cos 2x-\frac{1}{2}\displaystyle\sum_{k=1}^n \cos 2kx-\sin 2x -\frac{1}{2}\displaystyle\sum_{k=1}^n \sin 2kx\\
&=\cos 2x \cdot \cos (n+1)x \sin nx \csc x-\sin 2x \cdot \sin (n+1)x \sin nx \csc x\\
&\\
&=\{\cos (n+1)x \cos 2x-\sin (n+1)x \sin 2x\}\sin nx \csc x\\
&\\
&=\cos \{(n+1)+2\}x \sin nx \csc x=\cos (n+3)x \sin nx \csc x\\
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \sin kx \sin (k+2)x=\frac{n}{2}\cos 2x -\frac{1}{2} \cos (n+3)x \sin nx \csc x$$







\begin{alignat}{2}
&(3)  \displaystyle\sum_{k=1}^n \sin kx \cos (2k-1)y\\
&=\frac{1}{2}\displaystyle\sum_{k=1}^n [\sin \{kx+(2k-1)y\}+\sin\{kx-(2k-1)y\}]\\
&=\frac{1}{2}\displaystyle\sum_{k=1}^n [\sin \{-y+(x+2y)k\}+\sin\{y+(x-2y)\}]\\
&=-\frac{1}{2}\displaystyle\sum_{k=1}^n [\sin \{y-(x+2y)k\}-\sin\{y+(x-2y)\}]\\
\end{alignat}ここで次の \((D)\) を用います。$$\displaystyle\sum_{k=0}^{n-1} \sin (X+kY)=\sin \left(X+\frac{n-1}{2}Y\right)\sin \frac{nY}{2} \csc \frac{Y}{2}$$左辺を変形します。$$\displaystyle\sum_{k=0}^{n-1} \sin (X+kY)=\displaystyle\sum_{k=1}^{n} \sin \{X+(k-1)Y\}=\displaystyle\sum_{k=1}^n \sin (X-Y+kY)$$となるので$$\displaystyle\sum_{k=1}^n \sin (X-Y+kY)=\sin \left(X+\frac{n-1}{2}Y\right)\sin \frac{nY}{2} \csc \frac{Y}{2}  \cdots (E)$$

\((α)\) \((E)\) の式で \(X-Y=y,\,Y=-x-2y\) とします。\((X=-x-y)\)
\begin{alignat}{2}
\displaystyle\sum_{k=1}^n [\sin \{y-(x+2y)k\}&=\sin \left\{-x-y+\frac{n-1}{2}(-x-2y)\right\}\sin \frac{n(-x-2y)}{2}\csc \frac{x-2y}{2}\\
&=\sin \left(-x-y-\frac{nx}{2}+\frac{x}{2}-ny+y\right)\sin \frac{n(x+2y)}{2}\csc \frac{x+2y}{2}\\
&=\sin \left(-ny-\frac{nx}{2}-\frac{x}{2}\right)\sin \frac{n(x+2y)}{2}\csc \frac{x+2y}{2}\\
&=-\sin \left(ny+\frac{n+1}{2}x\right)\sin \frac{n(x+2y)}{2}\csc \frac{x+2y}{2}\\
\end{alignat}
\((β)\) \((E)\) の式で \(X-Y=y,\,Y=x-2y\) とします。\((X=x-y)\)
\begin{alignat}{2}
\displaystyle\sum_{k=1}^n [\sin \{y+(x-2y)k\}&=\sin \left\{x-y+\frac{n-1}{2}(x-2y)\right\}\sin \frac{n(x-2y)}{2}\csc \frac{x-2y}{2}\\
&=\sin \left(x-y+\frac{nx}{2}-\frac{x}{2}-ny+y\right)\sin \frac{n(2y-x)}{2}\csc \frac{2y-x}{2}\\
&=\sin \left(-ny+\frac{nx}{2}+\frac{x}{2}\right)\sin \frac{n(2y-x)}{2}\csc \frac{2y-x}{2}\\
&=-\sin \left(ny-\frac{n+1}{2}x\right)\sin \frac{n(2y-x)}{2}\csc \frac{2y-x}{2}\\
\end{alignat}これらの式を元の計算式に代入します。以上より$$\displaystyle\sum_{k=1}^n \sin kx \cos (2k-1)y=\frac{1}{2}\left\{\sin \left(ny+\frac{n+1}{2}x\right) \sin \frac{n(x+2y)}{2} \csc \frac{x+2y}{2}-\sin \left(ny-\frac{n+1}{2}x\right) \sin \frac{n(2y-x)}{2} \csc \frac{2y-x}{2}\right\}$$

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