Σ[k=1,n]sin^{3}kxなどの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^n \sin^2 kx=
\begin{cases}
\displaystyle \frac{1}{2}n-\frac{1}{2}\cos (n+1)x \sin nx \csc x\\
\displaystyle \frac{1}{4} \{(2n+1)\sin x-\sin (2n+1)x\}\csc x\\
\end{cases}\\
&(2)  \displaystyle\sum_{k=1}^n \cos^2 kx=
\begin{cases}
\displaystyle \frac{1}{2}n+\frac{1}{2}\cos (n+1)x \sin nx \csc x\\
\displaystyle \frac{n-1}{2}+\frac{1}{2}\sin (n+1)x \cos nx \csc x\\
\end{cases} \\
&(3)  \displaystyle\sum_{k=1}^n \sin^3 kx=\frac{3}{4}\sin \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}-\frac{1}{4} \sin \frac{3(n+1)x}{2}\sin \frac{3nx}{2}\csc \frac{3x}{2}\\
&(4)  \displaystyle\sum_{k=1}^n \cos^3 kx=\frac{3}{4}\cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}+\frac{1}{4} \cos \frac{3(n+1)x}{2}\sin \frac{3nx}{2}\csc \frac{3x}{2}\\
&(5)  \displaystyle\sum_{k=1}^n \sin^4 kx=\frac{1}{8}\{3n-4 \cos (n+1)x \sin nx \csc x+\cos 2(n+1)x \sin 2nx \csc 2x\}\\
&(5)  \displaystyle\sum_{k=1}^n \cos^4 kx=\frac{1}{8}\{3n+4 \cos (n+1)x \sin nx \csc x+\cos 2(n+1)x \sin 2nx \csc 2x\}\\
\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。)
\begin{alignat}{2}
&(A)  \displaystyle\sum_{k=1}^n \sin kx=\sin \frac{n+1}{2}x\sin \frac{nx}{2} \csc \frac{x}{2}\\
&(B)  \displaystyle\sum_{k=1}^n \cos kx=
\begin{cases}
\displaystyle \cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}\\
\displaystyle \sin \frac{n+1}{2}x \cos \frac{nx}{2} \csc \frac{x}{2}-1\\
\end{cases}
\end{alignat}







全て三角関数の次数を下げて \((A)(B)\) の式を用います。
\begin{alignat}{2}
(1)  \displaystyle\sum_{k=1}^n \sin^2 kx&=\frac{1}{2}\displaystyle\sum_{k=1}^n (1- \cos 2kx)\\
&=\frac{1}{2}n -\frac{1}{2}\displaystyle\sum_{k=1}^n \cos 2kx\\
&=\frac{1}{2}n-\frac{1}{2}\cos (n+1)x \sin nx \csc x\\
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \sin^2 kx=\frac{1}{2}n-\frac{1}{2}\cos (n+1)x \sin nx \csc x$$
そのまま上記の式を変形します。
\begin{alignat}{2}
\displaystyle\sum_{k=1}^n \sin^2 kx&=\frac{1}{2}\{n \sin x- \cos (n+1)x \sin nx\}\csc x\\
&=\frac{1}{4}\{2n \sin x- 2\cos (n+1)x \sin nx\}\csc x\\
&=\frac{1}{4}[2n \sin x-\{\sin (2n+1)x – \sin x\}]\csc x\\
&=\frac{1}{4}[2n \sin x-\sin (2n+1)x +\sin x\}]\csc x\\
&=\frac{1}{4} \{(2n+1)\sin x-\sin (2n+1)x\}\csc x\\
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \sin^2 kx=\frac{1}{4} \{(2n+1)\sin x-\sin (2n+1)x\}\csc x$$







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







\begin{alignat}{2}
(3)   \displaystyle\sum_{k=1}^n \sin^3 kx&=\frac{1}{4}\displaystyle\sum_{k=1}^n (3 \sin kx- \sin 3kx )\\
&=\frac{3}{4}\displaystyle\sum_{k=1}^n \sin kx-\frac{1}{4}\displaystyle\sum_{k=1}^n \sin 3kx\\
&=\frac{3}{4}\sin \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}-\frac{1}{4} \sin \frac{3(n+1)x}{2}\sin \frac{3nx}{2}\csc \frac{3x}{2}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \sin^3 kx=\frac{3}{4}\sin \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}-\frac{1}{4} \sin \frac{3(n+1)x}{2}\sin \frac{3nx}{2}\csc \frac{3x}{2}$$







\begin{alignat}{2}
(4)   \displaystyle\sum_{k=1}^n \cos^3 kx&=\frac{1}{4}\displaystyle\sum_{k=1}^n (3 \cos kx+ \cos 3kx )\\
&=\frac{3}{4}\displaystyle\sum_{k=1}^n \cos kx+\frac{1}{4}\displaystyle\sum_{k=1}^n \cos 3kx\\
&=\frac{3}{4}\cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}+\frac{1}{4} \cos \frac{3(n+1)x}{2}\sin \frac{3nx}{2}\csc \frac{3x}{2}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \cos^3 kx=\frac{3}{4}\cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}+\frac{1}{4} \cos \frac{3(n+1)x}{2}\sin \frac{3nx}{2}\csc \frac{3x}{2}$$







\begin{alignat}{2}
(5)  \displaystyle\sum_{k=1}^n \sin^4 kx&=\frac{1}{8}\displaystyle\sum_{k=1}^n (3-4 \cos 2kx+ \cos 4kx)\\
&=\frac{1}{8}\left(3n-4\displaystyle\sum_{k=1}^n \cos 2kx+\displaystyle\sum_{k=1}^n \cos 4kx \right)\\
&=\frac{1}{8}\{3n-4 \cos (n+1)x \sin nx \csc x+\cos 2(n+1)x \sin 2nx \csc 2x\}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \sin^4 kx=\frac{1}{8}\{3n-4 \cos (n+1)x \sin nx \csc x+\cos 2(n+1)x \sin 2nx \csc 2x\}$$







\begin{alignat}{2}
(6)  \displaystyle\sum_{k=1}^n \cos^4 kx&=\frac{1}{8}\displaystyle\sum_{k=1}^n (3+4 \cos 2kx+ \cos 4kx)\\
&=\frac{1}{8}\left(3n+4\displaystyle\sum_{k=1}^n \cos 2kx+\displaystyle\sum_{k=1}^n \cos 4kx \right)\\
&=\frac{1}{8}\{3n+4 \cos (n+1)x \sin nx \csc x+\cos 2(n+1)x \sin 2nx \csc 2x\}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \cos^4 kx=\frac{1}{8}\{3n+4 \cos (n+1)x \sin nx \csc x+\cos 2(n+1)x \sin 2nx \csc 2x\}$$

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