Σ[k=1,n]sin(4k-1)xなどの級数

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




全て、三角関数の加法定理で式を展開し、

\((A)(B)\) の式を \(x \to 4x\) または \(x \to 3x\) として用います。

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







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







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







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

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