Σ[k=0,n]coskxなどの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^n \sin kx=\sin \frac{n+1}{2}x\sin \frac{nx}{2} \csc \frac{x}{2}\\
&(2)  \displaystyle\sum_{k=0}^n \cos kx=
\begin{cases}
\displaystyle \cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}+1\\
\displaystyle \sin \frac{n+1}{2}x \cos \frac{nx}{2} \csc \frac{x}{2}\\
\displaystyle \frac{1}{2}\left\{1+\sin \left( n+\frac{1}{2}\right)x \csc \frac{x}{2}\right\}\\
\end{cases}
\end{alignat}









<証明>

オイラーの公式を用います。
\begin{alignat}{2}
\displaystyle\sum_{k=0}^n e^{ikx}&=\displaystyle\sum_{k=0}^n (\cos kx+i \sin kx)\\
&=\displaystyle\sum_{k=0}^n \cos kx+i\displaystyle\sum_{k=1}^n \sin kx  \cdots (A)\\
\end{alignat}左辺について、等比数列の和の公式を用います。
\begin{alignat}{2}
\displaystyle\sum_{k=0}^n e^{ikx}&=\frac{1-(e^{ix})^{n+1}}{1-e^{ix}}=\frac{1-e^{i(n+1)x}}{1-e^{ix}}\\
&=\frac{1-\{\cos (n+1)x +i \sin (n+1)x\}}{1-(\cos x+i \sin x)}\\
&=\frac{1-\cos (n+1)x -i \sin (n+1)x}{1-\cos x-i \sin x}\\
\end{alignat}分母の虚数を払い、実部と虚部に分けます。
\begin{alignat}{2}
&=\frac{\{1-\cos (n+1)x -i \sin (n+1)x\}(1- \cos x+i \sin x)}{(1-\cos x-i \sin x)(1- \cos x+i \sin x)}\\
&=\frac{\{1- \cos (n+1)x\}(1- \cos x)+\sin (n+1)x \sin x}{(1- \cos x)^2+\sin^2 x}+i \cdot \frac{\{1- \cos (n+1)x\}\sin x-(1-\cos x) \sin (n+1)x}{(1- \cos x)^2+\sin^2 x}\\
\end{alignat}\((A)\) の式と比較することで
\begin{alignat}{2}
\displaystyle\sum_{k=1}^n \sin kx&=\frac{\{1- \cos (n+1)x\}\sin x-(1-\cos x) \sin (n+1)x}{(1- \cos x)^2+\sin^2 x}\\
\\
\displaystyle\sum_{k=0}^n \cos kx&=\frac{\{1- \cos (n+1)x\}(1- \cos x)+\sin (n+1)x \sin x}{(1- \cos x)^2+\sin^2 x}\\
\end{alignat}を得ます。





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







\begin{alignat}{2}
(2)  \displaystyle\sum_{k=0}^n \cos kx&=\frac{\{1- \cos (n+1)x\}(1- \cos x)+\sin (n+1)x \sin x}{(1- \cos x)^2+\sin^2 x}\\
&=\frac{1-\cos x- \cos (n+1) x+\cos (n+1) \cos x+\sin (n+1) x \sin x}{1-2 \cos x+\cos^2 x+\sin^2 x}\\
&=\frac{1- \cos x- \cos (n+1) x+\cos \{(n+1)-1\}x}{2(1-\cos x)}\\
&=\frac{1- \cos x- \cos (n+1) x+\cos nx}{2(1-\cos x)}=-\frac{\cos (n+1) x-\cos nx}{2(1- \cos x)}+\frac{1}{2}\\
&=\frac{2 \sin \left(n+\frac{1}{2}\right)x \sin \frac{x}{2}}{4 \sin^2 \frac{x}{2}}+\frac{1}{2}=\frac{\sin \left(n+\frac{1}{2}\right)x}{2 \sin \frac{x}{2}}+\frac{1}{2}\\
&=\frac{1}{2}\left\{1+\frac{\sin \left(n+\frac{1}{2}\right)x}{\sin \frac{x}{2}}\right\}=\frac{1}{2}\left\{1+\sin \left( n+\frac{1}{2}\right)x \csc \frac{x}{2}\right\}
\end{alignat}以上より$$\displaystyle\sum_{k=0}^n \cos kx=\frac{1}{2}\left\{1+\sin \left( n+\frac{1}{2}\right)x \csc \frac{x}{2}\right\}  \cdots (B)$$





\((B)\) の式を変形します。
\begin{alignat}{2}
\displaystyle\sum_{k=0}^n \cos kx&=\frac{1}{2}\left\{1+\sin \left( n+\frac{1}{2}\right)x \csc \frac{x}{2}\right\}=\frac{1}{2}\cdot \frac{\sin \left(n+\frac{1}{2}\right)x+\sin \frac{x}{2}}{\sin \frac{x}{2}}\\
&=\frac{\sin \frac{n+1}{2}x \cos \frac{nx}{2}}{\sin \frac{x}{2}}=\sin \frac{n+1}{2}x \cos \frac{nx}{2} \csc \frac{x}{2}
\end{alignat}以上より$$\displaystyle\sum_{k=0}^n \cos kx=\sin \frac{n+1}{2}x \cos \frac{nx}{2} \csc \frac{x}{2}$$





\((B)\) の式を変形します。
\begin{alignat}{2}
\displaystyle\sum_{k=0}^n \cos kx&=\frac{1}{2}\left\{1+\sin \left( n+\frac{1}{2}\right)x \csc \frac{x}{2}\right\}=\frac{\sin \left(n+\frac{1}{2}\right)x}{2 \sin \frac{x}{2}}+\frac{1}{2}\\
&=\frac{\sin \left(n+\frac{1}{2}\right)x}{2 \sin \frac{x}{2}}-\frac{1}{2}+1=\frac{\sin \left(n+\frac{1}{2}\right)x – \sin \frac{x}{2}}{2 \sin \frac{x}{2}}+1\\
&=\frac{\cos \frac{n+1}{2}x \sin \frac{nx}{2}}{\sin \frac{x}{2}}+1=\displaystyle \cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}+1
\end{alignat}以上より$$\displaystyle\sum_{k=0}^n \cos kx=\displaystyle \cos \frac{n+1}{2}x \sin \frac{nx}{2} \csc \frac{x}{2}+1$$







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