$\sum_{k=1}^n \sin kx ,\quad \sum_{k=1}^n \cos kx$


1　正弦の和(1)

$\qquad \sin x +\sin 2x + \cdots + \sin nx = \cfrac{\sin \dfrac{n+1}{2}x \sin \dfrac{n}{2}x}{\sin \dfrac{x}{2}}$


$証明$

$\sin \cfrac{x}{2}\sin kx=-\cfrac{1}{2}\big\{\cos (\cfrac{1}{2}+k)x-\cos (\cfrac{1}{2}-k)x\big\} \quad より$

$2\sin \cfrac{x}{2}\sin kx=\cos (k-\cfrac{1}{2})x - \cos (k + \cfrac{1}{2})x　\quad において$

$\quad k=1 \quad とおくと \qquad 2\sin \cfrac{x}{2}\sin x=\cos \cfrac{1}{2} x - \cos \cfrac{3}{2}x$
$\quad k=2 \quad とおくと \qquad 2\sin \cfrac{x}{2}\sin 2x=\cos \cfrac{3}{2} x - \cos \cfrac{5}{2}x$
$\hspace{5em} \vdots $
$\quad k=n \quad とおくと \qquad 2\sin \cfrac{x}{2}\sin nx=\cos (n-\cfrac{1}{2}) x - \cos (n+\cfrac{1}{2})x$

$辺々加えて$

$\quad 2\sin \cfrac{x}{2}(\sin x +\sin 2x + \cdots + \sin nx )= \cos \cfrac{1}{2}x - \cos (n+\cfrac{1}{2})x = -2\sin \cfrac{n+1}{2}x \sin (-\cfrac{n}{2}x) $

$\quad \therefore \ \ \sin x +\sin 2x + \cdots + \sin nx = \cfrac{\sin \dfrac{n+1}{2}x \sin \dfrac{n}{2}x}{\sin \dfrac{x}{2}}$


2　余弦の和(1)

$\qquad \cos x +\cos 2x + \cdots + \cos nx = \cfrac{\sin (n+ \dfrac{1}{2})x}{2\sin \dfrac{x}{2}}-\cfrac{1}{2}= \cfrac{\cos \dfrac{n+1}{2}x \sin \dfrac{n}{2}x}{\sin \dfrac{x}{2}}$


$証明$

$\sin \cfrac{x}{2}\cos kx=\cfrac{1}{2}\big\{\sin (\cfrac{1}{2}+k)x + \sin (\cfrac{1}{2}-k)x\big\} \quad より$

$2\sin \cfrac{x}{2}\cos kx=-\sin (k-\cfrac{1}{2})x + \sin (k + \cfrac{1}{2})x　\quad において$

$\quad k=1 \quad とおくと \qquad 2\sin \cfrac{x}{2}\cos x=-\sin \cfrac{1}{2} x + \sin \cfrac{3}{2}x$
$\quad k=2 \quad とおくと \qquad 2\sin \cfrac{x}{2}\cos 2x=-\sin \cfrac{3}{2} x + \sin \cfrac{5}{2}x$
$\hspace{5em} \vdots $
$\quad k=n \quad とおくと \qquad 2\sin \cfrac{x}{2}\cos nx=-\sin (n-\cfrac{1}{2}) x + \sin (n+\cfrac{1}{2})x$

$辺々加えて$

$\quad 2\sin \cfrac{x}{2}(\cos x +\cos 2x + \cdots + \cos nx )= \sin (n+\cfrac{1}{2})x - \sin \cfrac{1}{2}x \hspace{5em}①$

$\quad \therefore \ \ \cos x +\cos 2x + \cdots + \cos nx = \cfrac{\sin (n+ \dfrac{1}{2})x}{2\sin \dfrac{x}{2}}-\cfrac{1}{2}$

$①で右辺の和を積に変形すると \quad 右辺=2\cos \cfrac{n+1}{2}x \sin \cfrac{n}{2}x \quad だから$

$\quad \therefore \ \ \cos x +\cos 2x + \cdots + \cos nx = \cfrac{\cos \dfrac{n+1}{2}x \sin \dfrac{n}{2}x}{\sin \dfrac{x}{2}}$


$(1),(2)は、複素数を用いて、次のように証明することもできます。$

$z=\cos \theta +i\sin \theta \quad とおくと$
\begin{eqnarray*} \sum _{k=1}^n z^k &=&\cfrac{z(1-z^n)}{1-z}\\ \\ &=&\cfrac{(\cos \theta +i\sin \theta )\{(1-\cos n\theta) - i\sin n\theta\}}{(1-\cos \theta)-i\sin \theta} \\ \\ &=&\cfrac{(\cos \theta +i\sin \theta )\big(2\sin ^2 \dfrac{n}{2}\theta - 2i\sin \dfrac{n}{2}\theta\cos \dfrac{n}{2}\theta\big)}{2\sin ^2 \dfrac{1}{2}\theta-2i\sin \dfrac{1}{2}\theta\cos \dfrac{1}{2}\theta} \\ \\ &=&\cfrac{\sin \dfrac{n}{2}\theta}{\sin \dfrac{1}{2}\theta} \cdot \cfrac{(\cos \theta +i\sin \theta )\big(\sin \dfrac{n}{2}\theta - i\cos \dfrac{n}{2}\theta\big)}{\sin \dfrac{1}{2}\theta -i\cos \dfrac{1}{2}\theta} \\ \\ & &\hspace{3em}分母・分子にiをかけて\\ \\ &=&\cfrac{\sin \dfrac{n}{2}\theta}{\sin \dfrac{1}{2}\theta} \cdot \cfrac{(\cos \theta +i\sin \theta )\big(\cos \dfrac{n}{2}\theta + i\sin \dfrac{n}{2}\theta\big)}{\cos \dfrac{1}{2}\theta +i\sin \dfrac{1}{2}\theta} \\ \\ &=&\cfrac{\sin \dfrac{n}{2}\theta}{\sin \dfrac{1}{2}\theta} \big(\cos (\theta + \dfrac{n}{2}\theta - \dfrac{1}{2}\theta) + i\sin (\theta + \dfrac{n}{2}\theta - \dfrac{1}{2}\theta) \big)\\ \\ &=&\cfrac{\sin \dfrac{n}{2}\theta}{\sin \dfrac{1}{2}\theta} \big(\cos \dfrac{n+1}{2}\theta + i\sin \dfrac{n+1}{2}\theta \big)\\ \end{eqnarray*} $一方左辺は、ド・モアブルの定理より$

\[\sum_{k=1}^n z^k=\sum_{k=1}^n (\cos \theta +i\sin \theta )^k=\sum_{k=1}^n (\cos k\theta +i\sin k\theta )=\sum_{k=1}^n \cos k\theta +i \sum_{k=1}^n \sin k\theta \] $実部と虚部を比べて$
\[\sum_{k=1}^n \cos k\theta =\cfrac{\sin \dfrac{n}{2}\theta}{\sin \dfrac{1}{2}\theta} \cos \dfrac{n+1}{2}\theta ,\qquad \sum_{k=1}^n \sin k\theta =\cfrac{\sin \dfrac{n}{2}\theta}{\sin \dfrac{1}{2}\theta} \sin \dfrac{n+1}{2}\theta \]

3　正弦の和(2)

$\qquad \sin x +\sin 3x + \cdots + \sin (2n-1)x = \cfrac{1-\cos 2nx}{2\sin x}$


$証明$

$\sin x \sin (2k-1)x=-\cfrac{1}{2}\{\cos 2kx- \cos (-2k+2)x \} \quad より$

$2\sin x \sin (2k-1)x=\cos (2k-2)x- \cos 2kx \quad において$

$\quad k=1 \quad とおくと \qquad 2\sin x \sin x=\cos 0 x - \cos 2x$
$\quad k=2 \quad とおくと \qquad 2\sin x \sin 3x=\cos 2x - \cos 4x$
$\hspace{5em} \vdots $
$\quad k=n \quad とおくと \qquad 2\sin x\sin (2n-1)x=\cos (2n-2) x - \cos 2nx$

$辺々加えて$

$\quad 2\sin x(\sin x +\sin 3x + \cdots + \sin (2n-1)x = 1 - \cos 2nx$

$\quad \therefore \ \ \sin x +\sin 3x + \cdots + \sin (2n-1)x = \cfrac{1-\cos 2nx}{2\sin x}$


4　余弦の和(2)

$\qquad \cos x +\cos 3x + \cdots + \cos (2n-1)x = \cfrac{\sin 2nx}{2\sin x}$


$証明$

$\sin x \cos (2k-1)x=\cfrac{1}{2}\{\sin 2kx + \sin (-2k+2)x \} \quad より$

$2\sin x \cos (2k-1)x=-\sin (2k-2)x + \sin 2kx \quad において$

$\quad k=1 \quad とおくと \qquad 2\sin x \cos x=-\sin 0 + \sin 2x$
$\quad k=2 \quad とおくと \qquad 2\sin x \cos 3x=-\sin 2x + \sin 4x$
$\hspace{5em} \vdots $
$\quad k=n \quad とおくと \qquad 2\sin x\cos (2n-1)x=-\sin (2n-2) x + \sin 2nx$

$辺々加えて$

$\quad 2\sin x(\cos x +\cos 3x + \cdots + \cos (2n-1)x ) = \sin 2nx$

$\quad \therefore \ \ \cos x +\cos 3x + \cdots + \cos (2n-1)x = \cfrac{\sin 2nx}{2\sin x}$


5　正弦の和(3)

$\qquad \sin 2x +\sin 4x + \cdots + \sin 2nx = \cfrac{\sin (n+1)x\sin nx}{\sin x}$


$証明$

$\sin x \sin 2kx=-\cfrac{1}{2}\{\cos (2k+1)x- \cos (2k-1)x \} \quad より$

$2\sin x \sin 2kx=\cos (2k-1)x- \cos (2k+1)x \quad において$

$\quad k=1 \quad とおくと \qquad 2\sin x \sin 2x=\cos x - \cos 3x$
$\quad k=2 \quad とおくと \qquad 2\sin x \sin 4x=\cos 3x - \cos 5x$
$\hspace{5em} \vdots $
$\quad k=n \quad とおくと \qquad 2\sin x\sin 2nx=\cos (2n-1) x - \cos (2n+1)x$

$辺々加えて$

$\quad 2\sin x(\sin 2x +\sin 4x + \cdots + \sin 2nx = \cos x - \cos (2n+1)x$

$\quad \therefore \ \ \sin 2x +\sin 4x + \cdots + \sin 2nx = \cfrac{\cos x-\cos (2n+1)x}{2\sin x}=\cfrac{\sin (n+1)x\sin nx}{\sin x}$


6　余弦の和(3)

$\qquad \cos 2x +\cos 4x + \cdots + \cos 2nx = \cfrac{\cos (n+1)x \sin nx}{\sin x}$


$証明$

$\sin x \cos 2kx=\cfrac{1}{2}\{\sin (2k+1)x - \sin (2k-1)x \} \quad より$

$2\sin x \cos 2kx=-\sin (2k-1)x + \sin (2k+1)x \quad において$

$\quad k=1 \quad とおくと \qquad 2\sin x \cos 2x=-\sin x + \sin 3x$
$\quad k=2 \quad とおくと \qquad 2\sin x \cos 4x=-\sin 3x + \sin 5x$
$\hspace{5em} \vdots $
$\quad k=n \quad とおくと \qquad 2\sin x\cos 2nx=-\sin (2n-1) x + \sin (2n+1)x$

$辺々加えて$

$\quad 2\sin x(\cos 2x +\cos 4x + \cdots + \cos 2nx )= -\sin x + \sin (2n+1)x$

$\quad \therefore \ \ \cos 2x +\cos 4x + \cdots + \cos 2nx = \cfrac{\sin (2n+1)x-\sin x}{2\sin x}=\cfrac{\cos (n+1)x\sin nx}{\sin x}$


$この公式の複素数の等比級数を使った求め方について($茨城大学(数学)2023年　問題3$)をご覧ください。$


7　累乗和


\[(1)\qquad \sum _{k=1}^n \sin ^2 kx=\sin ^2 x +\sin ^2 2x + \cdots + \sin ^2 nx \]
$\hspace{4em} \sin ^2\theta=\cfrac{1}{2}(1-\cos 2\theta) \quad だから$

\begin{eqnarray*} \quad \sum _{k=1}^n \sin ^2 kx &=&\cfrac{1}{2}\sum _{k=1}^n (1-\cos 2kx)\\ \\ &=&\cfrac{n}{2}-\cfrac{1}{2}\sum _{k=1}^n \cos k(2x)\\ \\ & & \quad 2の余弦の和(1)で　x \rightarrow 2x \quad とおいて\\ \\ &=&\cfrac{n}{2}-\cfrac{1}{2}\big(\cfrac{\sin (n+\dfrac{1}{2})2x}{2\sin x}- \cfrac{1}{2}\big)\\ \\ &=&\cfrac{2n+1}{4}-\cfrac{\sin (2n+1)x}{4\sin x}\\ \end{eqnarray*}

\[(2)\qquad \sum _{k=1}^n \cos ^2 kx=\cos ^2 x +\cos ^2 2x + \cdots + \cos ^2 nx \]
$\hspace{4em} \cos ^2\theta=\cfrac{1}{2}(1+\cos 2\theta) \quad だから$

\begin{eqnarray*} \quad \sum _{k=1}^n \cos ^2 kx &=&\cfrac{1}{2}\sum _{k=1}^n (1+\cos 2kx)\\ \\ &=&\cfrac{n}{2}+\cfrac{1}{2}\sum _{k=1}^n \cos k(2x)\\ \\ &=&\cfrac{n}{2}+\cfrac{1}{2}\big(\cfrac{\sin (n+\dfrac{1}{2})2x}{2\sin x}- \cfrac{1}{2}\big)\\ \\ &=&\cfrac{2n-1}{4}+\cfrac{\sin (2n+1)x}{4\sin x}\\ \end{eqnarray*}

\[(3)\qquad \sum _{k=1}^n \sin ^3 kx=\sin ^3 x +\sin ^3 2x + \cdots + \sin ^3 nx \]
$\hspace{4em} \sin 3\theta=3\sin \theta -4\sin ^3 \theta \quad より \quad \sin ^3 \theta=\cfrac{3}{4}\sin \theta -\cfrac{1}{4}\sin 3\theta \quad だから$
\begin{eqnarray*} \quad \sum _{k=1}^n \sin ^3 kx &=&\cfrac{3}{4}\sum _{k=1}^n \sin kx -\cfrac{1}{4}\sum _{k=1}^n \sin k(3x)\\ \\ &=&\cfrac{3\sin \dfrac{n+1}{2}x \sin \dfrac{n}{2}x}{4\sin \dfrac{x}{2}}-\cfrac{\sin \dfrac{3(n+1)}{2}x \sin \dfrac{3n}{2}x}{4\sin \dfrac{3x}{2}}\\ \end{eqnarray*}

\[(4)\qquad \sum _{k=1}^n \cos ^3 kx=\cos ^3 x +\cos ^3 2x + \cdots + \cos ^3 nx \]
$\hspace{4em} \cos 3\theta=4\cos ^3\theta -3\cos \theta \quad より \quad \cos ^3 \theta=\cfrac{3}{4}\cos \theta +\cfrac{1}{4}\cos 3\theta \quad だから$
\begin{eqnarray*} \quad \sum _{k=1}^n \cos ^3 kx &=&\cfrac{3}{4}\sum _{k=1}^n \cos kx +\cfrac{1}{4}\sum _{k=1}^n \cos k(3x)\\ \\ &=&\cfrac{3}{4}\big(\cfrac{\sin (n+ \dfrac{1}{2})x}{2\sin \dfrac{x}{2}}-\cfrac{1}{2}\big)+ \cfrac{1}{4}\big(\cfrac{\sin 3(n+\dfrac{1}{2})x}{2\sin \dfrac{3}{2}x}-\cfrac{1}{2}\big)\\ \\ &=&\cfrac{3\sin (n+ \dfrac{1}{2})x}{8\sin \dfrac{x}{2}} + \cfrac{\sin 3(n+\dfrac{1}{2})x}{8\sin \dfrac{3}{2}x}-\cfrac{1}{2}\\ \end{eqnarray*}


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