三角恒等式

lihpb

$\begin{aligned} & \frac{\sum_{i=1}^{k-1} \cos i x}{\sin k x}=\frac{1}{2}\left(\cot \frac{x}{2}-\cot \frac{k x}{2}\right), \\ & \frac{\sum_{i=1}^{k-1} \sin i x}{\sin k x}=\frac{1}{2}\left(\tan \frac{k x}{2} \cot \frac{x}{2}-1\right), \\ & 0<x<\frac{\pi}{k}, \quad k \text { 为正整数且 } k \geq 2\end{aligned}$
帮忙求第二个不定积分

战巡

\[\frac{\sum_{j=1}^{k-1}\cos(jx)}{\sin(kx)}+i\frac{\sum_{j=1}^{k-1}\sin(jx)}{\sin(kx)}=\frac{1}{\sin(kx)}\sum_{j=1}^{k-1}e^{ijx}=\frac{1}{\sin(kx)}\frac{e^{ikx}-e^{ix}}{e^{ix}-1}\]
\[=\frac{1}{\sin(kx)}\frac{(e^{ikx}-e^{ix})(e^{-ix}-1)}{2-2\cos(x)}\]
\[=\frac{1}{\sin(kx)}\frac{e^{i(k-1)x}+e^{ix}-e^{ikx}-1}{2-2\cos(x)}\]
\[=\frac{\cos((k-1)x)+\cos(x)-\cos(kx)-1}{\sin(kx)(2-2\cos(x))}+i\frac{\sin((k-1)x)+\sin(x)-\sin(kx)}{\sin(kx)(2-2\cos(x))}\]
最后化简你就自己想办法吧

kuing

战巡 发表于 2023-12-26 15:29
\[\frac{\sum_{j=1}^{k-1}\cos(jx)}{\sin(kx)}+i\frac{\sum_{j=1}^{k-1}\sin(jx)}{\sin(kx)}=\frac{1}{\sin ...

他好像是要求积分……🤔

lihpb

kuing 发表于 2023-12-26 15:37
他好像是要求积分……🤔

forum.php?mod=viewthread&tid=11827&extra=page=1

恒等式的证明我已经写了出来了，想问下你能看懂吗

就是不记得怎么求积分