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战巡
Post time 2023-12-26 15:29
\[\frac{\sum_{j=1}^{k-1}\cos(jx)}{\sin(kx)}+i\frac{\sum_{j=1}^{k-1}\sin(jx)}{\sin(kx)}=\frac{1}{\sin(kx)}\sum_{j=1}^{k-1}e^{ijx}=\frac{1}{\sin(kx)}\frac{e^{ikx}-e^{ix}}{e^{ix}-1}\]
\[=\frac{1}{\sin(kx)}\frac{(e^{ikx}-e^{ix})(e^{-ix}-1)}{2-2\cos(x)}\]
\[=\frac{1}{\sin(kx)}\frac{e^{i(k-1)x}+e^{ix}-e^{ikx}-1}{2-2\cos(x)}\]
\[=\frac{\cos((k-1)x)+\cos(x)-\cos(kx)-1}{\sin(kx)(2-2\cos(x))}+i\frac{\sin((k-1)x)+\sin(x)-\sin(kx)}{\sin(kx)(2-2\cos(x))}\]
最后化简你就自己想办法吧
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kuing
Post time 2023-12-26 15:37
