|
青青子衿
发表于 2021-2-24 12:08
试一下这个对齐方式
\begin{alignedat}{2}
\text{I}\,\,)&\enspace\enspace\enspace\enspace & &\sum_{k=1}^{n}\sin\left(\dfrac{k\pi}{n}\right)=\cot\left(\dfrac{2\pi}{n}\right)&\\
\text{II}\,)& & &\sum_{k=1}^{n}\sin\left[\dfrac{\left(2k-1\right)\pi}{n}\right]=0\\
\text{III)}& & &\sum_{k=1}^{n}\sin\left(\dfrac{k\pi}{2n-1}\right)=\dfrac{1}{2}\cot\left[\dfrac{\pi}{2(2n-1)}\right]-\cos\left[\dfrac{\pi}{2(2n-1)}\right]
\end{alignedat} |
|