923

1. 求 ${\displaystyle {\displaystyle {\tan }^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7}}}$ 的值;
2. 求 ${\displaystyle \sin {\displaystyle \frac{\pi }{14}-\sin \frac{3\pi }{14}+\sin \frac{5\pi }{14}}}$ 的值.

我们对每一问给一个代表性的处理方法.

1. ${\displaystyle {\displaystyle \theta =\frac{\pi }{7},\frac{2\pi }{7},\frac{3\pi }{7}}}$ 是 ${\displaystyle \tan 4\theta =-\tan 3\theta }$ 的三个解.将方程展开得$$\begin{array}{r}2\frac{\tan 2\theta }{1-{\tan }^{2}2\theta }=\frac{\frac{4\tan \theta }{1-{\tan }^2\theta }}{1-{\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)}^2}=-\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }.\end{array}$$
   消去${\displaystyle \tan \theta }$(我们不需要${\displaystyle \tan \theta =0}$)，令${\displaystyle x={\tan }^2\theta }$，上述方程整理得$$\begin{array}{r}{x}^3-21x^2+35x-7=0,\end{array}$$
   它的解 ${\displaystyle {\tan }^2{\displaystyle \frac{k\pi }{7},k=1,2,3}}$ 互不相等，因此它们是这个方程全部的解.由韦达定理，立得 ${\displaystyle {\displaystyle {\tan }^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7}=21}}$.
2. 记 ${\displaystyle {\displaystyle z=\cos \frac{\pi }{14}+\mathrm{i}\sin \frac{\pi }{14}}}$, 则 ${\displaystyle z^7=\mathrm{i},z^{14}=-1}$.这样$$\begin{array}{r}\sin \frac{k\pi }{14}=\frac{z^k-{\bar{z}}^k}{2\mathrm{i}}=\frac{z^k+z^{14-k}}{2\mathrm{i}},\end{array}$$
   从而$$\begin{array}{r}\sin \frac{\pi }{14}-\sin \frac{3\pi }{14}+\sin \frac{5\pi }{14}=\frac{1}{2\mathrm{i}}(z^{13}-z^{11}+z^9-z^7+z^5-z^3+z+\mathrm{i})=\frac{1}{2\mathrm{i}}(\frac{z(z^{14}+1)}{z^2+1}+\mathrm{i})=\frac{1}{2}.\end{array}$$