来自群的三角化简题$\tan(2\pi/13)+\tan(5\pi/13)+\tan(6\pi/13)$

爱好者-Salvation(13058978)  15:10:36
那这个怎么算比较好
tan(2pi/13)+tan(5pi/13)+tan(6pi/13)

求值 $\displaystyle \tan\frac{2\pi}{13}+\tan\frac{5\pi}{13}+\tan\frac{6\pi}{13}$

注意到三个角之和为 $\pi$，所以\[
\tan \frac{{2\pi }}{{13}} + \tan \frac{{5\pi }}{{13}} + \tan \frac{{6\pi }}{{13}} = \tan \frac{{2\pi }}{{13}}\tan \frac{{5\pi }}{{13}}\tan \frac{{6\pi }}{{13}}
\]利用欧拉公式，上式的右边可以写成\[
\frac{{i\left( {e^{ - \frac{{2i\pi }}{{13}}} - e^{\frac{{2i\pi }}{{13}}} } \right)}}{{e^{ - \frac{{2i\pi }}{{13}}} + e^{\frac{{2i\pi }}{{13}}} }} \cdot \frac{{i\left( {e^{ - \frac{{5i\pi }}{{13}}} - e^{\frac{{5i\pi }}{{13}}} } \right)}}{{e^{ - \frac{{5i\pi }}{{13}}} + e^{\frac{{5i\pi }}{{13}}} }} \cdot \frac{{i\left( {e^{ - \frac{{6i\pi }}{{13}}} - e^{\frac{{6i\pi }}{{13}}} } \right)}}{{e^{ - \frac{{6i\pi }}{{13}}} + e^{\frac{{6i\pi }}{{13}}} }}
\]展开可整理为\[
\frac{{i\left( {e^{ - \frac{{i\pi }}{{13}}} - e^{\frac{{i\pi }}{{13}}} } \right) + i\left( {e^{ - \frac{{3i\pi }}{{13}}} - e^{\frac{{3i\pi }}{{13}}} } \right) + i\left( {e^{ - \frac{{9i\pi }}{{13}}} - e^{\frac{{9i\pi }}{{13}}} } \right)}}{{ - 2 + e^{ - \frac{{i\pi }}{{13}}} + e^{\frac{{i\pi }}{{13}}} + e^{ - \frac{{3i\pi }}{{13}}} + e^{\frac{{3i\pi }}{{13}}} + e^{ - \frac{{9i\pi }}{{13}}} + e^{\frac{{9i\pi }}{{13}}} }}
\]再用欧拉公式化回三角式，为\[
\frac{{\sin \frac{\pi }{{13}} + \sin \frac{{3\pi }}{{13}} + \sin \frac{{9\pi }}{{13}}}}{{ - 1 + \cos \frac{\pi }{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{9\pi }}{{13}}}}
\]分母其实已经早有求过，看看分子如何，平方一下，利用两倍角公式，积化和差之类，有\[
\left( {\sin \frac{\pi }{{13}} + \sin \frac{{3\pi }}{{13}} + \sin \frac{{9\pi }}{{13}}} \right)^2 = \frac{1}{2}\left( {3 + 2\cos \frac{\pi }{{13}} + 2\cos \frac{{3\pi }}{{13}} + 2\cos \frac{{9\pi }}{{13}} + \cos \frac{{2\pi }}{{13}} + \cos \frac{{6\pi }}{{13}} + \cos \frac{{8\pi }}{{13}}} \right)
\]注意到\[
\cos \frac{{2\pi }}{{13}} + \cos \frac{{4\pi }}{{13}} + \cos \frac{{6\pi }}{{13}} + \cos \frac{{8\pi }}{{13}} + \cos \frac{{10\pi }}{{13}} + \cos \frac{{12\pi }}{{13}} = - \frac{1}{2}
\]于是易得\[
\left( {\sin \frac{\pi }{{13}} + \sin \frac{{3\pi }}{{13}} + \sin \frac{{9\pi }}{{13}}} \right)^2 = \frac{3}{2}\left( {\cos \frac{\pi }{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{9\pi }}{{13}}} \right) + \frac{5}{4}
\]又显然 $\sin \frac{\pi }{{13}} + \sin \frac{{3\pi }}{{13}} + \sin \frac{{9\pi }}{{13}}>0$，所以\[
\sin \frac{\pi }{{13}} + \sin \frac{{3\pi }}{{13}} + \sin \frac{{9\pi }}{{13}} = \sqrt {\frac{3}{2}\left( {\cos \frac{\pi }{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{9\pi }}{{13}}} \right) + \frac{5}{4}}
\]综上，最终得到\[
\tan \frac{{2\pi }}{{13}} + \tan \frac{{5\pi }}{{13}} + \tan \frac{{6\pi }}{{13}} = \frac{{\sqrt {\frac{3}{2}\left( {\cos \frac{\pi }{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{9\pi }}{{13}}} \right) + \frac{5}{4}} }}{{ - 1 + \cos \frac{\pi }{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{9\pi }}{{13}}}}
\]那 $\cos \frac{\pi }{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{9\pi }}{{13}}$ 是多少哇？时间关系，抄一段何版主的记录：

因为\[
\cos \frac{2\pi}{13} + \cos \frac{4\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{10\pi}{13} + \cos \frac{12\pi}{13} = -\frac{1}{2}
\]这样\begin{align*}
& \left(\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13}\right) - \left(\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{9\pi}{13}\right) \\
= & \cos \frac{2\pi}{13} + \cos \frac{4\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{10\pi}{13} + \cos \frac{12\pi}{13} \\
= & -\frac{1}{2}
\end{align*}即\[
\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} = \cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{9\pi}{13} - \frac{1}{2}
\]所以\begin{align*}
& \left(\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{9\pi}{13}\right)^2 \\
= & \cos^2\frac{\pi}{13} + \cos^2\frac{3\pi}{13} + \cos^2\frac{9\pi}{13} + 2\left(\cos \frac{\pi}{13}\cos \frac{3\pi}{13} + \cos \frac{\pi}{13}\cos \frac{9\pi}{13} + \cos \frac{3\pi}{13}\cos \frac{9\pi}{13}\right) \\
= & \frac{1}{2}\left(\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{18\pi}{13} + 3\right) \\
& {}+ \cos \frac{2\pi}{13} + \cos \frac{4\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{10\pi}{13} + \cos \frac{12\pi}{13} \\
= & \frac{1}{2}\left(\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13}\right) + 1 \\
= & \frac{1}{2}\left(\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{9\pi}{13}\right) + \frac{3}{4}
\end{align*}而\[
\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{9\pi}{13} = \cos \frac{\pi}{13} + \cos \frac{3\pi}{13} - \cos \frac{4\pi}{13} > 0
\]所以 $\cos \dfrac{\pi}{13} + \cos \dfrac{3\pi}{13} + \cos \dfrac{9\pi}{13}$ 是方程 $x^2 = \dfrac{1}{2}x + \dfrac{3}{4}$ 的正根，所以\[
\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{9\pi}{13} = \frac{\sqrt{13} + 1}{4}
\]

抄完，代入，即得\[
\tan \frac{{2\pi }}{{13}} + \tan \frac{{5\pi }}{{13}} + \tan \frac{{6\pi }}{{13}} = \frac{{\sqrt {\frac{3}{2} \cdot \frac{{\sqrt {13} + 1}}{4} + \frac{5}{4}} }}{{ - 1 + \frac{{\sqrt {13} + 1}}{4}}} = \sqrt {65 + 18\sqrt {13} }
\]

1# kuing
太牛了！

当年的转发连码都没打……