试证: $\dps{\cos\frac{2\pi}{5}}$ 为无理数.
证明: 设 $$\bex z=e^{i\frac{2\pi}{5}}, \eex$$ 则 $$\beex \bea z^5&=e^{i2\pi}=1,\\ (z-1)(z^4+z^3+z^2+z+1)&=0,\\ z^4+z^3+z^2+z+1&=0,\\ z^2+z+1+z^{-1}+z^{-2}&=0. \eea \eeex$$ 比较两端实部, 我们发现 $$\beex \bea 2\cos\frac{4\pi}{5}+2\cos\frac{2\pi}{5}+1&=0,\\ 2\sex{2\cos^2\frac{2\pi}{5}-1}+2\cos\frac{2\pi}{5}+1&=0,\\ 4\cos^2\frac{2\pi}{5}+2\cos\frac{2\pi}{5}-1&=0,\\ \cos\frac{2\pi}{5}&=\frac{\sqrt{5}-1}{4}\not\in\bbQ. \eea \eeex$$
原文：http://www.cnblogs.com/zhangzujin/p/4876000.html