求值：$\cos\dfrac{\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{9\pi}{13}$．
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【答案】
$\dfrac{1+\sqrt{13}}{4}$
【解析】
根据题意令$$\begin{split} &x=\cos\dfrac{\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{9\pi}{13}\\&y=\cos\dfrac{5\pi}{13}+\cos\dfrac{7\pi}{13}+\cos\dfrac{11\pi}{13}\end{split}$$则有$$x+y=\sum_{k=1}^6\cos\dfrac{(2k-1)\pi}{13}=\dfrac12.$$又因$$\begin{split} x\cdot y&=\left(\cos\dfrac{\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{9\pi}{13}\right)\cdot\left(\cos\dfrac{5\pi}{13}+\cos\dfrac{7\pi}{13}+\cos\dfrac{11\pi}{13}\right)\\
&=-\dfrac32\left(\cos\dfrac{\pi}{13}-\cos\dfrac{2\pi}{13}+\cos\dfrac{3\pi}{13}-\cos\dfrac{4\pi}{13}+\cos\dfrac{5\pi}{13}-\cos\dfrac{6\pi}{13}\right)\\
&=-\dfrac32\cdot\sum_{k=1}^6\cos\dfrac{(2k-1)\pi}{13}\\
&=-\dfrac34.\end{split}$$故 $x,y$ 是方程$$t^2-\dfrac12t-\dfrac34=0$$的两根，又 $x>0$，所以 $x=\dfrac{1+\sqrt{13}}{4}$，故$$\cos\dfrac{\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{9\pi}{13}=\dfrac{1+\sqrt{13}}{4}.$$
&=-\dfrac32\left(\cos\dfrac{\pi}{13}-\cos\dfrac{2\pi}{13}+\cos\dfrac{3\pi}{13}-\cos\dfrac{4\pi}{13}+\cos\dfrac{5\pi}{13}-\cos\dfrac{6\pi}{13}\right)\\
&=-\dfrac32\cdot\sum_{k=1}^6\cos\dfrac{(2k-1)\pi}{13}\\
&=-\dfrac34.\end{split}$$故 $x,y$ 是方程$$t^2-\dfrac12t-\dfrac34=0$$的两根，又 $x>0$，所以 $x=\dfrac{1+\sqrt{13}}{4}$，故$$\cos\dfrac{\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{9\pi}{13}=\dfrac{1+\sqrt{13}}{4}.$$
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