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2015年北京大学化学体验营数学试题
【标注】
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$\cos \dfrac {2\pi}{11}+\cos \dfrac {4\pi}{11}+\cos \dfrac {6\pi}{11}+\cos \dfrac {8\pi}{11}+\cos \dfrac {10\pi}{11}=-\dfrac1 2$；标注答案略解析根据题意，有\[\begin{split}LHS&=\cos \dfrac {2\pi}{11}+\cos \dfrac {4\pi}{11}+\cos \dfrac {6\pi}{11}+\cos \dfrac {8\pi}{11}+\cos \dfrac {10\pi}{11}\\
&=\dfrac{2\sin \dfrac{\pi}{11} \left (\cos \dfrac {2\pi}{11}+\cos \dfrac {4\pi}{11}+\cos \dfrac {6\pi}{11}+\cos \dfrac {8\pi}{11}+\cos \dfrac {10\pi}{11} \right )}{2\sin \dfrac {\pi}{11}}, \end{split}\]对上式中的分子应用积化和差公式，可得上式等于\[\dfrac {\sin \pi - \sin \dfrac {\pi}{11}}{2\sin{\dfrac {\pi}{11}}}=-\dfrac 1 2 .\] -
$\tan \dfrac {3\pi}{11}+4\sin \dfrac {2\pi}{11}=\sqrt {11}$．标注答案略解析因为\[\begin{split}&\left (\sin \dfrac {3\pi}{11} +4 \sin \dfrac {2\pi}{11} \cos \dfrac {3\pi}{11} \right ) ^2 -11 \cos^2 \dfrac {3\pi}{11}\\=&\left (\sin \dfrac {3\pi}{11} +2 \sin \dfrac {5\pi}{11}-2 \sin \dfrac {\pi}{11} \right ) ^2 -11 \cos^2 \dfrac {3\pi}{11}\\=&12\sin ^2 \dfrac {3\pi}{11}+4\sin ^2 \dfrac {5\pi}{11}+4 \sin ^2\dfrac{\pi}{11}\\ &\qquad+4\sin\dfrac{3\pi}{11}\sin\dfrac{5\pi}{11}-4\sin\dfrac{3\pi}{11}\sin\dfrac{\pi}{11}-8\sin\dfrac{5\pi}{11}\sin\dfrac{\pi}{11}-11\\=&6\left(1-\cos\dfrac{6\pi}{11}\right)+2\left(1-\cos\dfrac{10\pi}{11}\right)+2\left(1-\cos\dfrac{2\pi}{11}\right)\\ &\qquad +2\left(\cos\dfrac{2\pi}{11}-\cos\dfrac{8\pi}{11}\right)-2\left(\cos\dfrac{2\pi}{11}-\cos\dfrac{4\pi}{11}\right)-4\left(\cos\dfrac{4\pi}{11}-\cos\dfrac{6\pi}{11}\right)-11\\=&-1-2\left(\cos \dfrac {2\pi}{11}+\cos \dfrac {4\pi}{11}+\cos \dfrac {6\pi}{11}+\cos \dfrac {8\pi}{11}+\cos \dfrac {10\pi}{11}\right)\\=&0,\end{split}\]所以$$\left(\tan\dfrac{3\pi}{11}+4\sin\dfrac{2\pi}{11}\right)^2=11,$$易知 $\tan\dfrac{3\pi}{11}+4\sin\dfrac{2\pi}{11}>0$，故$$\tan\dfrac{3\pi}{11}+4\sin\dfrac{2\pi}{11}=\sqrt{11}.$$
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