每日一题[151] 三角求和

2015年高考数学江苏卷理科第14题:

设向量\(\vec a_k=\left(\cos\dfrac{k\pi}{6},\sin\dfrac{k\pi}{6}+\cos\dfrac{k\pi}{6}\right)\),\(k=0,1,2,\cdots,12\),则\(\sum\limits_{k=0}^{11}\left(\vec a_k\cdot \vec a_{k+1}\right)\)的值为_______.


cover正确答案是\(9\sqrt 3\).

法一

根据题意有\[\begin{split}\sum_{k=0}^{11}\left(\vec a_k\cdot \vec a_{k+1}\right)&=\sum_{k=0}^{11}\left[\cos\dfrac k6\pi\cos\dfrac{k+1}{6}\pi+\left(\sin\dfrac{k}{6}\pi+\cos\dfrac{k}{6}\pi\right)\left(\sin\dfrac{k+1}{6}\pi+\cos\dfrac{k+1}{6}\pi\right)\right]\\&=\sum_{k=0}^{11}\left(2\cos\dfrac{k}{6}\pi\cos\dfrac{k+1}{6}\pi+\sin\dfrac{k}{6}\pi\sin\dfrac{k+1}{6}\pi+\sin\dfrac{2k+1}6\pi\right),\end{split}\]

考虑到\[\sum_{k=0}^{11}\left(\cos\dfrac{k}{6}\pi\cos\dfrac{k+1}{6}\pi+\sin\dfrac{k}{6}\pi\sin\dfrac{k+1}{6}\pi\right)=\sum_{k=0}^{11}\cos\dfrac{\pi}6=6\sqrt 3,\]

且\[\begin{split}\sum_{k=0}^{11}\cos\dfrac{k}{6}\pi\cos\dfrac{k+1}{6}\pi&=\sum_{k=0}^{11}\sin\dfrac{k+3}{6}\pi\sin\dfrac{k+4}{6}\pi\\&=\sum_{k=3}^{14}\sin\dfrac{k}{6}\pi\sin\dfrac{k+1}{6}\pi\\&=\sum_{k=0}^{11}\sin\dfrac{k}{6}\pi\sin\dfrac{k+1}{6}\pi,\end{split}\]于是\[\sum_{k=0}^{11}\cos\dfrac{k}{6}\pi\cos\dfrac{k+1}{6}\pi=\sum_{k=0}^{11}\sin\dfrac{k}{6}\pi\sin\dfrac{k+1}{6}\pi=3\sqrt 3,\]又\[\begin{split}\sum_{k=0}^{11}\sin\dfrac{2k+1}6\pi&=\sum_{k=0}^{5}\sin\dfrac{2k+1}6\pi+\sum_{k=6}^{11}\sin\dfrac{2k+1}6\pi\\&=\sum_{k=0}^{5}\sin\dfrac{2k+1}6\pi+\sum_{k=0}^{5}\sin\left(2\pi+\dfrac{2k+1}6\pi\right)\\&=2\sum_{k=0}^{5}\sin\dfrac{2k+1}6\pi\\&=0,\end{split}\]因此所求和式的值为\(9\sqrt 3\).

法二(meiyun提供)

根据题意有\[\begin{split}\sum_{k=0}^{11}\left(\vec a_k\cdot \vec a_{k+1}\right)&=\sum_{k=0}^{11}\left[\cos\dfrac k6\pi\cos\dfrac{k+1}{6}\pi+\left(\sin\dfrac{k}{6}\pi+\cos\dfrac{k}{6}\pi\right)\left(\sin\dfrac{k+1}{6}\pi+\cos\dfrac{k+1}{6}\pi\right)\right]\\&=\sum_{k=0}^{11}\left[\cos\dfrac k6\pi\left(\dfrac{\sqrt 3}2\cos\dfrac k6\pi-\dfrac 12\sin\dfrac k6\pi\right)+\right.\\&\left.\qquad\qquad\left(\sin\dfrac k6\pi+\cos\dfrac k6\pi\right)\left(\dfrac{\sqrt 3-1}2\sin\dfrac k6\pi+\dfrac{\sqrt 3+1}2\cos\dfrac k6\pi\right)\right],\end{split}\]

注意到\[\begin{split}&\sum_{k=0}^{11}\cos^2\dfrac k6\pi=\sum_{k=0}^{11}\dfrac{1+\cos\dfrac k3\pi}{2}=6,\\&\sum_{k=0}^{11}\cos\dfrac k6\pi\sin\dfrac k6\pi=\sum_{k=0}^{11}\dfrac 12\sin\dfrac k3\pi=0,\\&\sum_{k=0}^{11}\sin^2\dfrac k6\pi=\sum_{k=0}^{11}\dfrac{1-\cos \dfrac k3\pi}{2}=6,\end{split}\]于是不难计算得所求和式的值为\(9\sqrt 3\).


   两种方法都刻意绕开了使用和差化积公式.

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