设向量 $\overrightarrow{a_k}=\left(\cos \dfrac{k{\mathrm \pi} }6,\sin \dfrac{k{\mathrm \pi} }6+\cos \dfrac{k{\mathrm \pi} }6\right)$（$k=0,1,2,\cdots,12$），则 $\displaystyle \sum\limits_{k=0}^{11}{\left(\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}\right)}$ 的值为 ．
【难度】
【出处】
2015年高考江苏卷
【标注】
【答案】
$9\sqrt 3$
【解析】
本题的方向是对 $\displaystyle \sum\limits_{k=0}^{11}{\left(\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}\right)}$ 进行化简，最后可借助单位圆求值．根据数量积的坐标式，可得\[\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}=\cos {\dfrac {k{\mathrm \pi} }6}\cos {\dfrac {\left(k+1\right){\mathrm \pi} }6}+\left(\sin {\dfrac {k{\mathrm \pi} }6}+\cos {\dfrac {k{\mathrm \pi} }6}\right)\left(\sin {\dfrac {\left(k+1\right){\mathrm \pi} }6}+\cos {\dfrac {\left(k+1\right){\mathrm \pi} }6}\right).\cdots\cdots ① \]根据积化和差公式可得\[\cos {\dfrac {k{\mathrm \pi} }6}\cos {\dfrac {\left(k+1\right){\mathrm \pi} }6}=\dfrac 12\left(\cos {\dfrac {2k+1}6{\mathrm \pi} }+\cos \dfrac {\mathrm \pi} 6\right).\cdots\cdots ② \]由和差角公式可得\[\begin{split}&\left(\sin {\dfrac {k{\mathrm \pi} }6}+\cos {\dfrac {k{\mathrm \pi} }6}\right)\left(\sin {\dfrac {\left(k+1\right){\mathrm \pi} }6}+\cos {\dfrac {\left(k+1\right){\mathrm \pi} }6}\right)\\=&\sin {\dfrac {k{\mathrm \pi} }6}\sin {\dfrac {\left(k+1\right){\mathrm \pi} }6}+\cos {\dfrac {k{\mathrm \pi} }6}\cos {\dfrac {\left(k+1\right){\mathrm \pi} }6}+\sin {\dfrac {k{\mathrm \pi} }6}\cos {\dfrac {\left(k+1\right){\mathrm \pi} }6}+\cos {\dfrac {k{\mathrm \pi} }6}\sin {\dfrac {\left(k+1\right){\mathrm \pi} }6}\\=&\cos {\dfrac {\mathrm \pi} 6}+\sin {\dfrac {\left(2k+1\right){\mathrm \pi} }6}.\cdots\cdots ③ \end{split}\]将 ②③ 带入 ① 可得\[\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}=\dfrac{3\sqrt3}{4}+\sin{\dfrac{2k+1}6}{\mathrm \pi} +\dfrac12\cos{\dfrac{2k+1}6}{\mathrm \pi} .\]当 $k$ 从 $0$ 取到 $11$，$\dfrac {\left(2k+1\right){\mathrm \pi} }6$ 对应的角的终边在单位圆中如图所示：
根据任意角三角函数的定义可知，\[\sum\limits_{k=0}^{11}{\sin {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}=0,\]\[\sum\limits_{k=0}^{11}{\dfrac 12\cos {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}=0.\]而\[\sum\limits_{k=0}^{11}{\dfrac {3\sqrt 3}4}=\dfrac {3\sqrt 3}4\cdot 12=9\sqrt 3;\]所以所求的代数式为\[\begin{split}&\sum\limits_{k=0}^{11}{\left(\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}\right)}\\=&\sum\limits_{k=0}^{11}{\dfrac {3\sqrt 3}4}+\sum\limits_{k=0}^{11}{\sin {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}+\sum\limits_{k=0}^{11}{\dfrac 12\cos {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}\\=&9\sqrt 3.\end{split}\]
根据任意角三角函数的定义可知，\[\sum\limits_{k=0}^{11}{\sin {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}=0,\]\[\sum\limits_{k=0}^{11}{\dfrac 12\cos {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}=0.\]而\[\sum\limits_{k=0}^{11}{\dfrac {3\sqrt 3}4}=\dfrac {3\sqrt 3}4\cdot 12=9\sqrt 3;\]所以所求的代数式为\[\begin{split}&\sum\limits_{k=0}^{11}{\left(\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}\right)}\\=&\sum\limits_{k=0}^{11}{\dfrac {3\sqrt 3}4}+\sum\limits_{k=0}^{11}{\sin {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}+\sum\limits_{k=0}^{11}{\dfrac 12\cos {\dfrac {\left(2k+1\right){\mathrm \pi} }6}}\\=&9\sqrt 3.\end{split}\]
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