已知函数 $ f\left( x \right) = 2\sin \left( {\dfrac{1}{3}x - \dfrac{{\rm{\pi }}}{6}} \right) $，$x \in {\mathbb{R}}$．
【难度】
【出处】
2011年高考广东卷（文）
【标注】
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求 $f\left( 0 \right)$ 的值；标注答案解析\[ f\left( 0 \right) = 2\sin \left( { - \frac{{\rm{\pi }}}
{6}} \right) = - 1. \] -
设 $\alpha , \beta \in \left[0,\dfrac {\rm{\pi}} 2 \right]$，$f\left( {3\alpha + \dfrac{{\mathrm{\pi }}}{2}} \right) = \dfrac{{10}}{{13}},f\left( {3\beta + 2{\mathrm{\pi }}} \right) = \dfrac{6}{5},$ 求 $\sin \left( {\alpha + \beta } \right)$ 的值．标注答案解析因为\[\begin{split}f\left( {3\alpha + \dfrac{{\mathrm{\pi }}}{2}} \right) &= 2\sin \left[ {\dfrac{1}{3}\left( {3\alpha + \dfrac{{\mathrm{\pi }}}{2}} \right) - \dfrac{{\mathrm{\pi }}}{6}} \right] \\& = 2\sin \alpha \\& = \dfrac{{10}}{{13}},\end{split}\]所以\[\sin \alpha = \dfrac{5}{{13}}.\]因为\[ \begin{split}f\left( {3\beta + 2{\rm{\pi }}} \right) & = 2\sin \left[ {\frac{1}{3}\left( {3\beta + 2{\rm{\pi }}} \right) - \frac{{\rm{\pi }}}{6}} \right] \\& = 2\sin \left( {\beta + \frac{{\rm{\pi }}}{2}} \right) \\&= 2\cos \beta \\&= \frac{6}{5}, \end{split}\]所以\[\cos \beta = \dfrac{3}{5}.\]因为 $\alpha ,\beta \in \left[ {0,\dfrac{{\mathrm{\pi }}}{2}} \right]$，所以\[\begin{split}\cos \alpha & = \sqrt {1 - \sin ^2 \alpha } = \frac{{12}}{{13}} , \\ \sin \beta & = \sqrt {1 - \cos ^2 \beta } = \frac{4}
{5},\end{split}\]所以\[\begin{split}\sin \left( {\alpha + \beta } \right) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\&= \dfrac{5}{{13}} \times \dfrac{3}{5} + \dfrac{{12}}{{13}} \times \dfrac{4}{5} \\&= \dfrac{{63}}{{65}}.\end{split}\]
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