设函数 $f\left( x \right) = 3\sin \left( {\omega x + \dfrac{{\mathrm{\pi}} }{6}} \right)$，$\omega >0$，$x \in \left( { - \infty , + \infty } \right)$，且以 $\dfrac{{\mathrm{\pi }}}{2}$ 为最小正周期．
【难度】
【出处】
2010年高考广东卷（文)
【标注】
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求 $f\left( 0 \right)$；标注答案解析因为函数 $f\left( x \right) = 3\sin \left( {\omega x + \dfrac{{\mathrm{\pi }}}{6}} \right),$ 所以\[f\left( 0 \right) = 3\sin \left( {\omega \times 0 + \dfrac{{\mathrm{\pi }}}{6}} \right) = 3\sin \dfrac{{\mathrm{\pi }}}{6} = \dfrac{3}{2}.\]
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求 $f\left( x \right)$ 的解析式；标注答案解析$\because$ 函数 $f\left( x \right) = 3\sin \left( {\omega x + \dfrac{{\mathrm{\pi}} }{6}} \right)$，$\omega >0$，$x \in \left( { - \infty , + \infty } \right)$，且以 $\dfrac{{\mathrm{\pi }}}{2}$ 为最小正周期．
$\therefore$ $\omega = 4$，$\therefore$ $f\left( x \right) = 3\sin \left( {4x + \dfrac{{\mathrm{\pi}} }{6}} \right)$． -
已知 $f\left( {\dfrac{\alpha }{4} + \dfrac{{\mathrm{\pi}} }{{12}}} \right) = \dfrac{9}{5}$，求 $\sin \alpha $ 的值．标注答案解析$\because$ $f\left(\dfrac{\alpha }{4} + \dfrac{{\mathrm{\pi }}}{{12}}\right) = \dfrac{9}{5}$，
$\therefore$ $3\sin \left(4\left(\dfrac{\alpha }{4} + \dfrac{{\mathrm{\pi}} }{{12}}\right) + \dfrac{{\mathrm{\pi}} }{6}\right) = \dfrac{9}{5}$，
$\therefore$ $\sin \left(\alpha + \dfrac{{\mathrm{\pi }}}{2}\right) = \dfrac{3}{5}$，
$\therefore$ $\cos \alpha = \dfrac{3}{5}$，$\therefore$ $1 - {\sin ^2}\alpha = \dfrac{9}{{25}}$，
$\therefore$ ${\sin ^2}\alpha = \dfrac{{16}}{{25}}$，$\therefore$ $\sin \alpha = \pm \dfrac{4}{5}$．
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