（本小题满分 12 分） 已知函数 f(x)=A sin…——2014 高考数学第 14 题答案解析

【解答】
（1）

$$ \begin{aligned} & \because f(x)=A \sin \left(x+\frac{\pi}{3}\right), \text { 且 } f\left(\frac{5 \pi}{12}\right)=\frac{3 \sqrt{2}}{2} . \\ & \therefore f\left(\frac{5 \pi}{12}\right)=A \sin \left(\frac{5 \pi}{12}+\frac{\pi}{3}\right)=A \sin \frac{3 \pi}{4}=A \cdot \frac{\sqrt{2}}{2}=\frac{3 \sqrt{2}}{2} . \\ & \therefore A=3 . \end{aligned} $$

②

$$ \begin{aligned} & \because f(x)=3 \sin \left(x+\frac{\pi}{3}\right), \text { 且 } f(\theta)-f(-\theta)=\sqrt{3} . \\ & \therefore f(\theta)-f(-\theta)=3 \sin \left(\theta+\frac{\pi}{3}\right)-3 \sin \left(-\theta+\frac{\pi}{3}\right) \\ & =3\left[\left(\sin \theta \cos \frac{\pi}{3}+\cos \theta \sin \frac{\pi}{3}\right)-\left(\sin \frac{\pi}{3} \cos \theta-\cos \frac{\pi}{3} \sin \theta\right)\right] \\ & =3 \cdot 2 \sin \theta \cos \frac{\pi}{3}=3 \sin \theta=\sqrt{3} . \\ & \therefore \sin \theta=\frac{\sqrt{3}}{3} . \text { 且 } \theta \in\left(0, \frac{\pi}{2}\right) \\ & \therefore \cos \theta=\sqrt{1-\sin ^{2} \theta}=\frac{\sqrt{6}}{3} . \\ & \because f\left(\frac{\pi}{6}-\theta\right)=3 \sin \left(\frac{\pi}{6}-\theta+\frac{\pi}{3}\right)=3 \sin \left(\frac{\pi}{2}-\theta\right)=3 \cos \theta=\sqrt{6} \end{aligned} $$