答案： 典型例题1. $-\frac{\sqrt{3}}{2}$ 解析:设$A\left(x_1,\frac{1}{2}\right),B\left(x_2,\frac{1}{2}\right)$,由$\vert AB\vert=\frac{\pi}{6}$可得$x_2 - x_1=\frac{\pi}{6}$,由$\sin x=\frac{1}{2}$可知,$x=\frac{\pi}{6}+2k\pi$或$x=\frac{5\pi}{6}+2k\pi,k\in Z$,由题图可知,$\omega x_2+\varphi-(\omega x_1+\varphi)=\frac{5}{6}\pi-\frac{\pi}{6}=\frac{2\pi}{3}$,即$\omega(x_2 - x_1)=\frac{2\pi}{3}$,所以$\omega=4$.因为$f\left(\frac{2}{3}\pi\right)=\sin\left(\frac{8\pi}{3}+\varphi\right)=0$,所以$\frac{8\pi}{3}+\varphi=k\pi$,即$\varphi=-\frac{8}{3}\pi+k\pi$,$k\in Z$.所以$f(x)=\sin\left(4x-\frac{8}{3}\pi+k\pi\right)=\sin\left(4x-\frac{2}{3}\pi+k\pi\right)$,所以$f(x)=\sin\left(4x-\frac{2}{3}\pi\right)$或$f(x)=-\sin\left(4x-\frac{2}{3}\pi\right)$.又因为$f(0)＜0$,所以$f(x)=\sin\left(4x-\frac{2}{3}\pi\right)$,所以$f(\pi)=\sin\left(4\pi-\frac{2}{3}\pi\right)=-\frac{\sqrt{3}}{2}$

答案： 变式训练1. C 解析:由题意可知,$A(2,0),B\left(2+\frac{\pi}{3},0\right)$,$C(0,A\sin\varphi)$,则$D\left(1+\frac{\pi}{2\omega},\frac{A\sin\varphi}{2}\right)$,有$\sqrt{3}\vert A\sin\varphi\vert=2+\frac{\pi}{\omega}$,$\sin(2\omega+\varphi)=0.\because\vert AD\vert=\frac{2\sqrt{21}}{3},\therefore\left(\frac{\pi}{2\omega}-1\right)^2+\frac{A^2\sin^2\varphi}{4}=\frac{28}{3}$,把$\vert A\sin\varphi\vert=\frac{1}{\sqrt{3}}\left(2+\frac{\pi}{\omega}\right)$代入上式,得$\left(\frac{\pi}{\omega}\right)^2-2×\frac{\pi}{\omega}-24=0$,解得$\frac{\pi}{\omega}=6$(负值舍去),$\therefore\omega=\frac{\pi}{6}$,$\therefore\sin\left(\frac{\pi}{3}+\varphi\right)=0$,由$\vert\varphi\vert\leq\frac{\pi}{2}$,解得$\varphi=-\frac{\pi}{3}$,$\therefore\sqrt{3}A\sin\left(-\frac{\pi}{3}\right)=8$,解得$A=\frac{16}{3}$,$\therefore f(x)=\frac{16}{3}\sin\left(\frac{\pi}{6}x-\frac{\pi}{3}\right)$,显然其周期$T=\frac{2\pi}{\frac{\pi}{6}}=12$,故$A$错误;当$x=8$时,$\frac{\pi}{6}x-\frac{\pi}{3}=\pi$,$f(x)=0$,故$B$错误;$f(2)=\frac{16}{3}\sin0=0$,$f(-4)=\frac{16}{3}\sin(-\pi)=0$,故$C$正确;$f(-x+2)=\frac{16}{3}\sin\left[\frac{\pi}{6}(-x+2)-\frac{\pi}{3}\right]=\frac{16}{3}\sin\left(-\frac{\pi}{6}x\right)$,显然是奇函数,故$D$错误.