(5分)已知函数$f(x)=\sin (\omega x+\varphi )$在区间$(\dfrac{\pi }{6}$,$\dfrac{2\pi }{3})$单调递增,直线$x=\dfrac{\pi }{6}$和$x=\dfrac{2\pi }{3}$为函数$y=f(x)$的图像的两条对称轴,则$f(-\dfrac{5\pi }{12})=($ $)$ A.$-\dfrac{\sqrt{3}}{2}$ B.$-\dfrac{1}{2}$ C.$\dfrac{1}{2}$ D.$\dfrac{\sqrt{3}}{2}$ 答案:$D$ 分析:先根据题意建立方程求出参数,再计算,即可得解. 解:根据题意可知$\dfrac{T}{2}=\dfrac{2\pi }{3}-\dfrac{\pi }{6}=\dfrac{\pi }{2}$, $\therefore T=\pi$,取$\omega > 0$,$\therefore \omega =\dfrac{2\pi }{T}=2$, 又根据“五点法“可得$2\times \dfrac{\pi }{6}+\varphi =-\dfrac{\pi }{2}+2k\pi$,$k\in Z$, $\therefore \varphi =-\dfrac{5\pi }{6}+2k\pi$,$k\in Z$, $\therefore f(x)=\sin (2x-\dfrac{5\pi }{6}+2k\pi )=\sin (2x-\dfrac{5\pi }{6})$, $\therefore f(-\dfrac{5\pi }{12})=\sin (-\dfrac{5\pi }{6}-\dfrac{5\pi }{6})=\sin (-\dfrac{5\pi }{3})=\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$. 故选:$D$. 点评:本题考查三角函数的性质,方程思想,属基础题.
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