$y=sinx$ đồng biến trên [tex][\frac{-\pi}{2};\frac{-\pi}{3}][/tex]
Nên $\displaystyle \min y _{x\epsilon [\frac{-\pi}{2};\frac{-\pi}{3}]}=y(\frac{-\pi}{2})=-1$
$\displaystyle \max y _{x\epsilon [\frac{-\pi}{2};\frac{-\pi}{3}]}=y(\frac{-\pi}{3})=\frac{-\sqrt{3}}{2} $
Chọn B