\( \sin \theta=-\frac{\sqrt{3}}{3}, \frac{3 \pi}{2}<\theta<2 \pi \) Find \( \sin (2 \theta) \)

To solve for \( \sin(2\theta) \) when \( \sin \theta = -\frac{\sqrt{3}}{3} \) and \( \frac{3 \pi}{2} < \theta < 2 \pi \), we first need to determine \( \theta \). Given that \( \sin \theta = -\frac{\sqrt{3}}{3} \), we know this occurs at angles in the fourth quadrant. The reference angle corresponding to \( \sin^{-1} \left(-\frac{\sqrt{3}}{3}\right) \) is \( \frac{\pi}{6} \). Hence, in the fourth quadrant, the value of \( \theta \) can be expressed as: \[ \theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \] Next, we can find \( \sin(2\theta) \) using the double angle identity for sine, which is given by: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \] We already have \( \sin \theta = -\frac{\sqrt{3}}{3} \). To find \( \cos \theta \), we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting for \( \sin \theta \): \[ \left(-\frac{\sqrt{3}}{3}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{3}{9} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{1}{3} = \frac{2}{3} \] Since \( \theta \) is in the fourth quadrant, \( \cos \theta \) is positive: \[ \cos \theta = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \] Now substituting back into the double angle formula: \[ \sin(2\theta) = 2 \left(-\frac{\sqrt{3}}{3}\right) \left(\frac{\sqrt{6}}{3}\right) \] \[ = 2 \left(-\frac{\sqrt{18}}{9}\right) = -\frac{2\sqrt{18}}{9} = -\frac{6\sqrt{2}}{9} = -\frac{2\sqrt{2}}{3} \] Thus, the value of \( \sin(2\theta) \) is \[ \boxed{-\frac{2\sqrt{2}}{3}} \]