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

To find \( \sin(2\theta) \) given that \( \sin \theta = -\frac{\sqrt{3}}{3} \) and \( \frac{3\pi}{2} < \theta < 2\pi \), we first need to determine the value of \( \theta \). In the interval \( \frac{3\pi}{2} < \theta < 2\pi \), the angle \( \theta \) corresponds to the sine value of \( -\frac{\sqrt{3}}{3} \). The reference angle where \( \sin x = \frac{\sqrt{3}}{3} \) is \( \frac{\pi}{6} \), thus: \[ \theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \] Now that we have \( \theta = \frac{11\pi}{6} \), we can use the double angle formula for sine, which is: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \] We already know \( \sin \theta = -\frac{\sqrt{3}}{3} \). Now we need to calculate \( \cos \theta \). Since \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \cos^2 \theta = 1 - \left(-\frac{\sqrt{3}}{3}\right)^2 = 1 - \frac{3}{9} = 1 - \frac{1}{3} = \frac{2}{3} \] \[ \cos \theta = -\sqrt{\frac{2}{3}} \quad \text{(since \(\theta\) is in the fourth quadrant)} \] Now we find \( \sin(2\theta) \): \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(-\frac{\sqrt{3}}{3}\right) \left(-\sqrt{\frac{2}{3}}\right) \] \[ = 2 \cdot \frac{\sqrt{3}}{3} \cdot \sqrt{\frac{2}{3}} = 2 \cdot \frac{\sqrt{6}}{9} = \frac{2\sqrt{6}}{9} \] Thus, the final result is: \[ \sin(2\theta) = \frac{2\sqrt{6}}{9} \]