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

To find \( \sin(2\theta) \), we can first determine the angle \( \theta \) using the given sine value. The sine function is negative in the fourth quadrant, and we know that: \[ \sin \theta = -\frac{\sqrt{3}}{3} \] The reference angle where \( \sin \theta = \frac{\sqrt{3}}{3} \) is \( \theta = \frac{\pi}{6} \). Therefore, in the fourth quadrant, we can express the angle \( \theta \) as: \[ \theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \] Now that we have \( \theta = \frac{11\pi}{6} \), we can use the double angle formula for sine: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \] We already have \( \sin \theta = -\frac{\sqrt{3}}{3} \). Next, we need to find \( \cos \theta \). Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting the known value of \( \sin \theta \): \[ \left(-\frac{\sqrt{3}}{3}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{3}{9} + \cos^2 \theta = 1 \quad \Rightarrow \quad \cos^2 \theta = 1 - \frac{1}{3} = \frac{2}{3} \] Taking the square root gives: \[ \cos \theta = \sqrt{\frac{2}{3}} \quad \text{(since} \ \cos \theta > 0 \ \text{in the fourth quadrant)} \] Now, substituting back into the double angle formula: \[ \sin(2\theta) = 2 \left(-\frac{\sqrt{3}}{3}\right) \left(\frac{\sqrt{2}}{\sqrt{3}}\right) \] Simplifying gives: \[ \sin(2\theta) = 2 \left(-\frac{\sqrt{3}}{3}\right) \left(\frac{\sqrt{2}}{\sqrt{3}}\right) = -\frac{2\sqrt{6}}{9} \] Thus, the final answer is: \[ \sin(2\theta) = -\frac{2\sqrt{6}}{9} \]