Use the information given about the angle \( \theta, \cos \theta=-\frac{\sqrt{3}}{3}, \frac{\pi}{2}<\theta<\pi \), to find the exact values of the foliowing, \( \begin{array}{llll}\text { (a) } \sin (2 \theta) & \text { (b) } \cos (2 \theta) & \text { (c) } \sin \frac{\theta}{2} & \text { (d) } \cos \frac{\theta}{2}\end{array} \) (a) \( \sin (2 \theta)=-\frac{2 \sqrt{2}}{3} \) (Type an exact answer, using radicals as needed.) (b) \( \cos (2 \theta)=\square \) (Type an exact answer, using radicals as needed.)

To find the values of \( \sin(2\theta) \) and \( \cos(2\theta) \), we can use the double angle identities and the given information: 1. Given \( \cos \theta = -\frac{\sqrt{3}}{3} \) and \( \frac{\pi}{2} < \theta < \pi \), we can determine \( \sin \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting for \( \cos \theta \): \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{3}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{1}{3} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3} \] Thus, \( \sin \theta = \sqrt{\frac{2}{3}} \) or \( \sin \theta = -\sqrt{\frac{2}{3}} \). Since \( \theta \) is in the second quadrant, we take the positive solution: \( \sin \theta = \frac{\sqrt{6}}{3} \). 2. Now we can find \( \sin(2\theta) \) and \( \cos(2\theta) \): - Using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \): \[ \sin(2\theta) = 2 \cdot \frac{\sqrt{6}}{3} \cdot \left(-\frac{\sqrt{3}}{3}\right) = -\frac{2\sqrt{18}}{9} = -\frac{6\sqrt{2}}{9} = -\frac{2\sqrt{2}}{3}. \] For (a): \( \sin(2 \theta) = -\frac{2 \sqrt{2}}{3} \). - Next, using the identity \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \): \[ \cos(2\theta) = \left(-\frac{\sqrt{3}}{3}\right)^2 - \left(\frac{\sqrt{6}}{3}\right)^2 = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}. \] For (b): \( \cos(2 \theta) = -\frac{1}{3} \). In summary: (a) \( \sin(2 \theta) = -\frac{2 \sqrt{2}}{3} \) (b) \( \cos(2 \theta) = -\frac{1}{3} \)