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) \( \sin (2 \theta)=\square \) (Type an exact answer, using radicals as needed.)

To solve this, we start with the given information where \( \cos \theta = -\frac{\sqrt{3}}{3} \) and \( \frac{\pi}{2} < \theta < \pi \). Since \( \theta \) is in the second quadrant, we can find \( \sin \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \] So, \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{3}\right)^2 = 1, \] \[ \sin^2 \theta + \frac{3}{9} = 1, \] \[ \sin^2 \theta + \frac{1}{3} = 1, \] \[ \sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}. \] Since \( \sin \theta \) is positive in the second quadrant, \[ \sin \theta = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}. \] Now we can find the values: **(a)** To find \( \sin(2\theta) \): \[ \sin(2\theta) = 2\sin \theta \cos \theta = 2 \left( \frac{\sqrt{6}}{3} \right) \left( -\frac{\sqrt{3}}{3} \right) = -\frac{2\sqrt{18}}{9} = -\frac{6\sqrt{2}}{9} = -\frac{2\sqrt{2}}{3}. \] So, (a) \( \sin(2\theta)=-\frac{2\sqrt{2}}{3} \). Now, we can use it to compute \( (b) \cos(2\theta) \), \( (c) \sin(\frac{\theta}{2}) \), and \( (d) \cos(\frac{\theta}{2}) \). **(b)** To find \( \cos(2\theta) \): \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta = \left(-\frac{\sqrt{3}}{3}\right)^2 - \left(\frac{\sqrt{6}}{3}\right)^2 = \frac{3}{9} - \frac{6}{9} = -\frac{3}{9} = -\frac{1}{3}. \] (b) \( \cos(2\theta)=-\frac{1}{3} \). **(c)** To find \( \sin(\frac{\theta}{2}) \) using the half-angle formula: \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{3}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{\frac{3 + \sqrt{3}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{3}}{6}}. \] (c) \( \sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{3 + \sqrt{3}}{6}} \). **(d)** To find \( \cos(\frac{\theta}{2}) \): \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{\frac{3 - \sqrt{3}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{3}}{6}}. \] (d) \( \cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{3 - \sqrt{3}}{6}} \). In conclusion, (a) \( \sin(2\theta) = -\frac{2\sqrt{2}}{3} \), (b) \( \cos(2\theta) = -\frac{1}{3} \), (c) \( \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{3 + \sqrt{3}}{6}} \), (d) \( \cos\left(\frac{\theta}{2}\right) =