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. \( \begin{array}{llll}\text { (a) } \sin (2 \theta) & \text { (b) } \cos (2 \theta) & \text { (c) } \sin \frac{\theta}{2} & \text { (d) } \cos \frac{\theta}{2} \\ \text { (a) } \sin (2 \theta)=-\frac{2 \sqrt{2}}{3} & \text { (Type an exact answer, using radicals as needed.) } \\ \text { (b) } \cos (2 \theta)=-\frac{1}{3} \text { (Type an exact answer, using radicals as needed.) } \\ \text { (c) } \sin \frac{\theta}{2}=\square \text { (Type an exact answer, using radicals as needed.) }\end{array} \)

To solve the problem, we start with the known condition \( \cos \theta = -\frac{\sqrt{3}}{3} \) and the range \( \frac{\pi}{2} < \theta < \pi \). This indicates that \( \theta \) is in the second quadrant. ### Step 1: Find \( \sin \theta \) Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting the value of \( \cos \theta \): \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{3}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{3}{9} = 1 \] \[ \sin^2 \theta + \frac{1}{3} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3} \] Taking the square root, since \( \sin \theta \) is positive in the second quadrant: \[ \sin \theta = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \] ### Step 2: Calculate \( \sin(2\theta) \) and \( \cos(2\theta) \) Using the double angle formulas: \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] #### (a) Calculate \( \sin(2\theta) \): \[ \sin(2\theta) = 2 \left(\frac{\sqrt{6}}{3}\right) \left(-\frac{\sqrt{3}}{3}\right) = 2 \cdot \frac{\sqrt{6} \cdot (-\sqrt{3})}{9} = -\frac{2\sqrt{18}}{9} = -\frac{6\sqrt{2}}{9} = -\frac{2\sqrt{2}}{3} \] #### (b) Calculate \( \cos(2\theta) \): \[ \cos(2\theta) = \left(-\frac{\sqrt{3}}{3}\right)^2 - \left(\frac{\sqrt{6}}{3}\right)^2 \] \[ = \frac{3}{9} - \frac{6}{9} = -\frac{3}{9} = -\frac{1}{3} \] ### Step 3: Calculate \( \sin\left(\frac{\theta}{2}\right) \) and \( \cos\left(\frac{\theta}{2}\right) \) Using the half-angle formulas: \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}} \] \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} \] #### (c) Calculate \( \sin\left(\frac{\theta}{2}\right) \): \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{3}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{\frac{3}{3} + \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{3}}{6}} \] #### (d) Calculate \( \cos\left(\frac{\theta}{2}\right) \): \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \left(-\frac{\sqrt{3}}{3}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{\frac{3}{3} - \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{3}}{6}} \] ### Final Answers - (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) \( \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{3 - \sqrt{3}}{6}} \)