Use the information given about the angle \( \theta, \cot \theta=-8, \sec \theta<0,0 \leq \theta<2 \pi \), to find the exact values of the following. \( \begin{array}{l}\text { (a) } \sin (2 \theta), \text { (b) } \cos (2 \theta), \text { (c) } \sin \frac{\theta}{2} \text {, and (d) } \cos \frac{\theta}{2} \\ \text { (a) } \sin (2 \theta)=\square \text { (Type an exact answer, using radicals as needed.) }\end{array} \)

To solve the problem, we start by using the information given about \( \cot \theta \) and \( \sec \theta \). Since \( \cot \theta = -8 \), we have: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = -8 \] This indicates that \( \cos \theta \) and \( \sin \theta \) have different signs. Additionally, \( \sec \theta < 0 \) tells us that \( \cos \theta < 0 \). Therefore, \( \sin \theta > 0 \), which places \( \theta \) in the second quadrant where sine is positive and cosine is negative. We can express \( \sin \theta \) and \( \cos \theta \) in terms of a right triangle where \( \cos \theta = -8k \) and \( \sin \theta = k \). Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ (k)^2 + (-8k)^2 = 1 \] \[ k^2 + 64k^2 = 1 \] \[ 65k^2 = 1 \implies k^2 = \frac{1}{65} \implies k = \frac{1}{\sqrt{65}} \] This gives: \[ \sin \theta = \frac{1}{\sqrt{65}}, \quad \cos \theta = -\frac{8}{\sqrt{65}} \] Now, we can find \( \sin 2\theta \) and \( \cos 2\theta \): \[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \left(\frac{1}{\sqrt{65}}\right) \left(-\frac{8}{\sqrt{65}}\right) = -\frac{16}{65} \] Thus, \[ \text{(a) } \sin (2 \theta) = -\frac{16}{65} \] Now, for \( \cos 2\theta \): \[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(-\frac{8}{\sqrt{65}}\right)^2 - \left(\frac{1}{\sqrt{65}}\right)^2 = \frac{64}{65} - \frac{1}{65} = \frac{63}{65} \] So, \[ \text{(b) } \cos (2 \theta) = \frac{63}{65} \] Next, we find \( \sin \frac{\theta}{2} \) and \( \cos \frac{\theta}{2} \) using the half-angle formulas: \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 + \frac{8}{\sqrt{65}}}{2}} = \sqrt{\frac{\frac{\sqrt{65} + 8}{\sqrt{65}}}{2}} = \frac{\sqrt{\sqrt{65} + 8}}{\sqrt{130}} \] And for \( \cos \frac{\theta}{2} \): \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 - \frac{8}{\sqrt{65}}}{2}} = \sqrt{\frac{\frac{\sqrt{65} - 8}{\sqrt{65}}}{2}} = \frac{\sqrt{\sqrt{65} - 8}}{\sqrt{130}} \] Thus, the final answers are: (a) \( \sin (2 \theta) = -\frac{16}{65} \) (b) \( \cos (2 \theta) = \frac{63}{65} \) (c) \( \sin \frac{\theta}{2} = \frac{\sqrt{\sqrt{65} + 8}}{\sqrt{130}} \) (d) \( \cos \frac{\theta}{2} = \frac{\sqrt{\sqrt{65} - 8}}{\sqrt{130}} \)