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. \( \begin{array}{l}\text { (a) } \sin (2 \theta),(b) \cos (2 \theta), \text { (c) } \sin \frac{\theta}{2} \text {, and (d) } \cos \frac{\theta}{2} \\ \text { (a) } \sin (2 \theta)=-\frac{16}{65} \text { (Type an exact answer, using radicals as needed.) } \\ \text { (b) } \cos (2 \theta)=\frac{63}{65} \text { (Type an exact answer, using radicals as needed.) } \\ \text { (c) } \sin \frac{\theta}{2}=\square \text { (Type an exact answer, using radicals as needed.) }\end{array} \) (2).

To find the exact values, we start from \( \cot \theta = -8 \). The positive cotangent value indicates that we are in the second quadrant where sine is positive and cosine is negative. This gives us: 1. From \( \cot \theta = \frac{\cos \theta}{\sin \theta} = -8 \): We can let \( \sin \theta = 1k = k \) and \( \cos \theta = -8k \). 2. Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ k^2 + (-8k)^2 = 1 \implies k^2 + 64k^2 = 1 \implies 65k^2 = 1 \implies k^2 = \frac{1}{65} \implies k = \frac{1}{\sqrt{65}}. \] Thus, we have: \[ \sin \theta = \frac{1}{\sqrt{65}}, \quad \cos \theta = -\frac{8}{\sqrt{65}}. \] Now we can find the required values: **(a)** For \( \sin (2\theta) \): \[ \sin (2\theta) = 2 \sin \theta \cos \theta = 2 \left(\frac{1}{\sqrt{65}}\right) \left(-\frac{8}{\sqrt{65}}\right) = -\frac{16}{65}. \] Thus, \( \sin (2\theta) = -\frac{16}{65}. \) **(b)** For \( \cos (2\theta) \): \[ \cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \left(-\frac{8}{\sqrt{65}}\right)^2 - \left(\frac{1}{\sqrt{65}}\right)^2 = \frac{64}{65} - \frac{1}{65} = \frac{63}{65}. \] Thus, \( \cos (2\theta) = \frac{63}{65}. \) **(c)** For \( \sin \frac{\theta}{2} \): Using the half-angle formula, \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - (-\frac{8}{\sqrt{65}})}{2}} = \sqrt{\frac{1 + \frac{8}{\sqrt{65}}}{2}}. \] Putting the fractions under a common denominator: \[ 1 + \frac{8}{\sqrt{65}} = \frac{\sqrt{65}}{\sqrt{65}} + \frac{8}{\sqrt{65}} = \frac{\sqrt{65} + 8}{\sqrt{65}}. \] Thus, \[ \sin \frac{\theta}{2} = \sqrt{\frac{\frac{\sqrt{65} + 8}{\sqrt{65}}}{2}} = \sqrt{\frac{\sqrt{65} + 8}{2\sqrt{65}}} = \frac{\sqrt{\sqrt{65} + 8}}{\sqrt{2\sqrt{65}}}. \] This method involves evaluating radical expressions accurately. **Final Answers:** \[ (a) \sin (2\theta) = -\frac{16}{65}, \quad (b) \cos (2\theta) = \frac{63}{65}, \quad (c) \sin \frac{\theta}{2} = \frac{\sqrt{\sqrt{65}+8}}{\sqrt{2\sqrt{65}}}. \] *Note: The (d) \( \cos \frac{\theta}{2} \) can be found similarly using \( \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \).*