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

To find the exact values of the trigonometric functions based on the given conditions, let's start with analyzing the information provided. Given \( \cot \theta = -8 \), we know that the tangent function is negative, indicating that \( \theta \) is in the second quadrant since \( \sec \theta < 0 \) (the cosine is also negative in this quadrant). From \( \cot \theta = \frac{\cos \theta}{\sin \theta} = -8 \), we can express this in terms of sine and cosine. We can assign values based on a right triangle: - Let \( \cos \theta = -8k \) and \( \sin \theta = k \) for some \( k > 0 \). - Then, using the Pythagorean identity, we find \( (-8k)^2 + (k)^2 = 1 \), which simplifies to: \[ 64k^2 + k^2 = 1 \implies 65k^2 = 1 \implies k^2 = \frac{1}{65} \implies k = \frac{1}{\sqrt{65}} \] Thus: \[ \sin \theta = \frac{1}{\sqrt{65}}, \quad \cos \theta = -\frac{8}{\sqrt{65}} \] Next, we can calculate the double angle identities. **(a)** To find \( \sin(2\theta) \): \[ \sin(2\theta) = 2 \sin \theta \cos \theta = 2 \left(\frac{1}{\sqrt{65}}\right) \left(-\frac{8}{\sqrt{65}}\right) = -\frac{16}{65} \] **(b)** To find \( \cos(2\theta) \): \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] First, calculate \( \cos^2 \theta \) and \( \sin^2 \theta \): \[ \cos^2 \theta = \left(-\frac{8}{\sqrt{65}}\right)^2 = \frac{64}{65} \] \[ \sin^2 \theta = \left(\frac{1}{\sqrt{65}}\right)^2 = \frac{1}{65} \] Now, substituting these into \( \cos(2\theta) \): \[ \cos(2\theta) = \frac{64}{65} - \frac{1}{65} = \frac{63}{65} \] Thus, the exact values are: (a) \( \sin(2\theta) = -\frac{16}{65} \) (b) \( \cos(2\theta) = \frac{63}{65} \) For parts (c) and (d), we can use the half-angle formulas: **(c)** Finding \( \sin \frac{\theta}{2} \): \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - \left(-\frac{8}{\sqrt{65}}\right)}{2}} = \sqrt{\frac{1 + \frac{8}{\sqrt{65}}}{2}} = \sqrt{\frac{\frac{65 + 8\sqrt{65}}{65}}{2}} = \frac{\sqrt{65 + 8\sqrt{65}}}{\sqrt{130}} \] **(d)** Finding \( \cos \frac{\theta}{2} \): \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 - \frac{8}{\sqrt{65}}}{2}} = \sqrt{\frac{\frac{65 - 8\sqrt{65}}{65}}{2}} = \frac{\sqrt{65 - 8\sqrt{65}}}{\sqrt{130}} \] Now we have the required values: - \( \sin(2 \theta) = -\frac{16}{65} \) - \( \cos(2 \theta) = \frac{63}{65} \)