Use the information given about the angle \( \theta, \cot \theta=-4, \sec \theta<0,0 \leq \theta<2 \pi \), to find the exact values \( 5 f \) the following. (a) \( \sin (2 \theta) \), (b) \( \cos (2 \theta) \), (c) \( \sin \frac{\theta}{2} \), and (d) \( \cos \frac{\theta}{2} \) (a) \( \sin (2 \theta)=\square \) (Type an exact answer, using radicals as needed.)

To solve this problem, we first need to determine the relevant trigonometric ratios based on the given values of \( \cot \theta \) and \( \sec \theta \). Since \( \cot \theta = -4 \), we have: \[ \tan \theta = -\frac{1}{4} \] This indicates that \( \theta \) is in the second quadrant (since \( \sec \theta < 0 \)). Let’s use the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \): 1. We can set \( \cos \theta = -4k \) and \( \sin \theta = k \) for some \( k \). 2. Thus, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{k}{-4k} = -\frac{1}{4} \). Next, we apply the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting in our expressions: \[ k^2 + (-4k)^2 = 1 \Rightarrow k^2 + 16k^2 = 1 \Rightarrow 17k^2 = 1 \Rightarrow k^2 = \frac{1}{17} \Rightarrow k = \frac{1}{\sqrt{17}} \] Hence, we compute: - \( \sin \theta = k = \frac{1}{\sqrt{17}} \) - \( \cos \theta = -4k = -\frac{4}{\sqrt{17}} \) Now, we calculate the values required: **Step 1: Find \(\sin(2\theta)\)**: Using the double angle formula: \[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \] Substituting the values: \[ \sin(2\theta) = 2 \cdot \frac{1}{\sqrt{17}} \cdot \left(-\frac{4}{\sqrt{17}}\right) = 2 \cdot \frac{-4}{17} = -\frac{8}{17} \] **The exact value is:** (a) \( \sin(2\theta) = -\frac{8}{17} \)