Given that \( \tan \theta=\frac{6}{5}, \pi<\theta<\frac{3 \pi}{2} \), find the exact value of each of the following. \( \begin{array}{lllll}\text { (a) } \sin (2 \theta) & \text { (b) } \cos (2 \theta) & \text { (c) } \sin \frac{\theta}{2} & \text { (d) } \cos \frac{\theta}{2} & \text { (e) } \tan (2 \theta) \\ \text { (f) } \tan \frac{\theta}{2}\end{array} \)

Given that \( \tan \theta = \frac{6}{5} \) and \( \pi < \theta < \frac{3\pi}{2} \), we know that both sine and cosine for \( \theta \) will be negative since \( \theta \) is in the third quadrant. ### Step 1: Find \( \sin \theta \) and \( \cos \theta \) Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \): Let \( \sin \theta = -6k \) and \( \cos \theta = -5k \) for some positive \( k \). Then: \[ \tan \theta = \frac{-6k}{-5k} = \frac{6}{5} \] To find \( k \), we use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \implies (-6k)^2 + (-5k)^2 = 1 \] \[ 36k^2 + 25k^2 = 1 \implies 61k^2 = 1 \implies k^2 = \frac{1}{61} \implies k = \frac{1}{\sqrt{61}} \] So: \[ \sin \theta = -\frac{6}{\sqrt{61}}, \quad \cos \theta = -\frac{5}{\sqrt{61}} \] ### Step 2: Find \( \sin(2\theta) \) and \( \cos(2\theta) \) Using the double angle formulas: \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] \[ = 2 \left(-\frac{6}{\sqrt{61}}\right) \left(-\frac{5}{\sqrt{61}}\right) = 2 \cdot \frac{30}{61} = \frac{60}{61} \] \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] \[ = \left(-\frac{5}{\sqrt{61}}\right)^2 - \left(-\frac{6}{\sqrt{61}}\right)^2 = \frac{25}{61} - \frac{36}{61} = -\frac{11}{61} \] ### Step 3: Find \( \sin\frac{\theta}{2} \) and \( \cos\frac{\theta}{2} \) Using the half-angle formulas: \[ \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} \] First, calculate \( 1 - \cos \theta \): \[ 1 - \cos \theta = 1 + \frac{5}{\sqrt{61}} = \frac{\sqrt{61} + 5}{\sqrt{61}} \] Then: \[ \sin\frac{\theta}{2} = \sqrt{\frac{\frac{\sqrt{61} + 5}{\sqrt{61}}}{2}} = \sqrt{\frac{\sqrt{61} + 5}{2\sqrt{61}}} = \frac{\sqrt{\sqrt{61} + 5}}{\sqrt{2\sqrt{61}}} \] Next, calculate \( \cos\frac{\theta}{2} \): \[ \cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \] Calculating \( 1 + \cos \theta \) gives: \[ 1 + \cos \theta = 1 - \frac{5}{\sqrt{61}} = \frac{\sqrt{61} - 5}{\sqrt{61}} \] Then: \[ \cos\frac{\theta}{2} = \sqrt{\frac{\frac{\sqrt{61} - 5}{\sqrt{61}}}{2}} = \sqrt{\frac{\sqrt{61} - 5}{2\sqrt{61}}} = \frac{\sqrt{\sqrt{61} - 5}}{\sqrt{2\sqrt{61}}} \] ### Step 4: Find \( \tan(2\theta) \) Using the identity: \[ \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \] Calculating \( \tan^2 \theta \): \[ \tan^2 \theta = \left(\frac{6}{5}\right)^2 = \frac{36}{25} \] So: