Find the exact value of each of the following under the given conditions below. \[ \tan \alpha=-\frac{12}{5}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{1}{2}, 0<\beta<\frac{\pi}{2} \] \( \begin{array}{llll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta) & \text { (d) } \tan (\alpha-\beta)\end{array} \)

To find the values of \(\sin(\alpha + \beta)\), \(\cos(\alpha + \beta)\), \(\sin(\alpha - \beta)\), and \(\tan(\alpha - \beta)\), we first need to find \(\sin \alpha\) and \(\cos \alpha\) based on the given condition for \(\tan \alpha\) and then find \(\sin \beta\) and \(\cos \beta\). **Step 1: Finding \(\sin \alpha\) and \(\cos \alpha\)** Given \(\tan \alpha = -\frac{12}{5}\) in the interval \(\frac{\pi}{2} < \alpha < \pi\): - Since \(\tan\) is negative in the second quadrant, we conclude that \(\sin \alpha > 0\) and \(\cos \alpha < 0\). - Let the opposite side be 12 and the adjacent side be -5 (negative because it's in the second quadrant). Now let's find the hypotenuse \(r\): \[ r = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \] Thus: \[ \sin \alpha = \frac{12}{13}, \quad \cos \alpha = -\frac{5}{13} \] **Step 2: Finding \(\sin \beta\) and \(\cos \beta\)** Given \(\cos \beta = \frac{1}{2}\) in the interval \(0 < \beta < \frac{\pi}{2\): - In this quadrant, both sine and cosine are positive. Thus: \[ \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] **Step 3: Calculating the values** (a) \(\sin(\alpha + \beta)\): \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] \[ = \frac{12}{13} \cdot \frac{1}{2} + \left(-\frac{5}{13}\right) \cdot \frac{\sqrt{3}}{2} \] \[ = \frac{12}{26} - \frac{5\sqrt{3}}{26} = \frac{12 - 5\sqrt{3}}{26} \] (b) \(\cos(\alpha + \beta)\): \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] \[ = \left(-\frac{5}{13}\right) \cdot \frac{1}{2} - \frac{12}{13} \cdot \frac{\sqrt{3}}{2} \] \[ = -\frac{5}{26} - \frac{12\sqrt{3}}{26} = -\frac{5 + 12\sqrt{3}}{26} \] (c) \(\sin(\alpha - \beta)\): \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] \[ = \frac{12}{13} \cdot \frac{1}{2} - \left(-\frac{5}{13}\right) \cdot \frac{\sqrt{3}}{2} \] \[ = \frac{12}{26} + \frac{5\sqrt{3}}{26} = \frac{12 + 5\sqrt{3}}{26} \] (d) \(\tan(\alpha - \beta)\): Using the formula \(\tan(\alpha - \beta)=\frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)}\) and knowing that \(\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\): \[ \cos(\alpha - \beta) = \left(-\frac{5}{13}\right) \cdot \frac{1}{2} + \frac{12}{13} \cdot \frac{\sqrt{3}}{2} \] \[ = -\frac{5}{26} + \frac{12\sqrt{3