Find the exact value of each of the following under the given conditions below. \[ \tan \alpha=-\frac{12}{5}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{1}{2}, 0<\beta<\frac{\pi}{2} \] \( \begin{array}{llll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta) & \text { (d) } \tan (\alpha-\beta)\end{array} \) (a) \( \sin (\alpha+\beta)=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To solve for \(\sin(\alpha+\beta)\), \(\cos(\alpha+\beta)\), \(\sin(\alpha-\beta)\), and \(\tan(\alpha-\beta)\), we first need to determine the values of \(\sin \alpha\), \(\cos \alpha\), and \(\sin \beta\), \(\cos \beta\) based on the given conditions. ### Step 1: Determine \(\sin \alpha\) and \(\cos \alpha\) Since \(\tan \alpha = -\frac{12}{5}\) and \(\alpha\) is in the second quadrant (where sine is positive and cosine is negative), we can use the identity: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \] Let \(\sin \alpha = 12k\) and \(\cos \alpha = -5k\) for some positive \(k\). Then we find \(k\) using the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] \[ (12k)^2 + (-5k)^2 = 1 \quad \Rightarrow \quad 144k^2 + 25k^2 = 1 \quad \Rightarrow \quad 169k^2 = 1 \quad \Rightarrow \quad k^2 = \frac{1}{169} \quad \Rightarrow \quad k = \frac{1}{13} \] Now we can find \(\sin \alpha\) and \(\cos \alpha\): \[ \sin \alpha = 12k = \frac{12}{13}, \quad \cos \alpha = -5k = -\frac{5}{13} \] ### Step 2: Determine \(\sin \beta\) and \(\cos \beta\) For \(\beta\), since \(\cos \beta = \frac{1}{2}\) and \(0 < \beta < \frac{\pi}{2}\), we can find \(\sin \beta\) using the Pythagorean identity: \[ \sin^2 \beta + \cos^2 \beta = 1 \] \[ \sin^2 \beta + \left(\frac{1}{2}\right)^2 = 1 \quad \Rightarrow \quad \sin^2 \beta + \frac{1}{4} = 1 \quad \Rightarrow \quad \sin^2 \beta = \frac{3}{4} \quad \Rightarrow \quad \sin \beta = \frac{\sqrt{3}}{2} \] ### Step 3: Calculate the required values Now we can calculate each part using the sine and cosine addition formulas: (a) \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\): \[ \sin(\alpha + \beta) = \left(\frac{12}{13}\right)\left(\frac{1}{2}\right) + \left(-\frac{5}{13}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{12}{26} - \frac{5\sqrt{3}}{26} = \frac{12 - 5\sqrt{3}}{26} \] (b) \(\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\): \[ \cos(\alpha + \beta) = \left(-\frac{5}{13}\right)\left(\frac{1}{2}\right) - \left(\frac{12}{13}\right)\left(\frac{\sqrt{3}}{2}\right) = -\frac{5}{26} - \frac{12\sqrt{3}}{26} = -\frac{5 + 12\sqrt{3}}{26} \] (c) \(\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\): \[ \sin(\alpha - \beta) = \left(\frac{12}{13}\right)\left(\frac{1}{2}\right) - \left(-\frac{5}{13}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{12}{26} + \frac{5\sqrt{3}}{26} = \frac{12 + 5\sqrt{3}}{26} \] (d) \(\tan(\alpha - \beta) = \frac{\sin(\alpha - \