Find the exact value of each of the following under the given conditions below. \[ \tan \alpha=-\frac{12}{5}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{1}{2}, 0<\beta<\frac{\pi}{2} \] (a) \( \sin (\alpha+\beta) \) (b) \( \cos (\alpha+\beta) \) (c) \( \sin (\alpha-\beta) \) (d) \( \boldsymbol{\operatorname { t a n }}(\alpha-\beta) \) (a) \( \sin (\alpha+\beta)=\frac{12-5 \sqrt{3}}{26} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) (b) \( \cos (\alpha+\beta)= \) \( \square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To solve for \( \sin (\alpha + \beta) \) and \( \cos (\alpha + \beta) \), we need to find the values of \( \sin \alpha \) and \( \cos \alpha \), and then we can use the angle addition formulas. Given: \[ \tan \alpha = -\frac{12}{5}, \quad \frac{\pi}{2} < \alpha < \pi \] This tells us that \( \alpha \) is in the second quadrant, where sine is positive and cosine is negative. We can represent \( \tan \alpha \) as: \[ \frac{\sin \alpha}{\cos \alpha} = -\frac{12}{5} \] Let: \[ \sin \alpha = 12k, \quad \cos \alpha = -5k \] Using the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \implies (12k)^2 + (-5k)^2 = 1 \implies 144k^2 + 25k^2 = 1 \implies 169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \frac{1}{13} \] Substituting back, we get: \[ \sin \alpha = 12 \left(\frac{1}{13}\right) = \frac{12}{13}, \quad \cos \alpha = -5 \left(\frac{1}{13}\right) = -\frac{5}{13} \] Next, we have: \[ \cos \beta = \frac{1}{2}, \quad 0 < \beta < \frac{\pi}{2 \Rightarrow \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Now, we can compute \( \sin (\alpha + \beta) \) and \( \cos (\alpha + \beta) \): (a) For \( \sin (\alpha + \beta) \): \[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] Substituting in the values: \[ \sin (\alpha + \beta) = \left(\frac{12}{13}\right) \left(\frac{1}{2}\right) + \left(-\frac{5}{13}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{12}{26} - \frac{5\sqrt{3}}{26} = \frac{12 - 5\sqrt{3}}{26} \] Thus, the exact value is: \[ \sin (\alpha + \beta) = \frac{12 - 5\sqrt{3}}{26} \] (b) For \( \cos (\alpha + \beta) \): \[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] Substituting our values: \[ \cos (\alpha + \beta) = \left(-\frac{5}{13}\right) \left(\frac{1}{2}\right) - \left(\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) = -\frac{5}{26} - \frac{12\sqrt{3}}{26} = -\frac{5 + 12\sqrt{3}}{26} \] Thus, the exact value is: \[ \cos (\alpha + \beta) = -\frac{5 + 12\sqrt{3}}{26} \]