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

To find the various trigonometric values based on the given conditions, we'll first determine the sine and cosine values of angles \(\alpha\) and \(\beta\). **For \(\alpha\)**: Given \(\tan \alpha = -\frac{5}{12}\) and \(\frac{\pi}{2} < \alpha < \pi\), we note \(\alpha\) is in the second quadrant where sine is positive and cosine is negative. Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \): 1. From \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{5}{12}\), we assign \(\sin \alpha = -5k\) and \(\cos \alpha = 12k\) for some \(k > 0\). 2. The Pythagorean identity gives: \[ (-5k)^2 + (12k)^2 = 1 \implies 25k^2 + 144k^2 = 1 \implies 169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \frac{1}{13} \] 3. Thus, we can compute: \[ \sin \alpha = -5 \cdot \frac{1}{13} = \frac{-5}{13} \quad \text{(since it's negative in the second quadrant)} \] \[ \cos \alpha = 12 \cdot \frac{1}{13} = \frac{12}{13} \quad \text{(it should be negative in the second quadrant, so it's } -\frac{12}{13} \text{)} \] **For \(\beta\)**: Given \(\cos \beta = \frac{\sqrt{3}}{2}\) and \(0 < \beta < \frac{\pi}{2\), which means \(\beta\) is in the first quadrant. The sine can easily be found using: \[ \sin^2 \beta + \cos^2 \beta = 1 \implies \sin^2 \beta = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4} \implies \sin \beta = \frac{1}{2} \] Now that we have: - \(\sin \alpha = \frac{-5}{13}\) - \(\cos \alpha = -\frac{12}{13}\) - \(\sin \beta = \frac{1}{2}\) - \(\cos \beta = \frac{\sqrt{3}}{2}\) We can proceed to solve each part. ### (a) Calculating \(\sin(\alpha + \beta)\): Using the sine addition formula: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] Substituting the values: \[ \sin(\alpha + \beta) = \left(\frac{-5}{13}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{12}{13}\right) \left(\frac{1}{2}\right) = \frac{-5\sqrt{3}}{26} - \frac{12}{26} = \frac{-5\sqrt{3} - 12}{26} \] Thus, \( \sin(\alpha + \beta) = \frac{-5\sqrt{3} - 12}{26} \). ### Final Answer: (a) \( \sin (\alpha+\beta)=\frac{-5\sqrt{3} - 12}{26} \)