(b) If \( 5 \cos \alpha+4=0 \) with \( \alpha \in\left[0^{\circ}: 180^{\circ}\right] \) and \( 13 \sin \beta+12=0 \) with \( \tan \beta>0 \), determine the value of the following without the use of a calculator and with the aid of a diagram: \( \begin{array}{lll}\text { (1) } \sin (\alpha-\beta) & \text { (2) } \cos (\alpha+\beta) \\ \text { (4) } \tan 2 \alpha & & \text { (3) } \cos 2 \alpha\end{array} \)

Given the equations \[ 5\cos \alpha + 4 = 0 \quad \text{and} \quad 13\sin \beta + 12 = 0, \] we first solve for \(\cos \alpha\) and \(\sin \beta\): \[ \cos \alpha = -\frac{4}{5}, \quad \sin \beta = -\frac{12}{13}. \] Since \(\alpha \in [0^\circ, 180^\circ]\) and \(\cos \alpha < 0\), it follows that \(\alpha\) is in the second quadrant, where \(\sin \alpha\) is positive. Using the Pythagorean identity: \[ \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}, \] \[ \sin \alpha = \frac{3}{5}. \] For \(\beta\), the condition \(\tan \beta > 0\) tells us that sine and cosine have the same sign. Since \(\sin \beta\) is negative, \(\beta\) must lie in the third quadrant. Then: \[ \cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}, \] \[ \cos \beta = -\frac{5}{13}. \] Now we proceed to compute each expression: --- 1. **Calculate \(\sin (\alpha-\beta)\):** Using the angle difference formula: \[ \sin (\alpha-\beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta, \] substitute the known values: \[ \sin (\alpha-\beta) = \frac{3}{5}\left(-\frac{5}{13}\right) - \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right). \] Simplify the products: \[ \frac{3}{5}\left(-\frac{5}{13}\right) = -\frac{15}{65}, \] \[ \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) = \frac{48}{65}. \] Thus, \[ \sin (\alpha-\beta) = -\frac{15}{65} - \frac{48}{65} = -\frac{63}{65}. \] --- 2. **Calculate \(\cos (\alpha+\beta)\):** Using the angle sum formula: \[ \cos (\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta, \] substitute in the values: \[ \cos (\alpha+\beta) = \left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right) - \frac{3}{5}\left(-\frac{12}{13}\right). \] Simplify: \[ \left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right) = \frac{20}{65}, \] \[ \frac{3}{5}\left(-\frac{12}{13}\right) = -\frac{36}{65}. \] Thus, \[ \cos (\alpha+\beta) = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}. \] --- 3. **Calculate \(\cos 2\alpha\):** Using the double-angle formula: \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha, \] substitute \(\cos^2 \alpha = \frac{16}{25}\) and \(\sin^2 \alpha = \frac{9}{25}\): \[ \cos 2\alpha = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}. \] --- 4. **Calculate \(\tan 2\alpha\):** First, find \(\tan \alpha\): \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}. \] Then use the tangent double-angle formula: \[ \tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2 \alpha}. \] Substitute \(\tan \alpha = -\frac{3}{4}\): \[ \tan 2\alpha = \frac{2\left(-\frac{3}{4}\right)}{1-\left(-\frac{3}{4}\right)^2} = \frac{-\frac{3}{2}}{1-\frac{9}{16}}. \] Simplify the denominator: \[ 1-\frac{9}{16} = \frac{16}{16}-\frac{9}{16} = \frac{7}{16}. \] Thus, \[ \tan 2\alpha = -\frac{3}{2} \cdot \frac{16}{7} = -\frac{48}{14} = -\frac{24}{7}. \] --- Final answers: \[ \sin (\alpha-\beta) = -\frac{63}{65}, \] \[ \cos (\alpha+\beta) = \frac{56}{65}, \] \[ \cos 2\alpha = \frac{7}{25}, \] \[ \tan 2\alpha = -\frac{24}{7}. \]