(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: \( \begin{array}{ll}\text { (1) } \sin (\alpha-\beta) & \text { (2) } \cos (\alpha+\beta) \\ \text { (4) } \tan 2 \alpha & \end{array} \)

To solve the equations step by step, let's first tackle the trigonometric equations provided. ### Finding \(\alpha\) Given the equation: \[ 5 \cos \alpha + 4 = 0 \] We can rearrange it to find \(\cos \alpha\): \[ 5 \cos \alpha = -4 \implies \cos \alpha = -\frac{4}{5} \] In the interval \([0^\circ, 180^\circ]\), \(\cos \alpha\) is negative in the second quadrant, implying that: \[ \alpha = 180^\circ - \cos^{-1}\left(\frac{4}{5}\right) \] From the known values of a right triangle with opposite side \(4\), adjacent side \(3\) (since \(3^2 + 4^2 = 5^2\)), the sine value becomes: \[ \sin \alpha = \frac{4}{5} \] ### Finding \(\beta\) Next, consider the equation: \[ 13 \sin \beta + 12 = 0 \] Rearranging gives: \[ 13 \sin \beta = -12 \implies \sin \beta = -\frac{12}{13} \] Since \(\tan \beta > 0\), \(\beta\) must be in the third quadrant (where sine is negative and cosine is also negative). Thus, we can find \(\cos \beta\) using the Pythagorean identity: \[ \cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \implies \cos \beta = -\frac{5}{13} \] Now we have: \[ \sin \beta = -\frac{12}{13}, \quad \cos \beta = -\frac{5}{13} \] ### Finding \( \sin(\alpha - \beta) \) Using the sine angle subtraction formula: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] Substituting the values obtained: \[ \sin(\alpha - \beta) = \left(\frac{4}{5}\right)\left(-\frac{5}{13}\right) - \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) \] Calculating each term: \[ \sin(\alpha - \beta) = -\frac{4}{13} - \frac{48}{65} \] Finding a common denominator yields: \[ = -\frac{20}{65} - \frac{48}{65} = -\frac{68}{65} \] ### Finding \( \cos(\alpha + \beta) \) Using the cosine angle addition formula: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] Substituting the known values: \[ \cos(\alpha + \beta) = \left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right) - \left(\frac{4}{5}\right)\left(-\frac{12}{13}\right) \] Calculating each term: \[ = \frac{20}{65} + \frac{48}{65} = \frac{68}{65} \] ### Finding \( \tan 2\alpha \) Using the double angle formula: \[ \tan(2\alpha) = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} \] Finding \(\tan \alpha\): \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{4}{5}}{-\frac{4}{5}} = -1 \] Substituting into the double angle formula: \[ \tan(2\alpha) = \frac{2(-1)}{1 - (-1)^2} = \frac{-2}{1 - 1} \text{ (indeterminate form, undefined)} \] Final answers: \[ \sin(\alpha - \beta) = -\frac{68}{65}, \quad \cos(\alpha + \beta) = \frac{68}{65}, \quad \tan(2\alpha) = \text{undefined} \]