Question \( 3 \quad[9] \) Find the exact values \( \begin{array}{l}\text { (a) } \sin \left(\frac{19 \pi}{12}\right) \\ \text { (Do not convert to degrees!!) } \\ \text { (b) } \sin \left(\cos ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)\right)\end{array} \)

We wish to find: \[ \text{(a) } \sin\left(\frac{19\pi}{12}\right) \] \[ \text{(b) } \sin\left(\cos^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{12}{13}\right)\right) \] ### Part (a) 1. First, express \(\frac{19\pi}{12}\) in terms of a known angle. Notice that \[ \frac{19\pi}{12} = \pi + \frac{7\pi}{12}. \] 2. Since \(\sin(\pi + \theta) = -\sin(\theta)\), we have \[ \sin\left(\frac{19\pi}{12}\right) = -\sin\left(\frac{7\pi}{12}\right). \] 3. Write \(\frac{7\pi}{12}\) as the sum of two angles whose trigonometric values are known. We can write \[ \frac{7\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}. \] 4. Using the sine addition formula: \[ \sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \sin\frac{\pi}{4}\cos\frac{\pi}{3} + \cos\frac{\pi}{4}\sin\frac{\pi}{3}. \] 5. Substitute the known values: \[ \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \] \[ \cos\frac{\pi}{3} = \frac{1}{2}, \quad \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}. \] 6. Compute the expression: \[ \sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \frac{\sqrt{2}}{2}\cdot\frac{1}{2} + \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{2}+\sqrt{6}}{4}. \] 7. Therefore, \[ \sin\left(\frac{19\pi}{12}\right) = -\sin\left(\frac{7\pi}{12}\right) = -\frac{\sqrt{2}+\sqrt{6}}{4}. \] ### Part (b) 1. Let \[ A = \cos^{-1}\left(\frac{3}{5}\right) \quad \text{and} \quad B = \sin^{-1}\left(\frac{12}{13}\right). \] Then we know: \[ \cos A = \frac{3}{5}, \quad \sin B = \frac{12}{13}. \] 2. Find \(\sin A\) using the Pythagorean identity: \[ \sin A = \sqrt{1-\cos^2A} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}. \] 3. Find \(\cos B\) similarly: \[ \cos B = \sqrt{1-\sin^2B} = \sqrt{1-\left(\frac{12}{13}\right)^2} = \sqrt{1-\frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}. \] 4. Now, use the sine addition formula: \[ \sin(A+B) = \sin A\cos B + \cos A\sin B. \] 5. Substitute the known values: \[ \sin(A+B) = \frac{4}{5}\cdot\frac{5}{13} + \frac{3}{5}\cdot\frac{12}{13} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}. \] ### Final Answers \[ \boxed{\sin\left(\frac{19\pi}{12}\right) = -\frac{\sqrt{2}+\sqrt{6}}{4}} \] \[ \boxed{\sin\left(\cos^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{12}{13}\right)\right) = \frac{56}{65}} \]