Question 1 Find the value of \( \sin \left(\frac{19 \pi}{12}\right) \)

1. Express the angle in a form that utilizes a known sine property: \[ \frac{19\pi}{12} = \pi + \frac{7\pi}{12} \] Since \[ \sin(\pi + \theta) = -\sin \theta, \] we have \[ \sin\left(\frac{19\pi}{12}\right) = -\sin\left(\frac{7\pi}{12}\right). \] 2. Decompose \(\frac{7\pi}{12}\) into the sum of two angles whose sine and cosine values are known: \[ \frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}. \] 3. Use the sine sum formula: \[ \sin\left(A+B\right) = \sin A \cos B + \cos A \sin B. \] Let \[ A = \frac{\pi}{3} \quad \text{and} \quad B = \frac{\pi}{4}. \] 4. Substitute the known values: \[ \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}, \quad \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}. \] Then, \[ \sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}. \] 5. Simplify the expression: \[ \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4}, \] \[ \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}. \] Therefore, \[ \sin\left(\frac{7\pi}{12}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}. \] 6. Substitute back to obtain the desired result: \[ \sin\left(\frac{19\pi}{12}\right) = -\sin\left(\frac{7\pi}{12}\right) = -\frac{\sqrt{6}+\sqrt{2}}{4}. \] Final answer: \[ \boxed{-\frac{\sqrt{6}+\sqrt{2}}{4}} \]