Calculate an exact answer using a formula learned in class. $\sin \frac{19 \pi}{12}$

Step 1 ：Given the expression \(\sin \frac{19 \pi}{12}\), we need to find its exact value.

Step 2 ：The angle \(\frac{19\pi}{12}\) is not a standard angle on the unit circle. However, it can be expressed as a sum or difference of angles that are standard angles on the unit circle.

Step 3 ：The standard angles on the unit circle are multiples of \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\).

Step 4 ：We can express \(\frac{19\pi}{12}\) as \(\frac{16\pi}{12} + \frac{3\pi}{12}\), which simplifies to \(\frac{4\pi}{3} + \frac{\pi}{4}\).

Step 5 ：Both \(\frac{4\pi}{3}\) and \(\frac{\pi}{4}\) are standard angles on the unit circle.

Step 6 ：We can use the sine addition formula, \(\sin(a + b) = \sin a \cos b + \cos a \sin b\), to find the exact value of \(\sin \frac{19\pi}{12}\).

Step 7 ：The exact values of \(\sin \frac{4\pi}{3}\) and \(\cos \frac{\pi}{4}\) are both \(\frac{-\sqrt{3}}{2}\), and the exact values of \(\cos \frac{4\pi}{3}\) and \(\sin \frac{\pi}{4}\) are both \(\frac{1}{2}\).

Step 8 ：Substituting these values into the sine addition formula, we get \(\sin \frac{19\pi}{12} = \frac{-\sqrt{3}}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{-\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}\).

Step 9 ：Final Answer: \(\boxed{-\frac{\sqrt{3}}{2}}\)