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: $\boxed{{\displaystyle -\frac{\sqrt{3}}{2}}}$