1. The problem is to find the value of $\sin 240^\circ$ using the unit circle. 2. Recall that the unit circle defines sine as the y-coordinate of the point on the circle at a given angle. 3. The angle $240^\circ$ is in the third quadrant, where sine values are negative. 4. The reference angle for $240^\circ$ is $240^\circ - 180^\circ = 60^\circ$. 5. The sine of $60^\circ$ is $\frac{\sqrt{3}}{2}$. 6. Since $240^\circ$ is in the third quadrant, $\sin 240^\circ = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$. 7. Therefore, the value of $\sin 240^\circ$ is $-\frac{\sqrt{3}}{2}$.