1. The problem is to find the value of $\sin 60^\circ$. 2. Recall that $60^\circ$ is a special angle in trigonometry. 3. From the unit circle or special triangle (30°-60°-90° triangle), the sine of $60^\circ$ is given by the ratio of the opposite side to the hypotenuse. 4. In a 30°-60°-90° triangle, the sides are in the ratio $1 : \sqrt{3} : 2$ where $\sqrt{3}$ is the side opposite to $60^\circ$. 5. Therefore, $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Hence, the value of $\sin 60^\circ$ is $\frac{\sqrt{3}}{2}$.