The exact value of $\sin 60^\circ$ is $\frac{\sqrt{3}}{2}$.

To understand why $\sin 60^\circ$ equals $\frac{\sqrt{3}}{2}$, we can analyze an equilateral triangle. An equilateral triangle has all sides of equal length and each angle measuring $60^\circ$. If we draw a height from one vertex to the midpoint of the opposite side, this action divides the equilateral triangle into two right-angled triangles. Each of these right-angled triangles will feature angles of $30^\circ$, $60^\circ$, and $90^\circ$.

Let’s assume the side length of the equilateral triangle is $2$ units. When we draw the height, it bisects the base into two segments, each measuring $1$ unit. This gives us a right-angled triangle with a hypotenuse of $2$ units, one leg measuring $1$ unit, and the height (which is opposite the $60^\circ$ angle) that we need to calculate.

We can apply the Pythagorean theorem, which states that $a^2 + b^2 = c^2$, where $a$ and $b$ are the lengths of the legs and $c$ is the length of the hypotenuse:

This simplifies to:

Subtracting $1$ from both sides gives:

Taking the square root of both sides results in:

In this right-angled triangle, the sine of an angle is defined as the ratio of the length of the opposite side to that of the hypotenuse. For the $60^\circ$ angle, we have:

Thus, we find that the exact value of $\sin 60^\circ$ is indeed $\frac{\sqrt{3}}{2}$.