The value of $\arcsin\left(\frac{\sqrt{3}}{2}\right)$ is the angle whose sine is $\frac{\sqrt{3}}{2}$. In the unit circle, the sine of $\frac{\sqrt{3}}{2}$ corresponds to an angle of:

$\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \text{ radians or } 60^\circ$

This is because $\sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$, and the principal value of $\arcsin$ is restricted to the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is $\arccos\left(\frac{\sqrt{3}}{2}\right)$?
2. How is the range of the arcsin function determined?
3. Can the sine of any angle ever be greater than 1 or less than -1?
4. How would you compute $\arcsin\left(\frac{1}{2}\right)$?
5. What is the general solution for $\sin(\theta) = \frac{\sqrt{3}}{2}$?

Tip: The $\arcsin$ function always returns an angle in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.