Arcsine Function

Approach: We need an angle $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ with $\sin \theta = \frac{\sqrt{3}}{2}$.

Steps:

1. Among the special angles, $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$;
2. $\frac{\pi}{3} \approx 1.047$, which lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$;
3. By uniqueness on the principal interval, $\arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.

Answer: $\arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.

Approach:

The arcsine is the inverse of the sine function restricted to its principal interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$: the domain comes from the range of $\sin x$, and the range is the principal interval itself.

Steps:

1. The range of the sine function $y = \sin x$ on $\mathbb{R}$ is $[-1, 1]$, and the domain of an inverse function is exactly the range of the original function, so the domain of $y = \arcsin x$ is $[-1, 1]$.
2. The sine function is monotonically increasing on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, running from $-1$ up to $1$, so this interval is chosen as the principal interval to make the inverse single-valued.
3. The inverse maps the angles of the principal interval back onto the domain of the original function, so the range of $y = \arcsin x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Answer:

Domain: $[-1, 1]$; Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$.