The given expression is:

$\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$

The value of $\sin^{-1}(x)$ gives the angle $\theta$ such that $\sin(\theta) = x$ and $\theta$ is in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$.

For $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$, the angle whose sine is $\frac{\sqrt{2}}{2}$ is $\frac{\pi}{4}$ (or $45^\circ$).

Thus, the exact value is:

$\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$

Final Answer: A. $\frac{\pi}{4}$

Would you like more explanation on this? Here are 5 follow-up questions:

1. What is the domain and range of $\sin^{-1}(x)$?
2. Why does $\frac{\pi}{4}$ satisfy the inverse sine definition?
3. How can this concept be extended to other inverse trigonometric functions?
4. What happens if the value of $\sin^{-1}(x)$ is outside its defined range?
5. How is $\sin^{-1}(x)$ related to arcsine?

Tip: Memorizing key sine, cosine, and tangent values for angles like $\frac{\pi}{4}, \frac{\pi}{6}, \frac{\pi}{3}$ will speed up solving problems like this!