Answer
$\dfrac{\pi}{4}$
Work Step by Step
$y = \sin^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \sin{y}$
$\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$
Let $\theta = \sin^{-1} {\left(\dfrac{\sqrt{2}}{2} \right)} \hspace{20pt} -\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$
Since $\theta = \sin^{-1} {\left(\dfrac{\sqrt{2}}{2} \right)}$, then
$\sin{\theta} = \dfrac{\sqrt{2}}{2}$
Referring to Table 1 on page 466:
The only angle in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$ whose sine is $\dfrac{\sqrt{2}}{2}$ is $\dfrac{\pi}{4}$.
Therefore, $\sin^{-1} {\left(\dfrac{\sqrt{2}}{2} \right)} = \dfrac{\pi}{4}$.