## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$-\dfrac{\pi}{4}$
Recall: $y = \sin^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \sin{y}$ $\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$ 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 $466$ of this textbook: 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, $\sin^{-1} {\left(-\dfrac{\sqrt{2}}{2} \right)} = -\dfrac{\pi}{4}$