1. **Problem Statement:** Evaluate the following inverse sine expressions and express the answers in radians within the interval $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$. 2. **Recall the definition:** The inverse sine function $$\sin^{-1}(x)$$ returns an angle $$\theta$$ such that $$\sin(\theta) = x$$ and $$\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$. 3. **Evaluate (a):** $$\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$. - We know $$\sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$. - Since $$-\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$, the answer is: $$\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}$$. 4. **Evaluate (b):** $$\sin^{-1}\left(-\frac{1}{2}\right)$$. - We know $$\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$$. - Since $$-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$, the answer is: $$\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$. 5. **Evaluate (c):** $$\sin^{-1}(1)$$. - We know $$\sin\left(\frac{\pi}{2}\right) = 1$$. - Since $$\frac{\pi}{2} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$, the answer is: $$\sin^{-1}(1) = \frac{\pi}{2}$$. **Final answers:** (a) $$-\frac{\pi}{4}$$ (b) $$-\frac{\pi}{6}$$ (c) $$\frac{\pi}{2}$$