Sec 6.6 Inverse Trig: Problen 5 (1 point) Evaluate the following expressions. Your answer must be an angle in radians and in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). (a) \( \sin ^{-1}(-1)=\square \) (b) \( \sin ^{-1}\left(\frac{\sqrt{2}}{2}\right)=\square \) (c) \( \sin ^{1}\left(\frac{\sqrt{3}}{2}\right)=\square \)

Let's evaluate each inverse sine expression within the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \): ### (a) \( \sin^{-1}(-1) \) The inverse sine of \(-1\) is the angle whose sine is \(-1\). \[ \sin\left(-\frac{\pi}{2}\right) = -1 \] \[ \sin^{-1}(-1) = -\frac{\pi}{2} \] ### (b) \( \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \) The inverse sine of \(\frac{\sqrt{2}}{2}\) is the angle whose sine is \(\frac{\sqrt{2}}{2}\). \[ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \] \[ \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \] ### (c) \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \) The inverse sine of \(\frac{\sqrt{3}}{2}\) is the angle whose sine is \(\frac{\sqrt{3}}{2}\). \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] ### **Final Answers:** - **(a)** \( \sin^{-1}(-1) = -\dfrac{\pi}{2} \) - **(b)** \( \sin^{-1}\left(\dfrac{\sqrt{2}}{2}\right) = \dfrac{\pi}{4} \) - **(c)** \( \sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3} \)