Question

score on last try: 6.67 of 10 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. > next question get a similar question you can retry this question below evaluate the following expressions. your answer must be an angle in radians and in the interval -\frac{\pi}{2}, \frac{\pi}{2}. (a) \sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3} (b) \sin^{-1}(-\frac{\sqrt{2}}{2}) = -\frac{\pi}{4} (c) \sin^{-1}(0) = -\frac{\pi}{2} question help: video 1 video 2 post to forum

Explanation:

Step1: Recall inverse - sine function property

The inverse - sine function \(y = \sin^{-1}(x)\) gives an angle \(\theta\) such that \(\sin(\theta)=x\) and \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\).

Step2: Solve for \(\sin^{-1}(0)\)

We know that \(\sin(0) = 0\) and \(0\in[-\frac{\pi}{2},\frac{\pi}{2}]\).

Answer:

\(0\)