find the exact value of the real number y.\n\n$y = \\sin^{-1}(\\frac{\\sqrt{2}}{2})$

Explanation:

Step1: Recall the range of (y = \sin^{-1}(x))

The range of (y=\sin^{-1}(x)) is (\left[-\frac{\pi}{2},\frac{\pi}{2}\right])

Step2: Recall the special - angle sine values

We know that (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}), (\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}), (\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2}), (\sin\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2})

But for (y = \sin^{-1}(x)), when (x=\frac{\sqrt{2}}{2}), and considering the range (\left[-\frac{\pi}{2},\frac{\pi}{2}\right])

Since (\frac{3\pi}{4}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), and (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}) with (\frac{\pi}{4}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right])

Answer:

C. (\frac{\pi}{4})