use the unit circle to find $sin^{-1}(\frac{sqrt{2}}{2})$ in radians. remember that the domain of inverse…

use the unit circle to find $sin^{-1}(\frac{sqrt{2}}{2})$ in radians. remember that the domain of inverse sine is limited to quadrants i and iv (the right side of the unit circle).\n\na. $\frac{pi}{4}$\nb. $\frac{7pi}{4}$\nc. $\frac{pi}{6}$\nd. $\frac{11pi}{6}$

use the unit circle to find $sin^{-1}(\frac{sqrt{2}}{2})$ in radians. remember that the domain of inverse sine is limited to quadrants i and iv (the right side of the unit circle).\n\na. $\frac{pi}{4}$\nb. $\frac{7pi}{4}$\nc. $\frac{pi}{6}$\nd. $\frac{11pi}{6}$

Answer

Explanation:

Step1: Recall sine - inverse sine relationship

The inverse - sine function (y = \sin^{-1}(x)) gives the angle (\theta) such that (\sin(\theta)=x) and (-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}) (quadrants I and IV).

Step2: Check sine values on unit - circle

We know that (\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}) and (\frac{\pi}{4}) is in the first quadrant ((0\lt\frac{\pi}{4}\lt\frac{\pi}{2})). Also, (\sin(\frac{7\pi}{4})=-\frac{\sqrt{2}}{2}), (\sin(\frac{\pi}{6})=\frac{1}{2}), and (\sin(\frac{11\pi}{6})=-\frac{1}{2}).

Answer:

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