Evaluate the expression. sin ⁡ − 1 ( 3 2 ) \sin^{-1}\left(\frac{\sqrt3}{2}\right) sin − 1 ( 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

Here we're asked to evaluate the expression the inverse sine of the square root of 3 over 2. So we can think of this as the sine of what angle is going to give me root 3 over 2, and we want to find the angle for which that is true. Now looking within my interval, negative pi over 2 to pi over 2, because I'm taking the inverse sine of a positive value, I know that my solution, my angle, has to be in quadrant 1. So of these angles in quadrant 1, which one gives me a sine of root 3 over 2? Now remember, sine is your y values when dealing with the unit circle, and here at pi over 3 I know that the sine is root 3 over 2, so that represents my solution. Pi over 3. And we are completely good to go here. Thanks for watching. And let me know if you have questions.