Evaluate the expression. cos ⁡ − 1 ( − 2 2 ) \cos^{-1}\left(-\frac{\sqrt2}{2}\right) cos − 1 ( − 2 2 <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"> )

this problem, we're asked to evaluate the expression the inverse cosine of negative root 2 over 2. Now, remember you can also think of this as the cosine of what angle is equal to negative root 2 over 2 and we're looking for the angle for which that is true. Now here because we're taking the inverse cosine of a negative value, I know that I have to be looking for an angle that is in quadrant 2 that is still within my interval from 0 to pi. So in this quadrant 2, which one of these angles gives me a cosine of negative root 2 over 2? Well, 3 pi over 4 will give me just that. So that represents my solution here, 3 pi over 4. And we're done here. Thanks for watching. And let me know if you have questions.