Section 4.7 [ATTACH=full]1124[/ATTACH] Question 54 sin (arccos sqrt{5}) Let u = arccos sqrt{5}. This means that cos u = sqrt{5}/1. The range of the arccos function is [0, pi]. I know that cos u is positive in quadrant 1. cos u = adj/hyp. I draw a right triangle in quadrant 1 to find the opposite side. I will use a^2 + b^2 = c^2 to find the opposite side. Let b = opposite side of right triangle. (sqrt{5})^2 + b^2 = 1^2 5 + b^2 = 1 b^2 = 1 - 5 b^2 = -4 Can't take the square of a negative numbers outside the complex number system. So, I cannot calculate the opposite side of the right triangle formed in quadrant 1 My conclusion is as follows: sin (arccos sqrt{5}) is undefined. You say?