Do you mean ? In that case it won't end up in a real number. is less than -1, hence you can't apply the real arccos function. The point would fall outside the circle, and I know, that is impossible. You will have to use complex numbers.
no, it's square root of 3 over 2.
and the answer is 5pi/6, why? i tried the unit circle thing and i dont get it. do you follow the x's ---> (x,y) coordinates of the unit circle? i used the ASTC, all students take calculus and i know it must be in either quadrant 2 or 3 because they are the only quads in which cos is negative. so why did they pick quad 2?
Oh, you mean . I think it is just a question of recognizing standard angles.
Say that , it means that . And since you know that , . Now, you know that , maybe that angle is easier to recognize. is in the interval , so , and then is either or . And since , the angle has to be .
The case when is extra special; if you draw a line from to , a line from to , and a line from that point to , you'll form a triangle. Since the lines beginning in are both radiuses, the have equal length. And since which is right between 0 and 1, then the distance from must be equal to and to , so all three lines must be of equal length and you have drawn a equilateral, which is known for having all angles being radians. Hence must be either or .