arccos( negative square root of 3 / 2)
i dont understand how you use the unit circle to figure out these kind of problems
Do you mean $\displaystyle arccos\left(-\sqrt{\frac{3}{2}}\right)$? In that case it won't end up in a real number. $\displaystyle -\sqrt{\frac{3}{2}}$ 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 $\displaystyle arccos\left(-\frac{\sqrt{3}}{2}\right)$. I think it is just a question of recognizing standard angles.
Say that $\displaystyle arccos\left(-\frac{\sqrt{3}}{2}\right)=\alpha$, it means that $\displaystyle cos(\alpha)=-\frac{\sqrt{3}}{2}$. And since you know that $\displaystyle cos^2(\alpha) + sin^2(\alpha) = 1$, $\displaystyle sin^2(\alpha) = 1 - \frac{3}{4} = \frac{1}{4}$. Now, you know that $\displaystyle sin(\alpha) = \pm 0.5$, maybe that angle is easier to recognize. $\displaystyle \alpha$ is in the interval $\displaystyle [0,\ \pi]$, so $\displaystyle sin(\alpha) \geq 0$, and then $\displaystyle \alpha$ is either $\displaystyle \frac{\pi}{6}$ or $\displaystyle \frac{5\pi}{6}$. And since $\displaystyle cos(\alpha) = -\frac{\sqrt{3}}{2} < 0$, the angle has to be $\displaystyle \frac{5\pi}{6}$.
The case when $\displaystyle sin(\alpha) = 0.5$ is extra special; if you draw a line from $\displaystyle (0,\ 0)$ to $\displaystyle (0,\ 1)$, a line from $\displaystyle (0,\ 0)$ to $\displaystyle (cos(\alpha),\ sin(\alpha))$, and a line from that point to $\displaystyle (1,\ 1)$, you'll form a triangle. Since the lines beginning in $\displaystyle (0, 0)$ are both radiuses, the have equal length. And since $\displaystyle sin(\alpha) = 0.5$ which is right between 0 and 1, then the distance from $\displaystyle (cos(\alpha),\ sin(\alpha))$ must be equal to $\displaystyle (0,\ 0)$ and to $\displaystyle (0,\ 1)$, so all three lines must be of equal length and you have drawn a equilateral, which is known for having all angles being $\displaystyle \frac{\pi}{3}$ radians. Hence $\displaystyle sin(\alpha)$ must be either $\displaystyle \frac{\pi}{2}+\frac{\pi}{3}$ or $\displaystyle \frac{\pi}{2}-\frac{\pi}{3}$.