# Thread: evaluating trig inverse functions

1. ## evaluating trig inverse functions

arccos( negative square root of 3 / 2)

i dont understand how you use the unit circle to figure out these kind of problems

2. Do you mean $arccos\left(-\sqrt{\frac{3}{2}}\right)$? In that case it won't end up in a real number. $-\sqrt{\frac{3}{2}}$ 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.

3. 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?

4. Oh, you mean $arccos\left(-\frac{\sqrt{3}}{2}\right)$. I think it is just a question of recognizing standard angles.

Say that $arccos\left(-\frac{\sqrt{3}}{2}\right)=\alpha$, it means that $cos(\alpha)=-\frac{\sqrt{3}}{2}$. And since you know that $cos^2(\alpha) + sin^2(\alpha) = 1$, $sin^2(\alpha) = 1 - \frac{3}{4} = \frac{1}{4}$. Now, you know that $sin(\alpha) = \pm 0.5$, maybe that angle is easier to recognize. $\alpha$ is in the interval $[0,\ \pi]$, so $sin(\alpha) \geq 0$, and then $\alpha$ is either $\frac{\pi}{6}$ or $\frac{5\pi}{6}$. And since $cos(\alpha) = -\frac{\sqrt{3}}{2} < 0$, the angle has to be $\frac{5\pi}{6}$.

The case when $sin(\alpha) = 0.5$ is extra special; if you draw a line from $(0,\ 0)$ to $(0,\ 1)$, a line from $(0,\ 0)$ to $(cos(\alpha),\ sin(\alpha))$, and a line from that point to $(1,\ 1)$, you'll form a triangle. Since the lines beginning in $(0, 0)$ are both radiuses, the have equal length. And since $sin(\alpha) = 0.5$ which is right between 0 and 1, then the distance from $(cos(\alpha),\ sin(\alpha))$ must be equal to $(0,\ 0)$ and to $(0,\ 1)$, so all three lines must be of equal length and you have drawn a equilateral, which is known for having all angles being $\frac{\pi}{3}$ radians. Hence $sin(\alpha)$ must be either $\frac{\pi}{2}+\frac{\pi}{3}$ or $\frac{\pi}{2}-\frac{\pi}{3}$.