# What are the coordinates for the point p((5pi)/6) where p(theta)=(x,y) is the point where the terminal arm of an angle theta intersects the unit circl?

Dec 31, 2017

$p \left(\frac{5 \pi}{6}\right) = \left(- \frac{\sqrt{3}}{2} , \frac{1}{2}\right)$

#### Explanation:

$\frac{5 \pi}{6}$ has reference angle $\frac{\pi}{6}$ because $\pi - \frac{5 \pi}{6} = \frac{\pi}{6}$.

So we know that:
$| \cos \left(\frac{5 \pi}{6}\right) | = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
$| \sin \left(\frac{5 \pi}{6}\right) | = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

Since $\frac{5 \pi}{6} \in$ QII, we know that $\cos \left(\frac{5 \pi}{6}\right) < 0$ and $\sin \left(\frac{5 \pi}{6}\right) > 0$.

Combining what we know, we have that $p \left(\frac{5 \pi}{6}\right) = \left(- \frac{\sqrt{3}}{2} , \frac{1}{2}\right)$.

I recommend memorizing the entire Unit Circle as it's not that difficult and is extremely useful.