I'm trying to understand how my teacher converted these angles. I'm not sure if my title is correct but I'm assuming that's what he was doing.
For a unit circle he had, \begin{align*} u & = \left(\cos\left(\frac{2\pi}{3}\right), \sin\left(\frac{2\pi}{3}\right)\right)\\ & = \left(-\cos\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right)\right) \end{align*}
How did he move from line 1 to 2, what is it called? Refer me to a youtube video or some other source for better understanding please?
Hmm, well, those two ordered pairs are not equal, so either your teacher made a mistake or you copied something down wrong.
Note that $$\Bigl(\cos\frac{2\pi}{3} , \sin\frac{2\pi}{3}\Bigr)=\Bigl(-\frac{1}{2},\frac{\sqrt{3}}{2}\Bigr)$$ whereas $$\Bigl(-\cos\frac{\pi}{6} , \sin\frac{\pi}{6}\Bigr)=\Bigl(-\frac{\sqrt{3}}{2},\frac{1}{2}\Bigr)\,.$$ Obviously these are not the same. Also, it's also not clear what your teacher might have meant, since $\frac{2\pi}{3}$ is coterminal with $-\frac{4\pi}{3}$, while $\frac{\pi}{6}$ is coterminal with $-\frac{11\pi}{6}$...
EDIT: Perhaps it is worth noting that $$\Bigl(\cos\frac{2\pi}{3} , \sin\frac{2\pi}{3}\Bigr)=\Bigl(-\sin\frac{\pi}{6} , -\cos\frac{\pi}{6}\Bigr)\,.$$