Answer
$\sin \frac{2\pi}{3}=\frac{\sqrt 3}{2}$,
$\cos \frac{2\pi}{3}=-\frac{1}{2}$
$\tan \frac{2\pi}{3}=-\sqrt {3}$
Work Step by Step
Reference angle $\theta' =\pi-\theta$ if $\theta $ is in quadrant II.
$\implies \theta'=\pi-\frac{2\pi}{3}=\frac{\pi}{3}$
Sine is positive, cosine and tangent are negative in quadrant II.
$\implies \sin \frac{2\pi}{3}=+\sin \frac{\pi}{3}=\frac{\sqrt 3}{2}$,
$\cos \frac{2\pi}{3}=-\cos \frac{\pi}{3}=-\frac{1}{2},$
$\tan \frac{2\pi}{3}=-\tan \frac{\pi}{3}=-\sqrt {3}$
