#### Answer

$\displaystyle \cos\theta=-\frac{\sqrt{30}}{6}$
$\displaystyle \sin\theta=\frac{\sqrt{6}}{6}$

#### Work Step by Step

$\theta$ terminates in Q.II, where
its cosine is negative, its sine is positive.
Double-Angle Identity:
$\cos 2\theta=2\cos^{2}\theta- \mathrm{l}$
$\displaystyle \frac{2}{3}+1=2\cos^{2}\theta$
$\displaystyle \frac{5}{3}=2\cos^{2}\theta$
$\displaystyle \cos^{2}\theta=\frac{5}{6}$
... $\cos\theta$ is negative,
$\displaystyle \cos\theta=-\sqrt{\frac{5}{6}}=-\frac{\sqrt{5}}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=-\frac{\sqrt{30}}{6}$
Pythagorean Identity ($\sin\theta$ is positive):
$\sin\theta=+\sqrt{1-\cos^{2}\theta}$
$=\displaystyle \sqrt{1-\frac{5}{6}}= \sqrt{\frac{1}{6}}=\frac{1}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}}{6}$