#### Answer

$\displaystyle \cos\theta=-\frac{\sqrt{42}}{12}$
$\displaystyle \sin\theta=\frac{\sqrt{102}}{12}$

#### Work Step by Step

$\theta$ terminates in Q.II, where
its cosine is negative, its sine is positive.
Double-Angle Identity:
$\cos 2\theta=2\cos^{2}\theta- \mathrm{l}$
$-\displaystyle \frac{5}{12}+1=2\cos^{2}\theta$
$\displaystyle \frac{7}{12}=2\cos^{2}\theta$
$\displaystyle \cos^{2}\theta=\frac{7}{24}$
... $\cos\theta$ is negative,
$\displaystyle \cos\theta=-\sqrt{\frac{7}{24}}=-\frac{\sqrt{7}}{2\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=-\frac{\sqrt{42}}{12}$
Pythagorean Identity ($\sin\theta$ is positive):
$\sin\theta=+\sqrt{1-\cos^{2}\theta}$
$=\displaystyle \sqrt{1-\frac{7}{24}}= \sqrt{\frac{17}{24}}=\frac{\sqrt{17}}{2\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{102}}{12}$