Answer
$\displaystyle \cos\theta=-\frac{\sqrt{70}}{10}$
$\displaystyle \sin\theta=-\frac{\sqrt{30}}{10}$
Work Step by Step
Double-Angle Identity:
$\cos 2\theta=2\cos^{2}\theta- \mathrm{l}$
$\displaystyle \frac{3}{4}+1=2\cos^{2}\theta$
$\displaystyle \frac{7}{5}=2\cos^{2}\theta$
$\displaystyle \cos^{2}\theta=\frac{7}{10}$
Since $\theta$ terminates in Q.III, its cosine (and sine) are negative
$\displaystyle \cos\theta=-\sqrt{\frac{7}{10}}=-\frac{\sqrt{7}}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=-\frac{\sqrt{70}}{10}$
Pythagorean Identity ($\sin\theta$ is negative):
$\sin\theta=-\sqrt{1-\cos^{2}\theta}$
$=-\sqrt{1-\frac{7}{10}}=- \sqrt{\frac{3}{10}}$
$=-\displaystyle \frac{\sqrt{3}}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=-\frac{\sqrt{30}}{10}$
