## Precalculus (6th Edition)

$\displaystyle \cos\theta=-\frac{\sqrt{70}}{10}$ $\displaystyle \sin\theta=-\frac{\sqrt{30}}{10}$
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): $\sin\theta=-\sqrt{1-\cos^{2}\theta}$ $=-\sqrt{1-\frac{7}{10}}=- \sqrt{\frac{3}{10}}$ $=-\displaystyle \frac{\sqrt{3}}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=-\frac{\sqrt{30}}{10}$