## Precalculus (6th Edition)

$\displaystyle \cos\theta=\frac{2\sqrt{5}}{5}$ $\displaystyle \sin\theta=\frac{\sqrt{5}}{5}$
Double-Angle Identity: $\cos 2\theta=2\cos^{2}\theta- \mathrm{l}$ $\displaystyle \frac{3}{5}+1=2\cos^{2}\theta$ $\displaystyle \frac{8}{5}=2\cos^{2}\theta$ $\displaystyle \cos^{2}\theta=\frac{8}{10}=\frac{4}{5}$ Since $\theta$ terminates in Q.I, its cosine (and sine) are positive $\displaystyle \cos\theta=+\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$ Pythagorean Identity ($\sin\theta$ is positive): $\sin\theta=+\sqrt{1-\cos^{2}\theta}$ $=\displaystyle \sqrt{1-\frac{4}{5}}= \sqrt{\frac{1}{5}}=\frac{\sqrt{5}}{5}$