## Precalculus (6th Edition)

$\displaystyle \cos\theta=-\frac{\sqrt{42}}{12}$ $\displaystyle \sin\theta=\frac{\sqrt{102}}{12}$
$\theta$ 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): $\sin\theta=+\sqrt{1-\cos^{2}\theta}$ $=\displaystyle \sqrt{1-\frac{7}{24}}= \sqrt{\frac{17}{24}}=\frac{\sqrt{17}}{2\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{102}}{12}$