## Precalculus (6th Edition)

$\displaystyle \cos 2x=-\frac{8}{17}$ $\displaystyle \sin 2x=\frac{15}{17}$
Plan: work out sinx and cosx, and then apply the double-angle formulas. $\tan x$ is positive and $\sin x$ is negative $\Rightarrow \cos x$ is negative, since $\displaystyle \tan x=\frac{\sin x}{\cos x}$. This also means that x is in quadrant III, where both its sine and cosine are negative. Pythagorean Identity: $\displaystyle \sec^{2}x=1+\tan^{2}x=1+(\frac{5}{3})^{2}=1+\frac{25}{9}=\frac{34}{9}$ ... cosine and secant are reciprocals, so $\displaystyle \cos^{2}x=\frac{9}{34}$ and, since $\cos x$ is negative, $\displaystyle \cos x=-\frac{3}{\sqrt{34}}\cdot\frac{\sqrt{34}}{\sqrt{34}}=-\frac{3\sqrt{34}}{34}$ Pythagorean Identity ($\sin x$ is negative): $\sin x=-\sqrt{1-\cos^{2}x}$ $=-\sqrt{1-(-\dfrac{3}{\sqrt{34}})^{2}}$ $=-\displaystyle \sqrt{1-\frac{9}{34}}=-\sqrt{\frac{25}{34}}=-\frac{5}{\sqrt{34}}\cdot\frac{\sqrt{34}}{\sqrt{34}}$ $=-\dfrac{5\sqrt{34}}{34}$ Finally, the Double-Angle Identities: $\displaystyle \cos 2x=2\cos^{2}x-1=2(\frac{9}{34})-1=\frac{9}{17}-1=-\frac{8}{17}$ $\displaystyle \sin 2x=2\sin x\cos x=2(-\frac{5\sqrt{34}}{34})(-\frac{3\sqrt{34}}{34})=\frac{15}{17}$