## Precalculus (6th Edition)

$\displaystyle \cos 2x=-\frac{3}{5}$ $\displaystyle \sin 2x=\frac{4}{5}$
$\tan x$ and $\cos x$ are positive $\Rightarrow \sin x$ is positive, since $\displaystyle \tan x=\frac{\sin x}{\cos x}$. This also means that x is in quadrant I, $0^{\mathrm{o}} < x < 90^{\mathrm{o}}\qquad/\times 2$ $0^{\mathrm{o}} < 2x < 180^{\mathrm{o}},$ .. so 2x is either in quadrant I or II (*) Double-Angle Identity for tan: $\displaystyle \tan 2x=\frac{2\tan x}{1-\tan^{2}x}=\frac{2\cdot 2}{1-2^{2}}=-\frac{4}{3}$ $\tan 2x$is negative, so one of $\sin 2x,\ \cos 2x$ is negative, and the other is positive. With (*), 2x is in quadrant II, where its sine is positive and cosine negative. Pythagorean Identity: $\sec^{2}2x=1+\tan^{2}2x=$ $=1+(-\displaystyle \frac{4}{3})=1+\frac{16}{9}=\frac{25}{9}$, ... cosine and secant are reciprocals, so $\displaystyle \cos^{2}2x=\frac{9}{25}$ and $\displaystyle \cos 2x=-\frac{3}{5}$ (cos2x is negative) Pythagorean Identity: $\sin^{2}\theta+\cos^{2}\theta=1,$ $\displaystyle \sin 2x=\sqrt{1-(-\frac{3}{5})^{2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}$