# How does one find sin2x, cos2x, and tan2x if x is in quadrant IV and equals -2/sqrt13?

May 17, 2018

$\sin 2 x = - \frac{12}{13}$
$\cos 2 x = \frac{5}{13}$
$\tan 2 x = - \frac{12}{5}$

#### Explanation:

$\sin x = - \frac{2}{\sqrt{13}}$ (x is in Quadrant 4). Find cos x
${\cos}^{2} x = 1 - {\sin}^{2} x = 1 - \frac{4}{13} = \frac{9}{13}$
$\cos x = \frac{3}{\sqrt{13}}$ (because x is in Q. 4)
$\sin 2 x = 2 \sin x . \cos x = 2 \left(- \frac{2}{\sqrt{13}}\right) \left(\frac{3}{\sqrt{13}}\right) = - \frac{12}{13}$
${\cos}^{2} 2 x = 1 - {\sin}^{2} 2 x = 1 - \frac{144}{169} = \frac{25}{169}$
$\cos 2 x = \frac{5}{13}$
$\tan 2 x = \frac{\sin 2 x}{\cos 2 x} = \left(- \frac{12}{13}\right) \left(\frac{13}{5}\right) = - \frac{12}{5}$