# If sinx = -3/5, x is in quadrant III, then how do you find tan2x?

Jul 10, 2016

$\frac{24}{7}$

#### Explanation:

$\sin x = - \frac{3}{5}$ . Find cos x by identity: ${\cos}^{2} x = 1 - {\sin}^{2} x$
${\cos}^{2} x = 1 - \frac{9}{25} = \frac{16}{25}$
$\cos x = \pm \frac{4}{5.}$
Since x is in Quadrant III, then cos x is negative.
$\tan x = \frac{\sin}{\cos} = \left(- \frac{3}{5}\right) \left(- \frac{5}{4}\right) = \frac{3}{4}$
Use trig identity:
$\tan 2 x = \frac{2 \tan x}{1 - {\tan}^{2} x}$
$\tan 2 x = \frac{\frac{6}{4}}{1 - \frac{9}{16}} = \left(\frac{6}{4}\right) \left(\frac{16}{7}\right) = \frac{24}{7}$