# If sin theta=3/5 and theta has its terminal side in quadrant II, how do you find the exact value of tan 2 theta?

Apr 17, 2016

$- \frac{24}{7}$

#### Explanation:

$\sin t = \frac{3}{5}$. Find cos t.
${\cos}^{2} t = 1 - {\sin}^{2} t = 1 - \frac{9}{25} = \frac{16}{25}$
$\cos t = \pm \frac{4}{5.}$
Since t is located in Quadrant II, cos t is negative.
$\cos t = - \frac{4}{5}$ --> $\tan t = \frac{\sin t}{\cos t} = \left(\frac{3}{5}\right) \left(- \frac{4}{5}\right) = - \frac{3}{4}$
$\tan 2 t = \frac{2 \tan t}{1 - {\tan}^{2} t} = \frac{- \frac{3}{2}}{1 - \frac{9}{16}} =$
$= - \frac{\frac{3}{2}}{\frac{7}{16}} = - \left(\frac{3}{2}\right) \left(\frac{16}{7}\right) = - \frac{24}{7}$
Check by calculator
$\tan t = - \frac{3}{4}$ -> $t = - {36}^{\circ} 87$ --> $\tan 2 t = \tan \left(- 73.74\right) = - 3.43$
tan 2t = - 24/7 = 3.43. OK