# How do you find tan2A, given sin A = 3/5 and A is in QII?

Dec 2, 2015

#### Answer:

Find tan 2A, given $\sin A = \frac{3}{5}$ and A in Quadrant II.

Ans: $- \frac{24}{7}$

#### Explanation:

Use the trig identity: $\tan 2 A = \frac{2 \tan A}{1 - {\tan}^{2} A} \left(1\right) .$
First, find $\tan A = \sin \frac{A}{\cos A} .$
$\sin A = \frac{3}{5}$ --> ${\sin}^{2} A = \frac{9}{25}$ --> ${\cos}^{2} A = 1 - \frac{9}{25} = \frac{15}{25}$ -
--> $\cos A = \pm \frac{4}{5}$.
Because A is in Quadrant II, its cos is negative. $\cos A = - \frac{4}{5}$
$\tan A = \sin \frac{A}{\cos A} = \left(\frac{3}{5}\right) \left(- \frac{5}{4}\right) = - \frac{3}{4}$
Replace value of tan A = -3/4 into identity (1) -->
$\tan 2 A = \frac{- \frac{3}{2}}{1 - \frac{9}{16}} = \left(- \frac{3}{2}\right) \left(\frac{16}{7}\right) = - \frac{24}{7}$
Check by calculator.
$\cos A = - \frac{4}{5} = - 0.8$ --> $A = 143.13$ --> $2 A = 286.26$.-->
-> $\tan 2 A = - 3.43$
$- \frac{24}{7} = - 3. 43.$ OK