# Chapter 5 - Section 5.3 - Double-Angle Formulas - 5.3 Problem Set - Page 296: 7

$\frac{24}{25}$

#### Work Step by Step

Given that $\sin A =\frac{-3}{5}$ and A is in quadrant 3 In Quadrant 3; $\cos A$ is negative. $\cos A$ = -$\sqrt {1-\sin^{2}A}$ $\cos A$ = -$\sqrt {1-(\frac{3}{5})^{2}}$ $\cos A$ = -$\sqrt {1-(\frac{9}{25}}$ $\cos A$ = -$\sqrt {\frac{16}{25}}$ $\cos A$ = -$\frac{4}{5}$ $\sin 2A$ = 2 $\sin A \cos A$ Plug in $\sin A$ and $\cos A$ values in above equation we get $\sin 2A$ = $2 \times\frac{-3}{5}\times\frac{-4}{5}$ $\sin 2A$ = $\frac{24}{25}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.