# If sinx = -3/5, x is in quadrant III, then how do you find sin2x?

Jun 5, 2016

$\sin 2 x = \frac{24}{25}$
$\sin 2 x = 2 \sin x \cos x$
cosx=-sqrt(1-sin^2x (cosx <0 in quadrant III)
so $\cos x = - \sqrt{1 - \frac{9}{25}}$
that's $\cos x = - \frac{4}{5}$
and $\sin 2 x = 2 \left(- \frac{3}{5}\right) \left(- \frac{4}{5}\right)$
that's $\frac{24}{25}$