# If sinx = -3/5, x is in quadrant III, then how do you find cos2x?

$\cos 2 x = \frac{7}{25}$

#### Explanation:

Given $\sin x = - \frac{3}{5}$

Using half-angle formulas

$\sin x = \pm \sqrt{\frac{1 - \cos 2 x}{2}}$

$- \frac{3}{5} = - \sqrt{\frac{1 - \cos 2 x}{2}}$

square both sides of the equation

${\left(- \frac{3}{5}\right)}^{2} = {\left(- \sqrt{\frac{1 - \cos 2 x}{2}}\right)}^{2}$

$\frac{9}{25} = \frac{1 - \cos 2 x}{2}$

$\frac{9}{25} = \frac{1}{2} - \frac{\cos 2 x}{2}$

$\frac{\cos 2 x}{2} = \frac{1}{2} - \frac{9}{25}$

$\cos 2 x = 1 - \frac{18}{25}$

$\cos 2 x = \frac{25 - 18}{25}$

$\cos 2 x = \frac{7}{25}$

God bless....I hope the explanation is useful.