How do you find the value for sin2theta, cos2theta, and tan2theta and the quadrant in which 2theta lies given sintheta=-sqrt10/10 and theta is in quadrant IV?

May 28, 2017

$\sin 2 t = - \frac{3}{5}$
$\cos 2 t = \frac{4}{5}$
$\tan 2 t = - \frac{3}{4}$

Explanation:

$\sin t = - \frac{1}{\sqrt{10}}$. Find cos t
${\cos}^{2} t = 1 - {\sin}^{2} t = 1 - \frac{1}{10} = \frac{9}{10}$
$\cos t = \frac{3}{\sqrt{10}}$ --> because t is in Quadrant 4
$\sin 2 t = 2 \sin t . \cos t = 2 \left(- \frac{1}{\sqrt{10}}\right) \left(\frac{3}{\sqrt{10}}\right) = - \frac{6}{10} = - \frac{3}{5}$
Find cos 2t.
$\cos 2 t = 2 {\cos}^{2} t - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$
2t is in Quadrant 4 because its sin is negative and its cos is positive.
$\tan 2 t = \frac{\sin 2 t}{\cos 2 t} = \left(- \frac{3}{5}\right) \left(\frac{5}{4}\right) = - \frac{3}{4}$