# How do you find the value for sin2theta, cos2theta, and tan2theta and the quadrant in which 2theta lies given tantheta=-5/12 and theta is in quadrant II?

Dec 23, 2016

#### Explanation:

First, find cos t by using trig identity:
${\cos}^{2} x = \frac{1}{1 + {\tan}^{2} x}$ -->
${\cos}^{2} t = \frac{1}{1 + \frac{25}{144}} = \frac{144}{169}$ --> $\cos t = \pm \frac{12}{13}$
Since t is in Quadrant II, cos t is negative. Take the negative value.
Find sin t.
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{144}{169} = \frac{25}{169}$ --> $\sin t = \pm \frac{5}{13}$
Since t is in Quadrant II, then sin t is positive.
$\sin 2 t = 2 \sin t . \cos t = 2 \left(\frac{5}{13}\right) \left(- \frac{12}{13}\right) = - \frac{120}{169}$
$\cos 2 t = 1 - 2 {\sin}^{2} t = 1 - \frac{50}{169} = \frac{119}{169}$
$\tan 2 t = \frac{\sin 2 t}{\cos 2 t} = \left(- \frac{120}{169}\right) \left(\frac{169}{119}\right) = - \frac{120}{119}$ (1)
You may check the results by calculator.
$\tan t = - \frac{5}{12}$ -->$t = - {22}^{\circ} 62 + {180}^{\circ} = {157}^{\circ} 38$ (Quadrant II)
$\tan 2 t = \tan {314}^{\circ} 76 = - 1.01$
$2 t = {314}^{\circ} 76$ is in Quadrant III. Compare to answer (1) -->
$\tan 2 t = - \frac{120}{119} = - 1.01$. OK