# Question #defee

Mar 15, 2016

$\sin 2 t = - \frac{6}{13}$

#### Explanation:

P is in the Quadrant II.
We have: $\angle t = \left(\frac{\pi}{2} + \angle a\right) .$
Angle a is defined by $\tan a = \frac{2}{3}$ --> $\sin a = \frac{2}{\sqrt{4 + 9}} = \frac{2}{\sqrt{13}}$ -->
${\cos}^{2} a = 1 - {\sin}^{2} a = 1 - \frac{4}{13} = \frac{9}{13}$ --> $\cos a = \frac{3}{\sqrt{13}}$.
Find the trig functions of t.
$\sin t = \sin \left(a + \frac{\pi}{2}\right) = \cos a = \frac{3}{\sqrt{13}}$
$\cos t = \cos \left(\frac{\pi}{2} + a\right) = - \sin a = - \frac{2}{\sqrt{13}}$
$\sin 2 t = 2 \sin t . \cos t = - \left(\frac{2}{\sqrt{13}}\right) \left(\frac{3}{\sqrt{13}}\right) = - \frac{6}{13}$
${\cos}^{2} 2 t = 1 - {\sin}^{2} t = 1 - \frac{36}{169} = \frac{133}{169}$
$\cos 2 t = - \frac{\sqrt{133}}{13}$. (2t is in Quadrant III)
sin (t - 150) = sin t.cos 150 - sin 150.cos t.
$= - \left(\frac{3}{\sqrt{13}}\right) \left(- \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right) \left(- \frac{2}{\sqrt{13}}\right) =$
$= - \left(\frac{3 \sqrt{3}}{2 \sqrt{13}}\right) + \left(\frac{2}{2 \sqrt{13}}\right) = \frac{2 - 3 \sqrt{3}}{2 \sqrt{13}}$
cos (225 - t) = cos 225.cos t + sin 225.sin t =
$\cos 225 = - \cos 45 = - \frac{\sqrt{2}}{2}$ ; and $\sin 225 = - \sin 45 = - \frac{\sqrt{2}}{2}$