# Given cos(t)=-7/9 and sin(t)= -sqrt32/9, both in the 3rd Quadrant how do you find cos(t/2) and sin(t/2)?

Jan 12, 2017

$\cos \left(\frac{t}{2}\right) = - \frac{1}{3}$
$\sin \left(\frac{t}{2}\right) = \frac{2 \sqrt{2}}{3}$

#### Explanation:

A:
$\cos \left(t\right) = \cos \left(\frac{t}{2} + \frac{t}{2}\right)$
$- 7 / 9 = {\cos}^{2} \left(\frac{t}{2}\right) - {\sin}^{2} \left(\frac{t}{2}\right)$
$- \frac{7}{9} = {\cos}^{2} \left(\frac{t}{2}\right) - \left(1 - {\cos}^{2} \left(\frac{t}{2}\right)\right)$
$- \frac{7}{9} = 2 {\cos}^{2} \left(\frac{t}{2}\right) - 1$
$- \frac{7}{9} + 1 = 2 {\cos}^{2} \left(\frac{t}{2}\right)$
$\frac{2}{9} = 2 {\cos}^{2} \left(\frac{t}{2}\right)$
$\frac{1}{9} = {\cos}^{2} \left(\frac{t}{2}\right)$
$\pm \sqrt{\frac{1}{9}} = \pm \frac{1}{3} = \cos \left(\frac{t}{2}\right)$
$\cos \left(\frac{t}{2}\right) = - \frac{1}{3}$ , it has a -ve value because t/2 lie at 2nd quadrant.

B:
$\sin \left(t\right) = \sin \left(\frac{t}{2} + \frac{t}{2}\right)$
$- \frac{\sqrt{32}}{9} = 2 \sin \left(\frac{t}{2}\right) \cos \left(\frac{t}{2}\right)$
$- \frac{\sqrt{32}}{9} = 2 \sin \left(\frac{t}{2}\right) \left(- \frac{1}{3}\right)$, from answer above $\cos \left(\frac{t}{2}\right) = - \frac{1}{3}$
$- \frac{\sqrt{32}}{9} = - \frac{2}{3} \sin \left(\frac{t}{2}\right)$
$- \frac{\sqrt{32}}{9} \cdot - \frac{3}{2} = \sin \left(\frac{t}{2}\right)$
$- \frac{\sqrt{16 \cdot 2}}{9} \cdot - \frac{3}{2} = \sin \left(\frac{t}{2}\right)$
$- 4 \frac{\sqrt{2}}{9} \cdot - \frac{3}{2} = \sin \left(\frac{t}{2}\right)$
$\frac{2 \sqrt{2}}{3} = \sin \left(\frac{t}{2}\right)$