# Question 8dcc0

Apr 10, 2017

$\sin \left(\theta\right) = \frac{- 2 \sqrt{2}}{3}$
$\sin \left(2 \theta\right) = \frac{- 4 \sqrt{2}}{9}$
$\cos \left(2 \theta\right) = \frac{7}{9}$
$\sin \left(\frac{\pi}{6} + \theta\right) = \frac{1 - 2 \sqrt{6}}{6}$
$\tan \left(\frac{x}{2}\right) = - \frac{\sqrt{2}}{2}$

#### Explanation:

Given: $\sec \left(\theta\right) = 3$

$\cos \left(\theta\right) = \frac{1}{\sec} \left(\theta\right) = \frac{1}{3}$

$\sin \left(\theta\right) = - \sqrt{1 - {\cos}^{2} \left(\theta\right)} \leftarrow$ we use the negative because we are told that it is the 4th quadrant.

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

$\sin \left(\theta\right) = - \sqrt{\frac{9}{9} - \frac{1}{9}}$

$\sin \left(\theta\right) = - \sqrt{\frac{8}{9}}$

$\sin \left(\theta\right) = \frac{- 2 \sqrt{2}}{3}$

$\sin \left(2 \theta\right) = 2 \sin \left(\theta\right) \cos \left(\theta\right)$

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

$\sin \left(2 \theta\right) = \frac{- 4 \sqrt{2}}{9}$

$\cos \left(2 \theta\right) = 1 - 2 {\cos}^{2} \left(\theta\right)$

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

$\cos \left(2 \theta\right) = \frac{7}{9}$

$\sin \left(\frac{\pi}{6} + \theta\right) = \sin \left(\frac{\pi}{6}\right) \cos \left(\theta\right) + \cos \left(\frac{\pi}{6}\right) \sin \left(\theta\right)$

$\sin \left(\frac{\pi}{6} + \theta\right) = \left(\frac{1}{2}\right) \left(\frac{1}{3}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{- 2 \sqrt{2}}{3}\right)$

$\sin \left(\frac{\pi}{6} + \theta\right) = \frac{1 - 2 \sqrt{6}}{6}$

tan(x/2) = sin(x)/(1+cos(x)#

$\tan \left(\frac{x}{2}\right) = \frac{\frac{- 2 \sqrt{2}}{3}}{1 + \frac{1}{3}}$

$\tan \left(\frac{x}{2}\right) = - \frac{\sqrt{2}}{2}$