If $ \cos(\theta) = - \frac{2}{3} $ and $ 450^{\circ} < \theta < 540^{\circ} $, find... If $ \cos(\theta) = - \frac{2}{3} $ and $ 450^{\circ} < \theta < 540^{\circ} $, find:


*

*The exact value of $ \cos \! \left( \frac{1}{2} \theta \right) $.

*The exact value of $ \tan(2 \theta) $.


What I’ve tried:


*

*I took the square root of $ \sqrt{\frac{1}{2} \left( 1 - \frac{2}{3} \right)} $, which equals the square root of $ \frac{1}{6} $.

*I considered the formula $ \sin(x) = \sqrt{1 - {\cos^{2}}(x)} $ and took the square root of $ 1 - \frac{4}{9} $, but I don’t know where to go from there.


Thanks in advance!

You were given that $\cos\theta = -\dfrac{2}{3}$, with $450^\circ < \theta < 540^\circ$. 
Recall that 
$$\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$
where the sign is determined by the measure of angle $\frac{\theta}{2}$.  Since $450^\circ < \theta < 540^\circ$, we may conclude that $225^\circ < \frac{\theta}{2} < 270^\circ$, so $\frac{\theta}{2}$ is a third-quadrant angle.  Hence, its cosine is negative.  Thus,
\begin{align*}
\cos\left(\frac{\theta}{2}\right) & = -\sqrt{\frac{1 + \cos\theta}{2}}\\
& = -\sqrt{\frac{1 - \frac{2}{3}}{2}}\\
& = -\sqrt{\frac{\frac{1}{3}}{2}}\\
& = -\sqrt{\frac{1}{6}}
\end{align*} 
so you found the correct magnitude but did not take into account into the sign of $\cos(\frac{\theta}{2})$.
To determine $\tan(2\theta)$, we can use the formula
$$\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}$$
together with the double angle formulas
\begin{align*}
\sin(2\theta) & = 2\sin\theta\cos\theta\\
\cos(2\theta) & = \cos^2\theta - \sin^2\theta\\
              & = 2\cos^2\theta - 1\\
              & = 1 - 2\sin^2\theta
\end{align*}
Since $450^\circ < \theta < 540^\circ$, $\theta$ is a second-quadrant angle, so $\sin\theta > 0$.  Hence,
\begin{align*}
\sin\theta & = \sqrt{1 - \cos^2\theta}\\
           & = \sqrt{1 - \left(-\frac{2}{3}\right)^2}\\
           & = \sqrt{1 - \frac{4}{9}}\\
           & = \sqrt{\frac{5}{9}}\\
           & = \frac{\sqrt{5}}{3}
\end{align*}
Therefore, 
\begin{align*}
\tan(2\theta) & = \frac{\sin(2\theta)}{\cos(2\theta)}\\
              & = \frac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta}\\
              & = \frac{2\left(\frac{\sqrt{5}}{3}\right)\left(-\frac{2}{3}\right)}{\left(-\frac{2}{3}\right)^2 - \left(\frac{\sqrt{5}}{3}\right)^2}\\
              & = \frac{\frac{-4\sqrt{5}}{9}}{\frac{4}{9} - \frac{5}{9}}\\
              & = \frac{-4\sqrt{5}}{4 - 5}\\
              & = 4\sqrt{5}
\end{align*}
