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

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

What I’ve tried:

  1. 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} $.
  2. 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!

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    $\begingroup$ Hint: Try fixing ranges for $\theta/2$ and $2\theta$ to determine if their cosine or tangent is positive or negative. $\endgroup$ – Arpan Apr 23 '15 at 4:01
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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*}

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