Problems

Solution:

1. Since \(\cos^2 \theta + \sin^2 \theta \equiv 1\), \(\sin \theta = \pm \frac45\) and since \(\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi\) it must be the case that \(\sin\) is negative, ie \(\sin \theta = -\frac45\). Therefore \(\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac35 \cdot (-\frac45) = -\frac{24}{25}\). \begin{align*} \cos 3 \theta &= \cos 2 \theta \cos \theta - \sin 2\theta \sin \theta \\ &= (\cos^2 \theta - \sin^2 \theta) \cos \theta - \sin 2 \theta \sin \theta \\ &= (\frac{9}{25} - \frac{16}{25}) \frac35 + \frac{24}{25} \cdot (-\frac{4}{5}) \\ &= -\frac{21}{125} - \frac{96}{125} \\ &= -\frac{117}{125} \end{align*}
2. \begin{align*} \tan 3 \theta &\equiv \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2 \tan \theta}{1- \tan^2 \theta} + \tan \theta}{1 - \frac{2 \tan^2 \theta}{1- \tan^2 \theta}} \\ &\equiv \frac{2\tan \theta + \tan \theta -\tan^3 \theta}{1 - \tan^2 \theta - 2 \tan^2 \theta} \\ &\equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \end{align*} Let \(t = \tan \theta\), then \begin{align*} && \frac{11}{2} &= \frac{3t - t^3}{1-3t^2} \\ \Leftrightarrow && 11 - 33t^2 &= 6t -2t^3 \\ \Leftrightarrow && 0 &= 2t^3-33t^2-6t+11 \\ \Leftrightarrow && 0 &= (2t-1)(t^2-16t-11) \end{align*} Therefore \(\tan \theta = \frac12, \tan \theta = \frac{16 \pm \sqrt{16^2+4 \cdot 1 \cdot 11}}{2} = \frac{16\pm10\sqrt{3}}{2} = 8 \pm 5 \sqrt{3}\). Since \(\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}\) we must have that \(\tan\) is both positive and \(\geq 1\), therefore \(\tan \theta = 8 + 5 \sqrt{3}\)