Problem source

\begin{questionparts}
\item Given that $\displaystyle \cos \theta = \frac35$ and that $\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi$, show that $\displaystyle \sin 2 \theta = -\frac{24}{25}$, and evaluate $\cos 3 \theta$.
\item Prove the identity $\displaystyle \tan 3\theta \equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$. 
Hence evaluate $\tan \theta$, given that $\displaystyle \tan 3\theta = \frac{11}{ 2}$ and that $\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}$.
\end{questionparts}

Solution source

\begin{questionparts}
\item 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*}

\item \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 $\tan \theta = 8 + 5 \sqrt{3}$
\end{questionparts}