Problem source

Show that $\displaystyle \tan 3\theta = \frac{3\tan\theta -\tan^3\theta}{1-3\tan^2\theta}$ .
Given that $\theta=  \cos^{-1} (2/\sqrt5)$ and
$0<\theta<\pi/2$, show that $\tan 3\theta =11/2$
Hence, or otherwise, find all solutions of the equations
\begin{questionparts}
\item $\tan(3\cos^{-1} x) =11/2$ , 
\item $\cos ({\frac13}\tan^{-1} y) = 2/\sqrt5$ .
\end{questionparts}

Solution source

Let $\tan \theta = t$
\begin{align*}
\tan 3 \theta &\equiv \tan (2 \theta + \theta) \\
&\equiv \frac{\tan 2 \theta +\tan \theta}{1 - \tan 2 \theta \tan \theta} \\
&\equiv \frac{\frac{2t}{1-t^2}+t}{1-\frac{2t^2}{1-t^2}} \\
&\equiv \frac{2t+t-t^3}{1-t^2-2t^2} \\
&\equiv \frac{3t-t^3}{1-3t^3} \\
&\equiv \frac{3\tan \theta - \tan^3 \theta}{1 - 3 \tan^3 \theta}
\end{align*}

If $\theta = \cos^{-1} (2/\sqrt{5})$, then $\sin \theta = 1/\sqrt{5}$ and $\tan \theta = 1/2$. Hence
\begin{align*}
\tan 3 \theta &= \frac{3 \cdot \frac12 - \frac18}{1 - \frac34} \\
&= \frac{11}{2}
\end{align*}

\begin{questionparts}
\item Since $\tan 3 y = 11/2$ has the solution $y = \cos^{-1} (2/\sqrt{5})$ it will also have the solutions $y = \cos^{-1}(2/\sqrt{5}) + \frac{\pi}{3}$ and $y = \cos^{-1}(2/\sqrt{5})+\frac{2\pi}{3}$, therefore 
\begin{align*}
&& \cos^{-1} x &= \cos^{-1} (2/\sqrt{5})\\
 \Rightarrow && x &= 2/\sqrt{5} \\
&& \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{\pi}{3}\\
 \Rightarrow && x &= \frac{2}{\sqrt{5}} \frac{1}{2} - \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\
&&&= \frac{2-\sqrt{3}}{2\sqrt{5}} \\
&& \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{2\pi}{3}\\
 \Rightarrow && x &= \frac{2}{\sqrt{5}} \left (-\frac{1}{2} \right)- \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\
&&&= \frac{-\sqrt{3}-2}{2\sqrt{5}} \\
\end{align*}
\item Since $\cos \frac13 x = \frac{2}{\sqrt{5}}$ has the solution $x = \tan^{-1} \frac{11}{2}$ it will also have the solutions $x = \tan^{-1} \frac{11}{2} + 2n \pi$ and $x = -\tan^{-1} \frac{11}{2} + 2n \pi$.

\begin{align*}
&& \tan^{-1} y &= \tan^{-1} \frac{11}{2} \\
\Rightarrow && y &= \frac{11}{2} \\
&& \tan^{-1} y &=  \tan^{-1} \frac{11}{2} + 2n \pi \\
\Rightarrow && y &= \frac{\frac{11}{2} + 0}{1-0} \\
&&&= \frac{11}{2} \\
&& \tan^{-1} y &=  -\tan^{-1} \frac{11}{2} + 2n \pi \\
\Rightarrow && y &= \frac{-\frac{11}{2} + 0}{1-0} \\
&&&= -\frac{11}{2} \\
\end{align*}

So our two solutions are $y = \pm \frac{11}{2}$
\end{questionparts}