A trigonometric value

Evaluate

$$\mathcal{V}= \tan \left( 2 \arcsin \frac{1}{\sqrt{5}}\right)$$

Solution



We are invoking two lemmata first.

Lemma 1: It holds that $\tan (2\arctan x)=\frac{2x}{1-x^2}$ since if we call $\arctan x = \theta$ then $\tan \theta =x$ and as a consequence:

$$\tan 2 \theta = \frac{2\tan \theta}{1-\tan^2 \theta}=\frac{2x}{1-x^2}$$

proving lemma 1.

Lemma 2: $\arcsin \frac{1}{\sqrt{5}}=\arctan \frac{1}{2}$. $\color{red}{^{(1)}}$. This holds because:

\begin{align*}
\arcsin \frac{1}{\sqrt{5}} &=-i \log \left ( \frac{i}{\sqrt{5}} +\sqrt{1- \left (\frac{1}{\sqrt{5}}  \right )^2 } \right ) \\
 &=-i \log \left ( \frac{i}{\sqrt{5}} + \sqrt{\frac{4}{5}} \right ) \\
 &= -i \log \left ( \frac{2+i}{\sqrt{5}} \right )\\
 &=-i \log \left ( 2+i \right ) + \frac{i}{2} \log 5 \\
 &= -i \left ( \frac{\log 5}{2} + i \arctan \frac{1}{2} \right ) + \frac{i \log 5}{2} \\
 &= \arctan \frac{1}{2}
\end{align*}

Now we are ready to evaluate $\mathcal{V}$. For that we have successively:

\begin{align*}
\mathcal{V}&=\tan \left ( 2 \arcsin \frac{1}{\sqrt{5}} \right ) \\
 &\overset{(2)}{=} \tan \left ( 2 \arctan \frac{1}{2} \right ) \\
 &\overset{(1)}{=} \frac{2\cdot \frac{1}{2}}{1- \left ( \frac{1}{2} \right )^2} \\
 &=\frac{1}{\frac{3}{4}} \\
 &= \frac{4}{3}
\end{align*}

$\color{red}{^{(1)}}$ In general it holds that $\displaystyle \arctan(x) = \arcsin \left (\frac{x}{\sqrt{1+x^2}}\right)$ and $\displaystyle \arcsin x = \arctan \left( \frac{x}{\sqrt{1-x^2}}\right)$.