A trigonometric value

Evaluate

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

Solution

We are invoking two lemmata first:

Lemma 1: It holds that $\tan \left ( 2 \arctan x \right ) = \frac{2x}{1-x^2}$ .

Proof:

If we call $\theta=\arctan x$ then $\tan \theta= x$ and as a consequence we have that

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

proving lemma 1. $\square$

Lemma 2: It holds that $\arcsin \frac{1}{\sqrt{5}}=\arctan \frac{1}{2}$ .

Proof:

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

proving Lemma 2. $\square$

Hence for the value of $\mathcal{V}$ we have successively

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