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*}

Note: In general it holds that

\displaystyle \arctan(x) = \arcsin \left (\frac{x}{\sqrt{1+x^2}}\right)

as well as

\displaystyle \arcsin x = \arctan \left( \frac{x}{\sqrt{1-x^2}}\right)

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