# What is Tan(2arcsin(1/3))?

Sep 12, 2015

$\frac{4 \sqrt{2}}{7}$

#### Explanation:

Let $\arcsin \left(\frac{1}{3}\right) = \theta$

$\implies \tan \left(2 \arcsin \left(\frac{1}{3}\right)\right)$ becomes $\textcolor{b r o w n}{\tan \left(2 \theta\right)}$

and tan(2theta)=(2tan(theta))/(1-tan^2(theta)) color(red)(larr"this is what we are looking for"

• So we need $\textcolor{b r o w n}{\tan} \left(\theta\right)$

Here's how we do:

$\textcolor{g r e e n}{\text{Recall:}} \arcsin$ is the inverse trig function of $\sin$
So since, $\arcsin \left(\frac{1}{3}\right) = \theta$
$\implies \sin \left(\theta\right) = \frac{1}{3}$

$\implies \frac{1}{\sin} \left(\theta\right) = 3 = \csc \left(\theta\right)$

$\implies \csc \left(\theta\right) = 3$

$\implies {\left(\csc \left(\theta\right)\right)}^{2} = {\left(3\right)}^{2}$

$\implies {\csc}^{2} \left(\theta\right) = 9$

Use the trig identity : ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$
Divide all through by ${\sin}^{2} \left(\theta\right)$
$\implies {\cot}^{2} \left(\theta\right) + 1 = {\csc}^{2} \left(\theta\right)$
$\implies \frac{1}{\tan} ^ 2 \left(\theta\right) + 1 = {\csc}^{2} \left(\theta\right)$
$\implies {\tan}^{2} \left(\theta\right) = \frac{1}{{\csc}^{2} \left(\theta\right) - 1}$
$\implies \tan \left(\theta\right) = \sqrt{\frac{1}{{\csc}^{2} \left(\theta\right) - 1}}$

$\implies \tan \left(\theta\right) = \sqrt{\frac{1}{9 - 1}} = \sqrt{\frac{1}{8}} = \frac{\sqrt{8}}{8} = \frac{2 \sqrt{2}}{8} = \textcolor{b r o w n}{\frac{\sqrt{2}}{4}}$

Finally,
$\implies \tan \left(2 \theta\right) = \frac{2 \tan \left(\theta\right)}{1 - {\tan}^{2} \left(\theta\right)} = \frac{2 \left(\frac{\sqrt{2}}{4}\right)}{1 - {\left(\frac{\sqrt{2}}{4}\right)}^{2}} = \frac{8 \sqrt{2}}{16 - 2} = \textcolor{b l u e}{\frac{4 \sqrt{2}}{7}}$