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Simplify $\tan(2 \arcsin(1/3))$

$\tan(2 \arcsin(1/3))$

$=\tan(2\sin^{-1}(1/3))$

$=\tan^{-1}(2 \sin(1/3))$

Let $\theta =\sin^{-1} 1/3$

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

$=\frac{2\tan(\sin 1/3))}{(1-\tan^2(\sin 1/3))}$

$=[\frac{(2\tan(\sin 1/3)}{((1-\tan( \sin 1/3))^2)}$

$=\frac{(2 \tan(\sin 1/3))}{((1-(\sin(\sin 1/3)/\cos(\sin 1/3))^2)}$

$=\frac{(2 \tan( \sin 1/3))}{((1-(1/3)/(\cos(\sin 1/3))^2)}$

$=\frac{(2 \tan( \sin 1/3))}{((1-(\cos(\sin 1/3)/(1/3))^2)}$

$=[9(2\tan(\sin1/3))*((9-(\cos(\sin1/3))^2)]^{-1}$

. . . I have dont this problem quite a few different ways and can not seem to get the answer. typed into wolfram the answer should be $(4 \sqrt{2})/7$

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1 Answer

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Let $ \theta = \sin^{-1} \frac{1}{3} $ so $ \sin \theta= \frac{1}{3}$ ,$\cos \theta =\frac{\sqrt{8}}{3}$ ,$\tan \theta = \frac{1}{\sqrt{8}}$ so \begin{eqnarray*} \tan (2 \sin^{-1} \frac{1}{3}) =\tan( 2 \theta) = \frac{2 \tan ( \theta)}{1 - \tan^2(\theta)} =\frac{\frac{2}{\sqrt{8}}}{1-\frac{1}{8}} = \color{blue}{\frac{4 \sqrt{2}}{7}}. \end{eqnarray*}

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