# Simplify Sin[2tan^-1(sqrt((1-x)/(1+x)))]?

Dec 24, 2016

$\sin \left[2 {\tan}^{-} 1 \left(\frac{\sqrt{1 - x}}{\sqrt{1 + x}}\right)\right] = - \sqrt{1 - {x}^{2}}$

#### Explanation:

$\sin \left[2 {\tan}^{-} 1 \left(\frac{\sqrt{1 - x}}{\sqrt{1 + x}}\right)\right]$

As $\sqrt{1 - x}$, is defined, we have $x < 1$ and hence let $x = \sin 2 A = 2 \sin A \cos A$

then $\frac{\sqrt{1 - x}}{\sqrt{1 + x}} = \frac{\sqrt{{\sin}^{2} A + {\cos}^{2} A - 2 \sin A \cos A}}{\sqrt{{\sin}^{2} A + {\cos}^{2} A - 2 \sin A \cos A}}$

= $\frac{\sin A - \cos A}{\sin A + \cos A}$

= $\frac{\sin \frac{A}{\cos} A - 1}{\sin \frac{A}{\cos} A + 1}$

= $\frac{\tan A - \tan \left(\frac{\pi}{4}\right)}{\tan A \tan \left(\frac{\pi}{4}\right) + 1}$

= $\tan \left(A - \frac{\pi}{4}\right)$

Hence $2 {\tan}^{-} 1 \left(\frac{\sqrt{1 - x}}{\sqrt{1 + x}}\right) = 2 \left(A - \frac{\pi}{4}\right) = 2 A - \frac{\pi}{2}$ and

$\sin \left[2 {\tan}^{-} 1 \left(\frac{\sqrt{1 - x}}{\sqrt{1 + x}}\right)\right]$

= $\sin \left(2 A - \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{2} - 2 A\right) = - \cos 2 A = - \sqrt{1 - {\sin}^{2} 2 A} = - \sqrt{1 - {x}^{2}}$

Dec 24, 2016

let $x = \cos 2 A$

So given expression becomes

$S \in \left[2 {\tan}^{-} 1 \left(\sqrt{\frac{1 - x}{1 + x}}\right)\right]$

$= S \in \left[2 {\tan}^{-} 1 \left(\sqrt{\frac{1 - \cos 2 A}{1 + \cos 2 A}}\right)\right]$

$= S \in \left[2 {\tan}^{-} 1 \left(\sqrt{\frac{2 {\sin}^{2} A}{2 {\cos}^{2} A}}\right)\right]$

$= S \in \left[2 {\tan}^{-} 1 \left(\tan A\right)\right]$

$= S \in \left(2 A\right)$

$= \sqrt{1 - {\cos}^{2} \left(2 A\right)}$

$= \sqrt{1 - {x}^{2}}$

We have taken $x = \cos 2 A$
And $\sqrt{\frac{1 - x}{1 + x}}$ to be real the domain of this function will be $1 \ge x > - 1$ $\implies 2 A \in \left[0 , \pi\right) \implies A \in \left[0 , \frac{\pi}{2}\right)$
So $\sin 2 A$ or $S \in \left[2 {\tan}^{-} 1 \left(\sqrt{\frac{1 - x}{1 + x}}\right)\right]$ should be $\textcolor{red}{P O S I T I V E}$