Question 27fc5

Dec 24, 2016

let $x = \cos 2 A$

So given expression becomes

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

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

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

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

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

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

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

Alternative

Let ${\tan}^{-} 1 \left(\sqrt{\frac{1 - x}{1 + x}}\right) = \theta$

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

$= S \in \left[2 \theta\right]$

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

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

$= \frac{2 \sqrt{\frac{1 - x}{1 + x}}}{1 + \frac{1 - x}{1 + x}}$

$= \frac{2 \sqrt{\frac{1 - x}{1 + x}}}{\frac{1 + x + 1 - x}{1 + x}}$

$= 2 \sqrt{\frac{1 - x}{1 + x}} \times \frac{1 + x}{2}$

$= \sqrt{1 - x} \times \sqrt{1 + x}$

=sqrt(1-x^2#