Question

The formula $2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$
holds for

• A. $xϵ[0,1]$
• B. $xϵ[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$
• C. $xϵ(-1,0)$
• D. $xϵ[-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}]$

Answer 1

The correct option is B $xϵ[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

We know, $|2x\sqrt{1-x^2}|\le 1$
$4x^2(1-x^2)\le 1$
$4x^4-4x^2+1\ge 0$
$(1-2x^2{)}^2\ge 0$
$1-2x^2\ge 0$
$2x^2\le 1$
$x\le |\frac{1}{\sqrt{2}}]$
Hence
$xϵ[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}|$