Question

If $xϵ[-1,0)$, then find the value of ${\cos }^{-1}(2x^2-1)-2{\sin }^{-1}x$

• A. $-\pi /2$
• B. $+\pi /2$
• C. $-\pi$
• D. $+\pi$

Answer 1

The correct option is D $+\pi$

Let $x=\cos \theta$, for $xϵ[-1,0)\Rightarrow \theta ϵ(\pi /2,\pi ]$
$\Rightarrow$ ${\cos }^{-1}(2x^2-1)-2{\sin }^{-1}x={\cos }^{-1}(2{\cos }^2\theta -1)-2{\sin }^{-1}\left(\cos \theta \right)$
$={\cos }^{-1}\left(\cos 2\theta \right)-2(\pi /2-{\cos }^{-1}\left(\cos \theta \right))$
$={\cos }^{-1}\left(\cos 2\theta \right)+2{\cos }^{-1}\left(\cos \theta \right)-\pi$
$=2\pi -2\theta +2\theta -\pi =\pi$