If cos^-1 x > sin^-1 x, then x belong to the internal

We know that $ cos^{-1}x$ and  $sin^{-1}x$ exist for  x ∈ [-1, 1].

Now, 

$cos^{-1}x > sin^{-1}x $

$⇒ cos^{-1}x> \frac{\pi}{2}- cos^{-1}x$

$⇒ 2cos^{-1} x > \frac{\pi}{2}⇒cos^{-1} x > \frac{\pi}{4} ⇒ x ∈ [-1, \frac{1}{\sqrt{2}})$