The question is -
$$sin^{-1}(x)>cos^{-1}x ∨\ x\ ∃ (0,1)$$
I am getting two different answers by two different methods.I know which method is wrong and which one is right, as i plotted the solution of this inequality on Desmos. I am only posting the wrong solution and wish to find out what's wrong in my method.
$$sin^{-1}(x)+cos^{-1}(x) = \frac {\pi}{2} $$
Substituting $$sin^{-1}(x)= - cos^{-1}(x) + \frac {\pi}{2} $$
and if we substitute this in the original question we get
$$ 2cos^{-1}(x) < \frac {\pi}{2} $$
or $$ cos^{-1}(x) < \frac {\pi}{4} $$
taking cos and we get $$ x <cos( \frac {\pi}{4} ) $$
and get the solution $$ x \ ∃\ (0,\frac{1}{\sqrt2}) $$
If i replace $$ cos^{-1}(x)\ with\ sin^{-1}(x)$$ then i am getting $$ x \ ∃\ (\frac{1}{\sqrt2},1) $$
which apparently is the correct answer. At which step i am wrong?
Because $\cos$ decreases.
I think, it's better to use another way.
We need $$x>\sin\arccos{x}$$ or $$x>\sqrt{1-x^2}$$ or $$x>\frac{1}{\sqrt2}.$$