I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$ I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$
The easiest way was to just look at the graph and I found out that the region is $x \in ({1\over \sqrt{2}} , 1]$
But I tried to prove the statement algebraically also but couldn't get it correctly.
For $$\sin ^{-1}x > \cos ^{-1}x$$
$$\Rightarrow \sin ^{-1}x -\cos ^{-1}x>0$$
$$\Rightarrow \sin ^{-1}x - \sin ^{-1} \sqrt{1-x^2}>0 \tag1$$ 
Using the identity $$\sin ^{-1}x -\sin ^{-1}y=\sin ^{-1} \left (x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)$$ Equation $(1)$ can be written as
$$\Rightarrow \sin ^{-1}(2x^2-1)>0$$
For this to be true $0<2x^2-1<1$ and also for the equation to be valid $-1<2x^2-1<1$
Hence we can take 
$0<2x^2-1<1$ as the intersection of both the conditions
$$\Rightarrow 0\le  2x^2-1\le 1$$
$$\Rightarrow 1\le 2x^2\le 2$$
$$\Rightarrow 1/2\le x^2\le 1$$
The solution set of the inequality would be $x \in ({1\over \sqrt{2}} , 1] \cup [-1,{-1\over \sqrt{2}}),$
Can anybody tell me why I am getting the wrong answer.
 A: $$\sin^{-1}x>\cos^{-1}x=\dfrac\pi2-\sin^{-1}x$$
$$\iff\sin^{-1}x>\dfrac\pi4\iff x>\sin\dfrac\pi4$$
In fact, $$\cos^{-1}x=\begin{cases} \sin^{-1}\sqrt{1-x^2} &\mbox{if } x\ge0 \\ 
\pi-\sin^{-1}\sqrt{1-x^2} & \mbox{if } x<0\end{cases}$$
A: It is only true that $\cos^{-1}(x) = \sin^{-1}(\sqrt{1-x^2})$ in the region $x \in [0, 1]$. If you compare the graphs of the two on Wolfram Alpha, you can see how they diverge. Or, if you're more of a numeric kind of guy, note that $\cos^{-1}(-1) = \pi$ while $\sin^{-1}(\sqrt{1-(-1)^2})=\sin^{-1}(0) = 0$.
A: $$\sin^{-1}x>\cos^{-1}x $$
$$\pi/2-\sin^{-1}x < \pi/2 -\cos^{-1}x $$
$$ 2 \cos^{-1}x < \pi/2 $$
$$   \cos^{-1}x < \pi/4 $$
$$ \frac{1}{\sqrt 2}< x < 1, $$
and we can also include co-termianal angles with $ 2 k \pi.$
A: From very basic facts:
Use this corollary of the Mean value theorem:

Let $I$ be an interval, $x_0\in I$, and $f,\,g\,$  functions differentiable on the interior of $I$, such that $\;f(x_0)=g(x_0)\;$ and $\;f'(x)>g'(x)\;$ on $\;\stackrel{\circ}{I}$. Then 
  $$f(x)>g(x)\quad\text{for}\enspace x>x_0,\quad f(x)<g(x)\quad\text{for}\enspace x<x_0.$$

In the present case:


*

*$\arcsin\dfrac{\sqrt2}2=\arccos\dfrac{\sqrt2}2=\dfrac\pi4$,

*$(\arcsin)'(x)=\dfrac1{\sqrt{1-x^2}}>-\dfrac1{\sqrt{1-x^2}}=(\arccos)'(x)$.


Hence  the sought for interval is $\;\Bigl(\dfrac{\sqrt 2}2, 1\Bigr]$.
A: About your method you have to note that, unfortunately, $\arccos x=\arcsin\sqrt{1-x^2}$ is a false identity in general; it's true only for $0\le x\le 1$.
However, for $-1\le x<0$ we have $-\pi/2\le\arcsin x<0$ and $\pi/2<\arccos x\le\pi$, so we can exclude this interval, where the inequality certainly doesn't hold.
So we have
$$
\begin{cases}
\arcsin x>\arcsin\sqrt{1-x^2} \\[4px]
0\le x\le 1
\end{cases}
$$
that becomes, since the arcsine is increasing
$$
\begin{cases}
x>\sqrt{1-x^2} \\[4px]
0\le x\le 1
\end{cases}
$$
Due to the second limitation, we can square the first inequality
$$
\begin{cases}
x^2>1-x^2 \\[4px]
0\le x\le 1
\end{cases}
$$
and easily get $1/\sqrt{2} < x \le 1$.
On the other hand, if $\alpha=\arccos x$, we have $0\le\alpha\le\pi$, so $-\pi/2\le\pi/2-\alpha\le\pi/2$ and from $x=\cos\alpha$ we get $x=\sin(\pi/2-\alpha)$, so
$$
\frac{\pi}{2}-\alpha=\arcsin x
$$
and therefore
$$
\arccos x=\frac{\pi}{2}-\arcsin x
$$
which makes your inequality very easy to solve.
A: You're not getting the wrong answer, when you take the square root you just have to remember that $x^2=(−x)^2$. But as ConMan points out, your calculations aren't valid for negative numbers, so this is mostly luck.
$]\frac{-1}{\sqrt 2},-1[$ is an unusual way of writing an interval involving negative numbers, as $-1<\frac{-1}{\sqrt 2}$, I would say that $-1$ should be the left endpoint, but I guess everybody will understand you.
