I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$

Question

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.

Answer 1

$$\sin^{-1}x>\cos^{-1}x=\dfrac\pi2-\sin^{-1}x$$

$$\iff\sin^{-1}x>\dfrac\pi4\iff x>\sin\dfrac\pi4$$

In fact, $$\cos^{-1}x=\begin{cases} \sin^{-1}\sqrt{1-x^2} &\mbox{if } x\ge0 \\ \pi-\sin^{-1}\sqrt{1-x^2} & \mbox{if } x<0\end{cases}$$

Answer 2

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 $\sin^{-1}(\sqrt{1-(-1)^2})=\sin^{-1}(0) = 0$.

Answer 3

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

$$\pi/2-\sin^{-1}x < \pi/2 -\cos^{-1}x $$

$$ 2 \cos^{-1}x < \pi/2 $$

$$ \cos^{-1}x < \pi/4 $$

$$ \frac{1}{\sqrt 2}< x < 1, $$

and we can also include co-termianal angles with $ 2 k \pi.$

Answer 4

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 $\;\Bigl(\dfrac{\sqrt 2}2, 1\Bigr]$.

Answer 5

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.

Answer 6

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.