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Find the upper and lower limit of

$$ (\sin^{-1}x)^2+(\cos^{-1}x)^2 $$

My Attempt:

$$ \frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&\quad(\cos^{-1}x)^2\leq\pi^2\\ 0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4} $$

Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.

Why am I not getting the lower limit in my approach ?

How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?

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By C-S $$\arcsin^2{x}+\arccos^2{x}=\frac{1}{2}(1^2+1^2)(\arcsin^2{x}+\arccos^2{x})\geq\frac{1}{2}(\arcsin{x}+\arccos{x})^2=\frac{\pi^2}{8}.$$ The equality occurs for $x=\frac{1}{\sqrt2}$, which says that we got a minimal value.

In another hand, for $x\geq0$ we obtain: $$\arcsin^2{x}+\arccos^2{x}\leq\left(\arcsin{x}+\arccos{x}\right)^2=\frac{\pi^2}{4}.$$ The equality occurs for $x=0$.

For $x<0$ let $t=-\arcsin{x}$.

Thus, $0<t\leq\frac{\pi}{2}$, $\arccos{x}=\pi-\arccos(-x)=\pi-\left(\frac{\pi}{2}-\arcsin(-x)\right)=\frac{\pi}{2}+t$ and $$\arcsin^2{x}+\arccos^2{x}=t^2+\left(\frac{\pi}{2}+t\right)^2\leq\frac{5\pi^2}{4}$$ with equality for $x=-1$, which says that $\frac{5\pi^2}{4}$ is a maximal value.

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  • $\begingroup$ thanx. i didnt think abt C-S inequality. I understand how u found the min using C-S inequality. But, how do u find the max ?. srry didnt get second step. $\endgroup$ – ss1729 Jan 19 '18 at 18:19
  • $\begingroup$ I added something. I hope now it's clear. $\endgroup$ – Michael Rozenberg Jan 19 '18 at 18:44
  • $\begingroup$ How is minimum attained at x=π/4 shouldn't it be arcsin x= π/4 which implies x=1/√2 $\endgroup$ – Darkrai Jan 31 '18 at 9:35
  • $\begingroup$ It was typo. I fixed. Thank you! $\endgroup$ – Michael Rozenberg Jan 31 '18 at 12:06
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Let $\sin^{-1}x=a,\cos^{-1}x=b$

$a+b=\dfrac\pi2$

$$a^2+b^2=\left(\dfrac\pi2-b\right)^2+b^2=\dfrac{\pi^2}4+2\left(b-\dfrac\pi4\right)^2-2\left(\dfrac\pi4\right)^2$$

Again , $0\le b\le\pi\implies0\le\left(b-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$

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  • $\begingroup$ @Travis, Agreed. Thanks $\endgroup$ – lab bhattacharjee Jan 19 '18 at 10:42
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You can do it with calculus: if $f(x)=(\arcsin x)^2+(\arccos x)^2$, then $$ f'(x)=\frac{2\arcsin x}{\sqrt{1-x^2}}-\frac{2\arccos x}{\sqrt{1-x^2}} $$ (for $-1<x<1$). The derivative vanishes for $\arcsin x=\arccos x$, that is, $x=1/\sqrt{2}$. Since $$ f(1/\sqrt{2})=\frac{\pi^2}{8} \qquad f(-1)=\frac{5\pi^2}{4} \qquad f(1)=\frac{\pi^2}{4} $$ you can easily draw the conclusion.

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$$(\arcsin x)^2+(\arccos x)^2=(\arccos x+\arcsin x)^2-2\arccos x\cdot \arccos x\leq \left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{4}.$$

Equality hold when $x=0$ or $\displaystyle x=1$

Similarly using $$\frac{(\arcsin x)^2+(\arccos x)^2}{2}\geq \bigg(\frac{\arcsin x+\arccos x}{2}\bigg)^2$$

so $$(\arcsin x)^2+(\arccos x)^2\geq \frac{\pi^2}{8}$$

equality hold when $\displaystyle x=\frac{1}{\sqrt{2}}$.

And at $x=-1$ value of $\displaystyle (\arcsin x)^2+(\arccos x)^2=\frac{5\pi^2}{4}$

$$\Rightarrow \frac{\pi^2}{8}\leq (\arcsin x)^2(\arccos x)^2\leq \frac{5\pi^2}{4}$$

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Given, differentiate $f(x)$ with respect to $x$ now, $f'(x) = 0$ then, We also know, domain of and $∈ [ -1 1] ∴$ find $f(x)$ at $x = 1$ , $-1$ and $\frac{1}{\sqrt{2}}$

$f(1) = (sin^{-1}(1))^2 + (cos^{-1}(1))^2 = \frac{π^2}{4} + 0 = \frac{π^2}{4}$ $f(-1) = (sin^{-1}(-1))^2 + (cos^{-1}(-1))^2 = \frac{π^2}{4} + π^2 = \frac{5π^2}{4}$ $f(\frac{1}{\sqrt{2}}) = (sin^{-1}(\frac{1}{\sqrt{2}}))^{2} + (cos^{-1}(\frac{1}{\sqrt{2}}))^2 = \frac{π^2}{16} + \frac{π^2}{16} = \frac{π^2}{8}$

Hence, maximum value of $f(x) = \frac{5π^2}{4}$ Minimum value of $f(x) = \frac{π^2}{8}$

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    $\begingroup$ This doesn't really add anything that the existing answers didn't have. Except poor formatting. $\endgroup$ – Henrik Apr 4 at 14:37

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