Question

find the greatest and least values:(sin-1x)2+(cos-1x)2

Answer 1

Given, $\bold{f(x)=(sin^{-1}x)^2+(cos^{-1}x)^2}$
differentiate f(x) with respect to x
$\bold{f'(x)=2sin^{-1}x\frac{1}{\sqrt{1-x^2}}+2cos^{-1}\frac{-1}{\sqrt{1-x^2}}}$
$\bold{f'(x)=\frac{2}{\sqrt{1-x^2}}[sin^{-1}x-cos^{-1}x]}$
now, f'(x) = 0
then, $\bold{[sin^{-1}x-cos^{-1}x]=0}\\\bold{sin^{-1}x=cos^{-1}x}\implies x=\frac{1}{\sqrt{2}}$

We also know, domain of $Sin^{-1}x$ and $Cos^{-1}x$ ∈ [ -1 1]
∴ find f(x) at x = 1 , -1 and 1/√2

f(1) = (sin⁻¹(1))² + (cos⁻¹(1))² = π²/4 + 0 = π²/4
f(-1) = (sin⁻¹(-1))² + (cos⁻¹(-1))² = π²/4 + π² = 5π²/4
f(1/√2) = (sin⁻¹(1/√2))² + (cos⁻¹(1/√2))² = π²/16 + π²/16 = π²/8

Hence, maximum value of f(x) = 5π²/4
Minimum value of f(x) = π²/8

Answer 2

$answer \: = > \frac{5 {\pi}^{2} }{4} and \: \frac{ {\pi}^{2} }{8}$

$\red{</p><p>step \: by \: step \: explanation </p><p>} \\ = >$

$( { { \sin }^{ - 1}x})^{2} + ({ { \cos }^{ - 1} x})^{2} \\ = {[ ({ \sin }^{ - 1}x + { \cos}^{ - 1} ) ]}^{2} - 2 { \sin }^{ - 1} x { \cos }^{ - 1} \\ = { \frac{\pi}{4} }^{2} - 2 { \sin }^{ - 1} x( \frac{\pi}{2} - { \sin }^{ - 1} x) \\ = \frac{ {\pi}^{2} }{4} - \pi { \sin }^{ - 1} x + 2 {( { \sin}^{ - 1} x)}^{2} \\ = 2 [ {( { \sin }^{ - 1} x)}^{2} - \frac{\pi}{2} { \sin}^{ - 1} x + { \frac{\pi}{8} }^{2} ] \\ = 2 [ {( { \sin }^{ - 1} x)}^{2} - \frac{\pi}{2} { \sin}^{ - 1} x + \frac{ {\pi}^{2} }{16} - \frac{ {\pi}^{2} }{16} + { \frac{\pi}{8} }^{2} ] \\ = 2[ {( { \sin }^{ - 1}x - \frac{ {\pi}^{2} }{4} )} + \frac{ {\pi}^{2} }{16} ] \\ now \frac{ - \pi}{2} \leqslant { \sin}^ {- 1} x \leqslant \frac{\pi}{2} \: \: for \: \: all \: x ∈[ - 1,1 ] \\ = > \frac{ - \pi}{2} - \frac{\pi}{4} \leqslant { \sin}^{ - 1} - \frac{\pi}{4} \leqslant \frac{\pi}{2} - \frac{\pi}{4} : for \: \: all \: x ∈[ - 1,1 ] \\ = > - \frac{3\pi }{4} \leqslant {( { \sin }^{ - 1} x - \frac{\pi}{4} )}^{2} \leqslant \frac{\pi}{4} : for \: \: all \: x ∈[ - 1,1 ] \\ = > 0 \leqslant {( { \sin }^{ - 1} x - \frac{\pi}{4} )}^{2} \leqslant \frac{9 {\pi}^{2} }{16} \\ = > \frac{ {\pi}^{2} }{16} \leqslant {( \sin }^{ - 1} x - \frac{\pi}{4})^{2} + \frac{ {\pi}^{2} }{16} \leqslant \frac{9 {\pi}^{2} }{16} + \frac{ {\pi}^{2} }{16} \\ \\ = > \frac{ {\pi}^{2} }{16} \leqslant {( \sin }^{ - 1} x - \frac{\pi}{4})^{2} + \frac{ {\pi}^{2} }{16} \leqslant \frac{5 {\pi}^{2} }{8} \\ = > \frac{ {\pi}^{2} }{8 } \leqslant 2[ {( { \sin }^{ - 1} - \frac{\pi}{4} )}^{2} + \frac{ {\pi}^{2} }{16} ] \leqslant \frac{5 {\pi}^{2} }{4} \\ = > \frac{ {\pi}^{2} }{8} \leqslant {( { \sin }^{ - 1} x)}^{2} + {({ \cos}^{ - 1} x)}^{2} \leqslant \frac{5 {\pi}^{2} }{4} \\ \blue{ hence \: the \: greatest \: and \: the \: least \: value \: of} \pink { { ({ \sin }^{ - 1}x) }^{2} + { ({ \cos }^{ - 1} x)}^{2} } \purple{are} \green{ \frac{5 {\pi}^{2} }{4} } and \red { \frac{ {\pi}^{2} }{8} } \: respectively$

by rakhithakur