AOD maximum?

GRAPHICAL METHOD -

The expression can be written as

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

(Because $|\cos^{-1} x| = \cos^{-1} x$ )

Sketching the graphs for both $|\sin^{-1} x|$ and $\cos^{-1} x$

If we observe the graphs of $|\sin^{-1} x|$ and $\cos^{-1} x$ , we see that $|\sin^{-1} x|$ attains maximum value at $x = -1$ and $x = 1$ whereas $\cos^{-1} x$ attains maximum value at $x = -1$ .

Thus, $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ must have maximum value at $x = -1$

Putting $x = -1$ , We get -

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

$= \Big(\frac{-\pi}{2}\Big)^2 + (\pi)^2$

$= \frac{5}{4} (\pi)^2 = \boxed{\color{#20A900}{12.337}}$

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

$\Rightarrow (sin^{-1}x)^2 + (cos^{-1}x)^2 = \frac{\pi ^2}{4} - 2sin^{-1}x\times cos^{-1}x$

For maximum value of $(sin^{-1}x)^2 + (cos^{-1}x)^2$ , $2sin^{-1}x\times cos^{-1}x$ should be minimum & its minimum value occurs at $x=-1$ .

Hence maximum value of $(sin^{-1}x)^2 + (cos^{-1}x)^2 = \dfrac{\pi ^2}{4} -2\times (\dfrac{-\pi}{2})\times \pi$

$= \dfrac{5\pi ^2}{4} \approx 12.32$

Use chain rule $[(sin^{-1} x)^2+(cos^{-1} x)^2]'=\frac{2(\sin^{-1}x-\cos^{-1}x)}{\sqrt{1-x^2}}$

To find maximum

$\frac{2(\sin^{-1}x-\cos^{-1}x)}{\sqrt{1-x^2}}=0$

Minimun $2(\sin^{-1}x-\cos^{-1}x)=0$

Maximum $\sqrt{1-x^2}=0$

Then maximum is when $x=-1$ $(sin^{-1} {-1})^2+(cos^{-1}{-1})^2=\frac{\pi^2}{4}+\pi^2=\boxed{\frac{5\pi^2}{4}}$

Concerning real numbers, there can be maximum values. Inverse trigonometric functions give angles (in Radians) which are at the same status as logarithmic functions with principal values. For well defined inverse functions, only principal values are concerned. Domain for both of them is from -1 via 0 to 1, while range is from -Pi/ 2 via 0 to Pi/ 2 and 0 via Pi/ 2 to Pi respectively, for a one to one mappings. They are something opposite to the "All Science Teachers Crazy" for positive quadrants. The negative sides are actually taking another quadrants.

Taking mirror images about y = x, y = Asin x is just a curve with infinite slopes at both x = -1 and x = 1 while x = 0 is a point of inflexion. Acos x = Pi/ 2 - Asin x is then another treatment which is quite unusual and implicit in Excel. All streams in mathematics do the same. Acos x simply follows Asin x but not doing own series and own graph! In fact, where insisted, the curve of Acos x is just a mirror about x = 0 plus a raise up of Asin x. From 0 to Pi, Acos x is above y = 0. The greatest difference is obviously on our L.H.S.

y^2 + (Pi/ 2 - y)^2

= 2 y^2 +(- Pi) y + Pi^2/ 4

= 2 (y - Pi/ 4)^2 + Pi^2/ 8

= 2 (-Pi/ 2 - Pi/ 4)^2 + Pi^2/ 8 {Since y = -Pi/ 2 is optimum at x = -1}

= 5 Pi^2/ 4

=12.3370055013617 {Where x = -1 is obviously makes a maximum to contribute.}

I actually simulated with Excel knowing this complicated features so that I shall not get anything wrong. Sqrt (1 - x^2) from derivative tells the location of infinite slopes but x = 1 gives 2.46740110027234 instead and therefore actually doesn't tell. Either justify from Excel or from the above, the maximum values for sum of inverse functions can hardly be derived from other ways, usually.

Derivatives only tell where the turning points are located and not a generally way for determining maximum or minimum.

simpler way to do this is USING THE FACT THAT SINE INVERSE X PLUS COS INVESE X IS Pi/2 NOW (SIN INV X)^2 + (COS INV X)^2 =(SIN INV X + COS INV X)^2 -2(SIN INVX)(COSINV X)

LET {COS INV X= K} to get: =(Pi/4) - (K)(Pi/2-K)

HENCE WE GET A QUADRATIC IN K WHOSE MAX VALUE IS AT 12.32.