1. ## Find range of...?

Find the range of (arcsin x)^2 + (arccos x)^2 ?

2. Originally Posted by fardeen_gen
Find the range of (arcsin x)^2 + (arccos x)^2 ?
The range of this function are all $y\in \mathbb{R}$ such that the equation $(\arcsin x)^2 + (\arccos x)^2 = y$ has a solution in the domain i.e. $-1\leq x\leq 1$. Now we use the identity that $\arccos x = \tfrac{\pi}{2} - \arcsin x$.

This gives,
$(\arcsin x)^2 + \left( \tfrac{\pi}{2} - \arcsin x\right)^2 = y$
This becomes,
$2(\arcsin x)^2 - \pi \arcsin x + \left( \tfrac{\pi^2}{4} - y\right) = 0$
It order for this to have a solution we require,
$\pi^2 - 4(2)\left( \tfrac{\pi^2}{4} -y\right) \geq 0 \implies y\geq \frac{\pi^2}{8}$

But once we have $y$ satisfy this inequality we have,
$\arcsin x = \frac{\pi \pm \sqrt{8y - \pi^2}}{4}$
Therefore,
$x = \sin \left( \frac{\pi \pm \sqrt{8y - \pi^2}}{4} \right)$

Certainly this is in the domain, but does it solve the equation?
For that we need to substitute but we for we do that note,
$\arcsin(\sin x) = \left\{ \begin{array}{c} x \text{ for }-\pi \leq x\leq \pi \\ f(x+2\pi) = f(x) \end{array} \right.$

This means,
$\arcsin \left[ \sin \left( \frac{\pi + \sqrt{8y - \pi^2}}{4} \right) \right] = \left\{ \begin{array}{c} \frac{\pi+\sqrt{8y-\pi^2}}{4} \text{ for }y\leq \frac{5}{4}\pi^2 \\ f(y+2\pi) = f(y)\end{array} \right.$ . . . .[*]

Now if $y\leq \tfrac{5}{4}\pi^2$ then,
$(\arcsin x)^2 + (\arccos x)^2 = (\arcsin x)^2 + (\tfrac{\pi}{2} - \arcsin x)^2$ would give,
$\left( \frac{\pi+\sqrt{8y - \pi^2}}{4} \right)^2 + \left( \frac{\pi - \sqrt{8y - \pi^2}}{4} \right)^2 = y$

What happens if $y > \tfrac{5}{4}\pi^2$?
Therefore the range is $\left[ \frac{\pi^2}{8}, \frac{5\pi^2}{4} \right]$.
$(\arcsin x )^2 + (\arccos x)^2 = (\arcsin x + \arccos x)^2 - 2 \arcsin x + \arccos x$ $= \left( \frac{\pi}{2}\right)^2 - 2 ( \arcsin x ) ( \arccos x )$.
So all you have to do is get the range of $( \arcsin x ) ( \arccos x )$.