Set of values of $a$ satisfying an inverse trigonometric equation.

Question

Find the set of values of a so that the equation $(\arcsin x)^3+ (\arccos x)^3= a\pi^3$ has a solution.

Attempt:

Let $y = \arcsin x$ (just for convenience to avoid typing/ writing $\arcsin$ over and over again )

Note that $y + \arccos x = \pi /2$

We have

$(y+\arccos x)(y^2 + (\frac\pi 2-y)^2 - y (\frac \pi2 - y)= a\pi^3$

$\implies (\sqrt 3y- \dfrac{\sqrt3}{4}\pi)^2+\dfrac{\pi^2}{16}= 2a\pi^2$

$-\frac \pi 2\le y \le \frac\pi 2$

$\implies -\dfrac{\sqrt 3 \pi}{2} - \sqrt{3}\pi/ 4\le \sqrt 3 y - \frac{\sqrt {3}}{4}\pi \le \dfrac{\sqrt 3 \pi}{2}-\sqrt{3}\pi/ 4\\ \implies 0\le 2a\pi^2- \dfrac{\pi^2}{16}\le \dfrac{3}{16}\pi^2 \implies a\in [\dfrac{1}{32}, \dfrac{1}{16}]$

I get the right lower limit of $a$ but the upper limit given in the answer is $7/8$.

I'll be grateful if someone could point out my error and if that has been done, provide alternative, easier ways to solve this problem.

Answer 1

The question is equivalent to finding the range of the function $$x\mapsto \frac{(\arcsin x)^3+ (\arccos x)^3}{\pi^3}$$

defined on $[-1,1]$.

You can do that by studying the variations of the function via its derivative, and the data of extremal values in $x=-1$ and $x=1$.

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Edit

The problem is here: $ -\dfrac{\sqrt 3 \pi}{2} - \sqrt{3}\pi/ 4\le \sqrt 3 y - \frac{\sqrt {3}}{4}\pi \le \dfrac{\sqrt 3 \pi}{2}-\sqrt{3}\pi/ 4$

from here, the largest value for the square will come from squaring the LHS, not the RHS.

Answer 2

write your equation in the form $$\left(\frac{\pi}{2}-\arcsin(x)\right)^3+(\arcsin(x))^3=\pi^3a$$ and this is $$\frac{\pi^3}{8}-\frac{3}{4}\pi^2\arcsin(x)+\frac{3}{2}\pi(\arcsin(x))^2=\pi^3a$$ and this is a quaqdratic equation, can you solve it?