Question

What is the value of x ? satisfying

`(sin^-1` `x)^3` `-` `(cos^-1` `x)^3` `+` `(sin^-1` `x)` `(sin^-1` `x` `-` `cos^-1` `x)` = `(pi^3)/16`

Answer 1

Solve `arcsin^3 x - arccos^3 x + arcsin x ( arcsin x - arccos x) = pi^3/16`

We'll assume we're referring to the principal values of the inverse functions here.

What's the relation between `arcsin x` and `arccos x` ? Obviously the sine of the first equals the cosine of the second; i.e. they're complementary angles, which add to `pi/2.` This is true even if `-1 le x < 0` as the principal value of the inverse sine will be in the fourth quadrant and the inverse cosine in the second.

Let `a=arcsin x`

`arccos x = pi/2-a`

`a^3 - (pi/2-a)^3 + a( a - (pi/2-a)) = pi^3/16`

`16a^3 - 2(pi-2a)^3 + (32a^2-8 pi a) - pi^3 = 0`

`16a^3 - 2(pi^3 - 6pi^2 a + 12 pi a^2-8a^3) + (32a^2-8 pi a) - pi^3 = 0`

`32a^3 + (-24 pi + 32)a^2 + ( 12 pi^2 -8pi ) a - 3 pi^3 = 0`

Well that's a pretty ugly cubic. I'll skip the cubic formula and let Alpha solve it numerically:

`a approx 1.10268`

`x = sin a approx 0.89242`

Approximations, not very satisfying. Probably a mistake somewhere.