It should be simple: it's only trig, right?

[benorin]

I'm trying to determine the angle between an edge of a unit cube and either space diagonal which shares an end point with said edge (to the end that I may represent a vertex of the unit cube in spherical coordinates.)

My work: A face diagonal of the unit cube is of length $$\sqrt{1^2+1^2}=\sqrt{2}$$ by the Pythagorean theorem, now the right triangle having an edge and a face diagonal for its legs and a space diagonal as its hypotenuse gives the space diagonal's length as being $$\sqrt{1^2+( \sqrt{2} )^2}=\sqrt{3}$$ again by the Pythagorean theorem. Hence the desired angle is given by $$\mbox{arcsin} \left( \sqrt{\frac{2}{3}}\right)$$ or, equivalently, by $$\mbox{arccos} \left( \frac{1}{\sqrt{3}}\right)$$, yet I cannot determine an exact value for this angle (which I expect to be able to find given the simple geometric nature of the problem,) a numeric approximation of this angle is 54.7357... degrees.

Did I make a mistake? Is there some trick to this?

 

[Hurkyl]

I doubt that there is a simple expression for that angle, aside from simply writing it as something like $\cos^{-1} (1/\sqrt{3})$.


p.s. what is the reason to express it in spherical coordinates? Maybe there is a better approach that avoids the need for real numbers!

 

[benorin]

I am trying to prove that a particular sequence of sets converges to the unit hypercube (see https://www.physicsforums.com/showthread.php?t=104727), in particular I was working the 3-d case, finding the points on the surface $$x^{2N}+y^{2N}+z^{2N}=1$$ which are furthest from (closest to) the origin using spherical coordinates. I have since found a much simpler method for determining said extrema in the n-D case using the method of Lagrange multipliers but the angle thing was still bugging me.