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

  1. Jul 15, 2006 #1


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    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 [tex]\sqrt{1^2+1^2}=\sqrt{2}[/tex] 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 [tex]\sqrt{1^2+( \sqrt{2} )^2}=\sqrt{3} [/tex] again by the Pythagorean theorem. Hence the desired angle is given by [tex] \mbox{arcsin} \left( \sqrt{\frac{2}{3}}\right)[/tex] or, equivalently, by [tex]\mbox{arccos} \left( \frac{1}{\sqrt{3}}\right)[/tex], 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?
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  3. Jul 15, 2006 #2


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

    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!
  4. Jul 15, 2006 #3


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    I am trying to prove that a particular sequence of sets converges to the unit hypercube (see old post), in particular I was working the 3-d case, finding the points on the surface [tex]x^{2N}+y^{2N}+z^{2N}=1[/tex] 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.
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