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Could anyone please help me find the EXACT value (not numerical value) of this by Mathematica or by mathematical reasoning? Thanks a lot.

Maximize[Min[Abs[Sin[a]], Abs[Cos[a]], Abs[Sin[a - b]], 
  Abs[Cos[a - b]], Abs[Sin[b]], Abs[Cos[b]], Abs[Sin[a - c]], 
  Abs[Cos[a - c]], Abs[Sin[b - c]], Abs[Cos[b - c]], Abs[Sin[c]], 
  Abs[Cos[c]]], {a, b, c}]
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    $\begingroup$ $\sin(\pi/8)$ would appear to be the winner. $\endgroup$
    – Daniel Lichtblau
    Commented May 10, 2020 at 15:00

1 Answer 1

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NMaximize[Min[Abs[Sin[a]], Abs[Cos[a]], Abs[Sin[a - b]], Abs[Cos[a - b]], 
  Abs[Sin[b]], Abs[Cos[b]], Abs[Sin[a - c]], Abs[Cos[a - c]], 
  Abs[Sin[b - c]], Abs[Cos[b - c]], Abs[Sin[c]], Abs[Cos[c]]], {a, b, c}]

(*    {0.382683, {a -> -0.392699, b -> 0.392699, c -> -0.785398}}    *)

These values are exactly

(*    {Sin[π/8], {a -> -π/8, b -> π/8, c -> -π/4}}    *)

There are many equivalent solutions, forming an interesting grid in space.

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  • $\begingroup$ Thank you for your answer. Just wondering, how do you know the maximizer that NMaximize finds is a global maximizer? and how to rigorously show that the exact value must be Sin[π/8]? Thanks a lot. $\endgroup$
    – Natalia C.
    Commented Apr 10, 2020 at 10:54
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    $\begingroup$ You can never know if NMaximize gets stuck in a local extremum or finds the global one. Some of the options, like Method, can help greatly; also, doing a ContourPlot3D can give you an idea if there are any regions of higher values. For a rigorous answer you need to go to the math stackexchange. $\endgroup$
    – Roman
    Commented Apr 10, 2020 at 12:59

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