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let $y(x) = \cos(3 \arccos x)$, then $y(x) = 4x^3 - 3x$. (Chebychev poly), My question is CAN $ \cos( \frac{1}{3} \arccos x )$ be expressed as an alternate function of $x$? This would change the face of the cubic formula forever. Making it a theorem, possibly!

Icarus 369
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scottwaldonhayes
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  • What's an alternate function? –  Aug 14 '16 at 20:38
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    @G.Sassatelli I think the OP is thinking that is there another way to express it, and still be equivalent. – Sigma6RPU Aug 14 '16 at 23:31
  • Correct, G. If cos(3arcosx) = 4x^3 - 3x, then My question is: What is cos(1/3 arcos x) = ? Say for example = square-root( x^3 - 2) + square-root (x^3 + 2) for instance.... WHAT is the answer? A Macclauren series kind of works, even a first order Macclauren is nice, but NOT EXACT! – scottwaldonhayes Aug 15 '16 at 20:01

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$\cos(3 \arccos y) = x$, and thus $x = 4y^3 - 3y$.

Vaneet
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