(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I found this equality in the thread https://www.physicsforums.com/showthread.php?t=407130"

[tex]

cos\left(\frac{1}{3}arccos(z)\right) = \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}

[/tex]

and I'd like to know how it works.

Note: I haven't studied this, but I do know about complex numbers and I got some hints from that thread on what to do.

3. The attempt at a solution

First of all, I assumed it was true.

For the left side of the equality, let [tex]y=cos\left(\frac{1}{3}arccos(z)\right)[/tex]

so [tex]z=cos\left(3arccos(y)\right)[/tex]

by trig identities, [tex]cos\left(3arccos(y)\right)=4y^3-3y[/tex]

So the solutions to [tex]y=cos\left(\frac{1}{3}arccos(z)\right)[/tex] are the solutions (not exactly sure which of the 3) to the cubic [tex]4y^3-3y-z=0[/tex]

and now taking the right side, [tex]y=\frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}} [/tex]

[tex]z+\sqrt{z^2-1}=cos(z)+isin(z)[/tex]

simplifying this gives [tex]y=cos(z/3)[/tex]

So hence for this equality to be true, [tex]4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-z=0[/tex] for all z, but this isn't the case.

Please help me understand this more

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# Homework Help: Equality math problem

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