# Homework Help: Equality math problem

1. Jun 4, 2010

### Mentallic

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

so $$z=cos\left(3arccos(y)\right)$$

by trig identities, $$cos\left(3arccos(y)\right)=4y^3-3y$$

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

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

$$z+\sqrt{z^2-1}=cos(z)+isin(z)$$

simplifying this gives $$y=cos(z/3)$$

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

Last edited by a moderator: Apr 25, 2017
2. Jun 4, 2010

### tiny-tim

Hi Mentallic!
You mean $4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-cos(z)=0$, which is true …

just expand cos(z/3 + 2z/3).

3. Jun 4, 2010

### Mentallic

Re: Equality?

But $$z=cos\left(3arccos(y)\right)$$

and $$cos\left(3arccos(y)\right)=4y^3-3y$$ where $$y=cos\left(\frac{1}{3}arccos(z)\right)$$

So shouldn't it then be $$4y^3-3y=z$$ ???

Yeah if I change it to cos(z) it works, but I still don't see where I've gone wrong.

4. Jun 4, 2010

### tiny-tim

You're interchangeably using z as an angle and as a cosine.

In particular, your …
isn't true.

(unless cosz = z)

5. Jun 4, 2010

### Mentallic

Re: Equality?

Oh of course...