Trigonometric result concerning DeMoivre's formula.

Given this question is rather long to answer, and I'm losing hope it'll ever be, I just want an answer to this particular claim:

Working on the unitary circle, let $x=1-\cos \theta$ and $t=1-\cos n \theta$. Then how can one produce the following equations:

$$1-2z^n+z^{2n}=-2z^nt$$

$$1-2z+z^{2}=-2zx \text{ ?}$$

These arise from setting $z^n = l+\sqrt{l^2-1}$ in

If $l$ and $x$ are the cosines of two arcs $A$ and $B$ of a circle of radius unity, and if the first arc is to the second as the number $n$ is to unity then:

$$x = \frac{1}{2}\root n \of {l + \sqrt {{l^2} - 1} } + \frac{1}{2}\frac{1}{{\root n \of {l + \sqrt {{l^2} - 1} } }}$$

How can this be proven?

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What is the cosine of an arc? As in the cosine of the angle it forms? Also is the statement "the first arc to the second..." as in, the angle of one arc is the second multiplied by $n$? If yes, this should be related to Chebyshev polynomials. –  Alex R. Apr 14 '12 at 1:21
@Sam Why would there by any doubt what the cosine of an arc is? The statement is saying $l = \cos n\theta$ and $x= \cos \theta$ if I'm not reading things backwards- –  Pedro Tamaroff Apr 14 '12 at 1:55

I'm setting $l=\cos(n\theta)$ and $x=\cos(\theta)$ to match your notation in the comments.

The $n$'th Chebyshev polynomial $T_n(x)$ can be defined by $T_n(\cos(\theta))=\cos(n\theta)$. Then the identity that is highlighted is

$l=T_n(x)=\frac{(x-\sqrt{x^2-1})^n+(x+\sqrt{x^2-1})^n}{2}$

We just need to derive this relation. It comes from

$l=\cos(n\theta)=Re [ e^{in\theta}] = Re[(\cos(\theta)+i\sin(\theta))^n]=Re[(x+i\sqrt{1-x^2})^n]$

thus, using $z=a+bi$ and $Re[z]=\frac{1}{2}(z+\bar{z})$,

$l=\frac{(x+i\sqrt{1-x^2})^n+(x-i\sqrt{1-x^2})^n}{2}=\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{2}$

So maybe $x$ and $l$ should have their roles reversed if you want the form that's highlighted.

I'm not quite following why you say that $x=1-\cos(\theta)$ and $l=1-cos(n\theta)$ in the first part. Are these meant to be different variables from the ones below?

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They'd be the versed sines. Remember that you cannot use DeMoivre's formula, since what I want is to understand DeMoivre's derivation. (I'm not sure if you are in the middle step from $e^{i\theta} \to \cos \theta +i \sin\theta$) –  Pedro Tamaroff Apr 14 '12 at 14:06
I'm a bit confused. If you want a derivation of DeMoivre's formula, there is one by induction here: en.wikipedia.org/wiki/De_Moivre's_formula. The problem is that this formula you are interested in is necessarily of complex analysis nature. Notice how you are taking the square root of $x^2-1$ which is something like $\sqrt{-\sin^2(\theta)}$ –  Alex R. Apr 14 '12 at 20:47
I know. See this. I'm trying to understand how DeMoivre got there, and he obivously didn't know the formula before he discovered it. –  Pedro Tamaroff Apr 14 '12 at 20:51
It sounds like this might give you some insight. books.google.com/… –  Alex R. Apr 14 '12 at 20:54
No way! I had that book in my hand some months ago! I was going to buy it! –  Pedro Tamaroff Apr 14 '12 at 20:58