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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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    @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-2012-04-14

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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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    No way! I had that book i$n$ my ha$n$d some months ago! I was going to buy it!2012-04-14