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For a function:

$$f(x)=\frac{(1-x^2)}{(1+x^2)-2 x \cos \omega}$$

Let $x=-\frac{1}{3}$

That means that

$$f(-\frac{1}{3})=\frac{(\frac{8}{9})}{(\frac{10}{9})+\frac{2}{3} \cos \omega}=\frac{4}{5+3 \cos \omega}$$

If we are instead given

$$\frac{4}{5+3 \cos \omega}=\frac{(1-\alpha^2)}{(1+\alpha^2)-2 \alpha \cos \omega},$$

is there a way to determine $\alpha$ by observation? Or, is there a way to find $\alpha$ without simply solving for $\alpha$ and using the quadratic formula?

An example in my textbook does it in a single step; just wondering If I'm missing something.

Astor Florida
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    If one knows that $$\frac{u}{v+w \cos \omega}=\frac{1-\alpha^2}{(1+\alpha^2)-2 \alpha \cos \omega}$$ for every $\omega$, then $$\frac{u+v}v=\frac{1-\alpha^2}{1+\alpha^2}+1=\frac2{1+\alpha^2}\qquad\frac{w}v=\frac{-2\alpha}{1+\alpha^2}$$ hence $$\alpha=\frac{-w}{u+v}$$ For example, $$(u,v,w)=(4,5,3)$$ yields $$\alpha=\frac{-3}{4+5}=-\frac13$$ – Did Mar 01 '17 at 19:34
  • @Did Wow, is that a comment? Very cool – Astor Florida Mar 01 '17 at 19:35

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