I have this system

$$\begin{align*} & \cos\theta\left(r+\frac 1r\right)=-2\\ & r^2+\frac 1{r^2}+4\cos^2\theta=10\end{align*}$$

I solved for $r$ and $\cos\theta$ by first squaring the first equation$$r^2+\frac 1{r^2}=\frac 4{\cos^2\theta}-2$$So$$\frac 4{\cos^2\theta}+4\cos^2\theta=12\quad\implies\quad\cos\theta=\frac {\pm1\pm\sqrt5}2$$

Question: However after this, I'm not sure which value of the cosine I should take. The result used is $\cos\theta=1/\phi$, but I'm not sure why the other three solutions were discarded. The full solutions are$$\cos\theta=\frac 1{\phi}\quad\qquad\quad r=\phi+\sqrt \phi$$where $\phi=\tfrac 12\left(1+\sqrt5\right)$.

For context, I'm working out this problem which was solved by Ron Gordon. I'm having a hard time figuring out why he chose $1/\phi$ and $\phi+\sqrt{\phi}$.

  • $\begingroup$ is this $$\cos(\theta)\left(r+\frac{1}{r}\right)=-2$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 31 '17 at 18:55
  • $\begingroup$ @Dr.SonnhardGraubner Yes.$$\left(r+\frac 1r\right)^2=\frac 4{\cos^2\theta}\implies r^2+\frac 1{r^2}=\frac 4{\cos^2\theta}-2$$ $\endgroup$ – Crescendo Dec 31 '17 at 18:58
  • $\begingroup$ If $2\cos\theta=a, r+\dfrac1r=b$ $$ab=-4,a^2+b^2=12$$ $\endgroup$ – lab bhattacharjee Dec 31 '17 at 18:58
  • $\begingroup$ Are you sure that solution is correct? You have $\cos \theta > 0$ and $r > 0$ but $\cos \theta \left(r + \frac{1}{r}\right) < 0$. $\endgroup$ – B. Mehta Dec 31 '17 at 19:03
  • $\begingroup$ I think the choice that $\cos\theta\gt 0$ is just convention, since there are so many symmetries among the roots; this forces the plus sign on $\sqrt{5}$. The fact that $\cos\theta\lt 1$ (for all real $\theta$) then forces the minus sign on $1$. $\endgroup$ – Steven Stadnicki Dec 31 '17 at 19:08

Starting from

$$ \frac{4}{\cos^2\theta} + 4\cos^2\theta - 12 = 0 $$

You get

$$ \cos^4\theta - 3\cos^2\theta + 1 = 0 $$

which gives $$ \cos^2\theta = \frac{3\pm\sqrt{5}}{2} $$

However $\cos^2\theta \le 1$, so we must take the negative sign. Furthermore you want $\cos\theta < 0$ since $$ \cos\theta \left(r + \frac{1}{r}\right) = -2 < 0 $$

This leaves

$$ \cos\theta = -\sqrt{\frac{3-\sqrt{5}}{2}} = \frac{1-\sqrt{5}}{2} = -\frac{1}{\varphi} $$

which means $$ r + \frac{1}{r} = 2\varphi $$

or $$ r = \varphi \pm \sqrt{\varphi} $$

both roots are equally valid as they're reciprocals of each other, and $r$ and $r^{-1}$ are interchangeable.


You made a mistake getting to $$r^2+\frac 1{r^2}=\frac 4{\cos\theta}-2$$ In particular the $\cos \theta$ should be squared as well.

  • $\begingroup$ That was just an error I made while typing the question, you should still get the same answer when you work it through $\endgroup$ – Crescendo Dec 31 '17 at 18:58

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