1

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 $\phi+\sqrt{\phi}$.

Crescendo
  • 4,089

2 Answers2

2

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.

Dylan
  • 16,575
0

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.

B. Mehta
  • 12,774
  • That was just an error I made while typing the question, you should still get the same answer when you work it through – Crescendo Dec 31 '17 at 18:58