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Help me please to understand how does the Golden Angle on this mathworld page is derived: https://mathworld.wolfram.com/GoldenAngle.html

I can't understand how does it transformed from $2\pi(1-\dfrac{1}{\phi})$ to $\pi(3 - \sqrt 5)$.

After parenthesis expansion by hands it becomes $\dfrac{2\pi\phi - 2\pi}\phi$, right?

How it becomes as $\dfrac{2\pi}{(1+\phi)}$ after this? And other steps.

ewokx
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Axel
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    Just use $\phi = \frac{1+\sqrt{5}}{2}$. – Accelerator Aug 18 '22 at 05:36
  • @Accelerator thanks. I know about the solution of the $x^2-x-1=0$. I just was read today answer on StackOverflow on Fibonacci sphere and I became more interested in learning more about the number phi. Just want to know how it has been derived as this path. – Axel Aug 18 '22 at 05:41
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    The Golden Ratio is fun to manipulate. For instance, $\phi^2=\phi+1$ and $\phi^{-1}=\phi-1$. So, following your steps, but not multiplying-through by $2\pi$, we have $$2\pi(1-1/\phi) = 2\pi\frac{\phi-1}{\phi}=2\pi\frac{1/\phi}{\phi}=2\pi\frac1{\phi^2}=\frac{2\pi}{1+\phi}$$ Given the goal, though, it's easier to write $$2\pi(1-1/\phi)=2\pi(1-(\phi-1))=2\pi(2-\phi)$$ Then substitute $\phi=\frac12(1+\sqrt{5})$ to get the desired final form. – Blue Aug 18 '22 at 05:42
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    Given the length of the above comment, I don't understand why this question is being so heavily downvoted. The OP indicated an attempt, and it's clearly not completely obvious what manipulation you should try. – A. Thomas Yerger Aug 18 '22 at 06:07
  • @Blue Thanks. I'm sorry for some stupid questions. I understand the second step but don't understand how $2\pi\frac{\phi-1}{\phi}=2\pi\frac{1/\phi}{\phi}$. How does the $ϕ−1$ becomes $1/ϕ$? Is it some algebraic property of $ф$? – Axel Aug 18 '22 at 06:08
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    @Axel: Divide the relation $\phi^2=\phi+1$ by $\phi$, and rearrange a little. (BTW: I don't see any stupid questions.) – Blue Aug 18 '22 at 06:11

1 Answers1

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We can use the fact that $\phi^2 - \phi - 1 = 0$ is a defining equation for $\phi$, and rearrange it:

$\begin{eqnarray} \phi^2 - \phi - 1 & = & 0 \\ \phi^2 - \phi & = & 1 & \textrm{adding } 1 \textrm{ to both sides} \\ \phi - 1 & = & \frac{1}{\phi} & \textrm{dividing both sides by } \phi\end{eqnarray}$

We can then apply that identity to the expression, giving $2 \pi(1 - \frac{1}{\phi}) = 2 \pi(1 - (\phi - 1)) = 2 \pi(2 - \phi) = 2 \pi (2 - \frac{1 + \sqrt{5}}{2}) = \pi(4 - (1 + \sqrt{5})) = \pi(3 - \sqrt{5})$.

Alternatively, if we immediately expand out $\phi$, we can rationalise the denominator of the resulting expression:

$\begin{eqnarray} 2 \pi \left(1 - \frac{1}{\phi}\right) & = & 2 \pi \left(1 - \frac{2}{\sqrt{5} + 1}\right) \\ & = & 2 \pi \left(1 - \frac{2}{\sqrt{5} + 1} \frac{\sqrt{5} - 1}{\sqrt{5} - 1} \right) \\ & = & 2 \pi \left(1 - \frac{2(\sqrt{5} - 1)}{(\sqrt{5})^2 - 1^2} \right) \\ & = & 2 \pi \left(1 - \frac{2(\sqrt{5} - 1)}{4} \right) \\ & = & 2 \pi \left(1 - \frac{\sqrt{5} - 1}{2} \right) \\ & = & \pi \left(2 - (\sqrt{5} - 1) \right) \\ & = & \pi(3 - \sqrt{5}) \end{eqnarray}$

ConMan
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