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