$2\arctan(\phi^{-n})=\arctan\frac{p}{q}$ or $\arctan\frac{p\sqrt{5}}{q}$, where $\phi$ is the Golden Ratio. Is there a pattern in the $\frac{p}{q}$s?

Question

It is very interesting to know that

$$\arctan\frac{1}{\phi} + \arctan\frac{1}{\phi^3}= \arctan 1 = \frac{\pi}{4}$$

where Golden ratio $\phi = \frac12(\sqrt5 +1)$ is in association with circle constant $\pi$.

More interesting phenomenon is evaluation of inverse tan functions of inverse of $\phi$ in its consecutive powers as follows

$$\begin{align} 2\arctan\frac{1}{\phi} &= \arctan 2 & 2\arctan\frac{1}{\phi^2} &= \arctan\frac{2\sqrt{5}}{5}\\ 2\arctan\frac{1}{\phi^3} &= \arctan\frac{1}{2} & 2\arctan\frac{1}{\phi^4} &= \arctan\frac{2\sqrt{5}}{15}\\ 2\arctan\frac{1}{\phi^5} &= \arctan\frac{2}{11} & 2\arctan\frac{1}{\phi^6} &= \arctan\frac{\sqrt5}{20} \\ 2\arctan\frac{1}{\phi^7} &= \arctan\frac{2}{29} & 2\arctan\frac{1}{\phi^8} &= \arctan\frac{2\sqrt5}{105} \\ 2\arctan\frac{1}{\phi^9} &= \arctan\frac{1}{38} & 2\arctan\frac{1}{\phi^{10}} &= \arctan\frac{2\sqrt5}{275} \\ 2\arctan\frac{1}{\phi^{11}} &= \arctan\frac{2}{199} \end{align}$$

Here are the observations

1. Odd powers of inverse $\phi$ in double arctan functions lead to arctan of well defined fractions
2. Even powers of inverse $\phi$ in double arctan functions lead to arctan of fractions involving $\sqrt5$.

My curiosity is to know, is there any pattern in these interesting series?

I will be grateful to understand more, if anyone has come across such evaluations.

Answer 1

Hint: $$ \arctan(x)\pm\arctan(y) = \arctan(z) $$ where $z$ is: $$ z = \frac{x\pm y}{1\mp xy} $$ We will take a look at this: $$ 2\arctan\phi^{-n} = \arctan \phi^{-n}+\arctan\phi^{-n} = \arctan(\frac{\phi^{-n}+\phi^{-n}}{1-\phi^{-n}\phi^{-n}}) = \arctan(\frac{2\phi^{-n}}{1-\phi^{-2n}}) = $$ $$ = \arctan(\frac{2}{\phi^{n}-\phi^{-n}}) = \arctan(\frac{2}{e^{\ln\phi^{n}}-e^{\ln\phi^{-n}}}) =\arctan(\frac{1}{\sinh(n\ln\phi)})$$

Answer 2

The explanation is that $$\tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)}$$ so, for $-1 < x < 1$, $$2 \arctan(x) = \arctan\left(\frac{2x}{1-x^2}\right)$$ Thus $$ 2 \arctan(1/\phi^n) = = \arctan \left( \frac{2 \phi^{-n}}{1-\phi^{-2n}}\right) = \arctan\left(\frac{2}{\phi^{n} - \phi^{-n}}\right)$$ Now the Fibonacci numbers $$F_n = \frac{\phi^n - (-1/\phi)^n}{\sqrt{5}}$$ so if $n$ is even, $$2 \arctan(1/\phi^n) = \arctan\left(\frac{2}{\sqrt{5} F_n}\right)$$ On the other hand, the Lucas numbers $$L_n = \phi^n + (-1/\phi)^n$$ so if $n$ is odd, $$2 \arctan(1/\phi^n) = \arctan\left(\frac{2}{L_n}\right)$$

Answer 3

Here is the pattern:

Given $$2\arctan\frac1{\phi^n}=\arctan A_n$$

the following recursive reciprocals hold, for both odd and even $n$'s,

$$\frac1{A_{n+2}}+\frac1{A_{n-2}}=\frac3{A_{n}}$$