Trigonometry with golden ratio

Let $\varphi$ denote the golden ratio. Prove that

$4 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{\varphi^6 - 1}} = \frac{\pi}{2}$

Solution

Since $\varphi^2 = \varphi + 1$ we deduce that $\varphi^3 = \varphi^2 + \varphi = 2 \varphi + 1$ . Furthermore,

$\varphi^6 - 1 = \left ( \varphi^3 \right )^2 -1 = \left ( 2 \varphi + 1 \right )^2 -1 = 4\varphi^2 + 4 \varphi = 4 \varphi^3$

Also, taking into consideration that $\displaystyle 2 \arctan x = \arctan \frac{2x}{1-x^2} \; ,\; x^2<1$ we deduce that:

$\begin{align*} 4 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{\varphi^6 - 1}} &= 2 \cdot 2 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{4 \varphi^3}} \\ &= 2 \arctan \sqrt{\varphi} - \arctan \frac{1}{2 \sqrt{\varphi^3}} \\ &= \arctan \sqrt{\varphi} + \arctan \sqrt{\varphi} - \arctan \frac{1}{2 \sqrt{\varphi^3}} \\ &= \arctan \sqrt{\varphi} + \arctan \frac{1}{\sqrt{\varphi}} \\ &= \frac{\pi}{2} \end{align*}$