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Trying to answer this question (five years too late), I found to the surprising identity $$4 \cot ^{-1}\left(\sqrt{\phi }\right)+\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17 \sqrt{5}}\right)=\pi$$

How to prove it ?

Claude Leibovici
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1 Answers1

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Use the formula $$\cot2\alpha=\frac{\cot^2\alpha-1}{2\cot\alpha}$$ twice to conclude that $$ \cot(4\mathrm{arccot}\sqrt{\phi})=-\frac18\sqrt{\phi}-\frac58\sqrt{5\phi}. $$ The quantity on the right is equal to $-\dfrac14\sqrt{22+17\sqrt5}$. Together with $\cot(\pi-\alpha)=-\cot\alpha$ this leads to an answer. Manipulations of nested square roots are left to the reader.

Jyrki Lahtonen
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  • Quite a bit of manipulations of square roots are needed. Not sure that the above is the cleanest formula for $\cot (4\mathrm{arccot}\sqrt{\phi})$ :-) Anyway, the basic idea is not very different from the verification of Machin's formula. Only the numbers are more cumbersome. – Jyrki Lahtonen Sep 19 '21 at 09:02
  • Oh, it's that famous (if not infamous) integral thread! I'm surprised to learn about a connection @ClaudeLeibovici :-) – Jyrki Lahtonen Sep 19 '21 at 09:07
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    I think that I should have asked : what is $\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17 \sqrt{5}}\right)$ ? since this is where I had been stuck. Cheers and thanks again :-) – Claude Leibovici Sep 20 '21 at 11:44
  • @ClaudeLeibovici I'm afraid I cannot point you at an algorithm simplifying nested square roots as much as possible. Of course, when you have a target, it is straight forward, but if your starting point is $\dfrac14\sqrt{22+17\sqrt5}$ it's not clear what will work. Algebraic number theory of quadratic fields will say a lot about the factors of $22+17\sqrt5$, but it is not entirely clear how much that helps in a given situation. Here the presence of a factor $(1+5\sqrt5)^2$ improves matters, but I only used that piece for reverse engineering. – Jyrki Lahtonen Sep 20 '21 at 12:39
  • I totally agree with you, for sure. You cannot conceive how much I was disappointed when I obtain this result ! As you wrote the problem is simple (after reading your answer !) but the reverse engineering is far from obvious. It is amusing, at least for me) to see how simple is $\cot \left(4 \cot ^{-1}(x)\right)=\frac{x^4-6x^2+1}{4x(x^2-1)}$ – Claude Leibovici Sep 20 '21 at 12:49