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In the course of doing scratchwork to answer this question, I had occasion to write the trigonometric identity $$ \arctan x- \arctan(1-x) = \arctan\left( \frac{1-2x}{x^2-x-1} \right). $$ Now notice that this is $$ \arctan\left( -\frac{d}{dx} \log(x^2-x-1) \right) $$ and $x^2-x-1$ is the monic polynomial of lowest degree with rational coefficients that has the golden ratio $(1+\sqrt{5})/2$ as a root, and that number is $\lim\limits_{n\to\infty} F_{n+1}/F_n$, where $F_n$ is the $n$th Fibonacci number.

Googling indicates that someone has found connections between arctangents and Fibonacci numbers: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibpi.html#section2

Is the fact I state above "known"? What related facts are "known"? (Scare quotes because obviously it is known to those who read this question.)

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  • $\begingroup$ I dont see any interesting "fact" just accidental coincidence. $\endgroup$ – Leox Oct 2 '14 at 19:36
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    $\begingroup$ @Leox : An accidental coincidence in mathematics may be an occasion to develop it into an interesting "fact". ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 2 '14 at 19:38
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    $\begingroup$ @Leox : BTW, did you name yourself after this guy?: en.wikipedia.org/wiki/Pope_Leo_X ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 2 '14 at 19:41
  • $\begingroup$ Of cource, but right now I still dont see such an interesting "fact". Moreover I like such "fact" $$ \arctan\left( -\frac{d}{dx} \log(x^2-x-1) \right)=\arctan\left( i^2\frac{d}{dx} \log(x^2-x-(\sin^2(x)+\cos^2(x))) \right)$$ $\endgroup$ – Leox Oct 2 '14 at 19:43
  • $\begingroup$ @MichaelHardy What about the domains? The identity is only true for a certain interval? $\endgroup$ – imranfat Oct 2 '14 at 19:51

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