2

Consider Machin's Formula

$$4\cot^{-1}{5}-\cot^{-1}{239}={\pi\over 4}\tag1$$

How does one show that

$$\cot^{-1}{\phi}-\cot^{-1}{\phi^3}-\cot^{-1}{\phi^5}-\cot^{-1}{\phi^7}=\cot^{-1}{5}\tag2$$ and $$2\cot^{-1}{\phi}-3\cot^{-1}{\phi^3}-3\cot^{-1}{\phi^5}-4\cot^{-1}{\phi^7}=\cot^{-1}{239}\tag3$$ $\phi$;Golden ratio

An attempt:

$$\cot^{-1}{\phi}-\cot^{-1}{\phi^3}=\cot^{-1}({\phi^2-\phi^{-2}})=\cot^{-1}{3}\tag4$$

$$\cot^{-1}{\phi^5}+\cot^{-1}{\phi^7}=\cot^{-1}{\phi^6-\phi^{-6}\over \sqrt{5}}=\cot^{-1}8\tag5$$

Then $(4)-(5)$

$$\cot^{-1}3-\cot^{-1}8=\cot^{-1}5\tag6$$

$(3)$ seems a bit messy if we apply as above, so I think there is an easy way. How else can we tackle $(3)?$

gymbvghjkgkjkhgfkl
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  • Use http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html#section3.1 and http://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function and http://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Feb 20 '17 at 12:55

1 Answers1

1

You may simply consider that $\cot^{-1}(\varphi^k)=\text{arg}(\varphi^k+i)$, hence $(2)$ boils down to checking that

$$ 65 \left(29+13 \sqrt{5}\right)\cdot(\varphi+i) = (5+i)(\varphi^3+i)(\varphi^5+i)(\varphi^7+i) \tag{2}$$ and $(3)$ boils down to checking that: $$ 7140250 \left(375125+167761 \sqrt{5}\right)\cdot(\varphi+i)^2=(239+i)(\varphi^3+i)^3(\varphi^5+i)^3(\varphi^7+i)^4.\tag{3} $$

Jack D'Aurizio
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