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1-8-10-6-tan-89-9999-pi-upto-9-decimal-places-can-i-have-some-explanations-how-it-is-worked-out-Thank-you-




Question Number 47239 by JDlix last updated on 06/Nov/18
((1.8×10^6 )/(tan(89.9999°))) ∼ π (upto 9 decimal places)  can i have some explanations how it is worked out ?  Thank you!
$$\frac{\mathrm{1}.\mathrm{8}×\mathrm{10}^{\mathrm{6}} }{{tan}\left(\mathrm{89}.\mathrm{9999}°\right)}\:\sim\:\pi\:\left({upto}\:\mathrm{9}\:{decimal}\:{places}\right) \\ $$$${can}\:{i}\:{have}\:{some}\:{explanations}\:{how}\:{it}\:{is}\:{worked}\:{out}\:? \\ $$$${Thank}\:{you}! \\ $$
Commented by MJS last updated on 07/Nov/18
I guess somebody just messed around a bit.  you can find many almost−identities  πln (π) =((143851)/(40000)) (error ≈−2.7×10^(−10) )  πln (πln (π)) =((100523)/(25000)) (error ≈−2.6×10^(−10) )  π=((6955380174−(√3))/(2213966271)) (error ≈−5×10^(−13) )
$$\mathrm{I}\:\mathrm{guess}\:\mathrm{somebody}\:\mathrm{just}\:\mathrm{messed}\:\mathrm{around}\:\mathrm{a}\:\mathrm{bit}. \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{find}\:\mathrm{many}\:\mathrm{almost}−\mathrm{identities} \\ $$$$\pi\mathrm{ln}\:\left(\pi\right)\:=\frac{\mathrm{143851}}{\mathrm{40000}}\:\left(\mathrm{error}\:\approx−\mathrm{2}.\mathrm{7}×\mathrm{10}^{−\mathrm{10}} \right) \\ $$$$\pi\mathrm{ln}\:\left(\pi\mathrm{ln}\:\left(\pi\right)\right)\:=\frac{\mathrm{100523}}{\mathrm{25000}}\:\left(\mathrm{error}\:\approx−\mathrm{2}.\mathrm{6}×\mathrm{10}^{−\mathrm{10}} \right) \\ $$$$\pi=\frac{\mathrm{6955380174}−\sqrt{\mathrm{3}}}{\mathrm{2213966271}}\:\left(\mathrm{error}\:\approx−\mathrm{5}×\mathrm{10}^{−\mathrm{13}} \right) \\ $$
Answered by MJS last updated on 07/Nov/18
tan ((90−(1/(10^d )))°)=(1/(tan (π/(180×10^d ))))≈((180×10^d )/π) with d∈N
$$\mathrm{tan}\:\left(\left(\mathrm{90}−\frac{\mathrm{1}}{\mathrm{10}^{{d}} }\right)°\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\pi}{\mathrm{180}×\mathrm{10}^{{d}} }}\approx\frac{\mathrm{180}×\mathrm{10}^{{d}} }{\pi}\:\mathrm{with}\:{d}\in\mathbb{N} \\ $$

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