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Question Number 151665 by Huy last updated on 22/Aug/21

prove 4arccot5−arccot239=(π/4)

$$\mathrm{prove}\:\mathrm{4arccot5}−\mathrm{arccot239}=\frac{\pi}{\mathrm{4}} \\ $$

Answered by puissant last updated on 22/Aug/21

tan(4x)=((4tanx−4(tanx)^3 )/(1−6(tanx)^2 +(tanx)^4 )) tan(4arctan(1/5))=((120)/(119)) β=4tan((1/5))−arctan((1/(239))) ⇒ tanβ=((tan(4arctan((1/5)))−tan(arctan((1/(239)))))/(1+tan(4arctan((1/5)))tan(arctan((1/(239)))))) ⇒ tanβ=((((120)/(119))−(1/(239)))/(1+((120)/(119))×(1/(239)))) = 1 ⇒ tanβ=1 ⇒ β=arctan(1)=(π/4) ∵ 4arctan((1/5))−arctan((1/(239))) = (π/4) ( MACHIN formula)..

$${tan}\left(\mathrm{4}{x}\right)=\frac{\mathrm{4}{tanx}−\mathrm{4}\left({tanx}\right)^{\mathrm{3}} }{\mathrm{1}−\mathrm{6}\left({tanx}\right)^{\mathrm{2}} +\left({tanx}\right)^{\mathrm{4}} } \\ $$$${tan}\left(\mathrm{4}{arctan}\frac{\mathrm{1}}{\mathrm{5}}\right)=\frac{\mathrm{120}}{\mathrm{119}} \\ $$$$\beta=\mathrm{4}{tan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right) \\ $$$$\Rightarrow\:{tan}\beta=\frac{{tan}\left(\mathrm{4}{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\right)−{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right)\right)}{\mathrm{1}+{tan}\left(\mathrm{4}{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\right){tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right)\right)} \\ $$$$\Rightarrow\:{tan}\beta=\frac{\frac{\mathrm{120}}{\mathrm{119}}−\frac{\mathrm{1}}{\mathrm{239}}}{\mathrm{1}+\frac{\mathrm{120}}{\mathrm{119}}×\frac{\mathrm{1}}{\mathrm{239}}}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:{tan}\beta=\mathrm{1}\:\Rightarrow\:\beta={arctan}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\because\:\mathrm{4}{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$\left(\:{MACHIN}\:{formula}\right).. \\ $$

Answered by qaz last updated on 22/Aug/21

∵ (((5+i)^4 )/(239+i))=((28561+28561i)/(60)) ∴ 4arccot 5−arccot 239 =4arctan (1/5)−arctan (1/(239)) =arctan ((((28561)/(60))/((28561)/(60)))) =(π/4)

$$\because\:\:\:\frac{\left(\mathrm{5}+\mathrm{i}\right)^{\mathrm{4}} }{\mathrm{239}+\mathrm{i}}=\frac{\mathrm{28561}+\mathrm{28561i}}{\mathrm{60}} \\ $$$$\therefore\:\:\mathrm{4arccot}\:\mathrm{5}−\mathrm{arccot}\:\mathrm{239} \\ $$$$\:\:\:\:\:\:\:=\mathrm{4arctan}\:\frac{\mathrm{1}}{\mathrm{5}}−\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{239}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{arctan}\:\left(\frac{\frac{\mathrm{28561}}{\mathrm{60}}}{\frac{\mathrm{28561}}{\mathrm{60}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}} \\ $$

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