prove-4arccot5-arccot239-pi-4-

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

prove 4 arccot 5 - arccot 239 = \frac{π}{4}

Answered by puissant last updated on 22/Aug/21

t a n (4 x) = \frac{4 t a n x - 4 {(t a n x)}^{3}}{1 - 6 {(t a n x)}^{2} + {(t a n x)}^{4}}

t a n (4 a r c t a n \frac{1}{5}) = \frac{120}{119}

β = 4 t a n (\frac{1}{5}) - a r c t a n (\frac{1}{239})

\Rightarrow t a n β = \frac{t a n (4 a r c t a n (\frac{1}{5})) - t a n (a r c t a n (\frac{1}{239}))}{1 + t a n (4 a r c t a n (\frac{1}{5})) t a n (a r c t a n (\frac{1}{239}))}

\Rightarrow t a n β = \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}} = 1

\Rightarrow t a n β = 1 \Rightarrow β = a r c t a n (1) = \frac{π}{4}

∵ 4 a r c t a n (\frac{1}{5}) - a r c t a n (\frac{1}{239}) = \frac{π}{4}

(M A C H I N f o r m u l a) . .

Answered by qaz last updated on 22/Aug/21

∵ \frac{{(5 + i)}^{4}}{239 + i} = \frac{28561 + 28561 i}{60}

∴ 4 arccot 5 - arccot 239

= 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}

= \arctan (\frac{\frac{28561}{60}}{\frac{28561}{60}})

= \frac{π}{4}

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