prove-that-4tan-1-1-5-tan-1-1-239-pi-4-

Question Number 17068 by chux last updated on 30/Jun/17

prove that

{4 \tan}^{- 1} (\frac{1}{5}) - \tan^{- 1} (\frac{1}{239}) = π / 4

Answered by ajfour last updated on 30/Jun/17

z_{1} = 5 + i; z_{2} = 239 + i

Arg (z_{1}) = \tan^{- 1} (1 / 5) = α

Arg (z_{2}) = \tan^{- 1} (1 / 239) = β

let z = \frac{z_{1}^{4}}{z_{2}}, then Arg (z) = 4 α - β

z = \frac{z_{1}^{4}}{z_{2}} = \frac{{(5 + i)}^{4}}{239 + i} = \frac{{(24 + 10 i)}^{2} (239 - i)}{239^{2} + 1}

= \frac{(476 + 480 i) (239 - i)}{239^{2} + 1}

= \frac{(476 \times 239 + 480) + i (480 \times 239 - 476)}{239^{2} + 1}

z = \frac{114244}{239^{2} + 1} (1 + i)

z has argument θ = π / 4

\Rightarrow 4 α - β = \frac{π}{4} + k π

or {4 \tan}^{- 1} (\frac{1}{5}) - \tan^{- 1} (\frac{1}{239}) = \frac{π}{4} + k π .

Commented by chux last updated on 01/Jul/17

mr Ajfour I really enjoy this

approach

Answered by mrW1 last updated on 30/Jun/17

α = \tan^{- 1} \frac{1}{5}

β = \tan^{- 1} \frac{1}{239}

\tan 2 α = \frac{2 \tan α}{1 - \tan^{2} α} = \frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}} = \frac{5}{12}

\tan 4 α = \frac{2 \tan 2 α}{1 - \tan^{2} 2 α} = \frac{2 \times \frac{5}{12}}{1 - {(\frac{5}{12})}^{2}} = \frac{120}{119}

\tan (4 α - β) = \frac{\tan 4 α - \tan β}{1 + \tan 4 α \tan β}

= \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}}

= \frac{120 \times 239 - 119}{119 \times 239 + 120}

= \frac{28561}{28561}

= 1

\Rightarrow 4 α - β = \tan^{- 1} 1 = \frac{π}{4} + k π with k \in Z

\Rightarrow 4 \tan^{- 1} (\frac{1}{5}) - \tan^{- 1} (\frac{1}{239}) = \frac{π}{4} + k π

Commented by chux last updated on 30/Jun/17

wow \dots . . i really appreciate .

Commented by Abbas-Nahi last updated on 01/Jul/17

v e r y v e r y Good \dots . . with my best wishes