Show-that-tan-1-1-3-sin-1-1-3-pi-4-

Question Number 155898 by Tawa11 last updated on 05/Oct/21

Show that \tan^{- 1} (\frac{1}{3}) + \sin^{- 1} (\frac{1}{3}) = \frac{π}{4}

Answered by immortel last updated on 05/Oct/21

L = t a n ({t a n}^{- 1} (\frac{1}{3}) + {s i n}^{- 1} (\frac{1}{3})) = \frac{t a n (a r c t a n (\frac{1}{3})) + t a n (a r c s i n \frac{1}{3})}{1 - t a n (a r c t a n (\frac{1}{3})) t a n (a r c s i n \frac{1}{3})}

o r t a n (a r c s i n x) = \frac{x}{\sqrt{1 - x^{2}}}

d o n c t a n (a r c s i n \frac{1}{3}) = \frac{\frac{1}{3}}{\sqrt{1 - \frac{1}{9}}}

= \frac{1}{\sqrt{8}} = \frac{\sqrt{2}}{4}

a i n s i L = \frac{\frac{1}{3} + \frac{\sqrt{2}}{4}}{1 - \frac{\sqrt{2}}{12}}

= \frac{(4 + 3 \sqrt{2})}{12} \times \frac{12}{12 - \sqrt{2}}

= \frac{(4 + 3 \sqrt{2}) (12 + \sqrt{2})}{(12 - \sqrt{2}) (12 + \sqrt{2})}

= \frac{48 + 4 \sqrt{2} + 36 \sqrt{2} + 6}{144 - 2}

= \frac{54 + 40 \sqrt{2}}{142} = \frac{27 + 20 \sqrt{2}}{71} < 1

d o n c L \neq 1 a i n s i

\tan^{- 1} (\frac{1}{3}) + \sin^{- 1} (\frac{1}{3}) \neq \frac{π}{4}

\dots \dots \dots . l^{'} i m m o r t e l \dots \dots . .

Commented by Tawa11 last updated on 05/Oct/21

God bless you sir .

Answered by mr W last updated on 05/Oct/21

α = \tan^{- 1} \frac{1}{3} \Rightarrow \tan α = \frac{1}{3}

β = \sin^{- 1} \frac{1}{3} \Rightarrow \sin β = \frac{1}{3} \Rightarrow \tan β = \frac{1}{2 \sqrt{2}}

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

= \frac{\frac{1}{3} + \frac{1}{2 \sqrt{2}}}{1 - \frac{1}{6 \sqrt{2}}} = \frac{3 + 2 \sqrt{2}}{6 \sqrt{2} - 1} = \frac{27 + 20 \sqrt{2}}{71}

\approx 0.778 \neq 1

\Rightarrow α + β \neq 45 °

i n f a c t \tan^{- 1} (\frac{1}{3}) + \sin^{- 1} (\frac{1}{3}) \approx 38 °

Commented by Tawa11 last updated on 05/Oct/21

God bless you sir

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