show-that-sin-1-i-ln-2-1-pi-2-

Question Number 86476 by DuDono last updated on 28/Mar/20

show that

\sin^{- 1} α = - i \ln (α \pm \sqrt{α^{2} - 1}) - \frac{π}{2}

Commented by MJS last updated on 28/Mar/20

the path is this :

y = \sin x

y = \frac{e^{i x} - e^{- i x}}{2 i}

2 i y = e^{i x} - \frac{1}{e^{i x}}

let e^{i x} = t \Rightarrow x = 2 n π - i \ln t

2 i y = t - \frac{1}{t} \Rightarrow t = i y \pm \sqrt{1 - y^{2}} = i (y \pm \sqrt{y^{2} - 1})

now insert into x = 2 n π - i \ln t

\dots

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