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Using $$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$ Prove $$\sin^{-1} z =\frac{1}{i}\ln(iz+\sqrt{1-z^2}) $$

Attempted solution: Let $\sin z = u$ and $e^{iz} = v$. \begin{align*}& 2iu = v - \frac{1}{v}\\&v^2 - 2iuv - 1 = 0\\&v = \frac{2iu \pm \sqrt{-4u^2 + 4}}{2} = iu \pm \sqrt{1-u^2}\\&e^{iz} = iu \pm \sqrt{1-u^2}\\&iz = \ln\left(iu \pm \sqrt{1-u^2}\right)\\&z = \frac{1}{i}\ln\left(iu + \sqrt{1-u^2}\right)\end{align*}

But I'm looking for $\sin^{-1}z$ not $\sin^{-1}u$, so I'm not sure where to go from here.

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  • $\begingroup$ You cannot get the answer because your formula show does not give the derivative of $\sin$ but the inverse $\sin^{-1}(z) = \arcsin(z)$ $\endgroup$ – gammatester Jun 14 '16 at 11:13
  • $\begingroup$ If, by "$Sin'(z)$, you mean the derivative, then you cannot prove this- it isn't true. At, for example, z= 0, the lefts side is 1 but the right side is 0, $\endgroup$ – user247327 Jun 14 '16 at 11:13
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    $\begingroup$ Welcome to Mathematics Stack Exchange! I've suggested an edit that typesets your working in MathJax - please try to do so yourself next time. If you're not sure how to use MathJax, there's this awesome guide here for future reference! $\endgroup$ – KoA Jun 14 '16 at 11:23
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As you defined $u = \sin z$, so $z = \arcsin u$. So,

$\arcsin u = \frac{1}{i}\ln\left(iu + \sqrt{1-u^2}\right)$.

Note $u$ is a dummy variable that can be renamed to other name without change the expression. So, $\arcsin z = \frac{1}{i}\ln\left(iz + \sqrt{1-z^2}\right)$.

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