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Q: Find the inverse sine $\sin^{-1} (-2)$ by writing $ \sin(w) = -2 $

ii) Using the Def'N $ \sin w = \frac {e^{iw} - e^{-iw}}{2i} $

This was actually a 2 part question the first part was to find the inverse using rectangular representation for $ \sin w $ and wasn't too bad.

My attempt: $\frac {e^{iw} - e^{-iw}}{2i} = -2 $

$ e^{iw} - e^{-iw} = -4i $

$ e^{iw} (e^{iw} - e^{-iw}) = e^{iw} (-4i) $

$ (e^{2iw} - e^{0}) = -4i e^{iw} $

$ (e^{2iw} + 4i e^{iw} - 1 ) = 0 $ let $a =e^{iw}$

$ (a^2 + 4ia -1)= (a^2 + 4ia -4 +4 -1) = (a+2i)^2 +4 -1 = (a+2i)^2 +3 $

$(a+2i) = \pm i \sqrt 3 $

$a = -i (2 \pm \sqrt 3 ) $

$e^{iw} = -i (2 \pm \sqrt 3 ) $

$iw = \ln[ {-i (2 \pm \sqrt 3 )} ] $

$w =\frac {\ln[ {-i (2 \pm \sqrt 3 )} ]}{i} $

Honestly that doesn't look like it makes any sense.

Any ideas how to do this?

2 Answers 2

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Here $$e^{2iw} + 4i e^{iw} - 1 = 0$$ then $e^{iw}=-2i+\sqrt{-3}$ and $$iw=\log[-2i+\sqrt{-3}]=\log[-i(2-\sqrt{3})]=\ln(2-\sqrt{3})+i\arg[-i(2-\sqrt{3})]=\ln(2-\sqrt{3})+i(-\dfrac{\pi}{2}+k\pi)$$ so $$w=-\dfrac{\pi}{2}+k\pi-i\ln(2-\sqrt{3})$$ for main branch $k=0$.

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rewrite $−i(2±√3)$ as $e^{ix}$ then $w=x$