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I see here a list of inverse trigonometric functions written in terms of logarithms. The ones I'm most interested in for the purposes of this question are $\arcsin{z}=-i\ln\left(iz+\sqrt{1-z^2}\right)$ and $\arccos{z}=-i\ln\left(z+\sqrt{z^2-1}\right)$. These look sort of reminiscent of certain properties of their inverses, namely: $e^{i\theta}=\cos\theta+i\sin\theta$ leading to the natural log being used, as well as $\sin$ being the imaginary component and $\cos$ the real component leading to an $iz$ term in the former's inverse's expansion, and a regular $z$ term in the latter's. Additionally, if you differentiate each of these and simplify, you get their expected derivatives. And, of course, if you plug in these definitions into their inverses' exponential forms you get $x$.

What I'm struggling with is proving that this is the case. How can I take $\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos{x}=\frac{e^{ix}+e^{-ix}}{2}$ and invert them to get the definitions above?

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Start with $w= \sin z=\frac{e^{iz}-e^{-iz}}{2i}$. Then,

$$e^{iz}-e^{-iz}-2iw=0 \implies e^{2iz}-2iwe^{iz}-1=0$$

which is a quadratic equation in $e^{iz}$. Solve to get,

$$e^{iz} = iw+\sqrt{1-w^2} \implies z =-i\ln\left(iw+\sqrt{1-w^2}\right)$$

or, the inverse function,

$$w=\arcsin{z}=-i\ln\left(iz+\sqrt{1-z^2}\right)$$

Similarly, $\arccos{z}=-i\ln\left(z+\sqrt{z^2-1}\right)$.

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  • $\begingroup$ What's the justification for taking the positive root when solving the quadratic? $\endgroup$ – A. Goodier Mar 19 '20 at 16:42
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Take your formula for $\sin x$, multiply the numerator and denominator of the right hand side both by $e^{ix}$, multiply both sides by the denominator of the new right hand side, rearrange to obtain a quadratic expression in $e^{ix}$. Solve this, e.g. by completing the square.

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