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Background

I recently set out to derive the exponential forms of the inverse trigonemtric functions using eulers identity and demoivres theorem, deciding to start with $arcsin(x)$ I first got that:

$$e^{ix}=\cos(x)+i\sin(x) \implies e^{-ix}=\cos(x)- i\sin(x) $$ $$\therefore \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$


Deriving arcsin(x)

Letting $\sin(x)=y$

$$y=\frac{e^{ix}-e^{-ix}}{2i} \implies 2iye^{ix}=e^{2ix}-1$$


$$\implies e^{2ix}-2iye^{ix}-1=0$$

Completing the square; $$(e^{ix}-iy)^2=1-y^2 \implies e^{ix}=iy \space ± \space \sqrt{1-y^2}$$

Then by taking the natural log and noting that it exists only for positive non zero numbers; $$x=\frac{\ln(iy + \sqrt{1-y^2})}{i} $$


$$\therefore \text{arcsin}(x)= \frac{\ln(ix + \sqrt{1-x^2})}{i}$$

The more I studied this the more it became apparent that if I were to rewrite the complex number within the natural log in its exponential form then would it not be true that:

$$\text{arcsin}(x)=\frac{(i)(\text{arg}(ix + \sqrt{1-x^2})}{i}$$


$$\therefore \text{arcsin}(x)=\text{arg}(ix + \sqrt{1-x^2})$$


Deriving arctan(x)

I am extremely fascinated by the fact that to find the $arcsin$ of some value, you merely have to find the argument of some arbitrary complex number, upon noting this I also derived the $\text{arctan}(x)$ logarithmic form in the same way:

$$\text{arctan}(x)= \frac{\ln \left(\frac{2i}{x+i} -1 \right)}{2i}$$

So that I may plug in the value of $\frac{x}{\sqrt{1-x^2}}$ into the expression for $\text{arctan}(x)$ and find the argument.

Upon making the same observation as with $arcsin(x)$ I noted that this expression was the same as

$$\text{arctan}(x)= \frac{\text{arg} \left(\frac{2i}{x+i} -1 \right)}{2}$$


Questions and query about the implications

Again, the fact that finding the $arctan$ or $arcsim$ of a value is the same as finding the argument of seemingly arbitrary complex numbers seems extremely interesting to me and I cant wait to find out more about this and what exactly these complex numbers in question represent.

However I have not been able to find any satisfactory answers during my searching and so my question is, what do the complex numbers $ix + \sqrt{1-x^2}$ and $\frac{2i}{x+i} -1$ represent, and why do their respective arguments give values for the inverse trigonometric functions in question?

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    $\begingroup$ Incidentally, it's usually stated as $\arctan x=\tfrac{1}{2i}\ln\frac{1+ix}{1-ix}$. $\endgroup$
    – J.G.
    Apr 13, 2021 at 17:15
  • $\begingroup$ @J.G. That was the first form I got but thought it would be simpler to rewrite it as such $\endgroup$ Apr 13, 2021 at 17:16
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    $\begingroup$ This is not immediately related, but the fact that you can express such functions in terms of logarithms directly correlates with the existence of an antiderivative in terms of elementary functions. This is an "implication" of the fact that these functions may be expressed as logarithms. (See Liouville's theorem for more information about this). $\endgroup$ Apr 15, 2021 at 17:18
  • $\begingroup$ Thank you for awarding your bounty to my answer! I noticed that you didn't mark it as the accepted answer though; are there any points that you are still unclear about and you'd like me to clarify? $\endgroup$ Apr 17, 2021 at 21:17
  • $\begingroup$ @A-LevelStudent Your answer was clear and concise I simply forgot to accept it $\endgroup$ Apr 18, 2021 at 1:04

2 Answers 2

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Point 1

If we look at a right-angled triangle with horizontal length being equal to $\sqrt{1-x^2}$, and the vertical length equal to $x$ then by Pythagoras' theorem we can see that the the length of the hypotenuse of this triangle is $1$. Consider the angle between the horizontal length and the hypotenuse, $\theta$. We can see that $$\sin\theta=\frac{x}{1}=x$$ and therefore $\theta=\arcsin x$. However, if we were to instead look at the complex plane and the complex number $ix+\sqrt{1-x^2}$, the imaginary value $ix$ corresponds to the vertical side of length $x$ and the real value $\sqrt{1-x^2}$ corresponds to the horizontal side of length $\sqrt{1-x^2}$.

Hence, it is thus easy to see that $$\theta=\arcsin x=\arg(ix+\sqrt{1-x^2})$$ Similar reasoning applies to your observation about $\arctan x$.

Point 2

The regular trigonometric functions are well known, as are their inverses, such as $\sin x$ and $\arcsin x$, $\tan x$ and $\arctan x$ and so on. However, there are also some slightly lesser-known functions known as the hyperbolic trigonometric functions, or just the hyperbolic functions. The main hyperbolic functions are: $$\sinh x=\frac{e^x-e^{-x}}{2}$$ $$\cosh x=\frac{e^x+e^{-x}}{2}$$ $$\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{2x}-1}{e^{2x}+1}$$ and their inverse functions are, respectively: $$\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$$ $$\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})$$ $$\operatorname{artanh}x=\frac{1}{2}\ln\frac{1+x}{1-x}$$

The hyperbolic functions and the trigonometric functions have an incredible amount of similar properties; many of them are listed here . In particular, $$\arcsin x= -i\operatorname{arsinh} ix$$ $$\arctan x= -i\operatorname{artanh} ix$$ $$\arccos x= -i\operatorname{arcosh} x$$ ie $$\arcsin x= -i\ln(ix+\sqrt{(ix)^2+1}=-i\ln(ix+\sqrt{1-x^2})$$ $$\arctan x= -i\frac{1}{2}\ln\frac{1+ ix}{1-ix}$$ $$\arccos x= -i\ln(x+\sqrt{x^2-1})$$ So these complex numbers are not random: they are produced by making the argument of the inverse hyperbolic functions completely imaginary in the case of $\operatorname{arsinh}x$ and $\operatorname{artanh} x$ .


I hope that helps. If you have any questions please don't hesitate to ask :)

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Ultimately, your question asks for a geometric interpretation of the angles with given sines and tangents. In the circle $|z|=1$, add radii with ends $1$ and $e^{i\theta}$, where either $x=\sin\theta$ or $x=\tan\theta$, depending on which half of your question we're addressing. (In both cases, I'm only addressing the case $x\in\Bbb R$.)

In the first case, $e^{i\theta}$ is reached from $0$ in two steps, a real step of length $\sqrt{1-x^2}$ by Pythagoras, followed by an imaginary step $ix$ of length $|x|$. So $\theta=\tfrac1i\ln(ix+\sqrt{1-x^2})$.

It can be hard to make sense of a theorem other than by repeating a proof of it. I think it's harder in the second case than in the first, but I'll try. Since $x:=\tan\theta$ is the gradient of the radius to $e^{i\theta}$,$$e^{\pm i\theta}=\tfrac{1\pm ix}{\sqrt{1-x^2}}\implies e^{\pm2i\theta}=\ln\tfrac{1+ix}{1-ix}.$$The intuition here is that complex conjugate pairs are related by a reflection in the real axis, separated by an angle equal to twice the phase of $z=e^{i\theta}$, so $e^{2i\theta}$ is the ratio.

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