Inverse trigonometric functions

Generalization

Because all of the inverse trigonometric functions output an angle of a right triangle, they can be generalized by using Euler's formula to form a right triangle in the complex plane. Algebraically, this gives us:

$ce^{i\theta }=c\cos(\theta )+ic\sin(\theta )$

or

$ce^{i\theta }=a+ib$

where $a$ is the adjacent side, $b$ is the opposite side, and $c$ is the hypotenuse. From here, we can solve for $\theta$.

${\begin{aligned}e^{\ln(c)+i\theta }&=a+ib\\\ln c+i\theta &=\ln(a+ib)\\\theta &=\operatorname {Im} \left(\ln(a+ib)\right)\end{aligned}}$

or

$\theta =-i\ln \left({\frac {a+ib}{c}}\right)$

Simply taking the imaginary part works for any real-valued $a$ and $b$, but if $a$ or $b$ is complex-valued, we have to use the final equation so that the real part of the result isn't excluded. Since the length of the hypotenuse doesn't change the angle, ignoring the real part of $\ln(a+bi)$ also removes $c$ from the equation. In the final equation, we see that the angle of the triangle in the complex plane can be found by inputting the lengths of each side. By setting one of the three sides equal to 1 and one of the remaining sides equal to our input $z$, we obtain a formula for one of the inverse trig functions, for a total of six equations. Because the inverse trig functions require only one input, we must put the final side of the triangle in terms of the other two using the Pythagorean Theorem relation

$a^{2}+b^{2}=c^{2}$

The table below shows the values of a, b, and c for each of the inverse trig functions and the equivalent expressions for $\theta$ that result from plugging the values into the equations $\theta =-i\ln \left({\tfrac {a+ib}{c}}\right)$ above and simplifying.

${\begin{aligned}&a&&b&&c&&-i\ln \left({\frac {a+ib}{c}}\right)&&\theta &&\theta _{a,b\in \mathbb {R} }\\\arcsin(z)\ \ &{\sqrt {1-z^{2}}}&&z&&1&&-i\ln \left({\frac {{\sqrt {1-z^{2}}}+iz}{1}}\right)&&=-i\ln \left({\sqrt {1-z^{2}}}+iz\right)&&\operatorname {Im} \left(\ln \left({\sqrt {1-z^{2}}}+iz\right)\right)\\\arccos(z)\ \ &z&&{\sqrt {1-z^{2}}}&&1&&-i\ln \left({\frac {z+i{\sqrt {1-z^{2}}}}{1}}\right)&&=-i\ln \left(z+{\sqrt {z^{2}-1}}\right)&&\operatorname {Im} \left(\ln \left(z+{\sqrt {z^{2}-1}}\right)\right)\\\arctan(z)\ \ &1&&z&&{\sqrt {1+z^{2}}}&&-i\ln \left({\frac {1+iz}{\sqrt {1+z^{2}}}}\right)&&=-{\frac {i}{2}}\ln \left({\frac {i-z}{i+z}}\right)&&\operatorname {Im} \left(\ln \left(1+iz\right)\right)\\\operatorname {arccot}(z)\ \ &z&&1&&{\sqrt {z^{2}+1}}&&-i\ln \left({\frac {z+i}{\sqrt {z^{2}+1}}}\right)&&=-{\frac {i}{2}}\ln \left({\frac {z+i}{z-i}}\right)&&\operatorname {Im} \left(\ln \left(z+i\right)\right)\\\operatorname {arcsec}(z)\ \ &1&&{\sqrt {z^{2}-1}}&&z&&-i\ln \left({\frac {1+i{\sqrt {z^{2}-1}}}{z}}\right)&&=-i\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z^{2}}}-1}}\right)&&\operatorname {Im} \left(\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z^{2}}}-1}}\right)\right)\\\operatorname {arccsc}(z)\ \ &{\sqrt {z^{2}-1}}&&1&&z&&-i\ln \left({\frac {{\sqrt {z^{2}-1}}+i}{z}}\right)&&=-i\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)&&\operatorname {Im} \left(\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)\right)\\\end{aligned}}$

The particular form of the simplified expression can cause the output to differ from the usual principal branch of each of the inverse trig functions. The formulations given will output the usual principal branch when using the $\operatorname {Im} \left(\ln z\right)\in (-\pi ,\pi ]$ and $\operatorname {Re} \left({\sqrt {z}}\right)\geq 0$ principal branch for every function except arccotangent in the $\theta$ column. Arccotangent in the $\theta$ column will output on its usual principal branch by using the $\operatorname {Im} \left(\ln z\right)\in [0,2\pi )$ and $\operatorname {Im} \left({\sqrt {z}}\right)\geq 0$ convention.

In this sense, all of the inverse trig functions can be thought of as specific cases of the complex-valued log function. Since these definition work for any complex-valued $z$, the definitions allow for hyperbolic angles as outputs and can be used to further define the inverse hyperbolic functions. It's possible to algebraically prove these relations by starting with the exponential forms of the trigonometric functions and solving for the inverse function.