Inverse Trigonometric Functions - Logarithmic Forms

Logarithmic Forms

These functions may also be expressed using complex logarithms. This extends in a natural fashion their domain to the complex plane.


\begin{align}
\arcsin x &{}= -i\,\ln\left(i\,x+\sqrt{1-x^2}\right) &{}= \arccsc \frac{1}{x}\\
\arccos x &{}= -i\,\ln\left(x+i\,\sqrt{1-x^2}\right) = \frac{\pi}{2}\,+i\ln\left(i\,x+\sqrt{1-x^2}\right) = \frac{\pi}{2}-\arcsin x &{}= \arcsec \frac{1}{x}\\
\arctan x &{}= \tfrac{1}{2}i\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) &{}= \arccot \frac{1}{x}\\
\arccot x &{}= \tfrac{1}{2}i\left(\ln\left(1-\frac{i}{x}\right)-\ln\left(1+\frac{i}{x}\right)\right) &{}= \arctan \frac{1}{x}\\
\arcsec x &{}= -i\,\ln\left(i\,\sqrt{1-\frac{1}{x^2}}+\frac{1}{x}\right) = i\,\ln\left(\sqrt{1-\frac{1}{x^2}}+\frac{i}{x}\right)+\frac{\pi}{2} = \frac{\pi}{2}-\arccsc x &{}= \arccos \frac{1}{x}\\
\arccsc x &{}= -i\,\ln\left(\sqrt{1-\frac{1}{x^2}}+\frac{i}{x}\right) &{}= \arcsin \frac{1}{x}
\end{align}

Elementary proofs of these relations proceed via expansion to exponential forms of the trigonometric functions.

Read more about this topic:  Inverse Trigonometric Functions

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