Exponentials, trigonometric functions, hyperbolics and their inverses.
(Lecture 3 of Mathematical Methods II.)
We are now familiar with the important Euler formula:
\begin{equation} \tag{1} e^{i\theta}=\cos\theta+i\sin\theta \end{equation}
whose Feynman called the most beautiful formula because it brings together all the fundamental Mathematical numbers, $0$, $1$, $e$, $\pi$ and~$i$ through
\begin{equation} e^{-i\pi}+1=0\,. \end{equation}
In Eq.~(1), supposedly $\theta\in\mathbf{R}$. However the formula remains valid for complex arguments:
\begin{equation} \tag{2} e^{iz}=\cos z+i\sin z\,,\quad z\in\mathbf{C}\,. \end{equation}
as can be seen again by serie expansion ($\exp$ is holomorphic, hence analytic):
\begin{equation} \tag{3} e^{iz}=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}+i\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!} \end{equation}
and defining the sine an cosine of the complex variable $z$ as the real and imaginary part of Eq.~(3). The notation is justified by the fact that, when $z$ is real, we recover the usual trigonometric functions.
From this follows:
\begin{equation} \tag{4} \cos(z)=\frac{e^{iz}+e^{-iz}}2\,,\qquad\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}\,. \end{equation}
We also define the complex tangent as $\tan(z)=\sin(z)/\cos(z)$.
Most properties extend to the complex realm, such as $\cos^2z+\sin^2z=1$, $\cos(-z)=\cos(z)$, $\sin(-z)=-\sin(z)$, $\tan(-z)=-\tan(z)$, etc. Others break down, e.g., $|\sin(z)|^2=\sin^2x+\sinh^2y$ and $|\cos(z)|^2=\cos^2x+\cosh^2y$. This shows that the complex sine and cosine are unbounded.
The hyperbolic functions also connect nicely with complex numbers. We give them directly in their complex form (merely by using $z$ a complex variable instead of $x$ a real one):
\begin{equation} \tag{5} \cosh(z)=\frac{e^z+e^{-z}}2\,,\qquad \sinh(z)=\frac{e^z-e^{-z}}2 \end{equation}
with also $\tanh(z)=\sinh(z)/\cosh(z)$, etc. We now have $\cosh^2z-\sinh^2z=1$. The cosh describes the catenary, that is, the curve that a hanging chain or cable assumes under its own weight when supported only at its ends.
The link between trigonometric and hyperbolic functions is through complex numbers:
\begin{align} \tag{6} \sin(iz)=i\sinh(z)\,,\quad\cos(iz)=\cosh(z)\,,\quad\tan(iz)=i\tanh(z)\,,\\ \sinh(iz)=i\sin(z)\,,\quad\cosh(iz)=\cos(z)\,,\quad\tanh(iz)=i\tan(z)\,. \end{align}
Notations exist that are important to know, such as $\sec(z)=1/\cos(z)$, $\csc(z)=1/\sin(z)$, $\cot(z)=1/\tan(z)$ for the secant, cosecant, cotangent. They also exist for the hyperbolic case where they can become monstruous, e.g., $\mathrm{csch}=1/\mathrm{sinh}$. More common is $\mathrm{sech}=1/\cosh$.
More important than this terminology are inverse functions in the sense that if
\begin{equation} \tag{7} z=\sin(w) \end{equation}
what is $w$ as a function of $z$? The answer is denoted $\arcsin(z)$. Because trigonometric and hyperbolic functions are all periodic, they are many-to-one; hence their inverses are necessarily multivalued. The most important ones are:
\begin{align} \arcsin(z)=-i\ln(iz+\sqrt{1-z^2})\,,\quad\arccos(z)=-i\ln(z+i\sqrt{1-z^2})\,,\quad\arctan(z)=-\frac{i}2\ln\left(\frac{1+iz}{1-iz}\right)\,. \end{align}
Let us prove the first one: Eq.~(7) reads $z={(e^{iw}-e^{-iw})/(2i)}$ which we rewrite as $e^{iw}-2iz-e^{-iw}=0$, or, multiplying all terms by $e^{iw}$:
\begin{equation} \tag{8} \chi^2-2iz\chi-1=0 \end{equation}
where we wrote~$\chi$ for $e^{iw}$. This is a quadratic equation in~$\chi$, with solutions:
\begin{equation} \tag{9} \chi=\frac{2iz\pm\sqrt{-4z^2+4}}{2}\,, \end{equation}
that is, keeping only one solution (the $\sqrt{}$ is multivalued anyway) $\chi=iz+\sqrt{1-z^2}$. Taking the logarithm of both sides, the result is obtained.
Finally, the derivatives of all these functions should also be known:
\begin{align} \tag{10} &(\sin(z))'=\cos(z)\,,\quad(\cos(z))'=\sin(z)\,,\quad(\tan(z))'=\sec(z)^2\,,\\ &(\cot(z))'=-(\mathrm{csc}(z))^2\,,\quad(\sec(z))'=\sec(z)\tan(z)\,,\\ &(\cosh(z))'=\sinh(z)\,,\quad(\sinh(z))'=\cosh(z)\,,\quad(\tanh(z))'=\mathrm{sech}(z)^2\,,\\ &(\mathrm{arcsin}(z))'=1/\sqrt{1-z^2}\,,\quad (\mathrm{arccos}(z))'=-1/\sqrt{1-z^2}\,,\quad (\mathrm{arctan}(z))'=1/(1+z^2)\,. \end{align}
Etc.
