FP3.1 - Hyperbolic functions

Hyperbolic functions are combinations and ratios of $e^x$ and $e^{-x}$. The two basic functions are$$\sinh x=\frac{e^x-e^{-x}}2,\qquad \cosh x=\frac{e^x+e^{-x}}2.$$The other four are ratios or reciprocals of these, so their domains and asymptotes follow from where $\sinh x$ or $\cosh x$ can be zero.

The fundamental identity follows directly from the exponential definitions:$$\cosh^2x-\sinh^2x=\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4}=1.$$Dividing by $\cosh^2x$ gives $1-\tanh^2x=\operatorname{sech}^2x$; dividing by $\sinh^2x$ gives $\coth^2x-\operatorname{cosech}^2x=1$. Also, $\cosh^2x+\sinh^2x=\cosh2x$.

These addition formulae are proved by replacing every hyperbolic function with its exponential definition, expanding, and collecting $e^{A+B}$ with $e^{-(A+B)}$. They also allow a linear combination to be written in a shifted form. For example, $P\sinh x+Q\cosh x=R\sinh(x+\alpha)$ when $R\cosh\alpha=P$ and $R\sinh\alpha=Q$.

A universal route for $$a\cosh x+b\sinh x=c$$ is to set $u=e^x>0$. Multiplying the exponential form by $2u$ gives$$(a+b)u^2-2cu+(a-b)=0.$$Solve the quadratic, reject every root with $u\le0$, and use $x=\ln u$. For example, $5\cosh x+3\sinh x=7$ gives $4u^2-7u+1=0$, so$$x=\ln\frac{7+\sqrt{33}}8\quad\text{or}\quad x=\ln\frac{7-\sqrt{33}}8.$$

An inverse hyperbolic function reverses a one-to-one branch of a hyperbolic function. Its graph is the reflection of that restricted branch in $y=x$. The prefix $\operatorname{ar}$ means inverse function; it does not mean reciprocal.

To prove the official example, let $y=\operatorname{arsinh}x$, so $x=\sinh y$. With $u=e^y>0$,$$x=\frac{u-u^{-1}}2\quad\Longrightarrow\quad u^2-2xu-1=0.$$Thus $u=x\pm\sqrt{x^2+1}$. Since $\sqrt{x^2+1}>|x|$, only $u=x+\sqrt{x^2+1}$ is positive. Therefore$$y=\ln u=\ln\!\left(x+\sqrt{x^2+1}\right).$$

The same quadratic method gives the other forms. From $x=\cosh y$ with the principal restriction $y\ge0$, choose $e^y=x+\sqrt{x^2-1}$. From $x=\tanh y$, rearrangement gives $e^{2y}=(1+x)/(1-x)$, hence the factor $\tfrac12$ in $\operatorname{artanh}x$.

Branch information matters. If $\cosh y=x$ with $x\ge1$, then both$$y=\pm\operatorname{arcosh}x$$solve the unrestricted equation. A condition $y<0$ selects$$y=-\operatorname{arcosh}x=\ln\!\left(x-\sqrt{x^2-1}\right),$$because the two positive exponential roots are reciprocals.

Because $\sinh$ and $\tanh$ are odd, $\operatorname{arsinh}$ and $\operatorname{artanh}$ are odd. The principal $\operatorname{arcosh}$ graph begins at $(1,0)$ and is increasing. The logarithmic forms make exact equation solutions possible without treating inverse values as decimal approximations.