Identities, Addition Formulas and Logarithmic Forms

Fundamental identities. From the definitions, $\cosh^2 x - \sinh^2 x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2 = \frac{(e^{2x}+2+e^{-2x}) - (e^{2x}-2+e^{-2x})}{4} = \frac{4}{4} = 1$ for every real $x$ — the hyperbolic analogue of $\cos^2\theta+\sin^2\theta=1$, but with a minus sign, which is exactly why these are called hyperbolic functions: the point $(\cosh x, \sinh x)$ traces the right branch of the hyperbola $u^2 - v^2 = 1$, just as $(\cos\theta,\sin\theta)$ traces the unit circle. Dividing through by $\cosh^2 x$ gives $1 - \tanh^2 x = \operatorname{sech}^2 x$, and dividing by $\sinh^2 x$ gives $\coth^2 x - 1 = \operatorname{csch}^2 x$.

Addition formulas (Theorem 9.3.1). For all real $x,y$:

Proof (for $\sinh(x+y)$). Writing everything in terms of $e^x$: $\sinh x\cosh y + \cosh x\sinh y = \frac{(e^x-e^{-x})(e^y+e^{-y}) + (e^x+e^{-x})(e^y-e^{-y})}{4}$. Expanding both products and adding, the cross-terms $e^{x-y}$ and $-e^{x-y}$ (and similarly $e^{y-x}$) cancel, leaving $\frac{2e^{x+y} - 2e^{-(x+y)}}{4} = \frac{e^{x+y}-e^{-(x+y)}}{2} = \sinh(x+y)$. The $\cosh$ and $\tanh$ addition formulas, and the minus-sign versions, follow by the same expansion or by substituting $y \to -y$ and using the parities from §9.1.

Double-angle corollaries. Setting $y=x$: $\sinh 2x = 2\sinh x\cosh x$, and $\cosh 2x = \cosh^2 x+\sinh^2x = 2\cosh^2x-1 = 1+2\sinh^2x$ (the last two forms follow by substituting $\cosh^2x-\sinh^2x=1$).

Logarithmic forms of the inverse functions. These theorems convert an inverse hyperbolic function into an explicit logarithm, which is what makes numerical evaluation possible.

Theorem: $\sinh^{-1}x=\log\left(x+\sqrt{x^2+1}\right)$ for all real $x$. Proof: let $y=\sinh^{-1}x$, so $x=\sinh y=\frac{e^y-e^{-y}}{2}$, i.e. $e^{2y}-2xe^y-1=0$, a quadratic in $e^y$. Solving, $e^y=\frac{2x\pm\sqrt{4x^2+4}}{2}=x\pm\sqrt{x^2+1}$; since $e^y>0$ and $\sqrt{x^2+1}>|x|$, only the $+$ sign is admissible, giving $e^y=x+\sqrt{x^2+1}$ and hence $y=\log\left(x+\sqrt{x^2+1}\right)$. …