Editing Gudermannian function (section)

== Argument-addition identities ==

By combining [[hyperbolic functions#Sums of arguments|hyperbolic]] and [[trigonometric functions#Sum and difference formulas|circular]] argument-addition identities,

:<math>\begin{align}
\tanh(z + w) &= \frac{\tanh z + \tanh w}{1 + \tanh z \, \tanh w }, \\[10mu]
\tan(z + w)  &= \frac{\tan z  + \tan w }{1 - \tan z  \, \tan w  },
\end{align}</math>

with the [[#Circular–hyperbolic identities|circular–hyperbolic identity]],
:<math>
\tan \tfrac12 (\operatorname{gd} z) = \tanh \tfrac12 z,
</math>

we have the Gudermannian argument-addition identities:

:<math>\begin{align}
\operatorname{gd}(z + w) &= 2 \arctan \frac
  {\tan \tfrac12(\operatorname{gd} z) + \tan\tfrac12(\operatorname{gd} w)}
  {1 + \tan\tfrac12(\operatorname{gd} z) \, \tan\tfrac12(\operatorname{gd} w)}, \\[12mu]
\operatorname{gd}^{-1}(z + w) &= 2 \operatorname{artanh} \frac
  {\tanh\tfrac12(\operatorname{gd}^{-1} z) + \tanh\tfrac12(\operatorname{gd}^{-1} w)}
  {1 - \tanh\tfrac12(\operatorname{gd}^{-1} z) \, \tanh\tfrac12(\operatorname{gd}^{-1} w)}.
\end{align}</math>

Further argument-addition identities can be written in terms of other circular functions,<ref>{{harvp|Cayley|1862}} [https://archive.org/details/londonedinburg4241862lond/page/21 p. 21]</ref> but they require greater care in choosing branches in inverse functions. Notably,

:<math>\begin{align}
\operatorname{gd}(z + w) &= u + v, \quad \text{where}\ \tan u = \frac{\sinh z}{\cosh w},\ \tan v = \frac{\sinh w}{\cosh z}, \\[10mu]
\operatorname{gd}^{-1}(z + w) &= u + v, \quad \text{where}\ \tanh u = \frac{\sin z}{\cos w},\ \tanh v = \frac{\sin w}{\cos z},
\end{align}</math>

which can be used to derive the [[Gudermannian function#Complex values|per-component computation]] for the complex Gudermannian and inverse Gudermannian.<ref>{{harvp|Kennelly|1929}} [https://archive.org/details/dli.ministry.19102/page/180 pp. 180–183]</ref>

In the specific case <math display=inline>z = w,</math> double-argument identities are

:<math>\begin{align}
\operatorname{gd}(2z)
&= 2 \arctan (\sin(\operatorname{gd} z)), \\[5mu]
\operatorname{gd}^{-1}(2z) &= 2 \operatorname{artanh}(\sinh(\operatorname{gd}^{-1}z)).
\end{align}</math>