Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/97

$\operatorname {Cos} .nx={\frac {1}{2}}\left(u^{n}+v^{n}\right)\,;$

de sorte qu’en posant

${\begin{aligned}&U={\frac {au}{1}}-{\frac {a^{3}u^{3}}{3}}+{\frac {a^{5}u^{5}}{5}}-{\frac {a^{7}u^{7}}{7}}+\ldots ,\\\\&V={\frac {av}{1}}-{\frac {a^{3}v^{3}}{3}}+{\frac {a^{5}v^{5}}{5}}-{\frac {a^{7}v^{7}}{7}}+\ldots ,\end{aligned}}$

nous aurons

$S={\frac {1}{2}}(U+V)\,;$

cela donne

$U=\operatorname {Arc} .(\operatorname {Tang} .=au),\qquad V=\operatorname {Arc} .(\operatorname {Tang} .=av)\,;$

mais, en renversant le théorème connu,

$\operatorname {Tang} .(p+q)={\frac {\operatorname {Tang} .p+\operatorname {Tang} .q}{1-\operatorname {Tang} .p\operatorname {Tang} .q}},$

on a

$\operatorname {Arc} .(\operatorname {Tang} .=p')+\operatorname {Arc} .(\operatorname {Tang} .=q')=\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {p'+q'}{1-p'q'}}\right),$

et par suite,

$\operatorname {Arc} .(\operatorname {Tang} .=au)+\operatorname {Arc} .(\operatorname {Tang} .=av)=\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {a(u+v)}{1-a^{2}uv}}\right)\,;$

remettant donc $2\operatorname {Cos} .r$ pour $u+v$ et $1$ pour $uv,$ il viendra finalement