Page:Joseph Louis de Lagrange - Œuvres, Tome 4.djvu/478

donc, si l’on fait

$\varphi '+\varphi ''=2\psi ,\quad \varphi ''-\varphi '=2\omega ,$

en sorte que $\omega$ soit la moitié de l’angle intercepté entre les deux rayons $r'$ et $r''$ qui renferment le secteur dont il s’agit, on aura d’abord

${\sqrt {\frac {r'}{r}}}={\frac {\cos {\cfrac {\varphi ''}{2}}}{\cos {\cfrac {\varphi '}{2}}}}={\frac {\cos {\cfrac {\psi +\omega }{2}}}{\cos {\cfrac {\psi -\omega }{2}}}}={\frac {1-\operatorname {tang} {\cfrac {\psi }{2}}\operatorname {tang} {\cfrac {\omega }{2}}}{1+\operatorname {tang} {\cfrac {\psi }{2}}\operatorname {tang} {\cfrac {\omega }{2}}}},$

d’où l’on tire

$\operatorname {tang} {\frac {\psi }{2}}={\frac {{\sqrt {r''}}-{\sqrt {r'}}}{\left({\sqrt {r''}}+{\sqrt {r'}}\right)\operatorname {tang} {\cfrac {\omega }{2}}}}.$

On a de plus

${\sqrt {r'r''}}={\frac {p}{\cos {\cfrac {\varphi '}{2}}\cos {\cfrac {\varphi ''}{2}}}},$

d’où l’on tire

$p={\sqrt {r'r''}}\cos {\frac {\varphi '}{2}}\cos {\frac {\varphi ''}{2}}={\frac {\sqrt {r'r''}}{2}}(\cos \psi +\cos \omega ).$

Mais

${\begin{aligned}\cos \psi =&{\frac {1-\operatorname {tang} ^{2}{\cfrac {\psi }{2}}}{1+\operatorname {tang} ^{2}{\cfrac {\psi }{2}}}}={\frac {\left({\sqrt {r''}}+{\sqrt {r'}}\right)^{2}\operatorname {tang} ^{2}{\cfrac {\omega }{2}}-\left({\sqrt {r''}}-{\sqrt {r'}}\right)^{2}}{\left({\sqrt {r''}}+{\sqrt {r'}}\right)^{2}\operatorname {tang} ^{2}{\cfrac {\omega }{2}}+\left({\sqrt {r''}}-{\sqrt {r'}}\right)^{2}}}\\=&{\frac {(r''+r')\left(\operatorname {tang} ^{2}{\cfrac {\omega }{2}}-1\right)+2{\sqrt {r'r''}}\left(\operatorname {tang} ^{2}{\cfrac {\omega }{2}}+1\right)}{(r''+r')\left(\operatorname {tang} ^{2}{\cfrac {\omega }{2}}+1\right)+2{\sqrt {r'r''}}\left(\operatorname {tang} ^{2}{\cfrac {\omega }{2}}-1\right)}}\\=&{\frac {2{\sqrt {r'r''}}-(r'+r'')\cos \omega }{r'+r''-2{\sqrt {r'r''}}\cos \omega }},\end{aligned}}$

en substituant la valeur précédente de $\operatorname {tang} {\frac {\varphi }{2}}\,;$ ajoutons à cette quantité $\cos \omega ,$ et l’on aura

$\cos \psi +\cos \omega ={\frac {2{\sqrt {r'r''}}\sin ^{2}\omega }{r'+r''-2{\sqrt {r'r''}}\cos \omega }}\,;$

donc

$p={\frac {r'r''\sin ^{2}\omega }{r'+r''-2{\sqrt {r'r''}}\cos \omega }}.$