Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/81

75

DU TRIANGLE SPHÉRIQUE.

${\begin{aligned}&=\mathrm {C} +{\frac {1}{\sqrt {c^{2}-a^{2}+b^{2}}}}Log.{\frac {{\sqrt {c^{2}-a^{2}+b^{2}}}+(c-a)y}{{\sqrt {c^{2}-a^{2}+b^{2}}}-(c-a)y}}\\&=\mathrm {C} +{\frac {1}{\sqrt {c^{2}-a^{2}+b^{2}}}}Log.{\frac {{\sqrt {c^{2}-a^{2}+b^{2}}}+(c-a)\operatorname {Tang} .{\tfrac {1}{2}}x}{{\sqrt {c^{2}-a^{2}+b^{2}}}-(c-a)\operatorname {Tang} .{\tfrac {1}{2}}x}}\\\end{aligned}}$

Si l’on veut, en particulier, que $z$ et $x$ soient zéro en même temps, la constante $\mathrm {C}$ devra être nulle.

Troisième supposition. Soit enfin $c<a,$ on aura

${\begin{aligned}{\frac {\operatorname {d} z}{\operatorname {d} y}}&={\frac {2}{(a+c)+2by+(a-c)y^{2}}}\\&={\frac {2}{(a-c)}}\cdot {\frac {1}{{\frac {(a+c)}{(a-c)}}+{\frac {2b}{(a-c)}}y+y^{2}}}\\&={\frac {2}{(a-c)}}\cdot {\frac {1}{\left\{{\frac {(a+c)}{(a-c)}}-{\frac {b^{2}}{(a-c)^{2}}}\right\}+\left\{{\frac {b}{a-c}}+y\right\}^{2}}}\\&={\frac {2}{(a-c)}}\cdot {\frac {1}{{\frac {a^{2}-b^{2}-c^{2}}{(a-c)^{2}}}+\left\{{\frac {b}{a-c}}+y\right\}^{2}}}\\&=2\cdot {\frac {a-c}{a^{2}-b^{2}-c^{2}}}\cdot {\frac {1}{1+\left\{{\frac {b}{\sqrt {a^{2}-b^{2}-c^{2}}}}+{\frac {a-c}{\sqrt {a^{2}-b^{2}-c^{2}}}}y\right\}^{2}}}\\&={\frac {2}{\sqrt {a^{2}-b^{2}-c^{2}}}}\cdot {\frac {\frac {a-c}{\sqrt {a^{2}-b^{2}-c^{2}}}}{1+\left\{{\frac {b}{\sqrt {a^{2}-b^{2}-c^{2}}}}+{\frac {a-c}{\sqrt {a^{2}-b^{2}-c^{2}}}}y\right\}^{2}}},\\\end{aligned}}$

d’où on conclura

$z=\mathrm {C} +{\frac {2}{\sqrt {a^{2}-b^{2}-c^{2}}}}.\operatorname {Arc} \left\{\operatorname {Tang} .={\frac {b+(a-c)y}{\sqrt {a^{2}-b^{2}-c^{2}}}}\right\}$