Page:Gauss - Théorie du mouvement des corps célestes, traduction Dubois, 1864.djvu/351

332

NOTES DU TRADUCTEUR.

en faisant $\mathrm {C} \!=\!0,$ et prenant les logarithmes, nous avons

$\log \operatorname {tang} {\frac {1}{2}}v=\log \operatorname {tang} {\frac {1}{2}}w-{\frac {1}{2}}\log \left(1-{\frac {4}{5}}\beta \operatorname {tang} ^{2}\!{\frac {1}{2}}w\right)+\log \gamma \,;$

d’où, en différentiant,

${\frac {dv}{\sin v}}={\frac {dw}{2\sin {\dfrac {1}{2}}w\cos {\dfrac {1}{2}}w\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)}}+{\frac {d\gamma }{\gamma }}+{\frac {{\dfrac {2}{5}}\mathrm {A} }{\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)}}{\frac {d\beta }{\beta }}$

ou,

${\frac {dv}{\sin v}}={\frac {dw}{2\cos ^{4}\!{\dfrac {1}{2}}w}}\times {\frac {\cos ^{2}\!{\dfrac {1}{2}}w}{\operatorname {tang} {\dfrac {1}{2}}w\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)}}+{\frac {d\gamma }{\gamma }}+{\frac {{\dfrac {2}{5}}\mathrm {A} }{\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)}}{\frac {d\beta }{\beta }}.$

Mais comme on a

$\gamma ={\sqrt {\frac {5+5{\overset {}{e}}}{1+9e}}},$  $\beta ={\frac {5-5e}{1+9e}},$

ou en déduit

${\frac {d\gamma }{\gamma }}={\frac {-2}{(1+e)(1+9e)}}\,de,$  ${\frac {d\beta }{\beta }}={\frac {-10}{(1+9e)(1-e)}}\,de.$

Il vient donc, en substituant ces valeurs dans ${\frac {dv}{\sin v}},$ et aussi l’expression trouvée pour ${\frac {dw}{2\cos ^{4}\!{\frac {1}{2}}w}}$

${\begin{aligned}{\frac {dv}{\sin v}}&={\frac {\alpha \cos ^{2}{\dfrac {1}{2}}w}{75\operatorname {tang} {\dfrac {1}{2}}w\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)}}\,dt-{\frac {3\alpha t\cos ^{2}\!{\dfrac {1}{2}}w}{2.q.75\operatorname {tang} {\dfrac {1}{2}}w\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)}}\,dq\\&+{\frac {t\cos ^{2}\!{\dfrac {1}{2}}w.9\alpha }{75\operatorname {tang} {\dfrac {1}{2}}w\left(1\!-\!{\dfrac {4}{5}}\mathrm {A} \right)2(1\!+\!9e)}}\,de-{\frac {4}{(1\!+\!e)(1\!+\!9e)}}\,de\\&-{\frac {{\dfrac {2}{5}}\mathrm {A} }{1\!-\!{\dfrac {4}{5}}\mathrm {A} }}.{\frac {10}{(1\!+\!9e)(1\!-\!e)}}\,de\end{aligned}}$