Page:Gauss - Théorie du mouvement des corps célestes, traduction Dubois, 1864.djvu/382

363

MÉTHODE D’OLBERS.

$q^{\frac {1}{2}}={\frac {\left(rr''\right)^{\frac {1}{2}}\sin {\dfrac {1}{2}}(v''\!-v)}{\left(r''\!+r-2\left(rr''\right)^{\frac {1}{2}}\cos {\dfrac {1}{2}}(v''\!-v)\right)^{\frac {1}{2}}}}$

Si nous portons cette valeur de $q^{\frac {1}{2}}$ dans $\mathrm {T} ''-\mathrm {T} ,$ nous aurons

${\begin{aligned}\mathrm {T} ''\!\!-\!\mathrm {T} &=\mathrm {C} \left(r''\!\!+\!r-2\left(rr''\right)^{\frac {1}{2}}\cos {\frac {1}{2}}(v''\!\!-\!v)\right)^{\!{\frac {1}{2}}}\left(r''\!\!+\!r+\left(rr''\right)^{\frac {1}{2}}\cos {\frac {1}{2}}(v''\!\!-\!v)\right)\\&=\mathrm {C} \left(r''\!\!+\!r\right)^{\frac {3}{2}}\!\left(\!1+{\frac {\left(rr''\right)^{\frac {1}{2}}\cos {\dfrac {1}{2}}(v''\!\!-\!v)}{(r+r'')}}\!\right)\!\left(\!1-{\frac {\left(2rr''\right)^{\frac {1}{2}}\cos {\dfrac {1}{2}}(v''\!\!-\!v)}{(r+r'')}}\!\right)^{\!{\frac {1}{2}}}\\&=\mathrm {C} \left(r''\!\!+\!r\right)^{\frac {3}{2}}\!\left(\!1+{\frac {\left({\dfrac {r}{r''}}\right)^{\frac {1}{2}}\cos {\dfrac {1}{2}}(v''\!\!-\!v)}{1+{\dfrac {r}{r''}}}}\!\right)\!\left(\!1-{\frac {2\left({\dfrac {r}{r''}}\right)^{\frac {1}{2}}\cos {\dfrac {1}{2}}(v''\!\!-\!v)}{1+{\dfrac {r}{r''}}}}\!\right)^{\!{\frac {1}{2}}}\end{aligned}}$

ou, en mettant $\operatorname {tang} ^{2}\!z$ à la place de ${\frac {r}{r''}},$ et réduisant,

$\mathrm {T} ''\!\!-\!\mathrm {T} =\mathrm {C} \left(r''\!\!+\!r\right)^{\frac {3}{2}}\!\left(\!1+\sin z\cos z\cos {\frac {1}{2}}(v''\!\!-\!v)\!\right)\!\left(\!1-2\sin z\cos z\cos {\frac {1}{2}}(v''\!\!-\!v)\!\right)^{\!{\frac {1}{2}}}$

Posons

$2\sin z\cos z\cos {\frac {1}{2}}(v''\!\!-\!v)=\sin x$

et remarquons que

$\left(1-\sin x\right)^{\frac {1}{2}}=\cos {\frac {1}{2}}x-\sin {\frac {1}{2}}x$

et

$\left(1+{\frac {1}{2}}\sin x\right)=1+\sin {\frac {1}{2}}x\cos {\frac {1}{2}}x$

on aura, toutes réductions faites,

${\begin{aligned}\mathrm {T} ''\!\!-\!\mathrm {T} =\mathrm {C} \left(r\!+\!r''\right)^{\frac {3}{2}}\left(\cos ^{3}\!{\frac {1}{2}}x-\sin ^{3}\!{\frac {1}{2}}x\right)\end{aligned}}$

ou

$\mathrm {T} ''\!-\!\mathrm {T} =\mathrm {C} \left[\left({\frac {(r\!+\!r'')+(r\!+\!r'')\cos x}{2}}\right)^{\frac {3}{2}}\!-\left({\frac {(r\!+\!r'')-(r\!+\!r'')\cos x}{2}}\right)^{\frac {3}{2}}\right].$