Page:Mémoires de l’Académie des sciences, Tome 7.djvu/526

$i=1$ et $i=2,$ on aura :

${\begin{aligned}\sum \left({\frac {e_{i}\cos .2iq-f_{i}\sin .2iq}{1+2{,}324\sin .^{2}2iq}}\right)&={\frac {{\underset {-2}{e}}\cos .4q+{\underset {-2}{f}}\sin .4q}{1+2{,}324\sin .^{2}4q}}+{\frac {{\underset {-1}{e}}\cos .2q+{\underset {-1}{f}}\sin .2q}{1+2{,}324\sin .^{2}2q}}+e_{0}\\&+{\frac {e_{1}\cos .2q-f_{1}\sin .2q}{1+2{,}324\sin .^{2}2q}}+{\frac {e_{2}\cos .4q-f_{2}\sin .4q}{1+2{,}324\sin .^{2}4q}}\,;\\\sum \left({\frac {1}{1+2{,}324\sin .^{2}2iq}}\right)\ \ \quad &=1+{\frac {2}{1+2{,}324\sin .^{2}4q}}+{\frac {2}{1+2{,}324\sin .^{2}2q}}\,;\end{aligned}}$

on trouvera de même que

${\begin{aligned}\sum \left({\frac {e_{i}\sin .2iq+f_{i}\cos .2iq}{1+2{,}324\cos .^{2}2iq}}\right)&={\frac {{\underset {-2}{f}}\cos .4q-{\underset {-2}{e}}\sin .4q}{1+2{,}324\cos .^{2}4q}}+{\frac {{\underset {-1}{f}}\cos .2q-{\underset {-1}{e}}\sin .2q}{1+2{,}324\cos .^{2}2q}}+f_{0}\\&+{\frac {f_{1}\cos .2q+e_{1}\sin .2q}{1+2{,}324\cos .^{2}2q}}+{\frac {f_{2}\cos .4q+e_{2}\sin .4q}{1+2{,}324\cos .^{2}4q}}\,;\\{\text{et}}\qquad \qquad \qquad \qquad &\\\sum \left({\frac {1}{1+2{,}324\cos .^{2}2iq}}\right)\ \ \quad &={\frac {1}{3{,}324}}+{\frac {2}{1+2{,}324\cos .^{2}4q}}+{\frac {2}{1+2{,}324\cos .^{2}2q}}\,;\end{aligned}}$

en substituant dans ces formules, les quantités connues, on trouve :

${\begin{alignedat}{2}\sum \left({\frac {e_{i}\cos .2iq-f_{i}\sin .2iq}{1+2{,}324\sin .^{2}2iq}}\right)&=+0{,}104697\,;&\sum \left({\frac {1}{1+2{,}324\sin .^{2}2iq}}\right)&=3{,}29667;\\\sum \left({\frac {e_{i}\sin .2iq+f_{i}\cos .2iq}{1+2{,}324\cos .^{2}2iq}}\right)&=+0{,}030367{\text{ et }}&\sum \left({\frac {1}{1+2{,}324\cos .^{2}2iq}}\right)&=1{,}97902.\end{alignedat}}$

d’où l’on tire :

$x=0{,}031758\quad {\text{et}}\quad y=0{,}015344.$

Maintenant, pour déterminer l’influence lunaire, nous reprendrons les deux équations

$x=4\mathrm {R} \sin .2\lambda \quad {\text{et}}\quad y=4\mathrm {R} \cos .2\lambda \,;$

d’où l’on tire :