Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/99

Ainsi, combinant cette équation avec la précédente

$\mathrm {A} =\mathrm {F} \cos n-\mathrm {G} ,$

on tirera

${\begin{aligned}\mathrm {F} =&{\frac {\mathrm {B-A} \cos n-{\dfrac {2\sin(n+\varphi )}{\sqrt {\cos n-\cos(n+\varphi )}}}}{\sin ^{2}n}},\\\mathrm {G} =&{\frac {\mathrm {B} \cos n-\mathrm {A} -{\dfrac {2\cos n\sin(n+\varphi )}{\sqrt {\cos n-\cos(n+\varphi )}}}}{\sin ^{2}n}}.\end{aligned}}$

Donc, substituant ces valeurs dans les expressions de $d\mathrm {A}$ et de $d\mathrm {B}$ trouvées ci-dessus, on aura

${\begin{aligned}d\mathrm {A} =&{\frac {dn+d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}-{\frac {\sin(n+\varphi )dn}{\sin n{\sqrt {\cos n-\cos(n+\varphi )}}}}+{\frac {(\mathrm {B-A} \cos n)dn}{2\sin n}},\\d\mathrm {B} =&{\frac {\cos(n+\varphi )(dn+d\varphi )}{\sqrt {\cos n-\cos(n+\varphi )}}}-{\frac {\cos n\sin(n+\varphi )dn}{\sin n{\sqrt {\cos n-\cos(n+\varphi )}}}}+{\frac {(\mathrm {B} \cos n-\mathrm {A} )dn}{2\sin n}}.\end{aligned}}$

Donc, remettant à la place de $n$ sa valeur $q+\alpha -m,$ et faisant $\varphi =m,$ on aura

${\begin{aligned}d\mathrm {A} =&{\frac {dq+d\alpha }{\sqrt {\cos(q+\alpha -m)-\cos(q+\alpha )}}}\\&-{\frac {\sin(q+\alpha )(dq+d\alpha -dm)}{\sin(q+\alpha -m){\sqrt {\cos(q+\alpha -m)-\cos(q+\alpha )}}}}\\&+{\frac {\mathrm {B-A} \cos(q+\alpha -m)}{2\sin(q+\alpha -m)}}(dq+d\alpha -dm),\\d\mathrm {B} =&{\frac {\cos(q+\alpha )(dq+d\alpha )}{\sqrt {\cos(q+\alpha -m)-\cos(q+\alpha )}}}\\&-{\frac {\cos(q+\alpha -m)\sin(q+\alpha )(dq+d\alpha -dm)}{\sin(q+\alpha -m){\sqrt {\cos(q+\alpha -m)-\cos(q+\alpha )}}}}\\&+{\frac {\mathrm {B} \cos(q+\alpha -m)-\mathrm {A} }{2\sin(q+\alpha -m)}}(dq+d\alpha -dm).\end{aligned}}$

Donc, faisant pour plus de simplicité $\alpha -m=\beta ,$ on aura enfin ces trois