Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/121

or

${\frac {1-\cos t}{\sin t}}=\operatorname {tang} {\frac {1}{2}}t,\quad \cos t={\frac {d\sin t}{dt}},$

de sorte que l’équation précédente deviendra

$\operatorname {tang} {\frac {1}{2}}u={\frac {m}{1+n}}\left[(1-n)\operatorname {tang} {\frac {1}{2}}t\right.$

$-\left({\frac {n^{2}}{2}}+{\frac {n^{3}}{2.3}}\right){\frac {d^{2}\sin t}{dt^{2}}}+\left({\frac {n^{3}}{2\times 2.3}}+{\frac {n^{4}}{2.3.4}}\right){\frac {d^{3}\sin ^{2}t}{dt^{3}}}$

$\left.-\left({\frac {n^{4}}{3\times 2.3.4}}+{\frac {n^{5}}{2.3.4.5}}\right){\frac {d^{4}\sin ^{3}t}{dt^{4}}}+\ldots \right].$

d’où l’on aura

$\operatorname {tang} {\frac {1}{2}}u={\frac {m}{1+n}}\left[(1-n)\operatorname {tang} {\frac {1}{2}}t\right.$

${\begin{aligned}&+{\frac {n^{2}}{2}}\left(1+{\frac {n}{3}}\right)\sin t-{\frac {n^{3}}{2\times 2.3}}\left({\frac {1}{2}}+{\frac {n}{4}}\right)2^{3}\sin 2t\\&-{\frac {n^{4}}{4\times 2.3.4}}\left({\frac {1}{3}}+{\frac {n}{5}}\right)\left(3\sin t-3^{4}\sin 3t\right)\\&+{\frac {n^{5}}{8\times 2.3.4.5}}\left({\frac {1}{4}}+{\frac {n}{6}}\right)\left(4.2^{5}.\sin 2t-4^{5}\sin 4t\right)\\&+{\frac {n^{6}}{16\times 2.3\ldots 6}}\left({\frac {1}{5}}+{\frac {n}{7}}\right)\left({\frac {5.4}{2}}\sin t-5.3^{6}\sin 3t+5^{6}\sin 5t\right)\\&{\biggl .}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots {\biggr ]}.\end{aligned}}$

VI.

Si l’on voulait avoir la valeur de l’angle même $u,$ il faudrait faire (Article II)

$\psi (x)=m\int {\frac {dx}{1+n\cos x}}=u\,;$

donc

$\psi (t)=\int {\frac {dt}{1+n\cos t}}\quad {\text{et}}\quad \psi '(t)={\frac {m}{1+n\cos t}}\,;$