Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/120

d’angles multiples de $t,$ et exécutant les différentiations indiquées, on aura

${\begin{aligned}r=&1+n\cos t-{\frac {n^{2}}{2}}(-1+\cos 2t)\\&-{\frac {n^{3}}{4\times 2}}\left(3\cos t-3\cos 3t\right)\\&+{\frac {n^{4}}{8\times 2.3}}\left(4.2^{2}.\cos 2t-4^{2}\cos 4t\right)\\&+{\frac {n^{5}}{16\times 2.3.4}}\left({\frac {5.4}{2}}\cos t-5.3^{3}.\cos 3t+5^{3}\cos 5t\right)\\&-{\frac {n^{6}}{32\times 2.3.4.5}}\left({\frac {6.5}{2}}2^{4}\cos 2t-6.4^{4}.\cos 4t+6^{4}\cos 6t\right)\\&-{\frac {n^{7}}{64\times 2.3\ldots 6}}\left({\frac {7.6.5}{2.3}}\cos t-{\frac {7.6}{2}}3^{5}\cos 3t+7.5^{5}.\cos 5t-7^{5}\cos 7t\right)\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

V.

De même, pour avoir la valeur de $\operatorname {tang} {\frac {1}{2}}u,$ on fera

$\psi (x)={\frac {m}{1+n}}\operatorname {tang} {\frac {1}{2}}x=\operatorname {tang} {\frac {1}{2}}u\,;$

par conséquent aussi

$\psi (t)={\frac {m}{1+n}}\operatorname {tang} {\frac {1}{2}}t,$

ce qui donnera

$\psi '(t)={\frac {m}{1+n}}{\frac {1}{2\cos ^{2}{\dfrac {1}{2}}t}}={\frac {m}{1+n}}{\frac {1}{1+\cos t}}={\frac {m}{1+n}}{\frac {1-\cos t}{\sin ^{2}t}}.$

Donc, substituant ces valeurs dans la formule de l’Article II, on aura

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

$-{\frac {n^{3}}{2.3}}{\frac {d^{2}(1-\cos t)\sin t}{dt^{2}}}+{\frac {n^{4}}{2.3.4}}{\frac {d^{3}(1-\cos t)\sin ^{2}t}{dt^{3}}}$

$\left.-{\frac {n^{5}}{2.3.4.5}}{\frac {d^{4}(1-\cos t)\sin ^{3}t}{dt^{4}}}+\ldots \right]\,;$