![{\displaystyle {\begin{aligned}-{\frac {n^{5}}{16.5}}&\left[{\frac {5.4}{2}}\left({\frac {6}{m^{5}}}+{\frac {8}{2.3.m^{3}}}+{\frac {10}{2.3.4.5.m}}\right)\sin t\right.\\&\,\ \ -5\left({\frac {6}{m^{5}}}+{\frac {8.3^{2}}{2.3.m^{3}}}+{\frac {10.3^{4}}{2.3.4.5.m}}\right)\sin 3t\\&\quad \ +\left.\left({\frac {6}{m^{5}}}+{\frac {8.5^{2}}{2.3.m^{3}}}+{\frac {10.5^{4}}{2.3.4.5.m}}\right)\sin 5t\right]\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57413ecd773e342389826ea0a31e703cd570467c)
![{\displaystyle -{\frac {n^{6}}{32.6}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/737206faec424361b5bedaea7abb0e94767975d3)
![{\displaystyle {\begin{aligned}\left[{\frac {6.6.5.4}{2.2.3.m^{7}}}t-{\frac {6.5}{2}}\left({\frac {6}{2m^{7}}}+{\frac {8.2}{2m^{5}}}+{\frac {10.2^{3}}{2.3.4.m^{3}}}+{\frac {12.2^{5}}{2.3\ldots 6.m}}\right)\sin 2t\right.\ \ &\\\quad +6\left({\frac {6}{4m^{7}}}+{\frac {8.4}{2m^{5}}}+{\frac {10.4^{3}}{2.3.4.m^{3}}}+{\frac {12.4^{5}}{2.3\ldots 6.m}}\right)\sin 4t\ \ &\\\quad -\left.\left({\frac {6}{6m^{7}}}+{\frac {8.6}{2m^{5}}}+{\frac {10.6^{3}}{2.3.4.m^{3}}}+{\frac {12.6^{5}}{2.3\ldots 6.m}}\right)\sin 6t\right]&\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e1632569c718684f12dbd52b9f9182d828cec1f)
. . . . . . . . . . . . . . . . . . . . . . . . .
VII.
Nous avons donné plus haut (Article V) la valeur de
si l’on voulait aussi avoir celle du logarithme de la même tangente, on la trouverait avec la même facilité en faisant
![{\displaystyle \psi (x)=\log \operatorname {tang} {\frac {1}{2}}u=\log \left({\frac {m}{1+n}}\operatorname {tang} {\frac {1}{2}}x\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a12ddadf8fcb0052a98d90b27db0f5349974c985)
et par conséquent
![{\displaystyle \psi (t)=\log {\frac {m}{1+n}}+\log \operatorname {tang} {\frac {1}{2}}t\quad {\text{et}}\quad \psi '(t)={\frac {1}{\sin t}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d451e537a94793fbb7ed073b1d1975c71c2ac6)
ce qui étant substitué dans la formule de l’Article II on aurait
![{\displaystyle {\begin{aligned}\log \operatorname {tang} {\frac {1}{2}}u=&\log {\frac {m}{1+n}}+\log \operatorname {tang} {\frac {1}{2}}t-n+{\frac {n^{2}}{2}}{\frac {d\sin t}{dt}}\\&-{\frac {n^{3}}{2.3}}{\frac {d^{2}\sin ^{2}t}{dt^{2}}}+{\frac {n^{4}}{2.3.4}}{\frac {d^{3}\sin ^{3}t}{dt^{3}}}-\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4189850b04daadd577bf7757fa61bc2bbbf3d16)
c’est-à-dire, en réduisant les puissances de
en sinus et cosinus de