Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/98

$S={\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right).$

II. Pour sommer la seconde série

$S={\frac {\operatorname {Cos} .x}{1}}+{\frac {1}{2}}{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}{\frac {\operatorname {Cos} .5x}{5}}+{\frac {1.3.5}{2.4.6}}{\frac {\operatorname {Cos} .7x}{7}}+\ldots$

nous poserons encore $\operatorname {Cos} .x={\frac {1}{2}}(u+v),$ avec la condition $uv=1\,;$ alors, en posant

${\begin{aligned}&U=u+{\frac {1}{2}}.{\frac {u^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {u^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {u^{7}}{7}}+\ldots =\operatorname {Arc} .(\operatorname {Sin} .=u),\\\\&V=v+{\frac {1}{2}}.{\frac {v^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {v^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {v^{7}}{7}}+\ldots =\operatorname {Arc} .(\operatorname {Sin} .=v),\end{aligned}}$

nous aurons

$S={\frac {1}{2}}(U+V)={\frac {1}{2}}\left\{\operatorname {Arc} .(\operatorname {Sin} .=u)+\operatorname {Arc} .(\operatorname {Sin} .=v)\right\}\,;$

donc

$\left.{\begin{aligned}&u=\operatorname {Sin} .U\\&v=\operatorname {Sin} .V\end{aligned}}\right\}\quad \mathrm {d'o{\grave {u}}} \quad \left\{{\begin{aligned}&{\sqrt {1-u^{2}}}=\operatorname {Cos} .U,\\&{\sqrt {1-v^{2}}}=\operatorname {Cos} .V,\end{aligned}}\right.$

donc aussi

$\operatorname {Sin} .2S=\operatorname {Sin} .(U+V)=u{\sqrt {1-v^{2}}}+v{\sqrt {1-u^{2}}},$

ou, en quarrant

$\operatorname {Sin} .^{2}2S=u^{2}\left(1-v^{2}\right)+v^{2}\left(1-u^{2}\right)+2uv{\sqrt {(1-u^{2})(1-v^{2})}},$