Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/118

114

QUESTIONS

$\operatorname {Log} .\left(1+2a\operatorname {Cos} .x+a^{2}\right)=$

$2\left({\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{2}\operatorname {Cos} .2x}{2}}+{\frac {a^{3}\operatorname {Cos} .3x}{3}}-\ldots \right).$

En faisant successivement $a=+1$ et $a=-1,$ il vient

$\operatorname {Log} .(1+\operatorname {Cos} .x)=$

$-\operatorname {Log} .2+2\left({\frac {\operatorname {Cos} .x}{1}}-{\frac {\operatorname {Cos} .2x}{2}}+{\frac {\operatorname {Cos} .3x}{3}}-{\frac {\operatorname {Cos} .4x}{4}}+\ldots \right),$

$\operatorname {Log} .(1-\operatorname {Cos} .x)=$

$-\operatorname {Log} .2-2\left({\frac {\operatorname {Cos} .x}{1}}+{\frac {\operatorname {Cos} .2x}{2}}+{\frac {\operatorname {Cos} .3x}{3}}+{\frac {\operatorname {Cos} .4x}{4}}+\ldots \right),$

développemens donnés par Euler (Voyez son Calcul intégral tom. I, à la fin du chap. VI) ; il en déduit ensuite

${\begin{aligned}&\operatorname {Log} .\operatorname {Cos} .{\tfrac {1}{2}}x=-\operatorname {Log} .2+\operatorname {Cos} .x-{\tfrac {1}{2}}\operatorname {Cos} .2x+{\tfrac {1}{3}}\operatorname {Cos} .3x-{\tfrac {1}{4}}\operatorname {Cos} .4x+\ldots \\&\operatorname {Log} .\operatorname {Sin} .{\tfrac {1}{2}}x=-\operatorname {Log} .2-\operatorname {Cos} .x-{\tfrac {1}{2}}\operatorname {Cos} .2x-{\tfrac {1}{3}}\operatorname {Cos} .3x-{\tfrac {1}{4}}\operatorname {Cos} .4x-\ldots \\&\operatorname {Log} .\operatorname {Tang} .{\tfrac {1}{2}}x=-2\left(\operatorname {Cos} .x+{\tfrac {1}{3}}\operatorname {Cos} .3x+{\tfrac {1}{5}}\operatorname {Cos} .5x+{\tfrac {1}{7}}\operatorname {Cos} .7x+\ldots \right)\end{aligned}}$