Page:Annales de mathématiques pures et appliquées, 1824-1825, Tome 15.djvu/56

${\frac {1}{4}}\left\{r^{2}+\left(r^{2}+k^{2}\right)\operatorname {Cos} .{\frac {2\varpi }{m}}\right\}\operatorname {Sin} .{\frac {2\varpi }{m}}$

$-{\frac {1}{4}}k\left\{4r\operatorname {Cos} .^{2}{\frac {2\varpi }{m}}\operatorname {Cos} .\left({\frac {2n\pi }{m}}-\alpha \right)-k\operatorname {Cos} .\left({\frac {4n\pi }{m}}-2\alpha \right)\right\}$

Il résulte de là que l’aire cherchée du polygone sera

${\frac {m}{4}}\left\{r^{2}+\left(r^{2}+k^{2}\right)\operatorname {Cos} .{\frac {2\varpi }{m}}\right\}\operatorname {Sin} .{\frac {2\varpi }{m}}$

$-{\frac {1}{4}}k\left\{4r\operatorname {Cos} .^{2}{\frac {2\varpi }{m}}\Sigma \operatorname {Cos} .\left({\frac {2n\pi }{m}}-\alpha \right)-k\Sigma \operatorname {Cos} .\left({\frac {4n\pi }{m}}-2\alpha \right)\right\}\,;$

les sommes étant prises depuis $n=0$ jusqu’à $n=m-1.$

Or, il est connu^[1] que

$\operatorname {Cos} .a+\operatorname {Cos} .(a+d)+\operatorname {Cos} .(a+2d)+\ldots +\operatorname {Cos} .\left[a+(m-1)d\right]=$

${\frac {\operatorname {Cos} .\left[a+{\frac {1}{2}}(m-1)d\right]\operatorname {Sin} .{\frac {1}{2}}md}{\operatorname {Sin} .{\frac {1}{2}}d}}$

Posant donc, 1.^o 

$a=-\alpha ,\qquad d={\frac {2\varpi }{m}},$

il viendra

$\Sigma \operatorname {Cos} .\left({\frac {2n\pi }{m}}-\alpha \right)={\frac {\operatorname {Cos} .\left[{\frac {(m-1)\pi }{m}}-\alpha \right]\operatorname {Sin} .\varpi }{\operatorname {Sin} .{\frac {\varpi }{m}}}}=0.$

Posant, en second lieu,

$a=-2\alpha ,\qquad d={\frac {4\varpi }{m}},$

nous trouverons

1. Euler, Introd. in analy. inf. (tom. I, chap. XIV, n.^o 260).