Page:Annales de mathématiques pures et appliquées, 1814-1815, Tome 5.djvu/18

14

PROBLÈMES.

de ces deux fractions la forme connue de binôme, savoir $F\operatorname {Cos} .\delta -G\operatorname {Sin} .\delta .$ Dans cette vue, nous ferons, pour abréger,

${\begin{aligned}&a=\operatorname {Cos} .A\operatorname {Cot} .B,&&a'=\operatorname {Sin} ,A\operatorname {Cot} .B,\\&b=\operatorname {Cos} .A'\operatorname {Cot} .B',&&b'=\operatorname {Sin} .A'\operatorname {Cot} .B',\\&c=\operatorname {Cos} .A''\operatorname {Cot} .B'',&&c'=\operatorname {Sin} .A''\operatorname {Cot} .B'',\\\end{aligned}}$

${\begin{aligned}m=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Sin} .\theta '-\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Sin} .\theta ,\\n=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Sin} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Sin} .\theta ',\\o=&\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Sin} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Sin} .\theta \,;\\\end{aligned}}$

${\begin{aligned}m'=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Cos} .\theta '-\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Cos} .\theta ,\\n'=&\operatorname {Sin} .(\theta -A)\operatorname {Cot} .B\operatorname {Cos} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Cos} .\theta ,\\o'=&\operatorname {Sin} .(\theta '-A')\operatorname {Cot} .B'\operatorname {Cos} .\theta ''-\operatorname {Sin} .(\theta ''-A'')\operatorname {Cot} .B''\operatorname {Cos} .\theta '.\\\end{aligned}}$

91. Enfin, proposons la dernière notation que la nature du problème exige, et qui paraît nécessaire pour présenter l’inconnue sous la forme la plus simple ; savoir,

${\begin{aligned}D=&hm-tn-c'h\operatorname {Sin} .t+b't\operatorname {Sin} .h,\\E=&hm'-tn'-ch\operatorname {Sin} .t+bt\operatorname {Sin} .h,\\F=&b'tn-c'hm,\\G=&b'tn'+btn-c'hm'-chm,\\H=&btn'-chm'\,;\\D'=&ho-t'n-a'h\operatorname {Sin} .t'+b't'\operatorname {Sin} .h,\\E'=&ho'-t'n'-ah\operatorname {Sin} .t'+bt'\operatorname {Sin} .h,\\F'=&b't'n-a'ho,\\G'=&bt'n+b't'n'-aho-a'ho',\\H'=&bt'n'-aho'.\\\end{aligned}}$

92. Les deux expressions (89) deviendront alors