Page:Annales de mathématiques pures et appliquées, 1824-1825, Tome 15.djvu/305

$\operatorname {Sin} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}=$

${\sqrt {\frac {\operatorname {Cos} .{\frac {1}{2}}s\operatorname {Cos} .{\frac {1}{2}}(s-a)\operatorname {Sin} .{\frac {1}{2}}(s-b)\operatorname {Sin} .{\frac {1}{2}}(s-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c}}},$(XXXVIII)

$\operatorname {Cos} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}=$

${\sqrt {\frac {\operatorname {Sin} .{\frac {1}{2}}s\operatorname {Sin} .{\frac {1}{2}}(s-a)\operatorname {Cos} .{\frac {1}{2}}(s-b)\operatorname {Cos} .{\frac {1}{2}}(s-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c}}},$(XXXVII)

d’où encore

$\operatorname {Tang} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}=$

${\sqrt {\operatorname {Cot} .{\frac {1}{2}}s\operatorname {Cot} .{\frac {1}{2}}(s-a)\operatorname {Tang} .{\frac {1}{2}}(s-b)\operatorname {Tang} .{\frac {1}{2}}(s-c)}}.$(XXXIX)

§. IV.

Recherches diverses.

Soit $d$ l’arc de grand cercle mené du sommet $A,$ au milieu $A'$ du côté $a\,;$ il divisera notre triangle en deux autres, pour lesquels on aura (i)

${\begin{aligned}&\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .d\operatorname {Cos} .AA'C=\operatorname {Cos} .b-\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .d,\\\\&\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .d\operatorname {Cos} .AA'B=\operatorname {Cos} .c-\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .d\,;\end{aligned}}$

en prenant la somme de ces deux équations et observant que

$\operatorname {Cos} .AA'C+\operatorname {Cos} .AA'B=0,$

il viendra, en transposant,

$2\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .d=\operatorname {Cos} .b+\operatorname {Cos} .c=2\operatorname {Cos} .{\frac {1}{2}}(b+c)\operatorname {Cos} .{\frac {1}{2}}(b-c)\,;$

et, par suite,