Page:Annales de mathématiques pures et appliquées, 1810-1811, Tome 1.djvu/156

152

QUESTIONS.

et $r$ respectivement, j’affirme que

$\mathrm {R} :r=1^{3}:4\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}A.\operatorname {Sin} .{\tfrac {1}{2}}B.\operatorname {Sin} .{\tfrac {1}{2}}C}$.

Démonstration.

${\begin{array}{rlll}\mathrm {AB} :&r&=1\times \mathrm {\operatorname {Cos} .{\tfrac {1}{2}}C} &:\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}A.\operatorname {Sin} .{\tfrac {1}{2}}B.{\text{,(§. I. Coroll.)}}} \\\mathrm {R} :&\mathrm {AB} &=1&:2\mathrm {\operatorname {Sin} .C{\text{ ; (§. II.)}}} \\{\text{donc,}}\ \mathrm {R} :&r&=1^{2}\times \mathrm {\operatorname {Cos} .{\tfrac {1}{2}}C} &:2\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}A.\operatorname {Sin} .{\tfrac {1}{2}}B.\operatorname {Sin} .C} ,\\&\quad &=1^{3}&:4\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}A.\operatorname {Sin} .{\tfrac {1}{2}}B.\operatorname {Sin} .{\tfrac {1}{2}}C} .\\\end{array}}$

SCHOLIE. Quatre fois le produit continuel du sinus des demi-angles d’un triangle, est égal au produit du quarré du sinus total par l’excès de la somme des cosinus des angles de ce triangle sur ce sinus total.

En effet,

$\mathrm {\operatorname {Cos} .A+\operatorname {Cos} .B+\operatorname {Cos} .C-1=\operatorname {Cos} .A+\operatorname {Cos} .B-2\operatorname {Sin} .^{2}{\tfrac {1}{2}}C}$

${\begin{aligned}\quad &=2\mathrm {\operatorname {Cos} .{\tfrac {1}{2}}(A+B)\operatorname {Cos} .{\tfrac {1}{2}}(A-B)-2\operatorname {Sin} .^{2}{\tfrac {1}{2}}C} \\\quad &=2\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}C\left\{\operatorname {Cos} .{\tfrac {1}{2}}(A-B)-\operatorname {Sin} .{\tfrac {1}{2}}C\right\}} \\\quad &=2\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}C\left\{\operatorname {Cos} .{\tfrac {1}{2}}(A-B)-\operatorname {Cos} .{\tfrac {1}{2}}(A+B)\right\}} \\\quad &=4\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}A.\operatorname {Sin} .{\tfrac {1}{2}}B.\operatorname {Sin} .{\tfrac {1}{2}}C.} \\\end{aligned}}$

On peut aussi exprimer ce produit continuel dans les côtés du triangle.

En effet,

${\begin{aligned}\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}A} &=\mathrm {\sqrt {\tfrac {{\tfrac {1}{2}}(AB-AC+BC)\times {\tfrac {1}{2}}(-AB+AC+BC)}{BA\times AC}}} \\\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}B} &=\mathrm {\sqrt {\tfrac {{\tfrac {1}{2}}(AB-BC+AC)\times {\tfrac {1}{2}}(-AB+BC+AC)}{AB\times BC}}} \\\mathrm {\operatorname {Sin} .{\tfrac {1}{2}}C} &=\mathrm {\sqrt {\tfrac {{\tfrac {1}{2}}(AC-BC+AB)\times {\tfrac {1}{2}}(-AC+BC+AB)}{AC\times CB}}} \\\end{aligned}}$