Page:Annales de mathématiques pures et appliquées, 1812-1813, Tome 3.djvu/12

8

FACULTÉS

5. Dans l’ouvrage déjà cité (chapitre III, 39) j’ai prouvé que, $h$ étant une fraction positive plus petite que ${\tfrac {1}{2}},$ on a

$\operatorname {Tang} .h={\frac {(+h)^{{\tfrac {1}{2}}|+1}}{(-h)^{{\tfrac {1}{2}}|-1}}}\;;$

mais, suivant les réductions enseignées ci-dessus, on a

$(+h)^{{\tfrac {1}{2}}|+1}={\frac {\left(h-{\tfrac {1}{2}}\right)!}{\left(h-1\right)!}}={\frac {2h}{1+2h}}\cdot {\frac {\left({\tfrac {1}{2}}+h\right)!}{h!}}\;;$

$(-h)^{{\tfrac {1}{2}}|-1}={\frac {(-h)!}{\left(-h-{\tfrac {1}{2}}\right)!}}={\frac {1-2h}{2-2h}}\cdot {\frac {(1-h)!}{\left({\tfrac {1}{2}}-h\right)!}}\;;$

d’où résulte

$\operatorname {Tang} .h\varpi ={\frac {h(1-h)}{\left({\tfrac {1}{2}}+h\right)\left({\tfrac {1}{2}}-h\right)}}\cdot {\frac {\left({\tfrac {1}{2}}+h\right)!\left({\tfrac {1}{2}}-h\right)!}{h!(1-h)!}}\;;$

Ainsi, si l’on demandait la tangente de $66^{\circ }.36',$ on aurait $h=0{,}37,$ $1-h=0{,}63,{\tfrac {1}{2}}+h=0{,}87,{\tfrac {1}{2}}-h=0{,}13\;;$ d’où

$\operatorname {Tang} .66^{\circ }.36'={\frac {0{,}37\times 0{,}63}{0{,}87\times 0{,}13}}\cdot {\frac {0{,}87!0{,}13!}{0{,}37!0{,}63!}}$

Voici le calcul :

${\begin{array}{rl}\operatorname {Log} .37&=1{,}56820\;17241,\\\operatorname {Log} .63&=1{,}79934\;05495,\\{\text{Comp. arith. }}\operatorname {Log} .87&=8{,}06048\;07474,\\{\text{Comp. arith. }}\operatorname {Log} .13&=8{,}88605\;66477,\\\operatorname {Log} .0{,}87!&=9{,}97856\;40362,\\\operatorname {Log} .0{,}13!&=9{,}97309\;61812,\\{\text{Comp. arith. }}\operatorname {Log} .0{,}37!&=0{,}05094\;51119,\\{\text{Comp. arith. }}\operatorname {Log} .0{,}62!&=0{,}04708\;93246,\\\end{array}}$

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$\qquad \qquad \operatorname {Log} .\operatorname {tang} .66^{\circ }.36'=0{,}36377\;43220.$

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