Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/93

87

TRIGONOMÉTRIQUES.

Éliminant, dans les équations des deux premières lignes, $\operatorname {Cos} .c$ et $\operatorname {Cos} .C,$ au moyen de celles de la dernière, en se rappelant que $1-\operatorname {Cos} .^{2}x=\operatorname {Sin} .^{2}x,$ et supprimant ensuite le facteur commun à tous les termes des équations résultantes, il viendra

$\operatorname {Sin} .c\operatorname {Cos} .A=\operatorname {Cos} .a\operatorname {Sin} .b-\operatorname {Sin} .a\operatorname {Cos} .b\operatorname {Cos} .C,\ \operatorname {Sin} .C\operatorname {Cos} .a$

$=\operatorname {Cos} .A\operatorname {Sin} .B+\operatorname {Sin} .A\operatorname {Cos} .B\operatorname {Cos} .c,$

$\operatorname {Sin} .c\operatorname {Cos} .B=\operatorname {Sin} .a\operatorname {Cos} .b-\operatorname {Cos} .a\operatorname {Sin} .b\operatorname {Cos} .C,\ \operatorname {Sin} .C\operatorname {Cos} .b$

$=\operatorname {Sin} .A\operatorname {Cos} .B+\operatorname {Cos} .A\operatorname {Sin} .B\operatorname {Cos} .c~;$

en ajoutant et retranchant successivement les équations de chaque colonne, les résultats qui en proviendront, pourront être écrits ainsi

$\operatorname {Sin} .c(\operatorname {Cos} .B+\operatorname {Cos} .A)=(1-\operatorname {Cos} .C)\operatorname {Sin} .(a+b),$

$\operatorname {Sin} .C(\operatorname {Cos} .b+\operatorname {Cos} .a)=(1+\operatorname {Cos} .c)\operatorname {Sin} .(A+B),$

$\operatorname {Sin} .c(\operatorname {Cos} .B-\operatorname {Cos} .A)=(1+\operatorname {Cos} .C)\operatorname {Sin} .(a-b),$

$\operatorname {Sin} .C(\operatorname {Cos} .b-\operatorname {Cos} .a)=(1-\operatorname {Cos} .c)\operatorname {Sin} .(A-B)\,;$

en observant que

$\operatorname {Cos} .y+\operatorname {Cos} .x=2\operatorname {Cos} .{\frac {1}{2}}(x+y)\operatorname {Cos} .{\frac {1}{2}}(x-y),$

$\operatorname {Cos} .y-\operatorname {Cos} .x=2\operatorname {Sin} .{\frac {1}{2}}(x+y)\operatorname {Sin} .{\frac {1}{2}}(x-y)$

$1-\operatorname {Cos} .x=2\operatorname {Sin} .^{2}{\frac {1}{2}}x,\quad 1+\operatorname {Cos} .x=2\operatorname {Cos} .^{2}{\frac {1}{2}}x,$

$\operatorname {Sin} .x=2\operatorname {Sin} .{\frac {1}{2}}x\operatorname {Cos} .{\frac {1}{2}}x,$

ces équations deviendront

${\begin{alignedat}{1}2\operatorname {Sin} .c.\operatorname {Cos} .{\frac {1}{2}}(A+B)\operatorname {Cos} .{\frac {1}{2}}(A-B)\ &=4\operatorname {Sin} .^{2}{\frac {1}{2}}C.\operatorname {Sin} .{\frac {1}{2}}(a+b)\operatorname {Cos} .{\frac {1}{2}}(a+b),\\2\operatorname {Sin} .c.\,\operatorname {Sin} .{\frac {1}{2}}(A+B)\,\operatorname {Sin} .{\frac {1}{2}}(A-B)\ &=4\operatorname {Cos} .^{2}{\frac {1}{2}}C.\operatorname {Sin} .{\frac {1}{2}}(a-b)\operatorname {Cos} .{\frac {1}{2}}(a-b),\\2\operatorname {Sin} .C.\operatorname {Cos} .{\frac {1}{2}}(a+b)\operatorname {Cos} .{\frac {1}{2}}(a-b)\ \ \ &=4\operatorname {Cos} .^{2}{\frac {1}{2}}c.\operatorname {Sin} .{\frac {1}{2}}(A+B)\operatorname {Cos} .{\frac {1}{2}}(A+B),\\2\operatorname {Sin} .C.\,\operatorname {Sin} .{\frac {1}{2}}(a+b)\,\operatorname {Sin} .{\frac {1}{2}}(a-b)\ \ \ &=4\operatorname {Cos} .^{2}{\frac {1}{2}}c.\operatorname {Sin} .{\frac {1}{2}}(A-B)\operatorname {Cos} .{\frac {1}{2}}(A-B)~;\\\end{alignedat}}$

divisant successivement les deux premières par chacune des deux dernières, il viendra

${\begin{aligned}{\frac {\operatorname {Sin} .c}{\operatorname {Sin} .C}}=\left\{{\frac {\operatorname {Sin} .{\frac {1}{2}}C.\operatorname {Cos} .{\frac {1}{2}}(a+b)}{\operatorname {Cos} .{\frac {1}{2}}c.\operatorname {Cos} .{\frac {1}{2}}(A+B)}}\right\}^{2}&\cdot {\frac {\operatorname {Sin} .{\frac {1}{2}}(a+b)\operatorname {Cos} .{\frac {1}{2}}(a-b)}{\operatorname {Sin} .{\frac {1}{2}}(A+B)\operatorname {Cos} .{\frac {1}{2}}(A-B)}},\\{\frac {\operatorname {Sin} .c}{\operatorname {Sin} .C}}=\left\{{\frac {\operatorname {Sin} .{\frac {1}{2}}C.\operatorname {Sin} .{\frac {1}{2}}(a+b)}{\operatorname {Sin} .{\frac {1}{2}}c.\operatorname {Cos} .{\frac {1}{2}}(A-B)}}\right\}^{2}&\cdot {\frac {\operatorname {Sin} .{\frac {1}{2}}(a-b)\operatorname {Cos} .{\frac {1}{2}}(a+b)}{\operatorname {Sin} .{\frac {1}{2}}(A-B)\operatorname {Cos} .{\frac {1}{2}}(A+B)}},\\{\frac {\operatorname {Sin} .c}{\operatorname {Sin} .C}}=\left\{{\frac {\operatorname {Cos} .{\frac {1}{2}}C.\operatorname {Cos} .{\frac {1}{2}}(a-b)}{\operatorname {Cos} .{\frac {1}{2}}c.\operatorname {Sin} .{\frac {1}{2}}(A+B)}}\right\}^{2}&\cdot {\frac {\operatorname {Sin} .{\frac {1}{2}}(a-b)\operatorname {Cos} .{\frac {1}{2}}(a+b)}{\operatorname {Sin} .{\frac {1}{2}}(A-B)\operatorname {Cos} .{\frac {1}{2}}(A+B)}},\\{\frac {\operatorname {Sin} .c}{\operatorname {Sin} .C}}=\left\{{\frac {\operatorname {Cos} .{\frac {1}{2}}C.\operatorname {Sin} .{\frac {1}{2}}(a-b)}{\operatorname {Sin} .{\frac {1}{2}}c.\operatorname {Sin} .{\frac {1}{2}}(A-B)}}\right\}^{2}&\cdot {\frac {\operatorname {Sin} .{\frac {1}{2}}(a+b)\operatorname {Cos} .{\frac {1}{2}}(a-b)}{\operatorname {Sin} .{\frac {1}{2}}(A+B)\operatorname {Cos} .{\frac {1}{2}}(A-B)}}~;\\\end{aligned}}$