Page:Joseph Louis de Lagrange - Œuvres, Tome 4.djvu/292

24. En second lieu, on aura

${\begin{aligned}&{\frac {c+b-a}{2}}\\&\quad =\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\sin a+{\frac {1}{2}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{2}\sin 2a\\&\qquad \qquad \qquad \qquad \qquad +{\frac {1}{3}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{3}\sin 3a+\ldots \\&\quad =\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\sin(b+c)-{\frac {1}{2}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{2}\sin 2(b+c)\\&\qquad \qquad \qquad \qquad \qquad +{\frac {1}{3}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{3}\sin 3(b+c)-\ldots .\end{aligned}}$

25. Enfin la cinquième équation du n^o 20, étant comparée à celle du n^o 10, donnera

$\varphi ={\frac {\beta -\gamma }{2}},\quad \omega ={\frac {\beta +\gamma }{2}},\quad x={\frac {b+c}{2}},\quad y={\frac {b-c}{2}}\,;$

donc

$\theta =-{\frac {\sin \gamma }{\sin \beta }},$

et par conséquent

${\begin{aligned}c=&{\frac {\sin \gamma }{\sin \beta }}\sin(b-c)+{\frac {1}{2}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{2}\sin 2(b-c)+{\frac {1}{3}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{3}\sin 3(b-c)+\ldots \\=&{\frac {\sin \gamma }{\sin \beta }}\sin(b+c)-{\frac {1}{2}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{2}\sin 2(b+c)+{\frac {1}{3}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{3}\sin 3(b+c)-\ldots .\end{aligned}}$

On peut aussi comparer d’une autre manière la même équation avec celle du numéro cité, en faisant

$\varphi ={\frac {b+c}{2}},\quad \omega ={\frac {b-c}{2}},\quad x={\frac {\beta -\gamma }{2}},\quad y={\frac {\beta +\gamma }{2}},$

ce qui donnera

$\theta ={\frac {\sin c}{\sin b}},$

et de là par le n^o 7

${\begin{aligned}\gamma =&{\frac {\sin c}{\sin b}}\sin(\beta +\gamma )-{\frac {1}{2}}\left({\frac {\sin c}{\sin b}}\right)^{2}\sin 2(\beta +\gamma )+{\frac {1}{3}}\left({\frac {\sin c}{\sin b}}\right)^{3}\sin 3(\beta +\gamma )-\ldots \\=&{\frac {\sin c}{\sin b}}\sin(\beta -\gamma )+{\frac {1}{2}}\left({\frac {\sin c}{\sin b}}\right)^{2}\sin 2(\beta -\gamma )+{\frac {1}{3}}\left({\frac {\sin c}{\sin b}}\right)^{3}\sin 3(\beta -\gamma )+\ldots .\end{aligned}}$