Page:Annales de mathématiques pures et appliquées, 1813-1814, Tome 4.djvu/147

139

DES GRANDEURS IMAGINAIRES.

${\frac {\mathrm {\overline {KA}} }{\mathrm {\overline {KA}} }}=1,\quad {\frac {\mathrm {\overline {KB}} }{\mathrm {\overline {KB'}} }}=u^{2},\quad {\frac {\mathrm {\overline {KC}} }{\mathrm {\overline {KC'}} }}=u^{4},\quad {\frac {\mathrm {\overline {KD}} }{\mathrm {\overline {KD'}} }}=u^{6},\ldots {\frac {\mathrm {\overline {KN}} }{\mathrm {\overline {KN'}} }}=u^{2n}.$

Et, si l’on prend, sur les rayons correspondans, $\mathrm {K\beta '=K\beta } ,\mathrm {K\gamma '=K\gamma } ,$$\mathrm {K\delta '=K\delta ,\ldots }$ les longueurs $\mathrm {K\beta } ,\mathrm {K\gamma ,K\delta ,\ldots }$ étant à volonté, on aura encore

${\frac {\mathrm {\overline {K\beta }} }{\mathrm {\overline {K\beta '}} }}=u^{2},\quad {\frac {\mathrm {\overline {K\gamma }} }{\mathrm {\overline {K\gamma '}} }}=u^{4},\quad {\frac {\mathrm {\overline {K\delta }} }{\mathrm {\overline {K\delta '}} }}=u^{6},\ldots$

Si sur des rayons $\mathrm {{\overline {KA}},{\overline {KM}},{\overline {KN}},\ldots ,}$ pris pour bases, on construit des figures semblables, et que ${\overline {a}},{\overline {m}},{\overline {n}},$ soient des lignes homologues de ces figures, on aura

(C)$\qquad {\overline {m}}={\overline {a}}\times \mathrm {\overline {KM}} ,\qquad {\overline {n}}={\overline {a}}\times \mathrm {\overline {KN}} ,\ldots$

9. Soient (fig. 5) $\operatorname {Arc} .\mathrm {AB=CD} =a,\ \operatorname {Arc} .\mathrm {AC} =b\,;$ on aura (5, 6, 7)

$\operatorname {Cos} .(a+b)+{\sqrt {-1}}\operatorname {Sin} .(a+b)={\overline {\mathrm {K\delta } }}+{\overline {\mathrm {\delta D} }}={\overline {\mathrm {KD} }}={\overline {\mathrm {KB} }}\times {\overline {\mathrm {KC} }}$

$=\left({\overline {\mathrm {K\beta } }}+{\overline {\mathrm {\beta B} }}\right)\times \left({\overline {\mathrm {K\gamma } }}+{\overline {\mathrm {\gamma C} }}\right)=\left(\operatorname {Cos} .a+{\sqrt {-1}}\operatorname {Sin} .a)\right)\left(\operatorname {Cos} .b+{\sqrt {-1}}\operatorname {Sin} .b\right)$

$=(\operatorname {Cos} .a\operatorname {Cos} .b-\operatorname {Sin} .a\operatorname {Sin} .b)+{\sqrt {-1}}(\operatorname {Sin} .a\operatorname {Cos} .b+\operatorname {Cos} .a\operatorname {Sin} .b)\,;$

donc, en séparant,

$\operatorname {Cos} .(a+b)=\operatorname {Cos} .a\operatorname {Cos} .b-\operatorname {Sin} .a\operatorname {Sin} .b,$

$\operatorname {Sin} .(a+b)=\operatorname {Sin} .a\operatorname {Cos} .b+\operatorname {Cos} .a\operatorname {Sin} .b.$

Soient (fig. 6) $\mathrm {AC} =a,\ \mathrm {AB} =b,\ \mathrm {BD} ={\tfrac {1}{2}}\mathrm {BC} ={\frac {a-b}{2}}\,;$ prenons $\mathrm {AE=BD}$ et tirons $\mathrm {KD}$ et $\mathrm {BC}$ se coupant en $d\,;$ nous aurons

$(\operatorname {Cos} .a-\operatorname {Cos} .b)+{\sqrt {-1}}(\operatorname {Sin} .a-\operatorname {Sin} .b)=\left(\operatorname {Cos} .a+{\sqrt {-1}}\operatorname {Sin} .a\right)$

$-\left(\operatorname {Cos} .b+{\sqrt {-1}}\operatorname {Sin} .b\right)=\left({\overline {\mathrm {K\gamma } }}+{\overline {\mathrm {\gamma C} }}\right)-\left({\overline {\mathrm {K\beta } }}+{\overline {\mathrm {\beta B} }}\right)=\mathrm {{\overline {KC}}-{\overline {KB}}}$

$=\mathrm {{\overline {KC}}+{\overline {BK}}={\overline {BC}}} =2{\overline {d\mathrm {C} }}=$ (n.^o 8. C) $2{\overline {\epsilon \mathrm {E} }}\times \mathrm {\overline {KD}}$

$=2{\overline {\epsilon \mathrm {E} }}\times \left({\overline {\mathrm {K\delta } }}+{\overline {\mathrm {\delta D} }}\right)=2{\sqrt {-1}}\operatorname {Sin} .{\tfrac {a-b}{2}}\left(\operatorname {Cos} .{\tfrac {a+b}{2}}+{\sqrt {-1}}\operatorname {Sin} .{\tfrac {a+b}{2}}\right)$

$=-2\operatorname {Sin} .{\frac {a-b}{2}}\operatorname {Sin} .{\frac {a+b}{2}}+2{\sqrt {-1}}\operatorname {Sin} .{\frac {a-b}{2}}\operatorname {Cos} .{\frac {a+b}{2}}\,;$