TRIGONOMÉTRIE SPHÉRIQUE.
Recherches sur les quadrilatères, tant rectilignes
que sphériques, inscrits au cercle ;
Par
M. Guéneau d’Aumont, professeur, secrétaire
et conservateur de l’observatoire de la faculté des sciences de Dijon.
≈≈≈≈≈≈≈≈≈
§. I. Quadrilatère rectiligne.
Soient
les côtés consécutifs d’un quadrilatère rectiligne inscrit au cercle ; soient
les deux diagonales, la première se terminant aux sommets des angles
et l’autre aux sommets des angles ![{\displaystyle (b,c),(a,d).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/567615226941ab4ee4ddb65b1f52e571f0bc2d38)
Par la nature du quadrilatère inscrit, on aura
![{\displaystyle {\begin{aligned}\operatorname {Cos} .(a,d)&=-\operatorname {Cos} .(b,c)={\frac {a^{2}+d^{2}-x^{2}}{2ad}}=-{\frac {b^{2}+c^{2}-x^{2}}{2bc}},\\\\\operatorname {Cos} .(a,b)&=-\operatorname {Cos} .(c,d)={\frac {a^{2}+b^{2}-y^{2}}{2ab}}=-{\frac {c^{2}+d^{2}-y^{2}}{2cd}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fae453d9a9ed6ce746f9d8e908f6137eb38f9bed)
donc
![{\displaystyle {\begin{aligned}&{\frac {a^{2}+d^{2}-x^{2}}{ad}}+{\frac {b^{2}+c^{2}-x^{2}}{bc}}=0,\\\\&{\frac {a^{2}+b^{2}-y^{2}}{ab}}+{\frac {c^{2}+d^{2}-y^{2}}{cd}}=0\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0231ed159304765f5b7b268acad5e7cc3a1ff09b)
équations d’où on tire
![{\displaystyle \left.{\begin{array}{rll}x^{2}&={\cfrac {bc\left(a^{2}+d^{2}\right)+ad\left(b^{2}+c^{2}\right)}{bc+ad}}&={\cfrac {(ac+bd)(ab+cd)}{bc+ad}},\\\\y^{2}&={\cfrac {ab\left(c^{2}+d^{2}\right)+cd\left(a^{2}+b^{2}\right)}{ab+cd}}&={\cfrac {(ac+bd)(bc+ad)}{ab+cd}}\,;\end{array}}\right\}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/00783d4b21f3bfe48936dc48d3f84072ef47a08e)
(1)
et par suite
![{\displaystyle xy=ac+bd,\qquad {\frac {x}{y}}={\frac {ab+cd}{bc+ad}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1c878449b5e6ad3f690b24d45d64c009286b38c)
(2)
Si, dans l’expression
![{\displaystyle \operatorname {Cos} .(a,d)={\frac {a^{2}+d^{2}-x^{2}}{2ad}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f645c47a3589b1a41de464fb1e646d1f0da99ad)
on met pour
la valeur que nous venons d’obtenir, il viendra, toutes réductions faites,
![{\displaystyle \operatorname {Cos} .(a,d)={\frac {a^{2}+d^{2}-b^{2}-c^{2}}{2(bc+ad)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50e9f924671f1c16e61d255fe5b5d9c4e8cf295a)
or, on a
![{\displaystyle 2\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=1-\operatorname {Cos} .(a,d),\quad 2\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=1+\operatorname {Cos} .(a,d)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41bb753f0b199bb56a884e4465361f30e233dec7)
donc, en substituant et divisant par ![{\displaystyle 2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/993d53082bc05c266933af9a892e1ce2128547cd)
![{\displaystyle {\begin{array}{rll}\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)&={\cfrac {(b+c)^{2}-(a-d)^{2}}{4(bc+ad)}}&={\cfrac {(b+c+d-a)(a+b+c-d)}{4(bc+ad)}},\\\\\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)&={\cfrac {(a+d)^{2}-(b-c)^{2}}{4(bc+ad)}}&={\cfrac {(a+c+d-b)(d+a+b-c)}{4(bc+ad)}},\,;\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4d888ce57f9cfdf4d6b008aa822de889fd253b1)
mais on a d’ailleurs
![{\displaystyle \operatorname {Sin} .(a,d)=2\operatorname {Sin} .{\frac {1}{2}}(a,d)\operatorname {Cos} .{\frac {1}{2}}(a,d)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10c207e303542e965371714576f17d3989a635d0)
substituant donc et posant, pour abréger,
![{\displaystyle {\begin{aligned}&b+c+d-a=A,\\&c+d+a-b=B,\\&d+a+b-c=C,\\&a+b+c-d=D\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f16281d7b530b89e1f25712228a9693aaf56c33)
il viendra finalement
![{\displaystyle \operatorname {Sin} .(a,d)=\operatorname {Sin} .(b,c)={\frac {\sqrt {ABCD}}{2(bc+ad)}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/04f6e183415f6924088a4b55e2c7fa7531ebcb11)
(3)
L’aire du quadrilatère dont il s’agit étant la somme des aires de deux triangles, dont l’un a pour ses trois côtés
et l’autre pour les siens
en représentant ce quadrilatère par
on aura
![{\displaystyle Q={\frac {1}{2}}bc\operatorname {Sin} .(b,c)+{\frac {1}{2}}ad\operatorname {Sin} .(a,d)={\frac {1}{2}}(bc+ad)\operatorname {Sin} .(a,d)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e813f973ca93b0f0375bf48c7603e3afa5eae934)
c’est-à-dire, en substituant,
![{\displaystyle Q{\frac {1}{4}}{\sqrt {ABCD}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/43e13edaf64e7201a38287e47badbf98b053573f)
(4)
Si, dans cette expression, on suppose
elle devient
![{\displaystyle {\frac {1}{4}}{\sqrt {(a+b+c)(b+c-a)(c+a-b)(a+b-c)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c359cd02272199e1642ca35b1f66cdd80a35498)
qui est précisément celle de l’aire d’un triangle, en fonction de ses trois côtés[1].
Le cercle auquel est inscrit le quadrilatère dont il s’agit se trouvant en même temps circonscrit au triangle dont les trois côtés sont
en représentant par
le rayon de ce cercle, il est aisé de voir qu’on aura
![{\displaystyle R={\frac {x}{2\operatorname {Sin} .(a,d)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2187dae66f7a61f19e3e3b5a6c451abf4dbf67cd)
mettant donc pour
et
les valeurs déterminées ci-dessus, et posant, pour abréger,
![{\displaystyle (ab+cd)(ac+bd)(bc+ad)=K,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7adc74f63d6f1f0dadbe3ba9c0043b52857b554)
il viendra
[2]![{\displaystyle \qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0a9ceb3f51a3855999d6bbee5f3b6a8d54ade22)
(5)
Si ensuite, dans cette formule, on suppose
on retombe
sur l’expression connue du rayon du cercle circonscrit à un triangle en fonction de ses trois côtés.
La plupart des résultats auxquels nous venons de parvenir sont connus depuis long-temps. Si donc nous les reproduisons ici, c’est uniquement dans la vue d’en déduire et de leur comparer ceux qui vont faire présentement le sujet de nos recherches.
§. II. Quadrilatère sphérique.
Soient
les côtés consécutifs d’un quadrilatère sphérique, inscrit à un cercle de la sphère ; soient
les deux diagonales de ce quadrilatère ; la première étant celle qui se termine aux sommets des angles
et la seconde celle qui se termine aux sommets des angles
Les cordes de ces quatre côtés et de ces deux diagonales, qui sont
![{\displaystyle 2\operatorname {Sin} .{\frac {1}{2}}a,\ 2\operatorname {Sin} .{\frac {1}{2}}b,\ 2\operatorname {Sin} .{\frac {1}{2}}c,\ 2\operatorname {Sin} .{\frac {1}{2}}d,\ 2\operatorname {Sin} .{\frac {1}{2}}x,\ 2\operatorname {Sin} .{\frac {1}{2}}y,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb9a6691320dd55e3b84fecd1a2982c4b3accf9d)
sont les quatre côtés et les deux diagonales d’un quadrilatère rectiligne inscrit au même cercle ; d’où il suit qu’on pourra les substituer à la place de
respectivement, dans les formules (1) précédemment obtenues. On aura ainsi, toutes réductions faites,
![{\displaystyle \left.{\begin{aligned}&\operatorname {Sin} .^{2}{\tfrac {1}{2}}x={\tfrac {\left(\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}d\right)\left(\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b+\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Sin} .{\frac {1}{2}}d\right)}{\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d}},\\\\&\operatorname {Sin} .^{2}{\tfrac {1}{2}}y={\tfrac {\left(\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}d\right)\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)}{\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b+\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Sin} .{\frac {1}{2}}d}}\,;\end{aligned}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2ff518b779a7854d493e1a4a0d6be653eb7092b)
(I)
d’où ensuite
![{\displaystyle \left.{\begin{aligned}\operatorname {Sin} .{\frac {1}{2}}x\operatorname {Sin} .{\frac {1}{2}}y&=\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}d,\\\\{\frac {\operatorname {Sin} .{\frac {1}{2}}x}{\operatorname {Sin} .{\frac {1}{2}}y}}&={\frac {\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Sin} .{\frac {1}{2}}d}{\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d}}\end{aligned}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/021e809c541ad1432ec6d5eb261a85ed85e21f39)
(II)
Dans le triangle sphérique dont les trois côtés sont
on a
![{\displaystyle \operatorname {Cos} .(a,d)={\frac {\operatorname {Cos} .x-\operatorname {Cos} .a\operatorname {Cos} .d}{\operatorname {Sin} .a\operatorname {Sin} .d}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23c72cd03a3a33598c3b582656390b35d22b96ed)
d’où
![{\displaystyle {\begin{array}{rll}2\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)&=1-\operatorname {Cos} .(a,d)&={\frac {\operatorname {Cos} .(a-d)-\operatorname {Cos} .x}{\operatorname {Sin} .a\operatorname {Sin} .d}},\\\\2\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)&=1+\operatorname {Cos} .(a,d)&={\frac {\operatorname {Cos} .x-\operatorname {Cos} .(a+d)}{\operatorname {Sin} .a\operatorname {Sin} .d}}\,;\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d224e3087bdc50a062441d53a25673423f411f88)
mais, on a généralement
![{\displaystyle \operatorname {Cos} .k=-2\operatorname {Sin} .^{2}{\frac {1}{2}}k\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9b8669d712a66cf610dab2bddbd340128d51dc2)
donc
![{\displaystyle {\begin{aligned}\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)&={\frac {\operatorname {Sin} .^{2}{\frac {1}{2}}x-\operatorname {Sin} .^{2}{\frac {1}{2}}(a-d)}{\operatorname {Sin} .a\operatorname {Sin} .d}},\\\\\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)&={\frac {\operatorname {Sin} .^{2}{\frac {1}{2}}(a+d)-\operatorname {Sin} .^{2}{\frac {1}{2}}x}{\operatorname {Sin} .a\operatorname {Sin} .d}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d3ceae597a92547f55fd618e8ea66a49b50b386)
En mettant, dans ces deux formules, pour
la valeur (I) que nous avons trouvée tout à l’heure, elles deviennent, en rassemblant d’une part tous les termes multipliés par
et de l’autre tous les termes multipliés par ![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44573bf5452d58b47c9adc9dd2f354291f1b42b9)
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&{\tfrac {\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\left\{\operatorname {Sin} .^{2}{\frac {1}{2}}b+\operatorname {Sin} .^{2}{\frac {1}{2}}c-\operatorname {Sin} .^{2}{\frac {1}{2}}(a-d)\right\}+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\left\{\operatorname {Sin} .^{2}{\frac {1}{2}}a+\operatorname {Sin} .^{2}{\frac {1}{2}}d-\operatorname {Sin} .^{2}{\frac {1}{2}}(a-d)\right\}}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Sin} .a\operatorname {Sin} .d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&{\tfrac {\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\left\{\operatorname {Sin} .^{2}{\frac {1}{2}}(a+d)-\operatorname {Sin} .^{2}{\frac {1}{2}}b-\operatorname {Sin} .^{2}{\frac {1}{2}}c\right\}+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\left\{\operatorname {Sin} .^{2}{\frac {1}{2}}(a+d)-\operatorname {Sin} .^{2}{\frac {1}{2}}a-\operatorname {Sin} .^{2}{\frac {1}{2}}d\right\}}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Sin} .a\operatorname {Sin} .d}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99500165cf66389329026fb42615c9a12ab921ff)
mais on trouve facilement
![{\displaystyle {\begin{aligned}\operatorname {Sin} .^{2}{\tfrac {1}{2}}a+\operatorname {Sin} .^{2}{\tfrac {1}{2}}d-\operatorname {Sin} .^{2}{\tfrac {1}{2}}(a-d)&=2\operatorname {Sin} .{\tfrac {1}{2}}a\operatorname {Sin} .{\tfrac {1}{2}}d\operatorname {Cos} .{\tfrac {1}{2}}(a-d),\\\\\operatorname {Sin} .^{2}{\tfrac {1}{2}}(a+d)-\operatorname {Sin} .^{2}{\tfrac {1}{2}}a-\operatorname {Sin} .^{2}{\tfrac {1}{2}}d&=2\operatorname {Sin} .{\tfrac {1}{2}}a\operatorname {Sin} .{\tfrac {1}{2}}d\operatorname {Cos} .{\tfrac {1}{2}}(a+d)\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7ce7513694801bc39901176171cd7d2a241f98a)
en substituant donc, mettant dans les dénominateurs pour
et ![{\displaystyle 2\operatorname {Sin} .d,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/391ed939027eb7c1636f5da1d94649c3d99262a7)
et
et supprimant, haut et bas, le facteur
il viendra
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Sin} .^{2}{\frac {1}{2}}b+\operatorname {Sin} .^{2}{\frac {1}{2}}c-\operatorname {Sin} .^{2}{\frac {1}{2}}(a-d)+2\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Cos} .{\frac {1}{2}}(a-d)}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Sin} .^{2}{\frac {1}{2}}(a+d)-\operatorname {Sin} .^{2}{\frac {1}{2}}b-\operatorname {Sin} .^{2}{\frac {1}{2}}c+2\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Cos} .{\frac {1}{2}}(a+d)}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/617ff36cc9845fd21974c70543c88a09560e174d)
ou, en changeant respectivement, dans les numérateurs,
et
en
et
,
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .^{2}{\frac {1}{2}}(a-d)+2\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Cos} .{\frac {1}{2}}(a-d)-\left(1-\operatorname {Sin} .^{2}{\frac {1}{2}}b-\operatorname {Sin} .^{2}{\frac {1}{2}}c\right)}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\left(1-\operatorname {Sin} .^{2}{\frac {1}{2}}b-\operatorname {Sin} .^{2}{\frac {1}{2}}c\right)-\left\{\operatorname {Cos} .^{2}{\frac {1}{2}}(a+d)-2\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Cos} .{\frac {1}{2}}(a+d)\right\}}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14dada38bcc87ca68cdf7a00c95ac0921988438f)
mais
![{\displaystyle \left(1-\operatorname {Sin} .^{2}{\tfrac {1}{2}}b-\operatorname {Sin} .^{2}{\tfrac {1}{2}}c\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21071ab6470c54a5e4779edd339e094fdceebf99)
![{\displaystyle =\left(1-\operatorname {Sin} .^{2}{\tfrac {1}{2}}b\right)\left(1-\operatorname {Sin} .^{2}{\tfrac {1}{2}}b\right)-\operatorname {Sin} .^{2}{\tfrac {1}{2}}b\operatorname {Sin} .^{2}{\tfrac {1}{2}}c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9188ccf77ff296e271fdae5c784cedba5ca81134)
![{\displaystyle =\operatorname {Cos} .^{2}{\tfrac {1}{2}}b\operatorname {Cos} .^{2}{\tfrac {1}{2}}c-\operatorname {Sin} .^{2}{\tfrac {1}{2}}b\operatorname {Sin} .^{2}{\tfrac {1}{2}}c\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d11b9e1ca35a6a9ab762483fe94d788f19ece57)
donc, en substituant
![{\displaystyle \operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)={\frac {\left\{\operatorname {Cos} .^{2}{\frac {1}{2}}(a-d)+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\right\}^{2}-\operatorname {Cos} .^{2}{\frac {1}{2}}b\operatorname {Cos} .^{2}{\frac {1}{2}}c}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3eb797a9dd0483c32c98d734397a69b163724464)
![{\displaystyle \operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)={\frac {\operatorname {Cos} .^{2}{\frac {1}{2}}b\operatorname {Cos} .^{2}{\frac {1}{2}}c-\left\{\operatorname {Cos} .^{2}{\frac {1}{2}}(a+d)-\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\right\}^{2}}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/921e6d8fe1893a327836d9d0cea618ede0a46e85)
Le numérateur de la première fraction se décompose en ces deux facteurs
![{\displaystyle {\begin{aligned}\operatorname {Cos} .{\tfrac {1}{2}}(a-d)+\operatorname {Cos} .{\tfrac {1}{2}}b\operatorname {Cos} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}b\operatorname {Sin} .{\tfrac {1}{2}}c&=\operatorname {Cos} .{\tfrac {1}{2}}(a-d)+\operatorname {Cos} .{\tfrac {1}{2}}(b+c),\\\\\operatorname {Cos} .{\tfrac {1}{2}}(a-d)-\operatorname {Cos} .{\tfrac {1}{2}}b\operatorname {Cos} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}b\operatorname {Sin} .{\tfrac {1}{2}}c&=\operatorname {Cos} .{\tfrac {1}{2}}(a-d)-\operatorname {Cos} .{\tfrac {1}{2}}(b+c),\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dcece73feeee1f6dd85e457f34c7b2e0acd0204)
et le numérateur de la seconde en ces deux-ci :
![{\displaystyle {\begin{aligned}\operatorname {Cos} .{\tfrac {1}{2}}b\operatorname {Cos} .{\tfrac {1}{2}}c-\operatorname {Sin} .{\tfrac {1}{2}}b\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Cos} .{\tfrac {1}{2}}(a+d)&=\operatorname {Cos} .{\tfrac {1}{2}}(b+c)+\operatorname {Cos} .{\tfrac {1}{2}}(a+d),\\\\\operatorname {Cos} .{\tfrac {1}{2}}b\operatorname {Cos} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}b\operatorname {Sin} .{\tfrac {1}{2}}c-\operatorname {Cos} .{\tfrac {1}{2}}(a+d)&=\operatorname {Cos} .{\tfrac {1}{2}}(b-c)-\operatorname {Cos} .{\tfrac {1}{2}}(a+d)\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2814363decbe5f2ca3de0b7c76f1e6843b73fe03)
de sorte qu’on a
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&\qquad {\frac {\left\{\operatorname {Cos} .{\frac {1}{2}}(a-d)+\operatorname {Cos} .{\frac {1}{2}}(b-c)\right\}\left\{\operatorname {Cos} .{\frac {1}{2}}(a-d)-\operatorname {Cos} .{\frac {1}{2}}(b+c)\right\}}{4\left(\ \operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\ \operatorname {Sin} .{\frac {1}{2}}a\ \operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&\qquad {\frac {\left\{\operatorname {Cos} .{\frac {1}{2}}(b+c)+\operatorname {Cos} .{\frac {1}{2}}(a+d)\right\}\left\{\operatorname {Cos} .{\frac {1}{2}}(b-c)-\operatorname {Cos} .{\frac {1}{2}}(a+d)\right\}}{4\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8b7647a33d01d004dba04bbfb38aac3948c0e16)
or, on a
![{\displaystyle {\begin{aligned}\operatorname {Cos} .{\tfrac {1}{2}}(a-d)+\operatorname {Cos} .{\tfrac {1}{2}}(b-c)&=2\operatorname {Cos} .{\tfrac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\tfrac {1}{4}}(a+c-b-d),\\\operatorname {Cos} .{\tfrac {1}{2}}(a-d)-\operatorname {Cos} .{\tfrac {1}{2}}(b+c)&=2\ \operatorname {Sin} .{\tfrac {1}{4}}(a+b+c-d)\ \operatorname {Sin} .{\tfrac {1}{4}}(b+c+d-a),\\\operatorname {Cos} .{\tfrac {1}{2}}(b+c)+\operatorname {Cos} .{\tfrac {1}{2}}(a+d)&=2\operatorname {Cos} .{\tfrac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\tfrac {1}{4}}(b+c-a-d),\\\operatorname {Cos} .{\tfrac {1}{2}}(b-c)-\operatorname {Cos} .{\tfrac {1}{2}}(a+d)&=2\ \operatorname {Sin} .{\tfrac {1}{4}}(d+a+b-c)\ \operatorname {Sin} .{\tfrac {1}{4}}(c+d+a-b)\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/902874e2c45c0d02221dabd8cb845b5958094df5)
donc finalement
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}.\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdf9c26c2e6adf1975bcb5e40b1e678453f32b35)
En faisant, pour abréger,
![{\displaystyle {\begin{aligned}&\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)=M\,;\\\\&\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)=N\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52a96d3646cd5f4f6309b258ab4fe56b5f139d0b)
auquel cas
et
seront des fonctions symétriques des quatre côtés
et en se rappelant que
![{\displaystyle \operatorname {Sin} .(a,d)=2\operatorname {Sin} .{\frac {1}{2}}(a,d)\operatorname {Cos} .{\frac {1}{2}}(a,d),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0691eb4f57ea9b41f5cd096f727c8107e090e12f)
on aura
![{\displaystyle \operatorname {Sin} .(a,d)={\frac {2{\sqrt {MN}}}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}\,;\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d2783282e7eddd7ae1689110c15cc793f08eeeb)
(III)
d’où l’on voit qu’ici les angles opposés ne sont pas supplément l’un de l’autre, comme dans le quadrilatère rectiligne inscrit.
En changeant respectivement
en
il vient
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(b,c)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(b,c)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4531a859d17c46fef6793a1e2131269b63592308)
mais on a
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\operatorname {Sin} .{\frac {1}{2}}(a,d)\operatorname {Cos} .{\frac {1}{2}}(b,c)+\operatorname {Cos} .{\frac {1}{2}}(a,d)\operatorname {Sin} .{\frac {1}{2}}(b,c),\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\operatorname {Cos} .{\frac {1}{2}}(a,d)\operatorname {Cos} .{\frac {1}{2}}(b,c)-\operatorname {Sin} .{\frac {1}{2}}(a,d)\operatorname {Sin} .{\frac {1}{2}}(b,c)\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bc28d2f5e96edc6823a198b6409ea1b5d61c0e6)
en substituant donc, et posant, pour abréger,
![{\displaystyle \operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c\operatorname {Cos} .{\frac {1}{2}}d=P,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4eb3c4fe760d4d77ba286b8865180f6f29f12d3b)
il viendra
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\\&{\frac {\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)+\operatorname {Sin} .{\frac {1}{4}}(c+d-a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)}{\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d}}{\sqrt {\frac {M}{P}}},\\\\&\operatorname {Cos} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)-\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)}{\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d}}{\sqrt {\frac {N}{P}}}\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/940367206abac2fe70dd20c7eac1a66acb6afe98)
cela revient à
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\\&{\frac {\left\{\operatorname {Cos} .{\frac {1}{2}}(a-d)-\operatorname {Cos} .{\frac {1}{2}}(b+c)\right\}+\left\{\operatorname {Cos} .{\frac {1}{2}}(b-c)-\operatorname {Cos} .{\frac {1}{2}}(a+d)\right\}}{2\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)}}{\sqrt {\frac {M}{P}}},\\\\&\operatorname {Cos} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\\&{\frac {\left\{\operatorname {Cos} .{\frac {1}{2}}(a+d)+\operatorname {Cos} .{\frac {1}{2}}(b+c)\right\}-\left\{\operatorname {Cos} .{\frac {1}{2}}(a-d)+\operatorname {Cos} .{\frac {1}{2}}(b-c)\right\}}{2\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)}}{\sqrt {\frac {N}{P}}}\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2ec2c1fb2653fc4d74895c98b2c93cc1659572e)
d’où, par un développement ultérieur,
![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=+{\sqrt {\frac {M}{P}}},\quad \operatorname {Cos} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=-{\sqrt {\frac {N}{P}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4686fbc6ffb542ce9f2bd71672dff140ca89bca4)
Ces deux fonctions étant symétriques, il en résulte que
![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\operatorname {Sin} .{\frac {1}{2}}\left\{(a,b)+(c,d)\right\}=+{\sqrt {\frac {M}{P}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55490390d1d024beafc52042263480d68a97bd0a)
![{\displaystyle \operatorname {Cos} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\operatorname {Cos} .{\frac {1}{2}}\left\{(a,b)+(c,d)\right\}=-{\sqrt {\frac {N}{P}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d700ad3e0c58e535e1c98e359e2e479f8e0a302f)
et, par conséquent
![{\displaystyle (a,d)+(b,c)=(a,b)+(c,d)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1cc2344f1b875f13ba38d82e8d2601bc93e97be)
c’est-à-dire que, dans tout quadrilatère sphérique inscrit, la somme de deux angles opposés est égale à la somme des deux autres angles ; propriété que le quadrilatère rectiligne inscrit partage, au surplus, avec le quadrilatère sphérique, avec cette différence seulement que, dans le premier, ces deux sommes sont constantes, tandis que, dans ce dernier, au contraire, elles sont variables.
Si l’on désigne par
l’aire du quadrilatère, l’aire du triangle sphérique trirectangle étant prise pour unité, on aura, comme l’on sait,
![{\displaystyle Q=(a,b)+(b,c)+(c,d)+(a,d)-2\varpi ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8abd3ad37f591f79df52bec1585f9a86cf15a01b)
ou bien, par ce qui vient d’être dit,
![{\displaystyle Q=2\left\{(a,d)+(b,c)-\varpi \right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06b5e5b8afb9c703776f400f94cbc226431cb23c)
donc
![{\displaystyle {\begin{array}{rll}\operatorname {Sin} .{\frac {1}{4}}Q&=-\operatorname {Cos} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}&={\sqrt {\frac {N}{P}}},\\\\\operatorname {Cos} .{\frac {1}{4}}Q&=+\operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}&={\sqrt {\frac {M}{P}}},\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3e1ef4c1c75afd63acce72c1d05db028d075b8f)
et de là encore
![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}Q=2\operatorname {Sin} .{\frac {1}{4}}Q\operatorname {Cos} .{\frac {1}{4}}Q={\frac {2{\sqrt {MN}}}{P}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddf9cc24f4d9b3da8b9fd057e770403e23470aa0)
![{\displaystyle \operatorname {Tang} .{\frac {1}{4}}Q={\frac {\operatorname {Sin} .{\frac {1}{4}}Q}{\operatorname {Cos} .{\frac {1}{4}}Q}}={\sqrt {\frac {N}{M}}}\,;\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8a3c7ae5d6fb994d3b28fa2a504557d7f82aa28)
(IV)
fonctions qui sont toutes symétriques.
Si l’on suppose
le quadrilatère devient un triangle ; de sorte qu’en représentant par
l’aire du triangle sphérique dont les trois côtés sont
on a
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .{\frac {1}{4}}T=\\&{\sqrt {\frac {\operatorname {Sin} .{\frac {1}{4}}(a+b+c)\operatorname {Sin} .{\frac {1}{4}}(b+c-a)\operatorname {Sin} .{\frac {1}{4}}(c+a-b)\operatorname {Sin} .{\frac {1}{4}}(a+b-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&\operatorname {Cos} .{\frac {1}{4}}T=\\&{\sqrt {\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c)\operatorname {Cos} .{\frac {1}{4}}(b+c-a)\operatorname {Cos} .{\frac {1}{4}}(c+a-b)\operatorname {Cos} .{\frac {1}{4}}(a+b-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&\operatorname {Sin} .{\frac {1}{2}}T=\\&{\sqrt {\frac {\operatorname {Sin} .{\frac {1}{2}}(a+b+c)\operatorname {Sin} .{\frac {1}{2}}(b+c-a)\operatorname {Sin} .{\frac {1}{2}}(c+a-b)\operatorname {Sin} .{\frac {1}{2}}(a+b-c)}{2\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&\operatorname {Tang} .{\frac {1}{4}}T=\\&{\sqrt {\operatorname {Tang} .{\tfrac {1}{4}}(a+b+c)\operatorname {Tang} .{\tfrac {1}{4}}(b+c-a)\operatorname {Tang} .{\tfrac {1}{4}}(c+a-b)\operatorname {Tang} .{\tfrac {1}{4}}(a+b-c)}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/262d59b45da152acb480809b738d85bea6629ce9)
formules connues, dont la dernière est due à M. Lhuilier, de Genève.
Si nous désignons par
l’arc de grand cercle qui joint le pôle du cercle auquel notre quadrilatère est inscrit à l’un quelconque des points de sa circonférence, cet arc aura pour sinus le rayon même de ce cercle, de sorte que, pour obtenir
il ne s’agira que de changer, dans la formule (5) précédemment obtenue,
![{\displaystyle a,b,c,d,R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f11e43d14c0be4cd367c5a50aa1e05c2848436c5)
respectivement en
![{\displaystyle 2\operatorname {Sin} .{\tfrac {1}{2}}a,\ 2\operatorname {Sin} .{\tfrac {1}{2}}b,\ 2\operatorname {Sin} .{\tfrac {1}{2}}c,\ 2\operatorname {Sin} .{\tfrac {1}{2}}d,\ \operatorname {Sin} .R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d17d00a0d9a54ad2fb557eff0b29ad352ac129ef)
Posant donc, pour abréger,
![{\displaystyle {\begin{aligned}\operatorname {Sin} .{\tfrac {1}{2}}b+\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}d-\operatorname {Sin} .{\tfrac {1}{2}}a&=\operatorname {Sin} .A,\\\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}d+\operatorname {Sin} .{\tfrac {1}{2}}a-\operatorname {Sin} .{\tfrac {1}{2}}b&=\operatorname {Sin} .B,\\\operatorname {Sin} .{\tfrac {1}{2}}d+\operatorname {Sin} .{\tfrac {1}{2}}a+\operatorname {Sin} .{\tfrac {1}{2}}b-\operatorname {Sin} .{\tfrac {1}{2}}c&=\operatorname {Sin} .C,\\\operatorname {Sin} .{\tfrac {1}{2}}a+\operatorname {Sin} .{\tfrac {1}{2}}b+\operatorname {Sin} .{\tfrac {1}{2}}c-\operatorname {Sin} .{\tfrac {1}{2}}d&=\operatorname {Sin} .D\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a56ec70f8a8ad276be4b3040c2b60bcbded846a4)
![{\displaystyle \left(\operatorname {Sin} .{\tfrac {1}{2}}a\operatorname {Sin} .{\tfrac {1}{2}}b+\operatorname {Sin} .{\tfrac {1}{2}}c\operatorname {Sin} .{\tfrac {1}{2}}d\right)\left(\operatorname {Sin} .{\tfrac {1}{2}}a\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}b\operatorname {Sin} .{\tfrac {1}{2}}d\right)\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d263ba85e8744ac3465fcdbcbe7e39796a43a5c)
![{\displaystyle \left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)=\operatorname {Sin} .K,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/095857b5005590e73ee699b92a25162533f7cd73)
il viendra
![{\displaystyle \operatorname {Sin} .R=2{\sqrt {\frac {\operatorname {Sin} .K}{\operatorname {Sin} .A\operatorname {Sin} .B\operatorname {Sin} .C\operatorname {Sin} .D}}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b26f0bb2aee8ef5f6826d159e0103dd82f48e23)
(V)
Il est presque superflu d’observer que les résultats que nous venons d’obtenir, en dernier lieu, s’appliquent littéralement à l’angle tétraèdre inscrit au cône droit.
Dijon, le 20 décembre 1821.