![{\displaystyle {\begin{aligned}&\operatorname {Sin} .{\frac {1}{2}}(s-a)={\sqrt {\frac {1-\operatorname {Cos} .(s-a)}{2}}}=\\&{\sqrt {-{\frac {1+\operatorname {Sin} .^{2}{\frac {1}{2}}A-\operatorname {Sin} .^{2}{\frac {1}{2}}B-\operatorname {Sin} .^{2}{\frac {1}{2}}C-2\operatorname {Sin} .{\frac {1}{2}}A\operatorname {Sin} .{\frac {1}{2}}B\operatorname {Sin} .{\frac {1}{2}}C}{4\operatorname {Sin} .{\frac {1}{2}}A\operatorname {Cos} .{\frac {1}{2}}B\operatorname {Cos} .{\frac {1}{2}}C}}}},\\\\&\operatorname {Cos} .{\frac {1}{2}}(s-a)={\sqrt {\frac {1+\operatorname {Cos} .(s-a)}{2}}}=\\&{\sqrt {\frac {1+\operatorname {Sin} .^{2}{\frac {1}{2}}A-\operatorname {Sin} .^{2}{\frac {1}{2}}B-\operatorname {Sin} .^{2}{\frac {1}{2}}C+2\operatorname {Sin} .{\frac {1}{2}}A\operatorname {Sin} .{\frac {1}{2}}B\operatorname {Sin} .{\frac {1}{2}}C}{4\operatorname {Sin} .{\frac {1}{2}}A\operatorname {Cos} .{\frac {1}{2}}B\operatorname {Cos} .{\frac {1}{2}}C}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e571a0c665d97302081235de33348df1ada5967e)
comparant ces formules aux formules (35) et (36), on pourra leur donner cette forme
![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}(s-a)=\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6746aeda51c1ad79c7540468eb6ba7b30fa40d4d)
(
xxxvii)
![{\displaystyle {\sqrt {-{\frac {\operatorname {Sin} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)\operatorname {Cos} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}\operatorname {Sin} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-B)\right\}\operatorname {Sin} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-C)\right\}}{\operatorname {Sin} .{\frac {1}{2}}A\operatorname {Sin} .{\frac {1}{2}}B\operatorname {Sin} .{\frac {1}{2}}C}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/690235125640c648a30ef5339c5aa8e261198312)
![{\displaystyle \operatorname {Cos} .{\frac {1}{2}}(s-a)=\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/270c72bf89c82705901c5c2a7f25cc6afbf062ca)
(
xxxviii)
![{\displaystyle {\sqrt {\frac {\operatorname {Cos} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)\operatorname {Sin} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}\operatorname {Cos} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-B)\right\}\operatorname {Cos} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-C)\right\}}{\operatorname {Sin} .{\frac {1}{2}}A\operatorname {Sin} .{\frac {1}{2}}B\operatorname {Sin} .{\frac {1}{2}}C}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac167e20fb0db4ce85845fc1e65351935f031c0b)
d’où encore
![{\displaystyle \operatorname {Tang} .{\frac {1}{2}}(s-a)=\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/45aea75ab89a4b68051f1b8df42c1228a8d261ef)
(
xxxix)
![{\displaystyle {\sqrt {-\operatorname {Tang} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)\operatorname {Cot} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}\operatorname {Tang} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-B)\right\}\operatorname {Tang} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-C)\right\}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e92809b333a13b696fc70f9e5bb7f1bc53518103)
Par les mêmes formules (5) et par la formule (XXX), on trouve
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}={\sqrt {\frac {1-\operatorname {Sin} .(S-A)}{2}}}=\\&{\sqrt {\frac {-1+\operatorname {Cos} .^{2}{\frac {1}{2}}a-\operatorname {Cos} .^{2}{\frac {1}{2}}b-\operatorname {Cos} .^{2}{\frac {1}{2}}c-2\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c}{4\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c}}},\\\\&\operatorname {Cos} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}={\sqrt {\frac {1+\operatorname {Sin} .(S-A)}{2}}}=\\&{\sqrt {\frac {1+\operatorname {Cos} .^{2}{\frac {1}{2}}a-\operatorname {Cos} .^{2}{\frac {1}{2}}b-\operatorname {Cos} .^{2}{\frac {1}{2}}c+2\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}{4\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfface675d9276c943a12f879c1d56fe66643020)
comparant ces formules aux formules (31) et (32), on pourra leur donner cette forme
de sorte que les formules (XIII), (XVII), (XX), (XXIV), (XXIX), (XXXI), (XXXIV), (XXXV), (XXXVI), offrent autant de moyens d’obtenir l’aire d’un triangle sphérique en fonction de trois de ses parties. On reconnaîtra en particulier la dernière pour celle de M. Lhuilier, donnés, par M. Legendre, dans ses Élémens de Géométrie.