![{\displaystyle \operatorname {Tang} .{\frac {1}{2}}s=\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eec3b604412de2a1c68eec6ea648de1f6b11776)
(
xxxvi)
![{\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 {Cot} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-B)\right\}\operatorname {Cot} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-C)\right\}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75464c7a9f2c71a6609245ae5b1f8058564af916)
Par les mêmes formules (5), et par la formule (XXIX), on a
![{\displaystyle {\begin{aligned}&-\operatorname {Sin} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)={\sqrt {\frac {1-\operatorname {Sin} .S}{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 {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&+\operatorname {Cos} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)={\sqrt {\frac {1+\operatorname {Sin} .S}{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 {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11696e3318b532a80a61245069d2c8e3a195f820)
comparant ces formules aux formules (29) et (30), on pourra leur donner cette forme
(XXXV)
(XXXIV)
d’où encore
![{\displaystyle -\operatorname {Tang} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adb87dedc0bcc9fbdf289f6588b0e75cb3582ecd)
[1](
XXXVI)
Par les formules (5) et par la formule (xxx), on trouve
- ↑ Si l’on désigne par
l’aire du triangle, on aura, comme l’on sait,
![{\displaystyle T=A+B+C-\varpi =2S-\varpi ,\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/de04b8a20a0be826aa73169c28e3b8401936a4f5)
d’où
![{\displaystyle \quad {\frac {1}{2}}T=-\left({\frac {1}{2}}\varpi -S\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86d9e88773b2fe3f88ca6107ebba1f838e3b73cc)
et par conséquent
![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}T=-\operatorname {Cos} .S,\quad \operatorname {Cos} .{\frac {1}{2}}T=\operatorname {Sin} .S,\quad \operatorname {Tang} .{\frac {1}{2}}T=-\operatorname {Cot} .S,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9ca47b2c1d79484cb04fd3076828af7a7cca59b)
on aura en outre
d’où
![{\displaystyle \operatorname {Sin} .{\frac {1}{4}}T=-\operatorname {Sin} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),\ \operatorname {Cos} .{\frac {1}{4}}T}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6684d92bfda877d455dcd26da1b592174c4b1455)
![{\displaystyle =\operatorname {Cos} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),\ \operatorname {Tang} .{\frac {1}{4}}T=-\operatorname {Tang} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb88d0a0758163cbf9aa3f2507f2c896c07eecb1)