étant toujours supposé très petit ; on a donc
![{\displaystyle 1-\sin ^{2}\mathrm {CRF} =\varphi ^{2}\cos ^{2}v\sin ^{2}\mathrm {CRF} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f44c2962504f56aa645b76590dc00e6d0a6dd009)
ce qui donne
![{\displaystyle \sin \mathrm {CRF} =1-{\frac {1}{2}}\varphi ^{2}\cos ^{2}v.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f14ceef77b5775ca621bc5bfde6780279053f0c)
Soit
![{\displaystyle \mathrm {CRF} =90^{\circ }-\gamma ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af6a54f4d0efe8d3e7d26407b410e05fd4cd35f7)
on aura
![{\displaystyle \sin \mathrm {CRF} =1-{\frac {\gamma ^{2}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a3fce42312af0ae4ae59a69b4afb7cb08d44a17)
Donc
![{\displaystyle \gamma =\varphi \cos v.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f52c3d87b798cae9c3b5a5158e4945c224485f7)
Partant,
![{\displaystyle \mathrm {MRF} =90^{\circ }+\gamma -\mathrm {V} \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a31e5c3cc069b0a56af90b9ef5eb32fbb9c7967)
or on a
![{\displaystyle \cos \mathrm {CRF} :\cos \mathrm {MRF} ::\cot v:\cot \mathrm {RM} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/41f315950ae8152d33f30bbd5519231f139931a1)
ou
![{\displaystyle \varphi \cos v:\sin(\mathrm {V} -\varphi \cos v)::\cot v:\cot \mathrm {RM} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0cd03257acf687f7046f29808863ac555dc9118)
Partant,
![{\displaystyle \cot \mathrm {RM} ={\frac {\cot v\sin(\mathrm {V} -\varphi \cos v)}{\varphi \cos v}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd28d08413b20679988267edf55337c959448667)
donc
![{\displaystyle dv={\frac {dz\sin(\mathrm {V} -\varphi \cos v)}{\sin \mathrm {V} }}=dz-\varphi dz\cos v\cot \mathrm {V} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03e405a87ffed4c5d176088bd4372bb65608243a)
2o Si l’on fait varier l’angle
on a
![{\displaystyle dv=-{\frac {d\varphi \sin v}{\operatorname {tang} \mathrm {V} }}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b142cc2da01e78f7a7a33bcabd5d6970bf5c1cb)
donc
![{\displaystyle dv=dz-{\frac {\varphi dz\cos v+d\varphi \sin v}{\operatorname {tang} \mathrm {V} }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/675f8f3f2539592dfad2fa345c50e92a264382d1)
LXIII.
Je suppose que l’on veuille déterminer la position de l’équinoxe pour un instant donné ; pour cela, je le rapporte au plan fixe, en abaissant des points
et
(fig. 4) les arcs
et
perpendiculaires sur
On aura très sensiblement
et
mais si l’on