Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/55

$${\begin{aligned}r^{2}+r'^{2}&-(q-q')^{2}-2r\cos \phi .r'\cos \phi '-2r\sin \phi .r'\sin \phi '\\&\quad =b^{2}+c^{2}-a^{2}-2bc\cos A\\&\quad =2bc(\cos A^{*}-\cos A),\end{aligned}}$$

combined with the development of the expression

$$r\sin \phi .r'\cos \phi '-r\cos \phi .r'\sin \phi '=bc\sin A,$$

gives the following formula:

$${\begin{aligned}\cos A^{*}-\cos A=-(q-q')p\sin A{\bigl (}{\tfrac {1}{3}}f^{\circ }&+{\tfrac {1}{6}}f'p+{\tfrac {1}{4}}g^{\circ }(q+q')\\&+({\tfrac {1}{10}}f''-{\tfrac {1}{45}}{f^{\circ }}^{2})p^{2}+{\tfrac {3}{20}}g'p(q+q')\\&+({\tfrac {1}{5}}h^{\circ }-{\tfrac {7}{90}}{f^{\circ }}^{2})(q^{2}+qq'+q'^{2})+{\text{etc.}}{\bigr )}\end{aligned}}$$

From this we have, to quantities of the fifth order,

$${\begin{aligned}A^{*}-A=+(q-q')p{\bigl (}{\tfrac {1}{3}}f^{\circ }&+{\tfrac {1}{6}}f'p+{\tfrac {1}{4}}g^{\circ }(q+q')+{\tfrac {1}{10}}f''p^{2}\\&+{\tfrac {3}{20}}g'p(q+q')+{\tfrac {1}{5}}h^{\circ }(q^{2}+qq'+q'^{2})\\&-{\tfrac {1}{90}}{f^{\circ }}^{2}(7p^{2}+7q^{2}+12qq'+7q'^{2}){\bigr )}\end{aligned}}$$

Combining this formula with

$$2\sigma =ap{\bigl (}1-{\tfrac {1}{6}}f^{\circ }(p^{2}+q^{2}+qq'+q'^{2})-{\text{etc.}}{\bigr )}$$

and with the values of the quantities ${\textstyle \alpha ,}$ ${\textstyle \beta ,}$ ${\textstyle \gamma }$ found in the preceding article, we obtain, to quantities of the fifth order,

[11]

$${\begin{aligned}A^{*}=A-\sigma {\bigl (}{\tfrac {1}{6}}\alpha &+{\tfrac {1}{12}}\beta +{\tfrac {1}{12}}\gamma +{\tfrac {2}{15}}f''p^{2}+{\tfrac {1}{5}}g'p(q+q')\\&+{\tfrac {1}{5}}h^{\circ }(3q^{2}-2qq'+3q'^{2})\\&+{\tfrac {1}{90}}{f^{\circ }}^{2}(4p^{2}-11q^{2}+14qq'-11q'^{2}){\bigr )}\end{aligned}}$$

By precisely similar operations we derive

[12]

$${\begin{aligned}B^{*}=B-\sigma {\bigl (}{\tfrac {1}{12}}\alpha &+{\tfrac {1}{6}}\beta +{\tfrac {1}{12}}\gamma +{\tfrac {1}{10}}f''p^{2}+{\tfrac {1}{10}}g'p(2q+q')\\&+{\tfrac {1}{5}}h^{\circ }(4q^{2}-4qq'+3q'^{2})\\&-{\tfrac {1}{90}}{f^{\circ }}^{2}(2p^{2}+8q^{2}-8qq'+11q'^{2}){\bigr )}\\\end{aligned}}$$

[13]

$${\begin{aligned}C^{*}=C-\sigma {\bigl (}{\tfrac {1}{12}}\alpha &+{\tfrac {1}{12}}\beta +{\tfrac {1}{6}}\gamma +{\tfrac {1}{10}}f''p^{2}+{\tfrac {1}{10}}g'p(q+2q')\\&+{\tfrac {1}{5}}h^{\circ }(3q^{2}-4qq'+4q'^{2})\\&-{\tfrac {1}{90}}{f^{\circ }}^{2}(2p^{2}+11q^{2}-8qq'+8q'^{2}){\bigr )}\end{aligned}}$$

From these formulæ we deduce, since the sum ${\textstyle A^{*}+B^{*}+C^{*}}$ is equal to two right angles, the excess of the sum ${\textstyle A+B+C}$ over two right angles, namely,

[14]

$${\begin{aligned}A+B+C=\pi +\sigma {\bigl (}{\tfrac {1}{3}}\alpha &+{\tfrac {1}{3}}\beta +{\tfrac {1}{3}}\gamma +{\tfrac {1}{3}}f''p^{2}+{\tfrac {1}{2}}g'p(q+q')\\&+(2h^{\circ }-{\tfrac {1}{3}}{f^{\circ }}^{2})(q^{2}-qq'+q'^{2}){\bigr )}\end{aligned}}$$

This last equation could also have been derived from formula [6].