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 $\alpha,$  $\beta,$  $\gamma$ 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 $A^{*} + B^{*} + C^{*}$ is equal to two right angles, the excess of the sum $A + B + C$  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].