Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/46

and also

$$N-M'=\psi .$$

Then the equation just found can be thrown into the following form:

$${\begin{alignedat}{1}{\sqrt {E}}.dp.\sin(M'&-\omega +\psi )+{\sqrt {G}}.dq.\sin(M'+\psi )\\&={\sqrt {E'}}.dp'.\sin M'+{\sqrt {G'}}.dq'.\sin(M'+\omega ')\end{alignedat}}$$

or

$${\begin{alignedat}{1}{\sqrt {E}}.dp.\sin(N'&-\omega -\omega '+\psi )-{\sqrt {G}}.dq.\sin(N'-\omega '+\psi )\\&={\sqrt {E'}}.dp'.\sin(N'-\omega ')+{\sqrt {G'}}.dq'.\sin N'\end{alignedat}}$$

And since the equation evidently must be independent of the initial direction, this direction can be chosen arbitrarily. Then, setting in the second formula ${\textstyle N'=0,}$ or in the first ${\textstyle M'=0,}$ we obtain the following equations:

$${\begin{aligned}{\sqrt {E'}}.\sin \omega '.dp'&={\sqrt {E}}.\sin(\omega +\omega '-\psi ).dp+{\sqrt {G}}.\sin(\omega '-\psi ).dq\\{\sqrt {G'}}.\sin \omega '.dq'&={\sqrt {E}}.\sin(\psi -\omega ).dp+{\sqrt {G}}.\sin \psi .dq\end{aligned}}$$

and these equations, since they must be identical with

$${\begin{alignedat}{2}dp'&=\alpha \,dp&&+\beta \,dq\\dq'&=\gamma \,dp&&+\delta \,dq\end{alignedat}}$$

determine the coefficients ${\textstyle \alpha ,}$ ${\textstyle \beta ,}$ ${\textstyle \gamma ,}$ ${\textstyle \delta .}$ We shall have

$${\begin{aligned}\alpha &={\sqrt {\frac {E}{E'}}}.{\frac {\sin(\omega +\omega '-\psi )}{\sin \omega '}},&\beta &={\sqrt {\frac {G}{E'}}}.{\frac {\sin(\omega '-\psi )}{\sin \omega '}}\\\gamma &={\sqrt {\frac {E}{G'}}}.{\frac {\sin(\psi -\omega )}{\sin \omega '}},&\delta &={\sqrt {\frac {G}{G'}}}.{\frac {\sin \psi }{\sin \omega '}}\end{aligned}}$$

These four equations, taken in connection with the equations

$${\begin{aligned}\cos \omega &={\frac {F}{\sqrt {EG}}},&\cos \omega '&={\frac {F'}{\sqrt {E'G'}}},\\\sin \omega &={\sqrt {\frac {EG-F^{2}}{EG}}},&\sin \omega '&={\sqrt {\frac {E'G'-F'^{2}}{E'G'}}},\end{aligned}}$$

may be written

$${\begin{aligned}\alpha {\sqrt {E'G'-F'^{2}}}&={\sqrt {EG'}}.\sin(\omega +\omega '-\psi )\\\beta {\sqrt {E'G'-F'^{2}}}&={\sqrt {GG'}}.\sin(\omega '-\psi )\\\gamma {\sqrt {E'G'-F'^{2}}}&={\sqrt {EE'}}.\sin(\psi -\omega )\\\delta {\sqrt {E'G'-F'^{2}}}&={\sqrt {GE'}}.\sin \psi \end{aligned}}$$

Since by the substitutions

$${\begin{alignedat}{2}dp'&=\alpha \,dp&&+\beta \,dq,\\dq'&=\gamma \,dp&&+\delta \,dq\end{alignedat}}$$