Page:Madras Journal of Literature and Science, series 1, volume 6 (1837).djvu/162

a function the error is given of the difference of longitude, by changing the formula.

In the diagram (Pl. 4 fig. 3) let A D be the meridian of the first station, B C the perpendicular of any other station, then A being the pole, D C will be the parallel of latitude of C, and A D will be the co-latitude of E, and not A B, and B D will be the error or the difference of latitude between the foot of perpendicular and the parallel.

${\begin{aligned}{\text{Take }}&\mathrm {A\ B} =90-\lambda \\&\mathrm {B\ C} =\pi \\&\mathrm {A\ C} =90-(\lambda -x)\\&\mathrm {D\ B} =x.{\text{ then by right angled spherics we have}}\\&\sin \ (\lambda -x)=\sin \ \lambda \ \cos \ \pi \\\sin \ \lambda .\ \cos \ x\ -\ &\sin \ x.\ \cos \ \lambda \ =\\&{\text{dividing by}}\ \sin \ \lambda \ {\text{we have}}\\&\cos \ x-\sin \ x.\ \cot \ \lambda =\cos \ \pi \\\end{aligned}}$

and the $\cos \ x$ being nearly $=1$, and substituting the value of $\cos \ \pi$

${\begin{aligned}&1-\sin \ x.\ \cot \ \lambda =1-2\ {\overline {\sin {\tfrac {1}{2}}}}^{2}\ \pi \\&\sin \ x.=2\ {\overline {\sin {\tfrac {1}{2}}}}^{2}\ \pi \ \operatorname {tang} \ \lambda \\\end{aligned}}$

and taking $x''$ and $\pi '$ for the sines

$x''={\tfrac {1}{2}}\pi ''^{2}.\operatorname {tang} \lambda .\sin l''$ which is the formula before given,—but the angle B A C is the difference of longitude between B and C, and in the right angled triangle A B C

${\begin{aligned}{\text{Tang }}\mathrm {A} &=\operatorname {tang} \ \pi .\ \sec \ \lambda \\\mathrm {A} ''&=\pi .\ \sec \ \lambda \ {\text{and transposing}}\\\pi ''&=\cos \ \lambda .\ \mathrm {A} ''\\&=\cos \ \lambda .\ {\text{diff. longitude}}\end{aligned}}$

substituting this in the last formula

${\begin{aligned}x''&={\tfrac {1}{2}}\ \delta ''\ ^{2}.{\overline {\cos \lambda .}}^{2}\operatorname {Tang} \ \lambda .\ \sin \ l''\\&={\tfrac {1}{2}}\ \delta ''\ ^{2}.\cos \lambda .\ \sin \ \lambda .\ \sin \ l''\end{aligned}}$

or become $\cos \ \lambda .\ \sin \ \lambda ={\tfrac {1}{2}}\ \sin \ 2\lambda$

$x''={\tfrac {1}{2}}\ \delta ''^{2}\sin .\ 2\ \lambda .\ \sin \ l''$ from this formula the following table is computed x always subtractive:—