£20 A NEW and UNIVERSAL SOLUTION 



Now, fin 2p fin (<p — n) -f cof 2<p cof (<p — ») — cof (<p + ;;) : 

 therefore 



cof Op + a) X cof 2<f/ = cof 20 X cof (<p — ff), 

 whence, as <p is now known, cof 2<J/ will be found by this pro- 

 portion, 



cof O + n) : cof (<p — n) : : cof 2<p : cof 2«k 

 Having thus found both the angles <p and ^, their difference 

 will give the angle » which is fought. 



In order to render the method of computation, derived from 

 the preceding analyfis, as fimple and as commodious for prac- 

 tice as the nature of the fubjeft will permit, we fhall change the 

 letters <p and <J> to denote the double of what they have hitherto 

 done ; and we fhall alfo put m for its equal 20 : This being ob- 

 ferved, we have the following rule : 



1. Let there be formed this cubic equation 



«* 1 I 1 J\ fin 70 



x* 4- (- — 1) x — , 



from which x is to be found : then * - fin f 



2. State this proportion, 



cof'^:cof^::cof>:cof+ > 



by which the angle ^ will be found. 



Then <p — ^ — 2v — arch of eccentric anomaly. 



This rule would be rigorous and exact if we could give to e 



fin 



711 (Jt, 



the value it has in the formula e = s X : and if, by 



means of the rule, we. compute a feries of arches,/', p', p" ', &c. 



tin £- 



by fucceffively fubftituting for e, the values ?, s X 2 



