£20 A NEW and UNIVERSAL SOLUTION 



Now, fin 2ip fin ((p — «) + cof 2(p cof ((p — «) — cof (ip + ") '• 

 therefore 



cof C<f> + ") X cof 24/ z= cof 2<p X cof {<p — «), 

 whence, as (p is now known, cof 2 4' will be found by this pro- 

 portion, 



cof ((p + «) : cof ((p — «) : : cof 2(p : cof 24'. 



Having thus found both the angles (p and 4', their difference 

 will give the angle v 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 fubjedl will permit, we Ihall change the 

 letters (p and 4' to denote the double of what they have hitherto 

 done ; and we fhall alfo put m for its eqxial 2« : This being ob- 

 ferved, we have the following rule : 



1. Let there be formed this cubic equation 



, , /I \ fin ff2 



from which x is to be found : then x zz Cm (p. 



2. State this proportion, 



^(p 4- ?!t r<P ''^ r /• I 



col : cof-!- : : col (p : cof 4', 



2 2 * 



by which the angle ^ will be found. 



Then (p — 4' = 2v =z arch of eccentric anomaly. 



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



^" "1 



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



?-(ro — ^) 



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



fin^^^^^ 



by fuccelfively fubflituting for e, the values s, e X ■ — 



I (;;/ - .a)' 



i X 



