2i8 A NEPV and UNIFERSAL SOLUTION 



" eccentric point in the diameter of a femicircle, that fliall di- 

 " vide the femicircle into two equal parts.'' 



II. Let us now proceed to confider the refolution of the 

 formula, 



— ^" '^ cof « 

 ~~ fini' cof II ' 

 fuppofing « to be any angle whatever not greater than a right 

 angle. 



From the point F, draw the ftraight line FKL, fo that the 

 part KL, infcribed in the angle ACL, may be equal to CA ; and 

 put q — angle FKA. Then, by proceeding in a manner fimilar 

 to what is done in Art. 9. we fliall eafily derive this equation, 



cof « fin n 



cof q fin q' 

 Further, let the fi:raight line FQ^be drawn to bifed: the 

 angle HFK : pvit <p =. angle FQA, and \// = angle KFQj= 

 angle QFH : then, fince it has already been fliewn, that 1/ =. 

 angle FH A, it is obvious that vzz (p — ij/ and q :zz (p -}- -^ : whence 

 we fliall have thefe two equations, for determining the two 

 angles (pand ^p, 



fin « cof « 



e rr 



e iz: 



fin (ip — 4') cof (<p — 4^)' 

 cof « fin n 



cof ((p + -4/) fin (<p + -vj/)' 

 By taking away the denominators, and remarking, that 

 fin ((p + 1^) X cof (ip + -4/) iz i fin (aip + 2-4') and fin ((p — i//) 

 X cof ((p — 4') — ^ fill (2(p — 2i^), there will refult 



- fin (29 — 24') = fin«Xcof ((p — -4) — cof«xfin((p — •4')jand 



e 



- fin (2(p + 2-^) = cof n X fin ((p + 4') — fin « X cof (<p + 4')- 



Take the fum and difference of thefe two equations, and, 



for the fums and differences of the fines and co-fines, having 



2 the 



