218 A NEW and UNIVERSAL SOLUTION 



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

 " vide the femicircle into two equal parts.'' 



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

 formula, 



— ^ n n cof « 

 ' "" fin* cof*' 

 fuppofing n 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 fhall eafily derive this equation, 



cof/? fin n 



' cof q fin q' 

 Further, let the ftraight line FQ^be drawn to bifect the 

 angle HFK : put <p = angle FCM, and ^ = angle KFQj= 

 angle QFH : then, fmce it has already been fhewn, that 4 =; 

 angle FHA, it is obvious that * — <p — -^ and q — <p -j- 4* : whence 

 we fhall have thefe two equations, for determining the two 

 angles pand •$/, 



fin n cof n 



~~ fin (<p — ^) cof (<f> — $)' 



cof n fin n 



~ cof (<p + ^) fin (<p + ^)* 

 By taking away the denominators, and remarking, that 

 fin (<p -f $) x cof (<p -f- ,/,) - ±. fm (29 + 2-40 and fin (<p — $) 

 X cof (<p — ^) - l fin. (29 — 2^), there will refult 



- 1111(29— 2^) = fin«Xcof(p — -4/) — cof«xfm(<p — 4,), and 



- fin (29 -f 21//) zr cof « x fin (9 + 4/) — fin n x cof (<p -f- -^). 



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

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



2 the 



