427 
Multiplying the two latter equations together, we have 
eo) +2(x+h) = eotr (p, x) cotr (h,k) + tres (p, x) tres (A, 2) 
+ tres (x, ¢) tres (A, h) 
+e {tres (x, ¢) tres (A, h) + cotr (d, x) tres (A, 4) 
+ tres (@, x) cotr (h, k)} 
+ {tres (p, x) tres (h, k) + tres(x, p) cotr (A, 2) 
+ cotr (p, x) tres (A, h)}. 
Now the left-hand member in this equation being also equal to 
cotr(pt+h, x+h)+etres(p+h, y+h)+etres(y +h, o+h), 
we may compare the similar parts of the two expressions, and 
thus get at once the three formule of which we were in search, 
Viz. : 
cotr (¢+h, x + k) = cotr (¢, x) cotr (A, k) 
+ tres (¢, x) tres (h, h) + tres (x, @) tres (h, k), 
tres (p+h, x +k) = tres (x; $) tres (A, h) 
+ cotr (, x) tres (A, &) + tres (, x) cotr (A, h), 
tres(y +h, p +h) = tres (¢, x) tres (A, 4) 
+ tres (x; p) cotr (h, &) + cotr (p, x) tres (A, A). 
From these equations, combined with (4), we obtain 
cotr (p+h, y+h) + cotr(p+ah, x+a°h) 
+cotr(¢+a°h, y+ ak) = 3 cotr(¢, x) cotr (A, 2), 
which is obviously analogous to the formula 
cos (0 + h) + cos (0 — h) = 2cos 0 cosh; 
and we might obtain similar formule for the tresines. 
‘‘ In many investigations great convenience arises from the 
peculiar way in which the functions of our new calculus are 
affected by differentiation. 
‘¢ From formule (1) or (3) we obtain the following : 
