384 Mr Glaisher, Addition to a previous paper on [May 17, 



this, by the theorem, gives 



sin (/3 — 7) sin (a — S) sin (/3 + 7) sin (a + S) 

 -I- sin (7 — a) sin (/3 — S) sin (7 + a) sin (/3 + 8) 



+ sin (a - /3) sin (7 - 8) sin (a + /3) sin (7 + g) = (3), 



and the proof is, that from (2) we have 



{/3-7)(a-S) + (7-a)(^-S) + (a-^)(7-8) = 0, 

 identically, and therefore 



(/3^ - 7^) (a^ - S^) + (7^ - a') (/3^ - S^) + (a^ - /S"^) (7^ - 8^) = 0, 



identically ; and this last equation on substituting sin a, sin yS, 

 sin 7, sin 8 for a, ^, 7, S becomes 



(sin^ /3 — sin^ 7) (sin^ a — sin^ S) + (sin^ 7 — sin^ a) (sin^ /3 — sin^ 8) 



+ (sin^ a — sin^ ^) (sin^ 7 — sin^ 8) = 0, 

 which is (3). 



The equation (3) is in fact {B) of § 7 (p. 823) in the form {B'), 

 (p. 324) ; and it thus appears that this formula {B), viz. 



n (sin a) = n (sin a') + 11 (sin a") 



is an immediate corollary from (2). 



§ 14. As another example of the theorem in § 12, take the 

 formula 



sin (/3 — 7) sin (/3 — 8) sin (7 — 8) 



+ sin (7 — 0) sin (7 — 8) sin (c. — 8) 

 + sin (a — yS) sin (a — 8) sin (/3 — 8) 

 + sin (/3 — 7) sin (7 — a) sin (a — y8) = 0, 

 which gives 



sin (^ - 7) sin {^ - 8) sin (7 - 8) sin (/3 + 7) sin (/3+ S) sin (7 + 8) 

 + sin (7 — a) sin (7 — 8) sin (a — 8) sin (7 + a) sin (7 + S) sin (a + 8) 

 + sin (a -/3) sin (a - 8) sin (/3- S) sin {a + /3) sin (a + S) sin (/3 + 8) 

 + sin (y8- 7) sin (7- a) sin (a - ^8) sin (/3+ 7) sin (7 + a) sin (a + y8) 



= 0. 

 This equation may, in the notation of § 7, be written 

 sin a sin 6 (sin a' sin &' sin a' sin 6" + sin c sin (^' sin c" sin d") 

 = sin c sin d (sin a' sin h' sin c" sin (i" + sin a" sin 6" sin c sin c?'). 



