CLASS OF ALTERNANTS WITH TRIGONOMETRICAL 

 ELEMENTS. 



Bv Sir Thomas Muih, LL.I)., F.R.S. 



(1) There is an interesting trigonometrical equality which I 

 have not hithei'to seen in print, but which deserves some little 

 attention, both on its own account and by reason of some fresh 

 results in alternants to which it readily gives rise. When the 

 number of angles is 4, a, [3, y, c, say it takes the form 



A .S7// a + B sin 3a + F ros a + G cos 3a = sin 5a 



wliere A, B. F, G are symmetrical functions of a, ft, y, c, namely, 



A = — i: ros (2o + 2/3) — i: COS (2a + 2ft + 2y). 



B = :S ros 2a + COS i;2o. 



F = — 2 sin (2a + 2ft) + :s sin (2o + 2/3 + 2y). 



G =:^ sin 2a — sin :2 2a. (I) 



To establish it, a simple course is to express A sin a + F cos a as 

 an aggregate of sines, namely. 



+ sin (o + 2/3 + 2y) 

 + sin (a + 2/3 + 2g) 

 + sin (o + 2y + 2c) 

 + sin (2/3 4- 2y + 2c - a), 



— sin (3a + 2/3) 



— sin (3o + 2y) 



— sin (3a + 2ci) 



— sin (a + 2/3 + 2y) 



— sin (a + 2/3 + 2c) 



— sin (a + 2y + 2^), 



and B sin 3a + G cos 3a in similar fashion, namely, 



sin 5a + sin (3a + 2/3) + sin (3a + 2y) + sin (3o + 2c) 

 -I- f^hi („ _ 2/3 — 2y — 2c). 



On addition the fifteen sines reduce to the one forming tlie right- 

 hand member of the equality. 



(2) Since A, B, F, G involve all the angles symmetrically, the 

 equality must hold when a is replaced by any one of the other 

 angles. W^e thus have a set of four equations involving A, B, F, 

 G : and viewing the latter as unknowns, we obtain, on solving for 

 them in the usual wav 



sin 3a sin 5a cos a cos 3a 



sin 3/3 sin 5/3 cos ft cos 3ft 



sin 3y sin 5y cos y cos 3y 



sin 3c sin 5o cos c cos 3c 



sin a sin 5a cos a cos 3a \ 

 sin a sin 3a cos a cos 3a | 



- A = V cos (2a + 2/3) 

 + ^ cos (2a + 2/3 + 2y), (II) 



= B = i: COS 2a + cos ^ 2a. (Ill) 



