328 G. B. AIRY, ESQ. SUPPLEMENT TO A PROOF OF THE THEOREM 



21. The division can be effected without difficulty by retaining the equation in the form 

 which I have given. But the subsidiary equations are much abbreviated by first multiplying 

 the equation by sin 9. Assume then that 



sinO .it^ + p. sind .of + q. sin .x^ + r. sin 9 .01^ + s .sin 9 ,01* + t . sin 9 .0^ 

 + V .sin 9 . ai^ + w .sind .w + X .sin 9 

 is equal to 



{a!^-2p.cos9 .x+f^] X {sin9.a^+P.a/^+Q.x*+R.a^+S.ar'+T.a!+V\ + Remainder: 



then upon actually performing the multiplication, and comparing coefficients of similar powers 

 of d?, we have the following equations : 



P m p.sin29 + p .sin9; 

 Q - 2pP .coa9~p^.sin9 + q.sin9i 

 R = 9,pQ . cos - |0=P + r . sin ; 

 S '=ZpR .cos9 - p^Q-\- s .sin9; 

 T=2pS. cos 9 - p^R + t .sin 9; 

 V=2pT. cos9 - p^S + v. sin 9; 

 and 



Remainder = (2jo7. cos 9 - p'^T + w .sin9) x + (- p'V + z . sin 9\ 



22. Solving the equations in order (constantly substituting each value in the two 

 following equations), we obtain, 



P = |0 . sin 20 + p . sin 9. 



Q = p* . sin 39 + pp . sin Z9 + q . sin 9. 



R = p^ . sin 40 + p'-p . sin 39 + pq . sin 29 + r . sin 9. 



S = p*. sin 59 + p^p . sin 4-9 + p^q . sin 39 + pr. sin 29 + s. sin 9. 



T = p^ . sin 69 + p*p . sin 59 + p'q . sin 40 + pV . sin S9 + ps . sin 29 + t . sin 9. 



V = p^ . sin 79 + p^p . sin 69 + p*q . sin 59 + f?r . sin 40 + p^s . sin 39 + pt . sin 29 + v . sin 9. 



Then substituting the values of T and V in the expression for Remainder, we find 

 Remainder = 

 {p' . sin 80+|O^p . sin 70+jo'g' . sin 60+|oV. sin 59 + p^s .sin 4!9+pH . sin S0+|O«.sin 20+2f;.sin0} x.v 

 + {-jo' . sin 79- pp . sin 69-p^q . sin 59-p^r . sin 40-p*s . sin 39-p^t . sin 29-p^v . sin 0+af . sin 0|. 



23. Use these symbols, for convenience, 



A-p^. cos 89 +p''p. cos 79 +p^q, cos 69+ p^r. cos 59+ p*s. cos 4-9+p^t. cos 39+ p%. cos 29+pto.cos9+x, 

 B=p^ . sin 80+p'p . sin 79+p^q . sin 69+p^r , sin 59+p'^s . sin4:9+pH . sin39+p^v . sin 29+pw . sin ; 

 then it is easily seen that 



Remainder = — x .r + {^ . sin - 5 . cos 0}. 

 P 



And now removing the multiplier sin which was introduced in article 21, 



Remainder of the original equation-function, after dividing by a? - 2p . cos 9 .x + p^, 



1 -^ < . B „, 



= - .—. — ■?, -x. X T \A : — ^ X cos 0J. 



p sin9 sin ' 



