306 On the Parallelogram of Forces. 
with the direction of 2, which equals 24; then z evidently bisects the 
angle 24, formed by the directions of R and S§, it also equals the sum 
of their components which act in its direction; hence and by (1), we 
_ have meee? <. 2=(R+8) . o(4)? =z(6)= the resultant re- 
solved in the direction of x; by resolving S in the direction of @, 
and adding R which acts in that direction, we have R+S)(2d)= 
the components resolved in the direction of 2, which must equal the 
resultant resolved in the same direction; hence we shall have 
(R+S) .¢(¢)?=R+89(28), or since R=S we shall have 29(6)? = 
1+¢(26), (3); which must evidently be an identical equation. 
It is manifest by (1), that if 6=0, (4) and 9(2¢) will each =135 
hence supposing 9(4) to be converted into a series, arranged accord- 
ing to the ascending powers of 4, its first term must = 1, and the 
powers of 6 must he positive, for should any of them be negative, 
(4) would be infinite when 6=0, instead of being = 1, as we have 
proved it must be; .°. o(8) must be of the form, o(8)=1-- A084 
Bs? +Cé°+D32-+, &c. (4), and by changing @ into 24, we shall 
have the expression for ¢(2d) ; by substituting the values of (4), o(20) 
in (3), we have 2(1+4 Aj7-+, &c.)? =242%. Ab?-+, &c., or 4AS% 
+ 2(A262¢+4 2B32) + 4(ABM+ 4 Coe) + 2(B2g2% 4+ 2DIe+ 
QACH240) 4, &e.= 24, Ad2 +2", Bir 13°. C9° +24, Dee +, Be. 
(5). Since (5) is to be identical, (so that 6 may be indeterminate,) 
it is evident that the coefficients of 6%, must be equal; ... 4=27%, 
which gives a=2, but A remains undetermined ; by substituting the 
value of a, and comparing the next higher powers of 4, we have 
2(A26*+2B)?)=2° Bd’, which requires that 6=4, .°. A?+2B 
= 8B, or oS in the same way we find c=6, d=8, &c. C= 
ae aia: oe PETA and so on. It is evident that (4) must gen- 
erally be less than 1, .*. put A= — a and we have Bay5 5 C= 
os ba D= a —___, and so on, where the law of contin- 
2.3.4.5.6 2.3.4.5.6.7.8 
uation is manifest; hence by substituting the values of a, 6, &c., A, 
hp2 62 hp364 kgs 
spies) 
B, &c., in (4), we have 9(9) 5 oT saaze te 
which is the well known expression for cos. kd; hence 9(8)=cos. ké, 
which substituted in (1) gives, =z cos. kd. ‘To determine f, let 
