—36.—

As an example of the type shown in Theorem IX, expand

& _2{(: 3 _}’»9;(3) il Ja Sile)

n kB 9o(2)  Ja Folz)

1 3 3
2¢° sine (1 - ¢*® II(1 +2¢3™3 4 ¢*"2) [I(1 - 20" cos 22 4 ¢*")
-.'T—-__‘-_-"——-‘-——-"—'———-——m__————“—‘
24° 4 II(I: Q") I1(1 + 2¢®" 4 ¢*%) II(1 - 2¢*"~%cos2z 4 ¢*"*%)
(t = O+ ¢*™Y)? (1 - %) II(1 - ¢3¢

e A e e e & e et whene t= ‘tfi.
2 II(]. + qan)e H(l = tiq2n—1) 11(1 = t-aqan-x)

Set H(1+¢* ™" )=(1+q+404 " )A+C+C+6% 40+ ***)(1 47204 B svo)eee
Here s'y=1 , $'05=0, s'oap=1, *eccce, st =s%y=0,
Set M1 ~t*"") =(1=C~t%*~0= """ )(1=0=0=0=t%* =0~ sere)ourer .
flere 8"; =0, s == 1)), eeese i g wisliiw 0,

For all the separate products we obtain the following s's:

5'1 =1 = S'OC‘ 23'00004 = cesens
3%03 O ® =% = s"ees = S"occoes = U
3"t o= T ("t—?).’ = " co0s =-5"cowes = °*CCC
sivOJ #1 = 3i'ocoj = si'mmd = esesen
s'1'. = ("ta)‘ = S'wi = s'ccoci = esesne
"“‘ - ("t‘z)‘ = S“m¢ = sv‘ccoc‘ = ereser

In accordance with Theorem IX, let 8= (2s' 4+ s" 4 s™ ) = (2s1Y 4 sV 4 s¥4),
Using the above table of s's, we find

sg=2+t*4t™® = 4cos’z =~ scq

832 2~1*~ 1t = 4 5in® Rz =~ s¢p
g2 +t%4 ¢t = 4cos®3z=~ 5.5
In general, s, == Son= Seon == Sccm = Sewocn = = Sececen = 20000

and all the other s's are zerc.

e alsc find by direct wultiplication that 5t = 48, ~ 35 ,
s,® = 15s, -—63',”-& &0y s;sq = 8, +28; -sg . If we let the C's be
related o;;url.y;tfék‘tfieAi‘hjfinit.e product—duotient without the factor
Fst Rt

--§--- , We obtain

1)