Practical  Method  of  Harmonic  Analysis.  35 
Fig.  1. 
Hence,  if  the  full  wave  or  C^  be  represented  by 
C1  =  a1  sin(W  —  Oi)  +as  sin  3(fttf—  ^)  +  a5  sin  5(W  —  05)  -f&c, 
its  third  component  C3  is 
=  ~"  92  [°1  Sm  (3®  * ~~  ^l)  +  a3  Sm  3  (3ft)/ — #3) 
+  a5  sin  5 (3ct)/—  -05)  +  &c.], 
and  its  fifth  component  C5  is 
^2  [«i  sin  (5ft)/ — #j )  +  a3  sin  3  (5ft)/  —  $s)  +  a5  sin  5  (5&)/  —  0-o)  +  &c] , 
and  so  on;  but  by  definition  C3  and  C5  are  also  given  by 
C3  =  6i3  sin 3 (cot — 03j  +a9  sin9(o>/— 09)  +  ... 
C5  =  a5  $m5{<0t  —  05)+ai5  sin  15(W  —  015)  + ... 
Hence,  identifying  the  expressions  for  the  same  components, 
we  lind  that 
ax=  —  32<33  =  52a5=  —  72a7  =  tfcc, 
so  that 
n           r*    t    4     a\      3in(3arf—  #,)       sin(5a>f— ft)       -     "1 
0!  =  ^  |j?iii(a>/— ^)~  32 ~  +  v       -—  &0.J. 
D 
