66 U. §, COAST AND GEODETIC SURVEY 
Evaluating these quantities we have 
sin (p—m)r __& COS (p—m)x 
@—m)=—0 7 19 \@—m)—o 
sin tm | _@ cos ta, | om 
SUN Sag Ploy aay COST aa. “Guiness 
In (275) it will be noted that when the integers p and m each equal 
, p=m 
sin ee 
n 
and 
> m must be an even number, and therefore cos nz is positive, while 
COS 7 1S negative. 
196. Assuming the condition that p and m are equal integers, each 
less than ~, we have by substituting (274) in (270), (271), and (272), 
Oy) 
a=(02=1) : a=(n—1) | 4 
De sin @ pwn Mm a— >) sina mi w— 4 (276) 
a=o0 a=0 
a=(n—1) a=(n—1) 
Dy) cos ar pw costa m a— — >a cost aim u— sin (277) 
a=0 a=0 
ae (n—1) ) a=(n-—1) , 
by sin @ p wicos am w—— >) sin am Wicostarmar—O0 (278) 
a=0 a=90 
197. Assuming the condition that p and m are each equal to 3 
we have by substituting (274) and (275) in (270), (271), and (272), 
a=(n-—1) 
sin? a m u=4n+3n cos r=0 (279) 
a=o 
a=(n—1) 2 
cos? a m u=4 n—}4 nN COS T=N (280) 
a=0 
a=(a—1) 
sin am u cos am u=0 (281) 
a=0 
198. Returning now to the solution of (258), by substituting the 
successive values of a from 0 to (n—1), we have 
h=H,+C; cos 0+ C, cos 04-_______- +0; cos 0 
+S, sin 0+8S, sin 0+ _______- +S; sin 0 
h,=H,+C, cos u+C, cos 2u+_______- +0, cos ku 
+S, sin ut S, sin 2u+______-- +S, sin lu 
hz =H,+ C0, cos 2u+C, cos 4u+_____-_- +0, cos 2ku 
JESh sin PHB SS sin Zs +§, sin 2lu/ (282) 
hea-y=H,+C; cos (n—1)u+C, cos 2(n—1)u+ ____-- 
+0, cos (n—1)ku 
+S, sin (n—1)u+ 8) sin 2(n—1)u-+-------- 
+S; sin (n—1)lu 
