262 Mr. Adams on Logarithmic and Circular hy [Ocr. 
1 ala | 
=]-— 7=— w+ ut — w + &e.; and ae 
ra utu + w+ qe 
— + - —_—- ++ - — &c. Subtract the second of these equa- 
ut 
tions from the first, and we get 0 = 1 — (w Bf -) ae (us ae -.) 
a8 (w + 5) + &c.; by transposition and dividing by 2, we 
1 1 1 
1 ut+— aha avers eT ee 
u u 
~ = — — &e. 
have 5 3 ee Pao > + &c. Then by 
De Moivre’s theorem, + = cos. uw — cos. 2 u + cos.3 u — cos. 
Ab UF CRESS RMN NEE. w 0 win}iayo.ese ia win, no n'a a: wcnis aelieye Riad lee (5) 
The same as equation (2), but independent of logarithms or dif- 
ferentiation. 
In like manner, — =l+utv+wuv+ut + &c. And 
LOTS EES Pe ae a eo! We &c. Subtract the second of 
—u+l u u? u3 ut 
these equations from the first, transpose and divide by 2, we 
1 ut 4 we u3 4 LY 
shall then have Ss ape Enmore = + &c. 
2 2 
Therefore 1 = — (cos.u + cos. 2u + cos.3 u + cos. 4u 
He fees JK pw’ aii in tobe. 'dle bn botal sibsbe: 5 ssilnig'n che nyse; 8h 040k ninicne sala paul (6) 
By taking the successive differential coefficients of equation 
(5), we have 
0= —sin.u + 2 sn. 2u—3 sn.3u+ 4 sin. 4u— &e. 
Q0= + sin.u — 2sin.2u + 33 sin.3 u — 4 sin. du + &e. 
Q0-= — sin.uw + 2’sin.2u— 3>sin. 3 u + 4 sin. 4u— Ke. 
0 = + sin.w — 27sin.2u + 37sin.3 u — 47 sin. 4u + &e. 
0. 
Bis 
eerre eee ee es ee eee eee eee se ee ets eeeseoseeeeeses Oe es eeessee 
ne TOL hee ee) ERR ES Bee i) 
0 = — cos.u + 22 cos.2 u — 32 cos.3 u + 42 cos.4u— Ke. 
0 = + cos. u — 24 cos. 2u + 3% cos.3 u — 4+ cos.4u4+ &e. 
0 = —cos.u + 2° cos.2u — 3% cos.3 u + 45 cos. 4 u— Ke. 
0 = + cos. u — 2% cos. 2 u + 3% cos. 3 u — 48 cos. 4u+ &e. 
evrecesr eee eee se se ee eee ese eseeeeseeee esses set se Oeeeseeeseesesn 
O=cos. u—2’" cos. 2 u+3°" cos. 3 u—4*" cos. 4u+ke.. .. (8) 
We shall have from equation (7), when uw = 90°, 
=—-1+3 —-5 +7 —9 + &e. 
O0O=+1—-3° +5 —7 + 9 — &e. 
=—14+3-—-5 + 7>— 9 + &c. 
=+1]-—37 + 57— 774 97 — &e. 
—e>) 
