222 W tsconstn Academy of Sciences, Arts and Letters. 



R n = R m = ^/l + E- D sin 5 cos St sin a 

 P n=:D cos 3^f cos 3^— l/l + E"- sin 5 sin a: J 



Pm— D cos 5l cos 5 +-1/I + E- sin 3 sinal 

 A,,?^:^ Pn^+A^^ = D cos 3Ll + .^'^sin2 :&-2'Vl+-E'"2 sin3cos3sina; J ' 

 A„,P=V Pm'^ + a„,m-=Dcos ^Ll+E"- sin'-* ^+2'\jl+E- sin5cos3sina: J 

 ^ A„n \1 + -EJ- sin3cosa 



Sin(5+u> 



A„p r 1 li 



Ll+^- sin- 5— 2 (1 + E-p sin^cos^sino: J 



Sin (5_y)— ^.uT^ _' \/l+^-^ sin3cos 



Ll + ^2 sin25+2 (!+£'-)* sin^cos^sina J 



o- .cv , ^ • /cv \ \1+jEJ^ /sin^3cos^a: 



Sm (3+u) sm (5— v)=^ — ■ =^ 



L(l+-E- sin25)2— 4 {l + E'^) sin^Scos^^sin^tx J' 



Expand and reduce this value for sin (3+u)sin (3 — v), and substitute 

 the result in expression (a) of this Art., 



Sin(u + v)=2 sina cosa: (sin23- + ^^ sin^^— -B'^ sin*5— ^* sin'^ + E* sin«5 

 +2 sin* 5sin^ a — 2 sin^ 3sin"a:+jE''' sin'*5 — „). 



Sin (u + v)=:2 sina (sin'-3- + -B'-* sin'^ ^—E'' sin'^^—E^ sin45+^-' sin«3— i 



sin-3 sin-a + 2 sin''^ sin^ a— 2 sin^3' sin-a + E'^ sia'^3- — „) 



tan (u+v)=3 sin a (sin^ ^+E'^ sin^ ^-E^ sin"* 3— -B* sin^ ^ + E* sin« 3--i sin* 

 3- sin^ a-\-2 sin* 3 sin^ a-f_E"= sin^ 5-"). 



The difference (4 sin^ 5 sin^ a) between tan u+v) and sin (u+v) is of the 

 order of the sixth power of eccentricity, therefore to the sixth power of 

 eccentricity, per this method of development, the sine, the arc and the 

 tangent of the angle (u-|-v) are of the same length. 



Diagram 9 is so constructed that triangles A„ P Rand A,„ P R are equal 

 to triangles like lettered in Dia. 8. In line P A,„ take P B equal to P A,„ 

 and let it be represented by unity. Produce P A,, to D, making P D equal 

 to P A,,„ and let it be represented by (1+a), Produce P D to F making 

 P F equal to (1+a)-. Complete parallograms P A,, C B, P D E B and 

 P F G B, and draw diagonals P C, P E and P G. 



