34 



Ellery W. Davis 



In particular, 



tan -1 oo i = 71-/2 — tan -1 00 = tan -1 ( 00^0). 



Thus, in fig. 32, letting 



<£" == tan" 1 BB' OB, <f>' = tt/2, 



we construct <£ as indicated. The red elements A and B, the 

 numerators of the tangent ratios, evidently belong to the same 



Fig. 32. 



elliptic line through the origin and making the angle </> with the 

 ;r-axis. 



One can write, of course, 



but this will give 



tan -1 ki = 6 + tan -1 .??, 



tan0 = 



(k — s)i 

 1 — ks 



so that 6 can be real only when s = k or i/k. This requires to 

 be an integral multiple of 71-/2. 



The functions cos 6 and sin 6, as well as the argument 6, can 

 be expressed as ratios of areas. 1 We take, in fact, for the de- 

 nominator of each ratio the triangle (o, o), (o, 1), (1,0). The 

 numerator of the cosine ratio is then (o, o), P, (o, 1) ; of the sine 

 ratio it is the triangle (0,0) ( 1, o), P, of 6 it is the complex sector, 

 whose initial line is the .r-axis and whose terminal line is (o, o)P. 



The ratio of all areas being unaltered by parallel projection, we 



1 Haskell, M. W., Bulletin Math. Soc, I, ISS- 



34 



