TRANSACTIONS OF SECTION A. 477 



the same problem lias been considered by Lindemann. I think, however, that a 

 purely geometrical method of lookuig at the question ma}' he of interest. 



The most general displacement of a rigid body is a rotation about an axis 

 combined with a rotation about the polar axis with regard to the absolute. These 

 two rotations form the unit of displacement. My problem is the determination of 

 the single imit of displacement, which is equivalent to the joint effect of two 

 displacements, all being small. This, it will be observed, includes every problem 

 of the composition of forces or rotations in non-EucUdian space. 



Each of the component imits involves a pair of conjugate polars, A,A' and B,B', 

 and we require to find a pair of conjugate polars C,C' the rotations around which 

 shall be equivalent to the given rotations about A,A' and B,B'. 



Draw the common transversals X and X' to the four rays A,B,A',B', then it 

 can easily be shown that the effect of the given displacements on four points 

 P Q R S on X will move those points on right of Imes directed towards P' Q'' 

 R' S' on X', so that the anharmonic ratio of P Q R S is equal to that of P' Q' 

 R'S'. 



On X there are two critical points, L and M, which are characterised by the 

 circumstance that they start in the same direction whether the displacement be 

 A, A' or B, B'. It is therefore necessary that C, 0' shall be such as to start L and 

 M in the same direction. This condition will enable C and C to be determined. 



Let the two given displacements convey L and M to L' and M', then and C 

 are two generators of the hyperboloid of which X, X', and L' M', are three genera- 

 tors of the other system. But when two hyperboloids are such that a pair of 

 generators of one system on one hyherboloid are conjugate polars of the other, 

 then a pair of generators of the other system are also conjugate polars. Obser- 

 ving that X and X' are conjugate polars of the absolute we therefore have C and 

 C completely determined. 



9. On the deduction of Trigonometrical from Elliptic Function Formulae. 

 By J. W. L. Glaisheb, M.A., F.B.8. 



In any elliptic function identity, connecting sn's, en's, and dn's, we may, of 

 course, as is well known, put k = 0, when the sn's and en's become respectively 

 sines and cosines and the dn becomes unity. But we may also expand the elliptic 

 functions in powers of /c^, and equate the coefficients of k'^, k*, Szc, to zero. 

 Considering only terms as far as k^, it can be shown that 



am u = u — \ k'^u + ■^k'' sin 2m 

 so that . . 



sn M = sin u — i k'^u cos u + -^ k- sin 2m cos m 

 en M = cos u + ^ k^u sin it — ^ A;' sin 2u sin u 

 dn M = 1 — ^k''' s\v? u 



Now the terms - ^ k-u cos u and ^ Fm sin u, in which the argument appeal's out- 

 side, will generally lead to terms which, on this account, are separately equal 

 to zero, so that in deducing trigonometrical from elliptic function formulae (in 

 which the arguments do not appear as external factors), we may put 



sn M = sin M (1 + 5 k^ cos'^ m) ^ 



en M = cos M (1 — ^k"^ sin'^ m) ^ . . . (1) 



duM = l — ^k'^ain'u ) 



and equate the powers of k"^ to zero. 



As an example, consider the elliptic function identity 



sn ^ sn y sn (/3 — y) + sn y sn a sn f y — a) + sn a sn /3 sn (a — ^) 

 + sn O — y) sn (y — a) sn (a — /3) =0 ; 



putting k = 0, we obtain the well-known trigonometrical identity 



sin /3 sin y sin (/3 — y) + sin y sin a sin (y — a) + sin a sin ^ sin (a - j3) 

 + sin (^ - y) sin (y — a) sin (a — /3) = 0, . . . (2"^ 



