Dr. Brewster on the laws of polarisation, &c. 247 
one negative axis of the same intensity as either of the other 
two, and placed at right angles to the plane of the positive 
axes. 
Let A,B,C, fig. 9, be the poles of the three axes of which 
B,C are the positive axis, and A the negative axis. 
The tint produced at any point E by the axis A alone, is of 
the same intensity, and the same character, as the tint that 
would be produced at the same point by the positive axes 
B,C acting jointly. Through E draw the great circles 
AEF, BE, and CE, and call EF=.r, BE=p, EC= 0 , BF=m, 
CF= n, BEF= tt, CEF= w. Then as the tint produced at 
E by the axes A alone is Sin. 3 AE or Cos. 2 .r, we must show 
that the resulting tint produced by the two axes B,C, is also 
equal to Cos.®.r. By spherical trigonometry we have 
Sin. 2 <p x Sin.V = Sin. 2 m 
Sin. 3 0 x Sin 8 w = Sin. 8 n 
But since m -f- n = go 0 Sin.® n = Cos. a m. Hence 
Sin. 2 m-J-Sin. 9 ra = 1 
and by adding together the two first equations we have 
Sin. 3 <p x Sin.® nr »j~ Sin.® 0 x Sin. 2 w = 1. 
Again , since Tang. <p = , and Tang, x =Tang. <p x Cos. tt 
and Cos. <p — Cos. x x Cos. m, we obtain by substitution 
Tang, x 
Sin. <p x Cos. <7t 
Cos. xx Cos. m 
and 
Tang, x x Cos. x x Cos. m = Sin. q> x Cos. nt ; but since 
Tang, x x Cos. uc = Sin. x v/e have 
Sin. x x Cos. m — Sin. <p x Cos.tt, and by the same reasoning 
Sin. x x Cos. n = Sin. 6 x Cos. u. 
Hence, after squaring both equations and adding them 
together, we have 
