486 
sin RR” sin R sin | WL sin W 
a sin 2 = eee 
Bim seer ons ; 
sn kh" LR = — 7 BL 
and consequently 
sin Rk” LR 
_ sin RR" sin WL sin ue 2sin NR cos NR sin WL sin W 
sn RL sin R’L sn RL sin R’L x 
or the expression is 
2 cos NR sinWLsnW sin NW 
~~ sin BR, RLsinR’L cos mas WR cos R'U' sin U’. 
But we have 
U'= sin NW 
SPOON iter yee 
and therefore 
sin NW cos R’U’ 
/ U’ = 
VR R'U' sin sin? NW —— Se 
* sin? NW COS VU- WR’) 
cos WR’ sin WU" 
= sin? NW (tan WR'+ cot WU’). 
But we have cot WU'= cot NW cos W, and the expression 
becomes 
WL W 
_2 cos — = eo 2N (tan WR’+ cot NWeos W): 
which is the expression previously found as the value of the 
left-hand side of the equation, and the theorem is therefore 
proved. 
‘It is obvious that the point Z might have been con- 
structed by taking on &’W, produced in the direction from 
R’ to W, a point K such that 
sin? NW 
ban TOW te toe yg sin RW sin R"W nM Ee 
and then taking the are KL in the reverse direction equal to 
90°. 
«¢ Passing now to the optical problem, it will be recollected 
