Strong’s Problems. | 425 — 3 
A: cosB+sinB. cosA ; which is the Siddin formula for ie “pd 
sine of the sum of two arcs, to the radius 1. -S 
Again, ifthrough O we draw the diameter DE perpendiet : 
lar to Aa, then will DC be the complement of (AB+BC.) 
Draw Cp, the sine of DC=cos (AB+ BC.) Through B draw 
diameter Bb; from 6, draw th@sines bz, br, of the arcs bc, 
bE respectively, and join z,r. Then by describing two cir- — 
cles, one on 4O as diameter, the other on OC, it may be prov- 
ed as before that the circle described on 6O passes through 
the points z and r, and that the circle described on CO pas- 
ses through p: and hence, by the same reasoning as before, 
zr=Cp=cos(AB+BC.) Now Obzr being a quadrilateral 
inscribed in the circle described on bO, we have (by the prop. 
before cited) bO- zr+-Or- bz=br. Oz; and hence 60: zr= 
br: Oz—Or- bz. But br= sine arc bE=sine arc BD; and 
since BD is the complement of AB, br=cosAB. In like 
manner Oz=cosBC, Or=sinAB, and 6z=sinBC; hence 
by substitution, bO- zr=cosAB: cosBC —sinAB: sinBC. By 
using the same notation as before, we have cos(A+B) 
_cosA. = sinB 
=(if r=1) cosA. cosB.—sinA. sin 
B, which is ikg known formula for the cosine of the sum 
of two ares. 
The same construction will answer for the two remaining 
cases: forif we suppose that bE and bc are two arcs, then will 
cE be their difference, and zr the sine of cE, as proved above; 
: br. Oz —Or. bz ; 
hence zr (=sin(OE —be))= b0 ~ But br=sin 
bE, and Or=its cosine; and bz=sine bc, and Oz=its cos. 
hence if bE be denoted by a, be by 6, and Ob as before, then 
sina. cosb—sinb, cosa 
r 
sinb: cosa. —... Again, AB++-BC is the complement of DC 
or cE; hence by the first part of the above investigation, 
xy=sin (AB+BC)=coscE: but zy or sin(A+B)=cos 
= = we A | 
ae abe + So ; and as sinA or AB 
will sin(a—b)= = (ifr=1) sina’ cosb — 
