sinAB · cosCB + sinCB · cosAB r .
If AB = A, BC = B, and the radius r = 1, sin(A + B) = sinA · CosB + sinB · cosA; which is the known formula for the sine of the sum of two arcs, to the radius 1.
Again, if through O we draw the diameter DE perpendicular to Aa, then will DC be the complement of (AB + BC). Draw Cp, the sine of DC = cos(AB + BC). Through B draw the diameter Bb; from b, draw the sines bz, br, of the arcs bc, bE respectively, and join z, r. Then by describing two circles, one on bO as diameter, the other on OC, it may be proved as before that the circle described on bO passes through the points z and r, and that the circle described on CO passes 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 bO · 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 bz = sinBC; hence by substitution, bO · zr = cosAB · cosBC – sinAB · sinBC. By using the same notation as before, we have cos(A + B) = cosA · cosB – sinA · sinB r = (if r = 1) cosA · cosB – sinA · sinB, which is the known formula for the cosine of the sum of two arcs.
The same construction will answer for the two remaining cases: for if we suppose that bE and bc are two arcs, then will cE be their difference, and zr the sine of cE, as proved above; hence zr (= sin(bE – bc)) = br · Oz – Or · bz bO . But br = sinbE, and Or = its cosine; and bz = sine bc, and Oz = its cos., hence if bE be denoted by a, bc by b, and Ob as before, then will sin(a – b) = sina · cosb – sinb · cosa r = (if r = 1) sina · cosb – 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 xy or sin(A + B) = cos(a – b) = sinA · cosB + sinB · cosA r ; and as sinA or AB = cosBD = cosbE, Ox = cosA or AB = sinBD = sinbE, By = bz = sinbc, and Oy = Oz = cosbc, we shall have, by substitution, cos(a – b) = cosa · cosb + sina · sinb r , = (if r = 1) cosa · cosb + sina · sinb .
From what has been said it appears, that if A and B be any two arcs, of which A is the greatest, then
Sin(A ± B) = sinA · cosB ± sinB · cosA r ;
Cos(A ± B) = cosA · cosB ∓ sinA · sinB r .
When the radius r is supposed = 1, the denominators in these formulæ disappear. In the latter, A and B are used for a and b, for the sake of homogeneity. The propriety of this is manifest; for as a and b denote two indefinite arcs, the same reasoning will apply to A and B, as to a and b, the first being supposed in each case the greatest.