22 Mr. Adams on the Sines of the Sum [JuLy, 
(3 and 30.3.¢e.) bisect at right angles the chords E B, E D, 
B D, in the points m, v, m; then wili (31.3 . e .) the triangle 
A BE be similar to C m E, and the triangle A D E similar to 
C vE; therefore (4.6.e.) A B is the double of Cm, and A D 
the ‘double of C v. 
We then have in fig. 10, by Prop. D, Simson’s Euclid, 
AS x DE +4bAcDix Bi =i AVE, Tools deagetshk (a) 
Or, 
2Cm x 2Ev+2Cvx 2Em=2 x 2 Bn (radius unity.) 
Or, 
Cm+Ev+CvxEm=Bn=Dn. 
Therefore, 
cos. Et sin. Er + cos. Ersin, Et = sin. Bw = sin. rt = sin. 
(Er+Et?). 
To find the Sine of the Difference of two Arcs. 
We have in fig. 2, by Prop. D, ibid. 
ADxBE+AExBD=AB~x DE. 
Therefore, 
ABxDE-—-ADxBE=AE~x BD. 
From whence, see equation (a), we have 
cos. E¢ sin. Er — cos. Ersin. Et = sin. B w = sin. r f= sin. 
(Er— Et). 
If from A and E, the extremities of the diameters A E, the 
perpendiculars A x, E s, be drawn to the chords B D, we shall, 
by a method equally as simple as the preceding, be able to find 
the cosines of the sum and difference of two arcs. 
For by Prop. C. 6, Simson’s Euclid, we have the following 
equations; viz.A B x AD=Awx AE,andEs x AE= 
EB x ED in both the figures, from whence we get 
AB x AD—EB x ED=AE (Ax—Es) =2Cu x AE,, fig. 1 .. (6) 
ABxAD+EB x ED=AE (Ar+ Es) =2Cn x AE, fig. 2 .. (c) 
Or, 
2Cmx2Cv—2Etx2Ev=2 x 2 Cn (radius unity). 
Or, 
Cmx Cv—Etx Ev=Cn. 
Therefore, 
cos. Et cos. Er — sin. Et sn. Er = cos. Bw = cos.rt = 
cos. (Er + Ed), by equation (6). 
In like manner, we have 
cos. Et cos. Er + sin. Et sin. Er = cos. Bw = cos. rt = 
cos. (Er + E 2), by equation (c). 
The arcs E B and E JD, as well as their halves E¢ and Er, 
are supposed to be of the same magnitude in both the figures. 
From the foregoing, we readily obtain the following equations ; 
viz. sin. (A + A) =sin. 2A = sin. Acos. A + sin. A cos. 
, A= 2sin. A cos. A. cos. (A + A) = cos. 2A =cos2A 
* Sei ie fees vs hs ted s shawls mid br Se poti's die 'We 6, Rites od e SN 
