Introductio in analysin infinitorum

218 DE MULTIPLICATIONE LIB. I. $s = \sin a + \sin(a+b) + \sin(a+2b) + \sin(a+3b) + \&c.$ & quia hæc Series est recurrens, cujus scala relationis est $2 \cos b, -1$, orietur hæc Series ex evolutione fractionis, cujus denominator est $1 - 2z \cos b + zz$, posito $z=1$. Ipsa vero fractio erit $=\frac{\sin a + z(\sin(a+b) - 2 \sin a \cos b)}{1 - 2z \cos b + zz}$, quare, facto $z=1$, crit $s = \frac{\sin a + \sin(a+b) - 2 \sin a \cos b}{2 - 2 \cos b} = \frac{\sin a - \sin(a-b)}{2(1 - \cos b)}$, ob $2 \sin a \cos b = \sin(a+b) + \sin(a-b)$. Cum autem fit $\sin f - \sin g = 2 \cos \frac{f+g}{2} \sin \frac{f-g}{2}$, erit $\sin a - \sin(a-b) = 2 \cos(a - \frac{1}{2}b) \sin \frac{1}{2}b$: & $1 - \cos b = 2(\sin \frac{1}{2}b)^2$, unde erit $s = \frac{\cos(a - \frac{1}{2}b)}{2 \sin \frac{1}{2}b}$. 259. Hinc itaque summa quotcunque Sinuum, quorum Arcus in arithmetica progressione incedunt, assignari poterit; quæratur nempe summa hujus progressionis $\sin a + \sin(a+b) + \sin(a+2b) + \sin(a+3b) + \dots\dots\dots\dots + \sin(a+nb)$. Quia summa hujus progressionis in infinitum continuatæ est $\frac{\cos(a - \frac{1}{2}b)}{2 \sin \frac{1}{2}b}$, considerentur termini ultimum sequentes in infinitum hi $\sin(a+(n+1)b) + \sin(a+(n+2)b) + \sin(a+(n+3)b) + \&c.$ quia horum Sinuum summa est $=\frac{\cos(a+(n+\frac{1}{2})b)}{2 \sin \frac{1}{2}b}$; si hæc a priori subtrahatur, remanebit summa quæsita. Scilicet, si fuerit $s = \sin a + \sin(a+b) + \sin(a+2b) + \dots\dots\dots\dots\dots + \sin(a+nb)$, erit $s = \frac{\cos(a - \frac{1}{2}b) - \cos(a+(n+\frac{1}{2})b)}{2 \sin \frac{1}{2}b} = \frac{\sin(a+\frac{1}{2}nb) \sin(\frac{1}{2}(n+1)b)}{\sin \frac{1}{2}b}$. 260. Pari