Introductio in analysin infinitorum

AC DIVISIONE ANGULORUM. 201 CAP. XIV. $\frac{3}{\sin 3z} = \frac{1}{\sin z} + \frac{1}{\sin(120^{\circ}+z)} + \frac{1}{\sin(240^{\circ}+z)} = \frac{1}{\sin z} + \frac{1}{\sin(60^{\circ}-z)} - \frac{1}{\sin(60^{\circ}+z)}$. $\sin 3z = -4 \sin z \sin(120^{\circ}+z) \sin(240^{\circ}+z) = 4 \sin z \sin(60^{\circ}-z) \sin(60^{\circ}+z)$. Erit ergo, uti jam supra notavimus, $\sin(60^{\circ}+z) = \sin z + \sin(60^{\circ}-z)$, & $3 \operatorname{cosec} 3z = \operatorname{cosec} z + \operatorname{cosec}(60^{\circ}-z) - \operatorname{cosec}(60^{\circ}+z)$. EXEMPLUM II. Ponamus esse $n=5$, atque prodibunt hæ æquationes: $0 = \sin z + \sin(\frac{2}{5}\pi+z) + \sin(\frac{4}{5}\pi+z) + \sin(\frac{6}{5}\pi [TRUNCATED]