Introductio in analysin infinitorum

FACTORUM TRINOMIALIUM. 121 nus ejus Arcus debet esse $= 0$, quod quidem ita patet, ut CAP. IX. hinc istos Factores a posteriori eruere licuisset. Simili modo, cum fit $\frac{e^{z\sqrt{-1}} + e^{-z\sqrt{-1}}}{2} = \cos z$, erit quoque $\cos z = (1 - \frac{4zz}{\pi\pi})(1 - \frac{4zz}{9\pi\pi})(1 - \frac{4zz}{25\pi\pi})$ $(1 - \frac{4zz}{49\pi\pi})$ &c., seu, his Factoribus in binos resolvendis, erit quoque $\cos z = (1 - \frac{2z}{\pi})(1 + \frac{2z}{\pi})(1 - \frac{2z}{3\pi})(1 + \frac{2z}{3\pi})$ $(1 - \frac{2z}{5\pi})(1 + \frac{2z}{5\pi})$ &c., ex qua pari modo patet, si fuerit $z = \pm \frac{(2k+1)}{2}\pi$, fore $\cos z = 0$, id quod etiam ex natura Circuli liquet. 159. Ex §. 152. etiam inveniri possunt Factores hujus expressionis $e^x - 2\cos g + e^{-x} = 2(1 - \cos g + \frac{xx}{1 \cdot 2} + \frac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \text{&c.})$. Transit enim haec expressio in hanc $( [TRUNCATED]