Inegalitate cu tangenta
Posted: Tue May 12, 2009 10:16 am
Sa se arate ca \( \tan\ \left(\frac{\pi}{4n}\right)\ \le\ \frac{1}{n}\ ,\ (\forall)\ n\ \in\ \mathbb{N}\ ,\ \ n\ \ge\ 1. \)
\( \mathrm{tg}\ \frac{\pi}{4n}\ <\ \frac{\pi}{4n}\ \cdot \ \frac{1}{\cos \ \frac{\pi}{4n}}\ \le \ \frac{\pi}{4n}\ \cdot \ \frac{1}{\cos \ \frac{\pi}{8}}\ =\ \frac{\pi}{4n}\ \cdot \ \frac{1}{\frac{\sqrt{2+\sqrt{2}}}{2}}\ <\ \frac{\pi}{4n}\ \cdot \ \frac{1}{\frac{\pi}{4}}\ =\ \frac{1}{n}\ ,\ (\forall)n\ge 2 \)
Daca \( n=1 \), avem egalitate.