Page 1 of 1
Inegalitate sin pi/2n>=1/n
Posted: Thu Jan 31, 2008 2:27 am
by Tudorel Lupu
Sa se arate ca daca \( n\in\mathbb{N}^{*} \) atunci are loc:
\( \sin\frac{\pi}{2n}\geq\frac{1}{n} \).
Marius Cavachi, Olimpiada judeteana Constanta, 1993
Posted: Mon Mar 10, 2008 11:53 am
by Wizzy
Sa luam numarul complex :
\( \omega=\cos\frac {\pi}{n}+i\sin\frac{\pi}{n} \) unde \( \omega ^n=-1 \) adica \( \omega ^n-1=-2 \)
Aplicand modul la ultima relatie,avem :
\( 2=|\omega^n-1|=|\omega-1| |\omega^{n-1}+\omega^{n-2}+...+\omega+1| \)
Din inegalitatea modulelor obtinem :
\( 2\leq n|\omega-1| \) , deci \( |\omega-1|\geq \frac{2}{n} \)
Dar \( \omega-1=\cos\frac {\pi}{n}-1+i\sin\frac{\pi}{n}=-2\sin^2 \frac{\pi}{2n}+2i \sin\frac{\pi}{2n}\cos\frac{\pi}{2n}=2i \sin\frac{\pi}{2n}(\cos\frac{\pi}{2n}+i\sin\frac{\pi}{2n}) \)
Asadar : \( |\omega-1|=2\sin\frac{\pi}{2n}\geq \frac{2}{n} \)
\( \sin\frac{\pi}{2n}\geq \frac{1}{n} \)