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O inegalitate trigonometrica
Posted: Sun Sep 30, 2007 2:44 pm
by Virgil Nicula
\( \{\ x\ ,\ y\ \}\subset\mathbb{R}\ \Rightarrow \ 1\ +\ \ 8\ \cos\ x\ \cos\ y\ \cos\ (\ x\ +\ y\ )\ \ge \ 0\ . \)
Caz particular. \( x\ +\ y\ +\ z\ =\ \pi\ \Rightarrow\ \cos\ x\ \cos\ y\ \cos\ z\ \le\ \frac 18\ . \)
Posted: Mon Oct 01, 2007 6:22 pm
by red_dog
Inegalitatea se scrie succesiv
\( 1+4[\cos(x+y)+\cos(x-y)]\cos(x+y)\geq 0 \)
\( 4\cos^2(x+y)+4\cos(x-y)\cos(x+y)+1\geq 0 \)
\( 4\cos^2(x+y)+4\cos(x-y)\cos(x+y)+\cos^2(x-y)+\sin^2(x-y)\geq 0 \)
\( [2\cos(x+y)+\cos(x-y)]^2+\sin^2(x-y)\geq 0 \)
ultima fiind evident adevarata.
Pentru cazul particular avem \( \cos(x+y)=\cos(\pi-z)=-\cos z \)
Atunci inegalitatea se scrie \( 1-8\cos x\cos y\cos z\geq 0\Leftrightarrow\cos x\cos y\cos z\leq\frac{1}{8} \)