Prove that among any four distinct numbers from the interval $ (0,\frac{\pi}{2})$ there are two, say $ x,y,$ such that: $ 8 \cos x \cos y \cos (x-y)+1>4(\cos ^2 x+\cos ^2 y).$
Problem
Source: Ukraine 1997 grade 11
Tags: trigonometry, algebra unsolved, algebra
Xantos C. Guin
23.07.2009 03:41
Some algebraic manipulation yields:
$ 8\cos(x)\cos(y)\cos(x-y) + 1 > 4(\cos^{2}(x)+\cos^{2}(y))$
$ 8\cos(x)\cos(y)[\cos(x)\cos(y) + \sin(x)\sin(y)] + 1 > 4\cos^{2}(x)+4\cos^{2}(y)$
$ 8\cos^2(x)\cos^2(y) - 4\cos^{2}(x) - 4\cos^{2}(y) + 2 + 8\cos(x)\cos(y)\sin(x)\sin(y) > 1$
$ 2(2\cos^2(x)-1)(2\cos^2(y)-1) + 2(2\sin(x)\cos(x))(2\sin(y)\cos(y)) > 1$
$ 2\cos(2x)\cos(2y) + 2\sin(2x)\sin(2y) > 1$
$ \cos(2x - 2y) > \dfrac{1}{2} = \cos\left(\dfrac{\pi}{3}\right)$
Since $ x,y \in \left(0,\dfrac{\pi}{2}\right)$, this is equivalent to:
$ |2x-2y| < \dfrac{\pi}{3} \Longleftrightarrow |x-y| < \dfrac{\pi}{6}$
Let the four distinct numbers be $ 0 < a < b < c < d < \dfrac{\pi}{2}$.
Suppose that no two of $ a,b,c,d$ satisfy $ |x-y| < \dfrac{\pi}{6}$.
Then we have:
$ |b-a| \ge \dfrac{\pi}{6} \Rightarrow b \ge a + \dfrac{\pi}{6}$
$ |c-b| \ge \dfrac{\pi}{6} \Rightarrow c \ge b + \dfrac{\pi}{6} \ge a + \dfrac{\pi}{3}$
$ |d-c| \ge \dfrac{\pi}{6} \Rightarrow d \ge c + \dfrac{\pi}{6} \ge a + \dfrac{\pi}{2}$
However, if $ a > 0$, then $ d > \dfrac{\pi}{2}$, a contradiction.
Therefore, two of $ a,b,c,d$ satisfy $ |x-y| < \dfrac{\pi}{6}$, and thus, the original inequality.
sqing
21.11.2013 13:55
when $x=\frac{5\pi}{12},y=\frac{\pi}{12}$, $ 2=8 \cos x \cos y \cos (x-y)+1<4(\cos ^2 x+\cos ^2 y)=4.$ ?
fractals
22.11.2013 03:15
moldovan wrote: Prove that among any four distinct numbers...