Prove that among any four distinct numbers from the interval (0,π2) there are two, say x,y, such that: 8cosxcosycos(x−y)+1>4(cos2x+cos2y).
Problem
Source: Ukraine 1997 grade 11
Tags: trigonometry, algebra unsolved, algebra
Xantos C. Guin
23.07.2009 03:41
Some algebraic manipulation yields:
8cos(x)cos(y)cos(x−y)+1>4(cos2(x)+cos2(y))
8cos(x)cos(y)[cos(x)cos(y)+sin(x)sin(y)]+1>4cos2(x)+4cos2(y)
8cos2(x)cos2(y)−4cos2(x)−4cos2(y)+2+8cos(x)cos(y)sin(x)sin(y)>1
2(2cos2(x)−1)(2cos2(y)−1)+2(2sin(x)cos(x))(2sin(y)cos(y))>1
2cos(2x)cos(2y)+2sin(2x)sin(2y)>1
cos(2x−2y)>12=cos(π3)
Since x,y∈(0,π2), this is equivalent to:
|2x−2y|<π3⟺|x−y|<π6
Let the four distinct numbers be 0<a<b<c<d<π2.
Suppose that no two of a,b,c,d satisfy |x−y|<π6.
Then we have:
|b−a|≥π6⇒b≥a+π6
|c−b|≥π6⇒c≥b+π6≥a+π3
|d−c|≥π6⇒d≥c+π6≥a+π2
However, if a>0, then d>π2, a contradiction.
Therefore, two of a,b,c,d satisfy |x−y|<π6, and thus, the original inequality.
sqing
21.11.2013 13:55
when x=5π12,y=π12, 2=8cosxcosycos(x−y)+1<4(cos2x+cos2y)=4. ?
fractals
22.11.2013 03:15
moldovan wrote: Prove that among any four distinct numbers...