# trigo proof 2

• Jun 14th 2011, 06:41 AM
hooke
trigo proof 2
Prove that among any four distinct numbers from the interval (0,pi/2) there are two, say, x and y such that

8 cos x cos y cos (x-y) + 1>4(cos^2 x + cos^2 y)
• Jun 14th 2011, 11:59 AM
Also sprach Zarathustra
Re: trigo proof 2
Quote:

Originally Posted by hooke
Prove that among any four distinct numbers from the interval (0,pi/2) there are two, say, x and y such that

8 cos x cos y cos (x-y) + 1>4(cos^2 x + cos^2 y)

I came up to this so far:

$8( cos x)( cos y)( cos (x-y)) + 1>4(cos^2 x + cos^2 y)$

$8( cos x )(cos y)( cos (x-y)) + 1-4(cos^2 x + cos^2 y)=$

$4(cos(x+y)+cos(x-y))cos(x-y)-4(cos^2 x + cos^2 y)+1=$

$4cos^2(x-y)+4cos(x-y)cos(x+y)-4(cos^2 x + cos^2 y)+1=$

$4cos^2(x-y)+2(cos(2x)+cos(2y))-4(cos^2 x + cos^2 y)+1=$

$4cos^2(x-y)+2(cos^2(x)-sin^2(x)+cos^2(y)-sin^2(y))-4(cos^2 x + cos^2 y)+1=$

$4cos^2(x-y)+2(cos^2(x)+cos^2(y))-2(sin^2(x)+sin^2(y))-4(cos^2 x + cos^2 y)+1=$

$4cos^2(x-y)-2(cos^2(x)+cos^2(y))-2(sin^2(x)+sin^2(y))+1=$

$4cos^2(x-y)-2(cos^2(x)+cos^2(y)+(sin^2(x)+sin^2(y))+1=$

$4cos^2(x-y)-2(1+1)+1=$

$4cos^2(x-y)-4+1=$

$4cos^2(x-y)-3$
• Jun 14th 2011, 12:15 PM
Opalg
Re: trigo proof 2
Quote:

Originally Posted by hooke
Prove that among any four distinct numbers from the interval (0,pi/2) there are two, say, x and y such that

8 cos x cos y cos (x-y) + 1>4(cos^2 x + cos^2 y)

$8 \cos x \cos y \cos (x-y) + 1>4(\cos^2 x + \cos^2 y)$
. . . . . . $\Longleftrightarrow \; 8 \cos x \cos y (\cos x\cos y-\sin x\sin y) + 1>4(\cos^2 x + \cos^2 y)$
. . . . . . $\Longleftrightarrow \; 2(2\cos^2x-1)(2\sin^2x-1) + 8\sin x\cos x\sin y\cos y -1>0$
. . . . . . $\Longleftrightarrow \; 2\bigl(\cos(2x)\cos(2y) + \sin(2x)\sin(2y)\bigr) -1>0$
. . . . . . $\Longleftrightarrow \; \cos(2|x-y|) > 1/2$
. . . . . . $\Longleftrightarrow \; 2|x-y| < \pi/3.$

If you have four numbers between 0 and $\pi/2$ then at least two of them must be less than distance $\pi/6$ apart.
• Jun 15th 2011, 06:31 AM
hooke
Re: trigo proof 2
Quote:

Originally Posted by Opalg
$8 \cos x \cos y \cos (x-y) + 1>4(\cos^2 x + \cos^2 y)$
. . . . . . $\Longleftrightarrow \; 8 \cos x \cos y (\cos x\cos y-\sin x\sin y) + 1>4(\cos^2 x + \cos^2 y)$
. . . . . . $\Longleftrightarrow \; 2(2\cos^2x-1)(2\sin^2x-1) + 8\sin x\cos x\sin y\cos y -1>0$
. . . . . . $\Longleftrightarrow \; 2\bigl(\cos(2x)\cos(2y) + \sin(2x)\sin(2y)\bigr) -1>0$
. . . . . . $\Longleftrightarrow \; \cos(2|x-y|) > 1/2$
. . . . . . $\Longleftrightarrow \; 2|x-y| < \pi/3.$

If you have four numbers between 0 and $\pi/2$ then at least two of them must be less than distance $\pi/6$ apart.

Thanks again Opalg.