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Math Help - trigo proof 2

  1. #1
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    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)
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: trigo proof 2

    Quote Originally Posted by hooke View Post
    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
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  3. #3
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    Re: trigo proof 2

    Quote Originally Posted by hooke View Post
    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.
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    Re: trigo proof 2

    Quote Originally Posted by Opalg View Post
    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.
    Last edited by hooke; June 15th 2011 at 06:43 AM.
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