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Math Help - 8*cosA*cosB*CosC

  1. #1
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    8*cosA*cosB*CosC

    Hi everyone.....
    I have a problem that say:

    if ( 1-8*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)

    **********************************

    I solved it
    1-8*cos(a)*cos(b)*cos(c)=

    1 - 8*cos A cos B cos C =
    1 - 4*cos A * [cos (B-C) + cos (B+C)]=
    [sin (B-C)]^2 + [cos (B-C)]^2 - 4*cos (A) cos (B - C) + 4*[cos A]^2 =
    [sin (B-C)]^2 +[cos (B-C) - 2*cos A]^2 >=0

    ********************************
    but they said that when 1-8*cos A * cos B * cos C=0
    that mean a=b=c=60 deg
    ?????
    how can i handle this problem.?
    Last edited by blue_blue; April 24th 2009 at 01:43 AM.
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  2. #2
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    Trigonometry

    Hello blue_blue
    Quote Originally Posted by blue_blue View Post
    Hi everyone.....
    I have a problem that say:

    if ( 1-8*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)

    **********************************

    I solved it
    1-8*cos(a)*cos(b)*cos(c)=

    1 - 8*cos A cos B cos C =
    1 - 4*cos A * [cos (B-C) + cos (B+C)]=
    [sin (B-C)]^2 + [cos (B-C)]^2 - 4*cos (A) cos (B - C) + 4*[cos A]^2 =
    [sin (B-C)]^2 +[cos (B-C) - 2*cos A]^2 >=0

    ********************************
    but they said that when 1-8*cos A * cos B * cos C=0
    that mean a=b=c=0
    ?????
    how can i handle this problem.?
    Welcome to Math Help Forum!

    I'm not sure what problem you are actually trying to solve here, but certainly a =b=c=60^o is a solution to the equation. Why? Because \cos 60^o = \tfrac12. So if a =b=c=60^o,

    1-8\cos a\cos b\cos c = 1 - 8\times \tfrac12\times \tfrac12\times \tfrac12 = 1 - 1=0

    But a=b=c=0 is not a solution, because that gives 1-8\cos a\cos b\cos c = 1 - 8\times 1\times 1\times 1 = 1 - 8=-7

    Grandad
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  3. #3
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    Tanks Grandad
    sorry about my mistake ....
    I mean
    if
    1-8*cos(a)*cos(b)*cos(c)=0
    that mean a=b=c=60 degree
    ***************************
    I know that cos (60)=0.5 ==>1-8*cos 60*cos 60* cos 60=0
    ***************************
    but i mean how can i improve that ??
    **************************
    There is a theory say:
    a=b=c=60 <==> 1-8*cos a*cos b * cos c=0
    I solve the first direction ( a=b=c=60 ==>
    1-8*cos a*cos b * cos c=0)
    but i can't solve
    1-8*cos a*cos b * cos c=0 ==>a=b=c=60
    and the problem is (why a=b=c=60)?
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  4. #4
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    Hello blue_blue
    Quote Originally Posted by blue_blue View Post
    Tanks Grandad
    sorry about my mistake ....
    I mean
    if
    1-8*cos(a)*cos(b)*cos(c)=0
    that mean a=b=c=60 degree
    ***************************
    I know that cos (60)=0.5 ==>1-8*cos 60*cos 60* cos 60=0
    ***************************
    but i mean how can i improve that ??
    **************************
    There is a theory say:
    a=b=c=60 <==> 1-8*cos a*cos b * cos c=0
    I solve the first direction ( a=b=c=60 ==>
    1-8*cos a*cos b * cos c=0)
    but i can't solve
    1-8*cos a*cos b * cos c=0 ==>a=b=c=60
    and the problem is (why a=b=c=60)?
    Unless you know that a = b = c, you can't go in the other direction. There will be infinitely many solutions. For example, you could have a = b = 0^o (so that \cos a = \cos b = 1) and c = \arccos (0.125) = 82.82^o. Any combination, in fact, where \cos a\cos b\cos c = \tfrac18.

    Grandad
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello blue_blueUnless you know that a = b = c, you can't go in the other direction. There will be infinitely many solutions. For example, you could have a = b = 0^o (so that \cos a = \cos b = 1) and c = \arccos (0.125) = 82.82^o. Any combination, in fact, where \cos a\cos b\cos c = \tfrac18.

    Grandad
    THX again....
    but the theory say that
    1-8*cos a*cos b * cos c=0 ==>a=b=c
    but a+b+c=180 deg==>a=b=c=60 deg


    I solve it:
    1-8*cos(a)*cos(b)*cos(c)= 0

    ==>1 - 8*cos A cos B cos C =0
    ==>1 - 4*cos A * [cos (B-C) + cos (B+C)]=0
    ==>[sin (B-C)]^2 + [cos (B-C)]^2 - 4*cos (A) cos (B - C) + 4*[cos A]^2 =0
    ==>[sin (B-C)]^2 +[cos (B-C) - 2*cos A]^2 =0
    ==>
    [cos (B-C) - 2*cos A]^2=-[sin (B-C)]^2
    ==>I have 2 solution
    either (first solution)
    cos (B-C) - 2*cos A=-sin (B-C)
    or (second solution)
    cos (B-C) - 2*cos A=sin (B-C)

    let's talk about first solution:
    cos (B-C) - 2*cos A=-sin (B-C)
    ==>
    cos (B-C)+sin(b-c)=2cos(a)

    ?????
    but i didn't know what i'll do after that
    so i can't reach that a=b=c ?

    let's talk about second solution:
    cos (B-C) - 2*cos A=sin (B-C)
    ==>
    cos (B-C)-sin(b-c)=2cos(a)

    I
    i didn't know what i'll do after that too?
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  6. #6
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    Hello blue_blue.

    Ah! I see. It's a pity you mention this sooner!
    Quote Originally Posted by blue_blue View Post
    a+b+c=180 deg
    I take it, then, that A, B and C are the angles of a triangle.

    In that case:

    1-8\cos A\cos B\cos C = 0

    \Rightarrow 1 - 8\cos(180-[B+C])\cos B \cos C = 0

    \Rightarrow 1+8\cos(B+C)\cos B \cos C = 0

    \Rightarrow 1 + 4\cos(B+C)[\cos(B+C) + \cos(B-C)]=0

    \Rightarrow 4\cos^2(B+C)+4\cos(B-C)\cos(B+C) + 1 = 0

    This can be considered as a quadratic in \cos(B+C), which has real roots only if

    16\cos^2(B-c) - 4.4.1 \ge 0

    i.e. \cos^2(B-C) \ge 1

    But, of course, \cos^2\theta \le 1 for all \theta.

    \Rightarrow \cos^2(B-C) = 1

    which is only possible, for 0^o<B<180^o and 0^o<C<180^o, if B = C.

    In the same way, we may eliminate B and prove A=C.

    Therefore the only solution is A=B=C=60^o

    Grandad
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  7. #7
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    Mr.Grandad...
    thanks so so so much....


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  8. #8
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    blue_blue wrote: "if ( 1-8*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)"

    suggests that, 1 - 8(cos 60)(cos 60)(cos 60) = 1 - 8(1/2)(1/2)(1/2) = 0.

    How about THIS:

    i) 1 - 8(cos 20)(cos 40)(cos 80) = 1 - 8(1/8) = 0?

    ii) 1 - 8(cos (pi/7) cos (2pi/7) cos (3pi/7)) = 1 - 8(1/8) = 0?

    beautiful isn't it?

    Now i know WHY A = B = C = 60 degrees is the solution because of the restriction A + B + C = 180 degrees! THANKS
    -
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  9. #9
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    Re: 8*cosA*cosB*CosC

    Found this nice thread on Google, so I will add one short solution using an advanced formula using Power of a Point:

    If O is the circumcentre, and H is the orthocentre, then it is well known that (OH)^2 = R^2(1-8\cos A \cos B \cos C).

    Then in this case, OH must be 0, and the only possible case for this is an equilateral triangle, so we are done.

    Best wishes,
    PowerofPoint
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