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Math Help - m^2 and n^2 type problem for a perfect Genius!!

  1. #1
    Newbie Lavlesh's Avatar
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    Question m^2 and n^2 type problem for a perfect Genius!!

    Given:-
    m^2 + m’^2 2mm’cosΘ = 1
    n^2 + n’^2 + 2nn’cosΘ = 1
    mn + m’n’ + (mn’ + m’n) cosΘ = 0

    To Prove:-
    m^2 + n^2 = cosec^2Θ

    Please help! Can anyone do this sum??


    And one more important thing i want to say, is that I'm a class 10th student....
    So please provide the easiest way round...
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    Last edited by Lavlesh; September 30th 2006 at 07:54 PM.
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  2. #2
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    Hello, Lavlesh!

    I haven't solved it yet, but I have some observations . . .


    Given:
    . . [1] . m≤ + mí≤ + 2mmícosθ .= .1
    . . [2] . n≤ + ní≤ + 2nnícosθ .= .1
    . . [3] .mn + míní + (mní + mín)cosθ .= .0

    Prove: .m≤ + n≤ .= .csc≤θ

    Add [1], [2], and 2 times [3]:
    . . . .[1] . m≤ + mí≤ + 2mmícosθ .= .1
    . . . .[2] . n≤ + ní≤ + 2nnícosθ .= .1
    . . 2x[3] .2mn + 2míní + 2(mní + mín)cosθ .= .0

    Their sum is:
    . . m≤ + 2mn + n≤ + m'≤ +2m'n' + n'≤ + 2(mm' + mn' + m'n + nn')cosθ .= .2

    which simplifies to:
    . . (m + n)≤ + (m' + n')≤ + 2(m + n)(m' + n')cosθ .= .2

    Did that help? . . . Not a bit!

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    The given equations look suspiciously like the Law of Cosines.

    I modified them:
    . . [1] . m≤ + mí≤ - 2mmícos(180į-θ) .= .1
    . . [2] . n≤ + ní≤ - 2nnícos(180į-θ) .= .1
    . . [3] .mn + míní - (mní + mín)cos(180į-θ) .= .0

    Adding [1], [2], and 2 times [3],

    . . we get: .(m + n)≤ + (m' + n')≤ - 2(m + n)(m' + n')cos(180į-θ) .= .2


    So I have a triangle . . .
    Code:
          *
           *    *
            *         *   2
       m + n *              *
              *                   *
            θ  *                        *
          - - - *  *  *  *  *  *  *  *  *  *  *
                            m' + n'

    But this too failed to inspire me.

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  3. #3
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    The equations are not independtant.
    Consider,
    x=n=n'=m=m'
    Then the first two equations give,
    2x^2+2x^2cosθ=0
    While the third gives,
    2x^2+2x^2cosθ=1
    This is an apparantet contradiction,
    Implying the set of numbers is empty, (no wonder the great Soroban failed).
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  4. #4
    Newbie Lavlesh's Avatar
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    Thumbs up Explain 'em!! Plz

    Good job done... Thanks...
    But I still cannot understand the second method Mr. Soroban discussed...
    But the first one given by him was pretty good!

    But can this be a little more easier to grasp?? B'coz I'm in 10th class and I couldn't get that cos(180į-θ) and I've learnt only cos(90į-θ)
    type methods....
    Can you further explain the second method, you gave... Please....
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  5. #5
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    Quote Originally Posted by Lavlesh View Post
    Good job done... Thanks...
    But I still cannot understand the second method Mr. Soroban discussed...
    But the first one given by him was pretty good!

    But can this be a little more easier to grasp?? B'coz I'm in 10th class and I couldn't get that cos(180į-θ) and I've learnt only cos(90į-θ)
    type methods....
    Can you further explain the second method, you gave... Please....
    He was reffering to Law of Cosines. In order to use it you need a negative in front of the cosine. Thus, he used the identity,
    cos(x)=-cos(180-x)
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  6. #6
    Newbie Lavlesh's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The equations are not independtant.
    Consider,
    x=n=n'=m=m'
    Then the first two equations give,
    2x^2+2x^2cosθ=0
    While the third gives,
    2x^2+2x^2cosθ=1
    This is an apparantet contradiction,
    Implying the set of numbers is empty, (no wonder the great Soroban failed).
    And can you explain what were you trying to do???
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  7. #7
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    Quote Originally Posted by Lavlesh View Post
    And can you explain what were you trying to do???
    That there is no solution to your problem, it leads to a contradiction (mathematically correct to say it leads to an empty set of numbers). Thus, your problem is wrong and has no solutions.

    EXAMPLE]
    Given that,
    x^2+y^2=2
    x^4+y^4=0
    Prove,
    x^3+y^3=1

    There is NO proof.

    Look ar the second equation, the only solution is (0,0) because some of positives is zero. But that is false because look at the first.

    Similarily what you are saying.
    You are asking to prove something which is not true, which is why it is not provable.
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