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Math Help - Please help me to do this problem!

  1. #1
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    Question Please help me to do this problem!

    Given a triangle ABC let
    X=tan[(B-C)/2]tan(A/2)
    Y=tan[(C-A)/2]tan(B/2)
    Z=tan[(A-B)/2]tan(C/2)
    Prove that X+Y+Z+XYZ=0
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  2. #2
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    Quote Originally Posted by mpgc_ac
    Given a triangle ABC let
    X=tan[(B-C)/2]tan(A/2)
    Y=tan[(C-A)/2]tan(B/2)
    Z=tan[(A-B)/2]tan(C/2)
    Prove that X+Y+Z+XYZ=0
    Some formulae that might help:
    If t = \tan \frac{\theta}{2}
    Then \tan \theta = \frac{2 t}{1 - t^2}
    And:
    tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1+\tan \theta \tan \phi}

    I tried, but the algebra is too much, there must be another way
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  3. #3
    Super Member malaygoel's Avatar
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    X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
    Similar are Y and Z
    Then solve (X+Y)/XY+1
    You will find it is equal to Z
    Your result immediately follows
    You can prove Napier Analogy by using Sine Law
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  4. #4
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    Quote Originally Posted by malaygoel
    X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
    Similar are Y and Z
    Then solve (X+Y)/XY+1
    You will find it is equal to Z
    Your result immediately follows
    You can prove Napier Analogy by using Sine Law
    Can you show a bit more detail? I havn't seen this method before...
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by chancey
    Can you show a bit more detail? I havn't seen this method before...
    Have you tried?
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  6. #6
    Super Member malaygoel's Avatar
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    Smile

    Quote Originally Posted by malaygoel
    X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
    Similar are Y and Z
    Then solve (X+Y)/XY+1
    You will find it is equal to Z
    Your result immediately follows
    You can prove Napier Analogy by using Sine Law
    X=(b-c)/(b+c)
    Y=(c-a)/(c+a)
    (X+Y)/(XY+1)={(b-c)/(b+c)+(c-a)/(c+a)}/[{(c-a)/(c+a)}*{(b-c)/(b+c)}+1]
    =(b-a)/(b+a)=-Z
    therefore,
    (X+Y)/(XY+1)=-Z
    multiply both sides by XY+1
    X+Y=-Z-XYZ
    X+Y+Z+XYZ=0
    Good Luck
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  7. #7
    Super Member malaygoel's Avatar
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    Proof of NApier Analogy

    Quote Originally Posted by malaygoel
    X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
    You can prove Napier Analogy by using Sine Law
    sine law states that
    a/sinA =b/sinB =c/sinC =2R
    (b-c)/(b+c)=(2RsinB-2RsinC)/(2RsinB+2RsinC)
    =(sinB-sinC)/(sinB+sinC)
    I can't write next step, use addition and subtraction fromulae and you will get X
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  8. #8
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    Quote Originally Posted by malaygoel
    sine law states that
    a/sinA =b/sinB =c/sinC =2R
    (b-c)/(b+c)=(2RsinB-2RsinC)/(2RsinB+2RsinC)
    =(sinB-sinC)/(sinB+sinC)
    I can't write next step, use addition and subtraction fromulae and you will get X
    Ahhh... I sort of get it, and I learnt a new sine rule, cool thanks.
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  9. #9
    Super Member malaygoel's Avatar
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    Quote Originally Posted by chancey
    Ahhh... I sort of get it, and I learnt a new sine rule, cool thanks.
    COULD YOU PROVE THE SINE RULE?
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  10. #10
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    Quote Originally Posted by malaygoel
    COULD YOU PROVE THE SINE RULE?
    In fact we can do better.
    The extended sines law
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