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Math Help - Cos Rule----- Sine Rule

  1. #1
    Member srirahulan's Avatar
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    Post Cos Rule----- Sine Rule

    Proof This a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)=0you can use sine rule and cos rule.


    Can You give me a hint to simplify this Problem>>><<<<<<
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    Re: Cos Rule----- Sine Rule

    Proof This a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)=0 you can use sine rule and cos rule.

    What the question doesn't mention is that ABC is a triangle.

    I only used the cosine rule. I converted CosA, cosB and cos c for its cos rule equivant.

    then I expanded each bracket.

    then I collected terms with a common denominator and simplified the numerators.

    then it all just fell out.
    Thanks from topsquark
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    Member srirahulan's Avatar
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    Re: Cos Rule----- Sine Rule

    A+B+C=\pi,that I did not mention>>>>>>
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    Re: Cos Rule----- Sine Rule

    Write out the law of cosines for each angle:

    a^2 = b^2+c^2-2bc\cos A Solve for \cos A

    b^2 = a^2+c^2-2ac\cos B Solve for \cos B

    c^2 = a^2+b^2-2ab\cos C Solve for \cos C

    Once you solve for all three, plug them back into the original expression.
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    Re: Cos Rule----- Sine Rule

    Just use sine rule and you will get sinerule : a/sinA = b/sin B = c/sin C = K ( Constant )
    LHS = K [ sin A ( cos B + cos C - 1 ) + sin B ( cos C + cos A - 1 ) + sin C ( cos A + cos B - 1 )]
    = K [ sin A cos B + sin A cos C - sin A + sin B cos C + sin B cos A - sin B + sin C cos A + sin C cos B - sin C] Rearrange terms
    = K [ sin A cos B + sin B cos A + sin A cos C + sin C cos A + sin B cos C + sin C cos B - sin A - sin B - sin C]
    = K [ sin ( A + B ) + sin ( A + C ) + sin ( B + C ) - sin A - sin B - sin C]
    Now we know that A + B + C = 2 pi OR 180 degree thus we will have sin ( A + B ) = sin C etc Thus we get
    LHS = K [ sin C + sin B + sin A - sin A - sin B - sin C ] = 0 = RHS
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