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Math Help - Trig Triangle Proof

  1. #1
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    Trig Triangle Proof

    Hello Everyone,

    I have another question again...

    In any acute-angle triangle ABC, prove that:
    b^2sin2C + c^2sin2B = 2bcsinA

    Does it have to do with sine law and cosine law? I can't figure it out.
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  2. #2
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    Quote Originally Posted by KelvinScale View Post
    Hello Everyone,

    I have another question again...

    In any acute-angle triangle ABC, prove that:
    b^2sin2C + c^2sin2B = 2bcsinA

    Does it have to do with sine law and cosine law? I can't figure it out.
    b^2\sin{2C} + c^2\sin{2B} = 2b^2\sin{C}\cos{C} + 2c^2\sin{B}\cos{B}


    Now from the cosine rule:

    c^2 = a^2 + b^2 - 2ab\cos{C}

    2ab\cos{C} = a^2 + b^2 - c^2

    \cos{C} = \frac{a^2 + b^2 - c^2}{2ab}.


    Similarly \cos{B} = \frac{a^2 + c^2 - b^2}{2ac}.



    So 2b^2\sin{C}\cos{C} + 2c^2\sin{B}\cos{B} = 2b^2\sin{C}\left(\frac{a^2 + b^2 - c^2}{2ab}\right) + 2c^2\sin{B}\left(\frac{a^2 + c^2 - b^2}{2ac}\right)

     = \frac{b(a^2 + b^2 - c^2)\sin{C}}{a} + \frac{c(a^2 + c^2 - b^2)\sin{B}}{a}

     = \frac{b(a^2 + b^2 - c^2)\sin{C} + c(a^2 + c^2 - b^2)\sin{B}}{a}



    We also know from the sine rule:

    \frac{\sin{C}}{c} = \frac{\sin{A}}{a}

    so \sin{C} = \frac{c\sin{A}}{a}.

    Similarly \sin{B} = \frac{b\sin{A}}{a}.


    So \frac{b(a^2 + b^2 - c^2)\sin{C} + c(a^2 + c^2 - b^2)\sin{B}}{a}

     = \frac{b(a^2 + b^2 - c^2)\left(\frac{c\sin{A}}{a}\right) + c(a^2 + c^2 - b^2)\left(\frac{b\sin{A}}{a}\right)}{a}

     = \frac{bc\sin{A}(a^2 + b^2 - c^2)}{a^2} + \frac{bc\sin{A}(a^2 + c^2 - b^2)}{a^2}

     = \frac{a^2bc\sin{A}}{a^2} + \frac{b^3c\sin{A}}{a^2} - \frac{bc^3\sin{A}}{a^2} + \frac{a^2bc\sin{A}}{a^2} + \frac{bc^3\sin{A}}{a^2} - \frac{b^3c\sin{A}}{a^2}

     = \frac{2a^2bc\sin{A}}{a^2}

     = 2bc\sin{A}.
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