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Thread: prove that a^2sin(B-C) =

  1. #1
    Super Member bigwave's Avatar
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    Talking prove that a^2sin(B-C) =

    prove that:

    $\displaystyle a^{2}sin\left(B-C\right) = \left(b^{2} - c^{2}\right)sinA$

    where $\displaystyle a,b,c$ are sides of triangle
    and $\displaystyle A,B,C$ are the corresponding angles
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  2. #2
    MHF Contributor red_dog's Avatar
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    I'll use the sine and cosine rules.

    $\displaystyle a^2\sin(B-C)=a^2(\sin B\cos C-\sin C\cos B)=$

    $\displaystyle =a^2\left(\frac{b}{2R}\cdot\frac{a^2+b^2-c^2}{2ab}-\frac{c}{2R}\cdot\frac{a^2+c^2-b^2}{2ac}\right)=$

    $\displaystyle =\frac{2a(b^2-c^2)}{4R}=\frac{a}{2R}(b^2-c^2)=(b^2-c^2)\sin A$
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