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Math Help - express in a symmetrical form??

  1. #1
    Super Member bigwave's Avatar
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    express in a symmetrical form??

    express in a symmetrical form \frac{a}{bc}+\frac{cosA}{a}

    basically i don't know what is meant by "symmetrical" since this is a triangle.
    the book did not give an example of this but think it was in the book prior to this Adv. Trig

    the answer to this is:

    \frac{a^2+b^2+c^2}{2abc}
    Last edited by bigwave; November 24th 2009 at 08:23 PM. Reason: spelling
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  2. #2
    MHF Contributor
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    Quote Originally Posted by bigwave View Post
    express in a symmetrical form \frac{a}{bc}+\frac{cosA}{a}

    basically i don't know what is meant by "symmetrical" since this is a triangle.
    the book did not give an example of this but think it was in the book prior to this Adv. Trig

    the answer to this is:

    \frac{a^2+b^2+c^2}{2abc}
    Symmetrical form means that the values of a,b and c could be interchanged without affecting the answer.

    Use law of cosines: a^2= b^2+c^2-2bc\cos(A)

    Add fractions: \frac{a^2+bc\cos(A)}{abc}
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  3. #3
    Super Member bigwave's Avatar
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    ok, can see how that was added but

    but from the law of cosines

    how is
    <br />
\frac{a^2+b^2+c^2}{abc}

    derived?

    know it is obvious but just don't see it.
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  4. #4
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    Solve law of cosines for cos(A) then substitute.

    The "bc" part will cancel but you will be left with everything over 2, thus multiply every term by 2. Simplify and you will get the final form.
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  5. #5
    Super Member bigwave's Avatar
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    as suggested

    <br />
a^2= b^2+c^2-2bc\cos(A)<br /> <br />

     <br />
\frac{a^2-b^2-c^2}{-2bc} = cos(A)<br />

    plug in cos(A)

     <br />
\frac{a^2+\frac{(bc)(a^2-b^2-c^2)}{-2bc}}{abc} . \frac{-2}{-2}<br />

     <br />
\frac{-2a^2+a^2-b^2-c^2}{-2abc}<br />

     <br />
\frac{a^2+b^2+c^2}{2abc}<br />
    Last edited by bigwave; November 24th 2009 at 10:46 PM. Reason: latex
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