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Math Help - Simplify Equation- Double Division with Radicals

  1. #1
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    Exclamation Simplify Equation- Double Division with Radicals

    Hi, I solved tan(5pi/12) using the sum/difference formulas and came up with the following:

    (1/sqrt(3) + 1) / (-1/sqrt(3) + 1)

    I know it simplifies further down into: 2 + sqrt(3), however I suppose I have forgotten some of the rules of simplification. I don't know how to get it down to that. If someone could explain how to simplify these sorts of expression, that would be much appreciated. Thanks!
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Use the method of rationalisation, that is multiplying by something such that the denominator becomes a difference of two squares.

    \dfrac{\frac{1}{\sqrt3} + 1}{-\frac{1}{\sqrt3} + 1} \times \dfrac{\frac{1}{\sqrt3} + 1}{\frac{1}{\sqrt3} + 1}
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  3. #3
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    Quote Originally Posted by MrCryptoPrime View Post
    Hi, I solved tan(5pi/12) using the sum/difference formulas and came up with the following:

    (1/sqrt(3) + 1) / (-1/sqrt(3) + 1)

    I know it simplifies further down into: 2 + sqrt(3), however I suppose I have forgotten some of the rules of simplification. I don't know how to get it down to that. If someone could explain how to simplify these sorts of expression, that would be much appreciated. Thanks!
    \dfrac{\frac1{\sqrt{3}}+1}{-\frac1{\sqrt{3}}+1}} = \dfrac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}

    \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\cdot \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{3+2\sqrt{3} +1}{3-1} = \sqrt{3} + 2
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    When I multiply do I use FOIL? (e.g. each term is multiplied by all the other terms) I already tried doing what you suggest using FOIL, but it does not seem to help. Could you please elaborate more on the specifics of multiplying that out? Thanks.
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  5. #5
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    Nevermind I understand now! Thanks you guys for refreshing my memory!
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