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Math Help - Equivalence of 2 expressions involving radicals

  1. #1
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    Equivalence of 2 expressions involving radicals

    Hi.

    I need help in seeing how the following two expressions are equal.

    \frac{\sqrt{6} + \sqrt{2}}{4} and \frac{\sqrt{2}\sqrt{1 + \frac{\sqrt{3}}{2}}}{2}

    I know they both evaluate to the same thing but I'm having a hard time seeing how they are related.
    If someone could, perhaps through algebraic manipulation, show how one is equivalent to the other,
    I would be immensely grateful.

    I don't think it matters but the first arrived at by evaluating cos \frac{\pi}{12} using a sum of angles
    formula while the second is what you get if you decide to use a half angle formula.
    Last edited by kodx; June 8th 2013 at 04:44 AM.
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  2. #2
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    Re: Equivalence of 2 expressions involving radicals

    Hint: Starting with \frac{\sqrt{6}+\sqrt{2}}{4} get the denominator the same as the other expression and then factorise \sqrt{2} out of the numerator
    Thanks from topsquark
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  3. #3
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    Re: Equivalence of 2 expressions involving radicals

    Or work the other way: start from \frac{\sqrt{2}\sqrt{1+ \frac{\sqrt{3}}{2}}}{4}, take the \sqrt{2} inside the other integral: \frac{\sqrt{2(1+ \frac{\sqrt{3}}{2}})}{4}, then take one of the "2"s in the denominator into it:
    \frac{\sqrt{\frac{2}{4}(1+ \frac{\sqrt{3}}{2}})}{2}= \frac{\sqrt{\frac{1}{2}(1+ \frac{\sqrt{3}}{2}})}{2}
    Last edited by HallsofIvy; June 8th 2013 at 05:01 AM.
    Thanks from kodx and topsquark
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  4. #4
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    Re: Equivalence of 2 expressions involving radicals

    Quote Originally Posted by kodx View Post
    I need help in seeing how the following two expressions are equal.
    \frac{\sqrt{6} + \sqrt{2}}{4} and \frac{\sqrt{2}\sqrt{1 + \frac{\sqrt{3}}{2}}}{2}
    Enough time has passed for me to give you a solution. I agree with reply #2.

    Note that {\left( {\frac{{\sqrt 6  + \sqrt 2 }}{4}} \right)^2} = \frac{1}{2} + \frac{{\sqrt 3 }}{4}, therefore \frac{{\sqrt 6  + \sqrt 2 }}{4} = \sqrt {\frac{1}{2} + \frac{{\sqrt 3 }}{4}}

    Thus
     \begin{align*} \frac{{\sqrt 6  + \sqrt 2 }}{4}  &= \sqrt {\frac{1}{2} + \frac{{\sqrt 3 }}{4}}\\ &= \frac{1}{{\sqrt 2 }}\left( {\sqrt {1 + \frac{{\sqrt 3 }}{2}} } \right)\\ &= \frac{{\sqrt 2 \left( {\sqrt {1 + \frac{{\sqrt 3 }}{2}} } \right)}}{2}\end{align*}
    Thanks from topsquark and kodx
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