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Math Help - Simplifying radicals in my trigonometry.

  1. #1
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    Simplifying radicals in my trigonometry.

    I am having a problem coming up with the correct solutions to work. The reason being I don't really understand how to work with radicals.

    A simple example of my problem is this.

    sqrt(33) * sqrt(32)

    I know the answer is 4sqrt(66) but I have no idea why.

    Another example of an answer I got in my problem that I cannot simplify is here (-4/-sqrt(33) + -sqrt(32)/2) /( 1 + -4/sqrt(33) + -sqrt(32)/2 ) - Wolfram|Alpha. (sorry for the link wolfram's makes it look pretty)

    I think I may have messed up some of the positives and negatives.
    I am using the sum formula for tangent.
    Tana = -4/-sqrt(33)
    Tanb = -sqrt(32)/2



    Can anybody point me in the direction of a radical tutorial somewhere or give me some pointers?

    Thanks!
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  2. #2
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    Quote Originally Posted by Jkeeb View Post
    I am having a problem coming up with the correct solutions to work. The reason being I don't really understand how to work with radicals.

    A simple example of my problem is this.

    sqrt(33) * sqrt(32)

    I know the answer is 4sqrt(66) but I have no idea why.

    Another example of an answer I got in my problem that I cannot simplify is here (-4/-sqrt(33) + -sqrt(32)/2) /( 1 + -4/sqrt(33) + -sqrt(32)/2 ) - Wolfram|Alpha. (sorry for the link wolfram's makes it look pretty)

    I think I may have messed up some of the positives and negatives.
    I am using the sum formula for tangent.
    Tana = -4/-sqrt(33)
    Tanb = -sqrt(32)/2



    Can anybody point me in the direction of a radical tutorial somewhere or give me some pointers?

    Thanks!
    Well
    \displaystyle \sqrt{33} \cdot \sqrt{32} = \sqrt{33 \cdot 32} = \sqrt{3 \cdot 11 \cdot 2 \cdot 16} = \sqrt{16} \cdot \sqrt{6 \cdot 11}

    \displaystyle = 4 \sqrt{66}

    As for your tangent formula, there is a mistake.
    \displaystyle tan(a + b) = \frac{tan(a) + tan(b)}{1 - tan(a) \cdot tan(b)}

    You subtracted the two tangents in Wolfram.

    -Dan
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