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Math Help - Trigonometry

  1. #1
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    Trigonometry

    If sin\;\beta = \frac{2}{7} find cos\;2\beta without using a calculator.

    First off I know that cos\;2\beta = 1 - 2\;sin^2\;\beta

    But I'm having trouble plugging in the sin\;\beta = \frac{2}{7}

    Anyone have any thoughts on this?
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  2. #2
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    Quote Originally Posted by OzzMan View Post
    If sin\;\beta = \frac{2}{7} find cos\;2\beta without using a calculator.

    First off I know that cos\;2\beta = 1 - 2\;sin^2\;\beta

    But I'm having trouble plugging in the sin\;\beta = \frac{2}{7}

    Anyone have any thoughts on this?
    Why not cos\;2\beta = 1 - 2\;sin^2\;\beta = 1 - 2 \left( \frac{2}{7}\right)^2 .....?

    Hmmmmmm ...... you do realise that \sin^2 \beta means (\sin \beta)^2, right ....?
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  3. #3
    Senior Member topher0805's Avatar
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    <br />
cos\;2\beta = 1 - 2\;sin^2\;\beta<br />

    Remember that:

    sin^2\;\beta = (sin \beta)^2

    So simply plug in \frac {2}{7} :

    cos\;2\beta = 1 - 2 \left( \frac{2}{7}\right)^2
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  4. #4
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    I thought you had to leave a sin in there because it was sin = \frac{2}{7} not sin^2 = \frac{2}{7}

    Alright so the answers cos\;2\beta = \frac{41}{49}
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  5. #5
    Senior Member topher0805's Avatar
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    Quote Originally Posted by OzzMan View Post
    I thought you had to leave a sin in there because it was sin = \frac{2}{7} not sin^2 = \frac{2}{7}
    You are right, it was sin = \frac{2}{7} not sin^2 = \frac{2}{7} .

    That is why you have to square the \frac {2}{7} .

    Quote Originally Posted by OzzMan View Post
    Alright so the answers cos\;2\beta = \frac{41}{49}
    Yes but it should be negative.
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  6. #6
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    1 - 2\left(\frac{2}{7}\right)^2

    1 - 2\left(\frac{4}{49}\right)

    1 - \frac{8}{49}

    \frac{49}{49} - \frac{8}{49}

    \frac{41}{49}
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  7. #7
    Senior Member topher0805's Avatar
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    Oops, sorry you are right.
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