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Math Help - Help with Trig Identities exercise.

  1. #1
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    Help with Trig Identities exercise. (unsolved)

    Ok, I'm supposed to find sin \theta when cos \theta = \frac{1}{3}. The answer is \frac{2\sqrt{2}}{3}. I got \sqrt{\frac{8}{9}. Does this somehow simplify to the correct answer? it seems like it might, but my algebra skills are terrible.
    Last edited by fogownz; November 15th 2012 at 10:15 AM.
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  2. #2
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    Re: Help with Trig Identities exercise.

    Note:

    \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{\sqrt{9}} = \frac{2\sqrt{2}}{3}.
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  3. #3
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    Re: Help with Trig Identities exercise.

    Awesome, thank you.

    Edit: Actually, I don't quite understand how it is determined that \sqrt{8}=2\sqrt{2} I know that it does, but how could I have gotten there.
    Last edited by fogownz; November 15th 2012 at 10:02 AM.
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  4. #4
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    Re: Help with Trig Identities exercise.

    Do you know that \sqrt{4}= 2? And 8= (4)(2).
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    Re: Help with Trig Identities exercise.

    Yes, but that doesn't really click for me. When I saw that, I tried applying the steps I saw to a different problem. (4)(4)=16, and \sqrt{4}=2, yet 4\sqrt{2} does not equal \sqrt{16}. There is something that I'm missing from this.
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  6. #6
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    Re: Help with Trig Identities exercise.

    note that \sqrt{ab} = \sqrt{a} \cdot \sqrt{b}

    \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}
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  7. #7
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    Re: Help with Trig Identities exercise.

    Ah, alright, I got it. Thank you everyone for your help.
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