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Math Help - Trig

  1. #1
    Junior Member R3ap3r's Avatar
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    Trig

    If cos\beta = -\frac{3}{11} and \beta is in the second quadrant, find the exact value of sin\;2\beta

    Well sin2\beta = 2sin\beta cos\beta

    My final answer is -5.77 is this correct?
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  2. #2
    Moo
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    Quote Originally Posted by R3ap3r View Post
    If cos\beta = -\frac{3}{11} and \beta is in the second quadrant, find the exact value of sin\;2\beta

    Well sin2\beta = 2sin\beta cos\beta

    My final answer is -5.77 is this correct?
    Hullo,

    A sine can barely be <-1 :/

    Here is how to do. You know that \beta is in the second quadrant, so \sin(\theta)>0 and \cos(\theta)<0

    We know that \cos^2(\beta)+\sin^2(\beta)=1

    --> \sin^2(\beta)=1-\left(-\frac{3}{11}\right)^2=\frac{112}{121}

    Since \sin(\beta)>0, \sin(\beta)=\sqrt{\frac{112}{121}} \approx \dots


    Hence, \sin(2 \beta)=\dots
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  3. #3
    Junior Member R3ap3r's Avatar
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    Well I know atleast what i did wrong. i forgot to put \sqrt{112} over 11 in the formula.

    New answer is -.525 hows this look?
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  4. #4
    Junior Member R3ap3r's Avatar
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    Quote Originally Posted by janvdl View Post
    Code:
       |\
       | \
       |  \  sqrt{130}
    11 |   \
       |__o_\
       
        3
    That o there represents Beta.

    We wish to find 2 Sin \beta Cos \beta

    = 2 \left( \frac{11}{ \sqrt{130} } \right) \left( \frac{3}{ \sqrt{130} } \right)

    = \frac{33}{65}
    I believe you made an error. For one its the \sqrt{112} and my math is 2\cdot \left(\frac{\sqrt{112}}{11}\right)\left(\frac{-3}{11}\right)

    whose right? lol
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  5. #5
    Moo
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    Quote Originally Posted by R3ap3r View Post
    I believe you made an error. For one its the \sqrt{112} and my math is 2\cdot \left(\frac{\sqrt{112}}{11}\right)\left(\frac{-3}{11}\right)

    whose right? lol
    I'm pretty sure it's this one
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  6. #6
    o_O
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    In Janvdl's diagram, the 11 should be the hypotenuse.
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by R3ap3r View Post
    I believe you made an error. For one its the \sqrt{112} and my math is 2\cdot \left(\frac{\sqrt{112}}{11}\right)\left(\frac{-3}{11}\right)

    whose right? lol
    Moo's. And I'm going to bed because I can't seem to concentrate anymore. Sorry for the mistake Reap3r.
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  8. #8
    Junior Member R3ap3r's Avatar
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    np
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