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Math Help - cot

  1. #1
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    cot

    Calculate . The answer may contain roots but not trigonometric functions.


    im really bad on the geometry so any1 give me tips?
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  2. #2
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    Re: cot

    Hint: \cot(\theta+k\pi)=\cot(\theta) where k\in\mathbb{Z}.
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  3. #3
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    Re: cot

    i want the range 0 ≤ x < 2π, so i have to choose k so it will be on tat range so k= 4 if i got right,sorry kkkeybord broken
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: cot

    -\frac{31}{6}=-5\frac{1}{6} so you have:

    -6\pi<-\frac{31}{6}\pi<-5\pi

    0<-\frac{31}{6}\pi+6\pi<\pi

    0<\frac{5}{6}\pi<\pi
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  5. #5
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    Re: cot

    ty again! but how can ik what cot 5pi/6 is on test
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  6. #6
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    Re: cot

    nvm solved it
    cot (5π/6) = 1 / tan (5π/6) = - 1/√3
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: cot

    \cot(x)=\frac{\cos(x)}{\sin(x)}

    You should know (or be able to quickly find using the unit circle):

    \cos\left(\frac{5\pi}{6} \right)=-\frac{\sqrt{3}}{2}

    \sin\left(\frac{5\pi}{6} \right)=\frac{1}{2}

    If you prefer to use a first quadrant angle, then you may write:

    \cot\left(-\frac{31\pi}{6} \right)=-\cot\left(\frac{31\pi}{6}-5\pi \right)=-\cot\left(\frac{\pi}{6} \right)
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: cot

    Quote Originally Posted by Petrus View Post
    nvm solved it
    cot (5π/6) = 1 / tan (5π/6) = - 1/√3
    You need to invert the result.
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