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Math Help - Need double checking.

  1. #1
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    Need double checking.

    Find the exact radian value of Arccos(\frac{-\sqrt{3}}{2})

    My answers: \theta = \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k<br />
\pi

    is this right?
    Last edited by c_323_h; March 26th 2006 at 02:21 PM.
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  2. #2
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    Quote Originally Posted by c_323_h
    Find the exact radian value of cos^-1(\frac-{\sqrt3}{2})

    My answers: \theta = \frac{5\pi}{6} + 2k\pi and \frac{7\pi}{6} + 2k\<br />
pi

    is the right?
    Notice that,
    -\frac{\sqrt{3}}{2} to get that you need,
    x=\frac{5\pi}{6} thus, all the solutions are,
    x=\frac{5\pi}{6}+\pi k,k\in\mathbb{Z}
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    Notice that,
    -\frac{\sqrt{3}}{2} to get that you need,
    x=\frac{5\pi}{6} thus, all the solutions are,
    x=\frac{5\pi}{6}+\pi k,k\in\mathbb{Z}
    why pi and not 2pi?
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  4. #4
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    Quote Originally Posted by c_323_h
    why pi and not 2pi?
    Your answers are right, by my answer is simpler.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    Your answers are right, by my answer is simpler.
    ThePerfectHacker, cos(5 \pi/6 + \pi) = + \sqrt3/2 so your form for the solution is not correct.

    Your original answer is correct, c_323_h.

    -Dan
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    Here are the two non-coterminal pairs.
    (\pi-\pi/6)+2\pi k
    (\pi+\pi/6)+2\pi k
    Which simplifies to,
    -\pi/6+\pi(2k+1)
    \pi/6+\pi(2k+1)
    Thus, all solutions are,
    \pm\frac{\pi}{6}+\pi k where k\in\{...-5,-3,-1,1,3,5,...\}
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  7. #7
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    Quote Originally Posted by ThePerfectHacker
    Here are the two non-coterminal pairs.
    (\pi-\pi/6)+2\pi k
    (\pi+\pi/6)+2\pi k
    Which simplifies to,
    -\pi/6+\pi(2k+1)
    \pi/6+\pi(2k+1)
    Thus, all solutions are,
    \pm\frac{\pi}{6}+\pi k where k\in\{...-5,-3,-1,1,3,5,...\}
    what does \in mean in layman's terms?
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  8. #8
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    Quote Originally Posted by c_323_h
    what does \in mean in layman's terms?
    It means "element of"

    That means k must be one of the "elements" of that set.
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  9. #9
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    wait a second. the range of arccos is [0,\pi] since it is the inverse of cos. This means that arccos is only negative in II. so the only answer would be  \frac{5\pi}{6}. is my reasoning correct? i was just thinking about the graph of arccos.
    Last edited by c_323_h; March 26th 2006 at 06:07 PM.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by c_323_h
    wait a second. the range of arccos is [0,\pi] since it is the inverse of cos. This means that arccos is only negative in II. so the only answer would be  \frac{5\pi}{6}. is my reasoning correct? i was just thinking about the graph of arccos.
    Technically you are correct here. However, in all cases that I have seen, when you are asked to solve this kind of equation the solution set is assumed to be over the whole real line, not just the range given by arccos.

    -Dan
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