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Homework Help: Computing and arccot

  1. Aug 12, 2009 #1
    1. The problem statement, all variables and given/known data
    compute arccot (-1/3 (3 in sqr. root))


    2. Relevant equations



    3. The attempt at a solution
    Sorry for posting one so soon after, but this is really annoying me.

    I have looked through my trig book but i cannot remember what an arccot is.

    I know that there cannot be a sqr. root as a denominator. So i moved it to be

    arccot(-sqr. root(3)/3)
     
  2. jcsd
  3. Aug 12, 2009 #2

    Cyosis

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    The arccot is the inverse of the cot function.

    [tex]
    \cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}
    [/tex]

    The arccot may be listed in your book as [itex]\cot^{-1} (x)[/itex], note that this means the inverse and not [itex]1/\cot x[/itex].

    This part I do not understand. Your original argument was just fine.
     
    Last edited: Aug 12, 2009
  4. Aug 12, 2009 #3
    I thought that you cannot have a sqr. root in the denominator? Like
    1/sqr. root 2

    and you would make it sqr. root 2/2 multiplying the bottom by sqr. root 2 and the top as well?
     
  5. Aug 12, 2009 #4

    Cyosis

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    Why do you think you can't have a square root in the denominator. You are correct that [itex]\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}[/itex], from which you can immediately conclude that there is nothing wrong with either of them. Do you perhaps mean that your teachers tell you to write a root in such a way that the root is in the denominator?

    Both methods are just fine and identical. So just pick one you're comfortable with. The real question is can you solve it now?
     
  6. Aug 12, 2009 #5

    tiny-tim

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    Hi Centurion1! :smile:

    (have a square-root: √ :tongue:)

    arccot(1/√3) = arctan(√3) :wink:
     
  7. Aug 12, 2009 #6
    Yes that is how they tell me to set it up.

    so newly set up it would be

    sin(x)/cos(x)= √ 3.
     
    Last edited: Aug 12, 2009
  8. Aug 12, 2009 #7

    tiny-tim

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    tan2(x) = 3, so sec(x) = … ? :smile:
     
  9. Aug 12, 2009 #8

    Mark44

    Staff: Mentor

    What your teacher probably said is that you shouldn't have a radical in the denominator, not that you must not have one there. There is a difference.

    Probably the most important reason for getting rid of the radical in the denominator is convenience.
     
  10. Aug 12, 2009 #9
    Sorry it took a while to respond

    sec(x)=1/cos right?
     
  11. Aug 12, 2009 #10

    Cyosis

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    Yep that's correct. Do you know for which value of x, [itex]\tan x=- \sqrt{3}[/itex]?
     
  12. Aug 12, 2009 #11
    is it -60
     
    Last edited: Aug 12, 2009
  13. Aug 12, 2009 #12

    Cyosis

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    Yep that's correct, if you mean degrees. So can you tell me what the [itex]\text{arccot} (-\frac{1}{\sqrt{3}})[/itex] is?
     
    Last edited: Aug 13, 2009
  14. Aug 12, 2009 #13
    wait what do you mean.

    tan(x)=-1/√3
     
  15. Aug 13, 2009 #14

    Cyosis

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    Sorry for the confusion I made a typo in the latex code that's why the arccot didn't display.
     
  16. Aug 13, 2009 #15
    Is it -30'? I am not sure though.

    i figure it is arccot(x)=.5773502692

    and since the arccot is the inverse of the cot function. Then doesn't that make it the tan function. As the cot is the inverse of the tan function?
     
    Last edited: Aug 13, 2009
  17. Aug 13, 2009 #16

    Cyosis

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    Cot is not the inverse of the tan version, but rather its reciprocal. Arccot is the inverse of the cot function and arctan the inverse of the tan function.

    Use the following fact about inverses:

    [tex]
    \begin{align*}
    &\text{arccot}(\frac{-1}{\sqrt{3}})=y
    \\
    &\cot(\text{arccot}(\frac{-1}{\sqrt{3}}))=\cot y
    \\
    &\frac{-1}{\sqrt{3}}=\cot y
    \\
    &\frac{-1}{\sqrt{3}}=\frac{1}{\tan y}
    \end{align*}
    [/tex]

    So to find your answer you need to determine y.
     
  18. Aug 13, 2009 #17
    99.29? I am pretty confused. You are explaining it well i just can't remember some of the basic procedures with sin, cos, tan, etc.
     
  19. Aug 13, 2009 #18

    Cyosis

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    In post #11 you got the correct answer all we need to do now is to make you see why. Do you understand the equations I listed if not which step is the problematic one?
     
  20. Aug 13, 2009 #19
    i understand the reciprocal. That makes it 1/tan. i just don't understand how to manipulate the tan in the final step.
     
  21. Aug 13, 2009 #20
    Multiply the whole equation by [itex]-tan(y)\sqrt{3}[/itex].
     
  22. Aug 13, 2009 #21

    tiny-tim

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    Hi Centurion1! :smile:
    This is getting needlessly complicated. :rolleyes:

    arccot(-1/√3) = arctan(-√3) = -arctan(√3) = … ?

    if you don't have tan tables or a tan button on your calculator (but only cos or sin tables or buttons), then use the standard trigonometric identities …

    tan2 = sec2 - 1 and sec = 1/cos :wink:

    (or you could have used cot2 = cosec2 - 1 and cosec = 1/sin)
     
  23. Aug 13, 2009 #22
    i underastand now. on my scientific calculator i was using tan-1. I just misunderstood what you meant.
     
  24. Aug 14, 2009 #23

    HallsofIvy

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    Draw an xy- coordinate system. Mark off 1 to the left (the -1) and there draw a line of length [itex]\sqrt{3}[/itex] up. You now have a right triangle in which the angle (measured from the positive x-axis) has cotangent [itex]-1/\sqrt{3}[/itex]. By the Pythagorean theorem, the hypotenuse has length [itex]\sqrt{(-1)^2+ (\sqrt{3})^2}= \sqrt{4}= 2[/itex]. In particular, note that one side is exactly half the length of the hypotenuse. If you copy that right triangle to the left of the vertical side, you have another right triangle with hypotenuse 2 and base 1. The two right triangles together give you a triangle in which all three sides (the two hypotenuses and the two bases together) have length 2. What are the angles in that triangle? Don't forget that the angle you want is measured from the positive x-axis. It is NOT one of the angles in the triangle but you can calculate it from that.
     
    Last edited by a moderator: Aug 15, 2009
  25. Aug 19, 2009 #24
    I'm under the impression that radicals in the denominators should always be rationalized (for stylistics and simplicity and unity) by multiplying the entire fraction by, essentially, 1. Am I foolish in thinking this is necessary?
     
  26. Aug 19, 2009 #25
    I read that it's mostly from convention that the denominator is rationalized, because, before hand-held calculators were common, it's much easier to divide an irrational number by an integer than vice versa. For example, √2/2 and 1/√2.
    Some textbooks don't bother to rationalize the denominator.
     
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