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How do I solve for x in cotx=2 by using a scientific calculator?

  1. Jan 20, 2010 #1
    1. The problem statement, all variables and given/known data

    How do I solve for x in cotx=2 by using a scientific calculator?
    What do I have to enter?

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I know that I can enter tan^-1 (1/2) to find x, but is there another way to do it on a calculator?
     
  2. jcsd
  3. Jan 20, 2010 #2

    Dick

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    Why do you need another way? If cot(x)=2 then 1/2=1/cot(x)=tan(x). The only other way to do it is to lobby the calculator companies for a cot^(-1) button.
     
  4. Jan 21, 2010 #3

    Char. Limit

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    There is another way:

    The arccotangent of x is equal to one half pi minus the arctangent of x. The arccotangent of x only equals the arctangent of 1/x if x is greater than zero.

    In short...

    [tex]arccot(x)=\frac{\pi}{2}-arctan(x)[/tex]

    [tex]arccot(x)=arctan(\frac{1}{x}), x>0[/tex]

    I don't know why this is true, my calculus book tells me so.
     
  5. Jan 21, 2010 #4

    Mentallic

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    Draw a right triangle with an angle [itex]\theta[/itex] and let the opposite side length be x and the adjacent side be 1 such that [itex]tan\theta=x[/itex]. Now, [itex]\theta=arctan(x)[/itex] and if we find the other angle in the triangle in terms of [itex]\theta[/itex], by sum of angles in a triangle, it is [itex]\pi/2 -\theta[/itex], so [itex]tan(\pi/2-\theta)=1/x[/itex] thus [itex]arccot(x)=\pi/2-\theta=\pi/2-arctan(x)[/itex].

    You can prove the second result by a similar method.

    @ the OP: you don't need an [itex]cot^{-1}[/itex] function on your calculator since it's very simple to take the [itex]tan^{-1}[/itex] of the reciprocal.
     
  6. Jan 21, 2010 #5

    Char. Limit

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    Thanks for the proof.
     
  7. Jan 21, 2010 #6

    vela

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    In fact, the "co" in "cosine" is there because the function gives the sine of the COmplementary angle. Same with the other co-functions.
     
  8. Jan 21, 2010 #7

    Mentallic

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    Aha thanks for that neat little info I didn't know :smile:
     
  9. Jan 21, 2010 #8
    Other proofs that arccot(x) = [itex]\pi[/tex]/2 - arctan(x) and arccot(x) = arctan(1/x):

    arccot(x) = [itex]\pi[/tex]/2 - arctan(x)

    Let x = tan(θ) = cot([itex]\pi[/itex]/2 - θ) (a trig identity)
    x = cot([itex]\pi[/itex]/2 - θ)
    arccot(x) = [itex]\pi[/itex]/2 - θ
    tan(θ) = x
    θ = arctan(x)
    arccot(x) = [itex]\pi[/itex]/2 - arctan(x)


    arccot(x) = arctan(1/x)

    Let θ = arccot(x)
    cot(θ) = x
    1/cot(θ) = tan(θ) = 1/x
    θ = arctan(1/x)
    arccot(x) = arctan(1/x)
     
    Last edited: Jan 21, 2010
  10. Jan 21, 2010 #9

    Mentallic

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    Well the trig identity was derived from the properties of a right-triangle so your proof hasn't really changed much, other than giving someone the opportunity to realize it can be done that way too.

    I'm curious as to why [tex]arccot(x)=arctan(\frac{1}{x}), x>0[/tex] as char.limit has posted. Why isn't this true for all non-zero x?
     
  11. Jan 21, 2010 #10

    Dick

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    arctan and arccot are defined with different angular ranges. Look it up and draw a graph. They only overlap for x>0 or resulting angle between 0 and pi/2.
     
  12. Jan 21, 2010 #11

    Char. Limit

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    Thanks, Dick. I didn't quite know myself.
     
  13. Jan 22, 2010 #12

    Mentallic

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    Sorry I didn't quit understand what "angular ranges" meant.
    I've drawn a graph for both:

    [tex]cot^{-1}x=\pi/2-tan^{-1}x[/tex]
    [tex]cot^{-1}x=tan^{-1}(1/x)[/tex]

    and it seems as though the first is only true for x>0 while the second is true for all x.

    Please elaborate so we can settle this.
     
  14. Jan 22, 2010 #13

    Char. Limit

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    Well, I believe that invtrig functions are only defined for certain angles. For example, I'm pretty sure that arcsin(x) is only defined from -pi/2 to pi/2 on the y axis. Thus, the angular range of arcsin(x) is [-pi/2,pi/2], but don't quote me on that.

    Arctangent I don't know about.
     
  15. Jan 22, 2010 #14

    Dick

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    I've just realized there is a another common convention for defining arccot. You can also define it to take values on (-pi/2,pi/2) but you pay the price of having it be discontinuous at 0. http://mathworld.wolfram.com/InverseCotangent.html Which 'arc' identities are true depends on what convention you are using. You HAVE to say which.
     
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