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Cyclometric Functions

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate [tex]arccot \left( \frac{1}{cot( \pi/5)} \right)[/tex] The answer may not contain any cyclometric functions.


    2. Relevant equations



    3. The attempt at a solution
    Can someone tell me where I went wrong? Cause I'm going insaaaane over this problem!

    [tex]arccot \left( \frac{1}{cot( \pi/5)} \right)[/tex]

    [tex]arccot(x) = \frac{arccos(x)}{arcsin(x)} = \frac{1}{arctan(x)} [/tex]


    [tex] \frac{1}{arctan\left( \frac{1}{cot( \pi/5)} \right)}[/tex]


    [tex] = \frac{1}{arctan\left( \frac{1}{ \frac{cos(\pi/5)}{sin(\pi/5)}} \right)}[/tex]


    [tex] = \frac{1}{arctan\left( \frac{sin(\pi/5)}{ cos(\pi/5)} \right)}[/tex]


    [tex] = \frac{1}{arctan\left(tan(\pi/5) \right)}[/tex]


    [tex] = \frac{1}{\pi/5} = \frac{5}{\pi}[/tex]


    And according to the practise test I'm doing, this is wrong.. help?
     
  2. jcsd
  3. Sep 15, 2009 #2

    Avodyne

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    [tex]{\rm arccot}(x) \ne \frac{\arccos(x)}{\arcsin(x)} [/tex]

    Try drawing a triangle.
     
  4. Sep 15, 2009 #3
    Ah that sucks.. why didn't my teacher say so? Is it true for arctan? cos I used arctanx = arcsinx / arccosx in class today (at the blackboard) and got the right answer and he didn't say anything?

    Anyway, I solved the problem (thanks for that tip, I never think of using triangles!) by saying that arccot(1/cotx) = arccot(tanx), but tanx=cot(90-x), i.e. 3π/10. Thanks!
     
  5. Sep 15, 2009 #4
    Actually, scratch that ^. I do get the right answer, but I didn't use that with tan, sorry.. I confused it, the other way around (tan(arccosx)= sin(arccosx)/cos(arccosx). My bad, weird/stressful day.
     
  6. Sep 15, 2009 #5

    Avodyne

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    Correct! And as you've figured out, the ratios like tan=sin/cos apply to the trig functions, and not to their inverses.
     
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