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Inverse trigonometric equation

  1. Jan 5, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Solve

    [itex]cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)[/itex]

    2. Relevant equations

    3. The attempt at a solution
    I can write the above equation as θ+α=β where the symbols represent the respective inverse functions. Now I take tan of both sides. Simplifying I get

    [itex](n^2+1-2x)(n-1) = x(n^2-x+1)[/itex]
     
  2. jcsd
  3. Jan 5, 2013 #2
    Are you asking how to solve that equation?
     
  4. Jan 5, 2013 #3

    SammyS

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    I get nearly the same thing:

    [itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]
     
  5. Jan 6, 2013 #4

    utkarshakash

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    Yes I have to find the value of x
     
  6. Jan 6, 2013 #5

    utkarshakash

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    But what should be the next step?
     
  7. Jan 6, 2013 #6

    SammyS

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    Multiply each side out.

    Get everything to the left side.

    You have a quadratic equation in x.
     
  8. Jan 6, 2013 #7

    utkarshakash

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    But its discriminant is way too big!
     
  9. Jan 6, 2013 #8

    SammyS

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    Too big in what sense ?

    What do you get for the discriminant ?
     
  10. Jan 7, 2013 #9

    utkarshakash

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    How are you getting that 'extra' 1 in the RHS. OK I'm only posting what I got from my answer i.e. without that '1'.

    D=[itex]n^4+6n^2-8n+2[/itex]
     
  11. Jan 7, 2013 #10

    SammyS

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    Wow! I showed that "extra" 1 way back in post #3. Now you mention it?

    I got it because,

    [itex]\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .[/itex]

    How did you not get it?
     
  12. Jan 8, 2013 #11

    utkarshakash

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    Hey there is a much better solution than this. Instead of taking tan if I take cot of both sides I get a very simple result and here's it

    [itex] x= \dfrac{n(n+1)}{2} [/itex]

    And yes you said right. There was a calculation error in my solution. It should have that extra 1
     
  13. Jan 8, 2013 #12

    SammyS

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    That is interesting. I'll have to take a look at it.
     
  14. Jan 9, 2013 #13

    SammyS

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    I still get a quadratic equation in x, so that can't be the solution.

    [itex]\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}[/itex]
     
  15. Jan 9, 2013 #14

    utkarshakash

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    OMG I made the silliest mistake ever. :bugeye: Instead of taking cot of both sides I just removed the 'inverse thing' and simply calculated the value of x from there. OK sticking to my previous solution I take tan of both sides and get the same thing what you got(with that extra 1) and get the discriminant as

    [itex] -4n^3+6n^2-8n+9[/itex]
     
  16. Jan 9, 2013 #15

    SammyS

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    What I got for a discriminant was

    [itex]n^4+6n^2-8n+9\ .[/itex]

    I just rechecked it using WolframAlpha on

    [itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]

    and it confirmed my result.
     
  17. Jan 9, 2013 #16

    utkarshakash

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    OK so what should I do after getting discriminant?
     
  18. Jan 9, 2013 #17

    haruspex

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    Nothing special. It's just a quadratic equation in x; solve it with the usual formula. If it's messy, it's messy.
     
  19. Jan 10, 2013 #18

    utkarshakash

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    You can't simply say that :cry:I cannot get rid of that square root. Please help me. Its not even a perfect square. Should I leave it this way?
     
  20. Jan 10, 2013 #19

    SammyS

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    If n=0, then the discriminant is a perfect square. I found no other cases. Also, the discriminant is positive for all values of n.

    So as haruspex said, it's a messy result.
     
  21. Jan 10, 2013 #20

    utkarshakash

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    OK so you want me to continue with that square root.
     
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