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Tan x = cos x

  1. Sep 13, 2010 #1
    If tan x = cos x, then what is x ? The answer includes the golden ratio !
     
  2. jcsd
  3. Sep 13, 2010 #2

    Char. Limit

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    You obviously know the answer, so why are you asking?
     
  4. Sep 13, 2010 #3
    So let s = sin x, c = cos x; then s/c = c, so s = c2 = 1 - s2. Solve for s, then take arcsin. Easy.
     
  5. Sep 13, 2010 #4

    Char. Limit

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    To be exact...

    [tex]Answer = sin^{-1}\left(\phi\right)[/tex]
     
  6. Sep 13, 2010 #5
    You forgot the other solution: [tex]\sin^{-1}\bigl[-\tfrac12(1+\sqrt5)\bigr][/tex]. (Not a real solution, though.)
     
  7. Sep 13, 2010 #6

    Char. Limit

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    I tend to use the real numbers.

    Complex numbers are not often the solutions to questions found in the general math section.

    EDIT: I just realized that [itex]sin^{-1}(\phi)[/itex] isn't real either... wow, so there are actually no real solutions.
     
  8. Sep 13, 2010 #7
    Tan(x) = Cos(x)
    Real solution are :
    x = ArcSin((Sqrt(5)-1)/2) + 2*k*Pi
    and
    x = Pi - ArcSin((Sqrt(5)-1)/2) + 2*k*Pi

    k = any n├ęgative, nul or positive integer.
     
  9. Sep 13, 2010 #8

    Char. Limit

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    So, in other words, any integer?
     
  10. Sep 13, 2010 #9
    Yes, indeed ! :wink:
     
  11. Sep 13, 2010 #10

    Mentallic

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    So cos(x) doesn't cross tan(x)?
     
  12. Sep 13, 2010 #11

    Char. Limit

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    Mistake number two...

    The real solution is in fact:

    [tex]sin^{-1}\left(\frac{-1}{\phi}\right)[/tex]

    I think, assuming that

    [tex]\frac{-1}{\phi} = \frac{1-\sqrt{5}}{2}[/tex]

    Which I seem to remember a similar property about phi...
     
  13. Sep 13, 2010 #12
    note that

    tan( x ) = cos( x ) = 1 / sqr( phi )

    so the solution can be made in terms of arctan or arccos.
     
  14. Sep 15, 2010 #13
    Since tan(x) increases monotonically from 0 to infinity in the domain (0, pi/1) and
    cos (x) decreases from 1 to 0 in the same domain, there should be a solution x between 0 and pi/2 and 0< sin(x) < 1. There's a solution in the second quadrant as well. cos(x) and tan(x) have different signs in the 3rd and fourth quadrant, so a solution where sin(x) < 0 is not possible

    Once you get all the signs right in the quadratic formula, you'll get

    [tex] sin(x) = \frac {1} {\phi} = \phi - 1 [/tex]

    from one of the solutions. The other solution has sin(x) >1 so that isn't a valid solution.

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