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Sin x =cosh x

  1. Aug 12, 2011 #1
    According to Wolfram Alpha, if sin x =cosh x then [tex]x=-(\frac{3\pi}{4}+\frac{3i\pi}{4})[/tex]

    But how do you derive this?

    Also, what is the solution to cos x = sinh x in exact form?


    Thanks.
     
  2. jcsd
  3. Aug 12, 2011 #2

    SteamKing

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    Express sin and cosh in their exponential form.
     
  4. Aug 12, 2011 #3

    Fredrik

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    This is not true. What's true is that if [tex]x=-(\frac{3\pi}{4}+\frac{3i\pi}{4}),[/tex] then sin x=cosh x. Note that "if A, then B" (where A and B are statements, not numbers) is not equivalent to "if B, then A". For example, "if x=dimension10, then x is human" is not equivalent to "if x is human, then x=dimension10". "if A, then B" is however equivalent to "if not B, then not A". This fact is very useful in proofs.

    I would start with the formulas \begin{align}
    \sin x=\frac{e^{ix}-e^{-ix}}{2i}\\
    \cosh x=\frac{e^{x}-e^{-x}}{2}
    \end{align} This is what SteamKing had in mind. I'm including them in this post because I suspect that you don't know them. I realize that you're going to need more than this, but I don't have time to think about this now.

    If you type "solve sin x=cosh x" into Wolfram Alpha, it will tell you all the solutions in the complex plane.
     
    Last edited: Aug 12, 2011
  5. Aug 13, 2011 #4
    I do know the exponential forms. However, they do not seem to be helpful in finding x. I even used Euler's formula, but nothing works.
     
  6. Aug 13, 2011 #5

    I like Serena

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    I suggest substituting y=ex...
     
  7. Aug 13, 2011 #6

    HallsofIvy

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    But that is still going to give you a messy equation. With the substitution [itex]y= e^x[/itex], I get [itex]y^{1+ i}- y^{1- i}= iy[/itex].

    What next?
     
  8. Aug 13, 2011 #7

    gb7nash

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    This might work:

    Dividing out y, we get:

    [itex]y^{i}- y^{-i}= i[/itex]. (since y != 0)

    Setting [itex]z = y^{i}[/itex], you'll get [itex]z - \frac{1}{z} = i[/itex], so:

    [tex]z^2 - 1 = iz[/tex] and [itex]z^2 - iz - 1[/itex]

    From here, you should get:

    [tex]z = \frac{i \pm \sqrt{-1 + 4}}{2}[/tex]

    Now, we managed to solve for y^i, but this is where I get stuck.
     
  9. Aug 13, 2011 #8

    D H

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    Yes, but it does so in a way that shows lack of insight (Mathematica's results ofttimes are messy; but then again, so are Maple's and Maxima's) . Mathematica doesn't recognize that there is a unifying way to represent these solutions:

    [tex]x = (n+1/4)(1\pm i)\pi,\quad n\in\mathbb{Z}[/tex]
     
  10. Aug 13, 2011 #9

    D H

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    I suggest setting [itex]x=u+iv[/itex].

    After just a bit of work you should arrive at
    [tex]\begin{align}
    \sin x &= \sin u \cosh v + i \cos u \sinh v \\
    \cosh x &= \cos v \cosh u + i \sin v \sinh u
    \end{align}[/tex]
    Equating the real and imaginary parts yields
    [tex]\begin{align}
    \sin u \cosh v &= \cos v \cosh u \\
    \cos u \sinh v &= \sin v \sinh u
    \end{align}[/tex]

    And that is more than enough of a hint for now.
     
    Last edited: Aug 13, 2011
  11. Aug 13, 2011 #10

    uart

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    Another way to do it is to recognize that cosh(x) = cos(ix)

    Then you have sin(x) = cos(ix) and from there you can use the complementary relationship between sin and cos.
     
    Last edited by a moderator: Aug 13, 2011
  12. Aug 13, 2011 #11

    Fredrik

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    It's not homework. dimension10 is too young to get this sort of problems in school.
     
    Last edited: Aug 13, 2011
  13. Aug 13, 2011 #12
    That's a nice way. Probably wouldn't get any easier.
     
  14. Aug 13, 2011 #13
    Thanks. Strangely, [tex] i \arcsin z [/tex] is the solution for cosh x = tanh x.

    But then x would pi/6 which is false.
     
    Last edited: Aug 13, 2011
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