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Trignometric and hyperbolic equalities: Why the golden ratio?

  1. Jul 14, 2011 #1
    1.

    [tex] \sin \theta = \cos \theta [/tex]

    \theta=\frac{\pi}{4}

    2.

    [tex] \sin \theta = \tan \theta [/tex]

    [tex]\theta = 0[/tex]

    3.

    [tex] \cos \theta = \tan \theta [/tex]

    [tex] \theta =\arcsin (\varphi -1) [/tex]

    4.

    [tex]\sin \theta = \csc \theta[/tex]

    [tex]\theta = \frac{\pi}{2}[/tex]

    5.

    [tex]\sin \theta =\sec \theta[/tex]

    [tex]\theta[/tex] does not exist.

    6.

    [tex]\sin \theta =\cot \theta[/tex]

    [tex]\theta = \arccos (\varphi -1)[/tex]

    7.

    [tex]\cos \theta =\csc \theta[/tex]

    [tex]\theta[/tex] does not exist.

    8.

    [tex]\cos \theta =\sec \theta[/tex]

    [tex]\theta=0[/tex]

    9.

    [tex]\cos \theta = \cot \theta [/tex]

    [tex]\theta=\frac{\pi}{2}[/tex]

    10.

    [tex]\tan \theta =\csc \theta[/tex]

    [tex]\theta =\arccos(\varphi-1)[/tex]

    11.

    [tex]\tan \theta = \sec \theta [/tex]


    [tex]\theta=\frac{\pi}{2}[/tex]

    12.

    [tex]\tan \theta = \cot \theta [/tex]

    [tex]\theta=\frac{\pi}{4}[/tex]


    13.

    [tex]\csc \theta =\sec \theta[/tex]

    [tex]\theta=\frac{\pi}{4}[/tex]

    14.

    [tex]\csc \theta =\cot \theta [/tex]

    [tex]\theta = \arccos (\varphi -1) [/tex]

    15.

    [tex] \sec \theta =\cot \theta [/tex]

    [tex]\theta=\arcsin (\varphi - 1) [/tex]

    I used quadratic equation for some equalities. Which showed that the golden ration was involved. But my question is "geometrically, why?"
     
  2. jcsd
  3. Jul 14, 2011 #2
    Hi!
    Interesting question. I have always been fascinated by the golden ratio which keeps appearing at places where you least expect :smile: Will have to think about this one.
     
  4. Jul 15, 2011 #3
    Oops. There was a LaTeX error.

    1.

    [tex] \sin \theta = \cos \theta [/tex]

    [tex]\theta=\frac{\pi}{4}[/tex]

    2.

    [tex] \sin \theta = \tan \theta [/tex]

    [tex]\theta = 0[/tex]

    3.

    [tex] \cos \theta = \tan \theta [/tex]

    [tex] \theta =\arcsin (\varphi -1) [/tex]

    4.

    [tex]\sin \theta = \csc \theta[/tex]

    [tex]\theta = \frac{\pi}{2}[/tex]

    5.

    [tex]\sin \theta =\sec \theta[/tex]

    [tex]\theta[/tex] does not exist.

    6.

    [tex]\sin \theta =\cot \theta[/tex]

    [tex]\theta = \arccos (\varphi -1)[/tex]

    7.

    [tex]\cos \theta =\csc \theta[/tex]

    [tex]\theta[/tex] does not exist.

    8.

    [tex]\cos \theta =\sec \theta[/tex]

    [tex]\theta=0[/tex]

    9.

    [tex]\cos \theta = \cot \theta [/tex]

    [tex]\theta=\frac{\pi}{2}[/tex]

    10.

    [tex]\tan \theta =\csc \theta[/tex]

    [tex]\theta =\arccos(\varphi-1)[/tex]

    11.

    [tex]\tan \theta = \sec \theta [/tex]


    [tex]\theta=\frac{\pi}{2}[/tex]

    12.

    [tex]\tan \theta = \cot \theta [/tex]

    [tex]\theta=\frac{\pi}{4}[/tex]


    13.

    [tex]\csc \theta =\sec \theta[/tex]

    [tex]\theta=\frac{\pi}{4}[/tex]

    14.

    [tex]\csc \theta =\cot \theta [/tex]

    [tex]\theta = \arccos (\varphi -1) [/tex]

    15.

    [tex] \sec \theta =\cot \theta [/tex]

    [tex]\theta=\arcsin (\varphi - 1) [/tex]

    I used quadratic equation for some equalities. Which showed that the golden ration was involved. But my question is "geometrically, why?"
     
  5. Jul 15, 2011 #4

    pwsnafu

    User Avatar
    Science Advisor

    IIRC, [itex]\cos\frac{\pi}{5} = \frac{\phi}{2}[/itex]. That's related to the pentagon. Hmm...
     
  6. Jul 15, 2011 #5
    Here are the hyperbolic equalities.

    1.

    [tex] \sinh x = \cosh x [/tex]

    [tex] x= \infty [/tex]

    2.

    [tex] \sinh x =\tanh x [/tex]

    [tex]x=0[/tex]

    3.

    [tex] \cosh x =\tanh x [/tex]

    [tex]{x}_{1}=\frac{- \arcsin (\frac{\sqrt{3}}{2}+\frac{i}{2})}{i}[/tex]


    [tex]{x}_{2}=\frac{- \arcsin (\frac{\sqrt{3}}{2}-\frac{i}{2})}{i}[/tex]

    4.

    [tex]\sinh x = csch x [/tex]

    [tex]x=arcsinh 1 [/tex]

    5.

    [tex] \sinh x = sech x [/tex]

    [tex] x=\frac{\ln (2 \varphi +1)}{2}[/tex]

    6.

    [tex] \sinh x =\coth x [/tex]

    [tex]x=\frac{\arccos(1-\varphi)}{i}[/tex]

    7.

    [tex] \cosh x =csch x [/tex]

    [tex]x=arcsinh \sqrt{\varphi-1}[/tex]

    8.

    [tex] \cosh x =sech x [/tex]

    [tex]x=0[/tex]

    9.

    [tex] \cosh x =\coth x [/tex]

    [tex]x=arcsinh (\varphi - 1)[/tex]

    10.

    [tex] \tanh x = csch x [/tex]

    [tex] x=arccosh (\varphi - 1) [/tex]

    11.

    [tex]\tanh x =sech x [/tex]

    [tex] x=arcsinh 1 [/tex]

    12.

    [tex] \tanh x = \coth x [/tex]

    [tex] x=\infty[/tex]

    13.

    [tex] csch x =sech x [/tex]

    [tex] x=arctanh 1[/tex]

    14.

    [tex]csch x =\coth x [/tex]

    [tex]x=0[/tex]

    15.

    [tex]sech x = \coth x[/tex]

    [tex]x=arcsinh (\varphi -1 )[/tex]

    Still a lot of golden ratios.
     
  7. Jul 15, 2011 #6
    Ok, so thats why its related...
     
  8. Jul 15, 2011 #7

    uart

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    Science Advisor

    This one doesn't look right. I'm pretty sure there should be a real solution there.
     
  9. Jul 15, 2011 #8

    uart

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    Science Advisor

    After just doing some calculations I'm pretty sure that the solution to 6. should be

    [tex]x = \pm \, \cosh^{-1} \phi[/tex]

    BTW. Inverse hyperbolics can usually be alternatively represented using logs.
     
    Last edited: Jul 15, 2011
  10. Jul 15, 2011 #9
    You mean arccosh right?
     
  11. Jul 15, 2011 #10
    That works too. But my solution also works. I converted sinh x to - i sin i x and it works.
     
  12. Jul 15, 2011 #11
    And by the way, what is the LaTeX code for arcsinh, arccosh, sech, cosech?
     
  13. Jul 15, 2011 #12

    Mute

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    Homework Helper

    Two things:

    1) The inverse hyperbolic functions are not "arc(whatever)". They are actually "ar(whatever)". i.e., arsinh(x), arcosh(x), artanh(x), etc.

    2) I don't believe there is latex commands for most of these. You typically have to use \mbox{arsinh}(x), etc.

    Tests:

    [itex]\arsinh(x), \arcosh(x), \artanh(x), \sech(x), \csch(x)[/itex]

    Check to make sure latex doesn't use the misnamed versions:

    [itex]\arcsinh(x), \arccosh(x)[/itex]
     
  14. Jul 15, 2011 #13
    Ok, thanks.

    Oh. So it can be written as [tex]\mbox{arsinh}(x)[/tex]?
     
  15. Jul 15, 2011 #14

    Mute

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    Homework Helper

    Yep. If you're writing in an actual latex document, you can always define new commands so that you don't always have to use mbox.

    For example, writing

    \newcommand{\arsinh}{\mbox{arsinh}}

    in the top before the document begins would let you use \arsinh as a command.
     
  16. Jul 16, 2011 #15

    uart

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    Science Advisor

    Yes the "f^{-1}" notation is still very commonly used for both trig and hyp-trig functions. See for example : http://mathworld.wolfram.com/InverseHyperbolicFunctions.html


    Yes I know that it works but I was considering being consistent with your original post in which you were clearly only considering real solutions.

    For example :
    There are no real solutions to that equation, but there definitely are complex solutions. Basically I was pointing out that the expression in question doesn't evaluate to a real number, whereas previously you seemed to be only considering reals.
     
  17. Jul 16, 2011 #16
    I was just clarifying to make sure it was an inverse hyperbolic function and not something like sin^2...




    Are you sure there are complex solutions?

    [tex]\sin \theta =\sec \theta[/tex]

    [tex]\sin \theta =\frac{1}{\cos \theta}[/tex]

    [tex]\sin \theta \cos \theta =1[/tex]

    As the maximum of both [tex]\sin \theta[/tex] and [tex]\cos \theta[/tex] is 1, both [tex]\sin \theta[/tex] and [tex]\cos \theta[/tex] should be 1. Then as [tex]\cos \theta=1[/tex], [tex]\frac{\theta}{\pi}[/tex] is a whole number, thus [tex]\sin \theta=0[/tex]. But we have earlier said that [tex]\sin\theta=1[/tex] and [tex]1 \neq 0[/tex] so there is no value of theta.
     
  18. Jul 16, 2011 #17
    Last edited by a moderator: Apr 26, 2017
  19. Jul 16, 2011 #18
    Last edited by a moderator: Apr 26, 2017
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