Trignometric and hyperbolic equalities: Why the golden ratio?

  • #1
371
0
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?"
 

Answers and Replies

  • #2
39
0
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.
 
  • #3
371
0
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?"
 
  • #4
pwsnafu
Science Advisor
1,080
85
IIRC, [itex]\cos\frac{\pi}{5} = \frac{\phi}{2}[/itex]. That's related to the pentagon. Hmm...
 
  • #5
371
0
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.
 
  • #6
371
0
IIRC, [itex]\cos\frac{\pi}{5} = \frac{\phi}{2}[/itex]. That's related to the pentagon. Hmm...
Ok, so thats why its related...
 
  • #7
uart
Science Advisor
2,776
9
Here are the hyperbolic equalities.


6.

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

[tex]x=\frac{\arccos(1-\varphi)}{i}[/tex]
This one doesn't look right. I'm pretty sure there should be a real solution there.
 
  • #8
uart
Science Advisor
2,776
9
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:
  • #9
371
0
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.
You mean arccosh right?
 
  • #10
371
0
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.
That works too. But my solution also works. I converted sinh x to - i sin i x and it works.
 
  • #11
371
0
And by the way, what is the LaTeX code for arcsinh, arccosh, sech, cosech?
 
  • #12
Mute
Homework Helper
1,388
10
And by the way, what is the LaTeX code for arcsinh, arccosh, sech, cosech?
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]
 
  • #13
371
0
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.
Ok, thanks.

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]
Oh. So it can be written as [tex]\mbox{arsinh}(x)[/tex]?
 
  • #14
Mute
Homework Helper
1,388
10
Oh. So it can be written as [tex]\mbox{arsinh}(x)[/tex]?
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.
 
  • #15
uart
Science Advisor
2,776
9
You mean arccosh right?
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


That works too. But my solution also works. I converted sinh x to - i sin i x and it works.
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 :
5.

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

[tex]\theta[/tex] does not exist.
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.
 
  • #16
371
0
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
I was just clarifying to make sure it was an inverse hyperbolic function and not something like sin^2...




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.
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.
 

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