Trignometric and hyperbolic equalities: Why the golden ratio?

[dimension10]

1.

\sin \theta = \cos \theta

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

2.

\sin \theta = \tan \theta

\theta = 0

3.

\cos \theta = \tan \theta

\theta =\arcsin (\varphi -1)

4.

\sin \theta = \csc \theta

\theta = \frac{\pi}{2}

5.

\sin \theta =\sec \theta

\theta does not exist.

6.

\sin \theta =\cot \theta

\theta = \arccos (\varphi -1)

7.

\cos \theta =\csc \theta

\theta does not exist.

8.

\cos \theta =\sec \theta

\theta=0

9.

\cos \theta = \cot \theta

\theta=\frac{\pi}{2}

10.

\tan \theta =\csc \theta

\theta =\arccos(\varphi-1)

11.

\tan \theta = \sec \theta\theta=\frac{\pi}{2}

12.

\tan \theta = \cot \theta

\theta=\frac{\pi}{4}13.

\csc \theta =\sec \theta

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

14.

\csc \theta =\cot \theta

\theta = \arccos (\varphi -1)

15.

\sec \theta =\cot \theta

\theta=\arcsin (\varphi - 1)

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

 

[Avijeet]

Hi!
Interesting question. I have always been fascinated by the golden ratio which keeps appearing at places where you least expect Will have to think about this one.

 

[dimension10]

Oops. There was a LaTeX error.

1.

\sin \theta = \cos \theta

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

2.

\sin \theta = \tan \theta

\theta = 0

3.

\cos \theta = \tan \theta

\theta =\arcsin (\varphi -1)

4.

\sin \theta = \csc \theta

\theta = \frac{\pi}{2}

5.

\sin \theta =\sec \theta

\theta does not exist.

6.

\sin \theta =\cot \theta

\theta = \arccos (\varphi -1)

7.

\cos \theta =\csc \theta

\theta does not exist.

8.

\cos \theta =\sec \theta

\theta=0

9.

\cos \theta = \cot \theta

\theta=\frac{\pi}{2}

10.

\tan \theta =\csc \theta

\theta =\arccos(\varphi-1)

11.

\tan \theta = \sec \theta


\theta=\frac{\pi}{2}

12.

\tan \theta = \cot \theta

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


13.

\csc \theta =\sec \theta

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

14.

\csc \theta =\cot \theta

\theta = \arccos (\varphi -1)

15.

\sec \theta =\cot \theta

\theta=\arcsin (\varphi - 1)

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

 

[pwsnafu]

IIRC, \cos\frac{\pi}{5} = \frac{\phi}{2}. That's related to the pentagon. Hmm...

 

[dimension10]

Here are the hyperbolic equalities.

1.

\sinh x = \cosh x

x= \infty

2.

\sinh x =\tanh x

x=0

3.

\cosh x =\tanh x

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


{x}_{2}=\frac{- \arcsin (\frac{\sqrt{3}}{2}-\frac{i}{2})}{i}

4.

\sinh x = csch x

x=arcsinh 1

5.

\sinh x = sech x

x=\frac{\ln (2 \varphi +1)}{2}

6.

\sinh x =\coth x

x=\frac{\arccos(1-\varphi)}{i}

7.

\cosh x =csch x

x=arcsinh \sqrt{\varphi-1}

8.

\cosh x =sech x

x=0

9.

\cosh x =\coth x

x=arcsinh (\varphi - 1)

10.

\tanh x = csch x

x=arccosh (\varphi - 1)

11.

\tanh x =sech x

x=arcsinh 1

12.

\tanh x = \coth x

x=\infty

13.

csch x =sech x

x=arctanh 1

14.

csch x =\coth x

x=0

15.

sech x = \coth x

x=arcsinh (\varphi -1 )

Still a lot of golden ratios.

 

[dimension10]

IIRC, \cos\frac{\pi}{5} = \frac{\phi}{2}. That's related to the pentagon. Hmm...

Ok, so that's why its related...

 

[uart]

Here are the hyperbolic equalities.


6.

\sinh x =\coth x

x=\frac{\arccos(1-\varphi)}{i}

This one doesn't look right. I'm pretty sure there should be a real solution there.

 

[uart]

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

x = \pm \, \cosh^{-1} \phi

BTW. Inverse hyperbolics can usually be alternatively represented using logs.

 

[dimension10]

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

x = \pm \, \cosh^{-1} \phi

BTW. Inverse hyperbolics can usually be alternatively represented using logs.

You mean arccosh right?

 

[dimension10]

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

x = \pm \, \cosh^{-1} \phi

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.

 

[dimension10]

And by the way, what is the LaTeX code for arcsinh, arccosh, sech, cosech?

 

[Mute]

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:

\arsinh(x), \arcosh(x), \artanh(x), \sech(x), \csch(x)

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

\arcsinh(x), \arccosh(x)

 

[dimension10]

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:

\arsinh(x), \arcosh(x), \artanh(x), \sech(x), \csch(x)

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

\arcsinh(x), \arccosh(x)

Oh. So it can be written as \mbox{arsinh}(x)?

 

[Mute]

Oh. So it can be written as \mbox{arsinh}(x)?

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.

 

[uart]

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.

\sin \theta =\sec \theta

\theta 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.

 

[dimension10]

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?

\sin \theta =\sec \theta

\sin \theta =\frac{1}{\cos \theta}

\sin \theta \cos \theta =1

As the maximum of both \sin \theta and \cos \theta is 1, both \sin \theta and \cos \theta should be 1. Then as \cos \theta=1, \frac{\theta}{\pi} is a whole number, thus \sin \theta=0. But we have earlier said that \sin\theta=1 and 1 \neq 0 so there is no value of theta.

 

[pmsrw3]

As the maximum of both \sin \theta and \cos \theta is 1,...

Only if theta is real.

Here are the solutions of \sin(\theta) = \sec(\theta) (courtesy of Wolfram Alpha): http://www.wolframalpha.com/input/?i=Solve+sin(theta)+==+sec(theta)"

 

[dimension10]

Only if theta is real.

Here are the solutions of \sin(\theta) = \sec(\theta) (courtesy of Wolfram Alpha): http://www.wolframalpha.com/input/?i=Solve+sin(theta)+==+sec(theta)"

So there is an arcsine of numbers greater than 1. Wow.