# Trignometric and hyperbolic equalities: Why the golden ratio?

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?"

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.

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...

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.

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

Ok, so thats 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.

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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?

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.

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

Mute
Homework Helper
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:

$\arcsinh(x), \arccosh(x)$

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
Homework Helper
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.

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.

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