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.