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Using Desmos graphing, I made an equation $\sin(\cos x)=\cos(\sin y)$ (here) which resulted in a strange lattice of symbols. I know that the trigonometric functions relate a triangle's sides to its angles (SOHCAHTOA), so how does this equation create circles?

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    $\begingroup$ It's no wonder that you obtain a lattice. If you have a solution $(x^\ast, y^\ast)$ then for any integer $k$ and $m$ the $(x^\ast + 2 \pi k, y^\ast + 2 \pi m)$ would be a solution too. $\endgroup$ – Evgeny Dec 1 '14 at 22:52
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    $\begingroup$ You can watch several videos on youtube searching for, say, unit circle sin cos and see how those functions are intimately related to circles. Doesn't really answer your question about your function, but about how those trig functions relate to circles as well as triangles. $\endgroup$ – turkeyhundt Dec 1 '14 at 22:53
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    $\begingroup$ Note that you can take $tan^{-1}$ of both sides and it reduces to $\sin(\cos(x)) = \cos(\sin(y))$. $\endgroup$ – Chantry Cargill Dec 1 '14 at 22:54
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    $\begingroup$ but isn't $SOHCAHTOA$ equal to $SO^2H^2CA^2T$? $\endgroup$ – user153330 Feb 1 '16 at 19:19
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Each of the following steps destroys some solutions, but we get them all back via Evgeny's comment above

$ \begin{align*} \sin(\cos(x)) &= \cos(\sin(y))\\ \cos(x) &= \sin^{-1}\cos(\sin(y))\\ \cos(x) &= \frac{\pi}{2}-\sin(y)\\ \cos(x) + \sin(y)&=\frac{\pi}{2} \end{align*} $

you can graph this and see that you get a row of "circles" like what you started with.

Tellingly, graphing $\cos(x) + \sin(y) =\frac{\pi}{n}$ for $n=2,3,4,5,..$ the graphs are no longer circles, but seem to approximate squares as $n \to \infty$.

This leads me to believe that $\cos(x) + \sin(y) =\frac{\pi}{2}$ is also not a circle, but just looks kind of like one. I will return to this post once I have found a proof that it is not circle, but hopefully this bit of progress will be helpful in the mean time.


Here is a proof:

$$x^2+\left(y-\frac{\pi }{2}\right)^2=\arccos \left(\frac{\pi }{2}-1\right)^2$$

is a circle which agrees with the "circle" given by OP at the $4$ points $$(\arccos(\frac{\pi}{2})-1,\pi/2)$$ $$(-\arccos(\frac{\pi}{2})-1,\pi/2)$$ $$(0,\frac{\pi}{2} + \arccos(\frac{\pi}{2}-1))$$ $$(0,\frac{\pi}{2} - \arccos(\frac{\pi}{2}-1))$$

but not at other points, as can be easily seen. Since two circles sharing 4 points must be identical, OP's "psuedocircle" cannot actually be a circle. It is curious that it is so close to one though.

The graphs of $\cos(x)+\sin(y)=t$ for $t \to 2$ get smaller and smaller, but also more and more perfectly circular. It seems that OP's example is just somewhere along that continuum.

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  • $\begingroup$ what's OP? I see it a lot and don't know what it means... $\endgroup$ – Conor O'Brien Dec 1 '14 at 23:54
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    $\begingroup$ OP = original poster, i.e. @ConorO'Brien in this case. $\endgroup$ – Robert Israel Dec 2 '14 at 0:00
  • $\begingroup$ Oooooooooh thanks! $\endgroup$ – Conor O'Brien Dec 2 '14 at 0:05
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If we write $y = \pi/2 + s$, then near $(x =0, y=\pi/2)$ we have $$\cos(x) + \sin(y) = \cos(x) + \cos(s) = 2 - (x^2 + s^2)/2 + (x^4 + s^4)/4! - \ldots$$ When $t > 0$ is small, and $x$ and $s$ are small, the terms in fourth and higher powers of $x$ and $s$ are negligible, so the equation $\cos(x) + \sin(y) = 2 - t$ is well approximated by $x^2 + s^2 = 2 t$, which is a circle.

Including more terms of the series, we get

$$ x = \sqrt{2t} u(t,\theta) \cos(\theta), s = \sqrt{2t} u(t,\theta) \sin(\theta)$$ where $$ u(t,\theta) = 1+ \left( \dfrac{\cos \left( 4\,\theta \right)}{48} +\dfrac{1}{16} \right) t + \left( {\frac {7\,\cos \left( 8\,\theta \right) }{9216}}+{\frac {9\,\cos \left( 4\,\theta \right) }{1280}}+{\frac {101}{9216}} \right) {t}^{2} + \ldots $$

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  • $\begingroup$ This is very insightful. $\endgroup$ – Steven Gubkin Dec 2 '14 at 0:19
  • $\begingroup$ yeah great answer +1 $\endgroup$ – user153330 Feb 1 '16 at 19:18
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I do like Desmos. I want to like it more, but it does have limitations. In this particular case, you can see how the loops arise quite clearly by plotting the 3D graphs of $z=\sin(\cos(x))$ and $z=\cos(\sin(y)$ on the same set of axes. The result looks like so:

enter image description here

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