Why does $\sin(\cos x)=\cos(\sin y)$ result in a lattice of circles?

Question

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?

Answer 1

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.

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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.

Answer 2

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

Answer 3

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: