# Explaining the graph of $\sin(x^2) + \sin(y^2) = 1$

I had to plot the graph of the implicitly defined function $$\sin^2 x + \sin^2 y = 1$$ in an exam. This is not particularly difficult, but it got me wondering what the graph would look like when the exponent is taken inside, viz.

$$\sin(x^2) + \sin(y^2) = 1$$

I found it difficult to figure this out, so I resorted to Desmos' graphing calculator. It looks like this:

I can explain some parts of this picture, but others elude me, and I think someone with more experience will do a better job of saying why this thing looks like it does.

I'd be particularly interested to know whether the figure in the middle is a special case of some other function, and similarly with the curlicues on the axes. (I have a good idea of what the polka dots are.)

Pre-emptive note: I had no trouble plotting the other thing mentioned ($$\sin^2 x + \sin^2 y = 1$$), so you needn't bother including that in your answer.

EDIT: After looking at @Jean Marie's answer below, I plotted the graph of $$(x^2 + y^2) - \frac{x^6 + y^6}{6} = 1,$$ reasoning that near the origin a few terms of the Taylor series might help. The resulting graph was quite similar to the strange shape near the origin in thing above. Somewhat unexpectedly (at least for a callow neophyte like me), a much closer approximation (shapewise) was offered by $$x^2 + y^2 - \frac{x^4 + y^4}{4} = 1$$

In case it is of any use, here is a picture (from Desmos) of the two plots I mentioned.

In the figure above, the blue is the sextic and the red the quartic.

• wolframalpha.com/input/…
– mwt
Sep 22, 2020 at 6:52
• @mwt That is not the OP's function... Sep 22, 2020 at 6:53
• @arvenka You have two different questions: in your title, you wrote $\;\sin x^2+\sin y^2=1\;$ ,whereas in the body of your questis it's written $\;\sin^2x+\sin^2 y=1\;$ . Pay attention to the fact that $\;\sin x^2\neq \sin^2 x\;$ ...! Sep 22, 2020 at 6:54
• @DonAntonio, I am aware of that. The first thing was just to explain the motivation for asking the question. As I said, I posted the question by mistake before it was finished. I hope there are no typos now? Sep 22, 2020 at 6:56
• In any manner, this is beautiful ! $\to +1$ Sep 22, 2020 at 7:15

The surface of equation $$z=\sin x+\sin y$$ has the shape of an "egg tray". It has maxima $$z=2$$ on a periodic grid, and this is why the level curves $$z=1$$ are regularly spaced approximate circles.

Now if we replace $$x$$ by $$x^2$$, we deform space horizontally so that $$x^2$$ increases faster and faster, giving a "compression effect".

By replacing $$y$$ with $$y^2$$, we get the effect on both axis.

Of course, the picture is symmetric by reflection, as the square function is even.

Remains to explain the "spikes" against the axis. If $$y$$ is small,

$$\sin x^2+\sin y^2=1\to y\approx\sqrt{1-\sin x^2}=\sqrt2\left|\cos\frac{x^2}2\right|$$

shows them.

• I like this! As it happens, I tried reasoning from the graph $\sin x + \sin y = 1$ also (that's why I was not too confused by the appearance of the circlish things that I called polka dots), but I couldn't account for the airfoil shapes near the axes or funny thing near the origin. And Desmos' plot of $\sin x^2 + \sin y = 1$ looked a bit odd, so I dropped this line. Clearly I forgot to account for the way $x^2$ is sublinear for small values and only later dominates, and your double stretching procedure explains the second. Sep 23, 2020 at 8:31
• Can you also explain the similarity to the graph of $(x^2 + y^2) - \frac{x^4 + y^4}{4} = 1$ that I described in an edit? Or is that perhaps serendipitous? Sep 23, 2020 at 8:33
• @arvenka:this is because $x^2-\frac{x^4}4$ peaks at $1$, while $x^2-\frac{x^6}6$ does not reach it.
– user65203
Sep 23, 2020 at 8:42

The picture you gave can be seen as the contour line at $$z=1$$ of the surface with equation:

$$z=\sin(x^2)+\sin(y^2)=2\sin\left(\frac{x^2+y^2}2\right)\cos\left(\frac{x^2-y^2}2\right)\tag{1}$$

(the RHS has been given by DonAntonio).

Here is a representation of this surface together with a certain number of contour lines:

This 3D surface (similar to a compressed eggbox) provides a help for understanding what's happening.

For example, in the vicinity of $$(0,0)$$, we see an almost circular contour line, explained by the fact that, using the rightmost expression of (1) $$z$$ is equivalent to $$x^2+y^2$$ in this area.

Another example: the different lemniscate-shaped level lines are classical phenomena in the vicinity of saddle points.

• I'm quite happy with your explanation of the lemniscates. It didn't strike me that thinking of the plot as the cross section of a surface would present advantages, but I can understand that saddle points (a somewhat uncomfortable saddle, though!) would lead to this. As for the appearance of the graph near (0,0), I felt that it was not exactly circular near the origin because $x$ is large enough that $\sin x = x$ is no longer good, so I added another term from the Taylor series. I have incorporated this into an edit of my question. Sep 23, 2020 at 7:48
• About the behavior near $0$, the yellow level line is almost circular... Sep 23, 2020 at 20:28

An idea: since $$\;\sin x+\sin y=2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)\;$$, the equation you have is equivalent to

$$\sin x^2+\sin y^2=1\iff\sin\left(\frac{x^2+y^2}2\right)\cos\left(\frac{x^2-y^2}2\right)=\frac12$$

Now, for example: from the left form above we can have $$\;x^2=y^2=\frac\pi2+2k\pi\;,\;\;k\in\Bbb N\cup\{0\}\;$$ (it must be $$\;k\ge 0\;$$ , of course...), or also

$$x^2=y^2=\begin{cases}\cfrac\pi6\\{}\\\cfrac{5\pi}6\end{cases}\;\;\;+\;2k\pi\;,\;\;k\in\Bbb N\cup\{0\}$$

and etc. From the right form of the equation above, we could as well get, for example,

$$\begin{cases}\cfrac{x^2+y^2}2=\cfrac\pi2\\{}\\\cfrac{x^2-y^2}2=\pm\cfrac\pi3\end{cases}\;\;\;+2k\pi\;,\;\;\;k\in\Bbb N\cup\{0\}$$

and etc. Solving the above infinite systems of equations gives the weird plot you got.

• Hi, mon ami ! How are you in this terrible time ? Take care. Sep 22, 2020 at 7:16
• @ClaudeLeibovici Salut cher ami ! We're doing fine, than you for asking...and I hope you're doing fine there. It's been a tough time since we haven't been able to travel (and we wanted to go to Paris...ah!), but at least we didn't lose our jobs, our families are fine, the kids and etc. I hope you and your beloved ones are trés bien. Please do take care. Sep 22, 2020 at 10:16
• @Claude Leibovici Hi, Claude! Happy to see that you are well! Not a good time, but some knowledge of history shows that there have been much worst times in the past... Sep 22, 2020 at 13:52