Why is $\sin(x+y)=c$ not a smooth curve for $c = \pm 1$?

Question

I understand that, for $c= 1$ and $c=-1$, the derivative of the function $f(x,y)=\sin(x+y)-c$ is a zero vector, and hence the locus which is given by $\sin(x+y)-c=0$ fails to satisfy the definition of the smooth curve.

However, I have the following doubt:

Suppose $c = 1$.

Then,

$$ \sin(x+y) = 1 \implies y = -x + \arcsin(1)$$

So, $$y = -x + (4n+1)\pi/2, \quad\forall n\in\mathbb{Z}$$

satisfies the equation. These are a set of parallel lines. But, why is it failing to satisfy the definition of a smooth curve?

I tried to plot the function for $c=1$ in WolframAlpha, but the plot did not contain any points. Most likely, I am doing some silly mistake in the above calculations. Can anyone help in identifying the flaw in my thinking?

Answer 1

Presumably you started out by using the implicit function theorem to the zero locus of the function. $$ F(x,y)=\sin(x+y)-c=0.\qquad(*) $$ Assume that $P=(x_0,y_0)$ satisfies $(*)$. IFT promises $y$ as a smooth function of $x$ in a neighborhood of $P$, if $F_y(x_0,y_0)\neq0$.

When $c=1$ and, for example $(x_0,y_0)=(\pi/2,0)$, we see that the partial derivative $$ F_y(x,y)=\cos(x+y) $$ vanishes at $(x_0,y_0)$ because$\cos(\pi/2)=0$.

What goes wrong even though you have those lines?

The solutions of $\sin (x+y)=c$ are $$x+y=\arcsin c+n\cdot2\pi$$ and $$x+y=\pi-\arcsin c+n\cdot 2\pi.$$ All because $\sin(\pi-\alpha)=\sin\alpha$. Supplementary angles and all that.

[Comment: originally I fumbled this and misguidedly thought that the other family of lines share the slope +1, thanks to @anon for setting this straight.]

So we have two families of parallel lines. Within a family the vertical separation of any two lines is an integer multiple of $2\pi$. Furthermore, you get one family from the other with a vertical (or horizontal) translation by $\pi-2\arcsin c$. So we see that the two families actually coincide when $\arcsin c=\pm\pi/2$, i.e. when $c=\pm1$.

IFT diagnoses it when the lines describing the set of solutions "double on themselves".

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In other words, the flow of the logic in IFT is one-way-traffic only:

• A non-vanishing Jacobian promises that locally within the set of solutions the prescribed subset of coordinates is a smooth function of the remaining coordinates.
• But, the set of solutions can be the graph of a smooth function even when the Jacobian vanishes.

My favorite example of this is the smooth curve of intersection formed by a donut lying flat on the table. The points of contact of the two surfaces form a perfectly smooth circle. But, because the surface of the table is the shared plane of tangency of the donut and the table, the IFT breaks down there as the Jacobian determinant vanishes at all points of intersection.

Answer 2

It seems that Alpha is stuck because it uses a numerical solver to get the curves. As all the roots are double, it has a hard time finding changes of sign.

You can try $$\sin(x+y)=1-\epsilon$$ for smaller and smaller values of $\epsilon$. You will first see pairs of parallel lines, then they fuse with each other, then become lacunar and in the end vanish.

Anyway, the analytical expressions are quite correct.