Let $f: \mathbb R \to \mathbb R$. A nonregular value $y$ of $f$ is any value such that not all $x \in f^{-1}(y)$ are regular. A point is regular if the Jacobian at it is surjective, in this case, has rank one.
If we tilt the map $f(x) = \sin x$ by some angle about the origin then the resulting map has countably infinitely many nonregular values.
My problem is: I don't understand why this is the case.
Please could someone give me some geometric insight into why tilting the sine functions yields nonregular values?
It seems to me that the untilted sine function has just as many nonregular values: The Jacobian is the derivative $J(x) = \cos x$ which vanishes infinitely many times...?
I am thinking about this because of the following text I read:
Note that without the tilting, all of the zeros of the derivative satisfy $\sin(x) = \pm 1$. Thus, even though there are infinite points $x$ with $f'(x) = 0$, the only nonregular values (referring to elements of the codomain rather than elements of the domain) of $\sin x$ are $\pm 1$.