Is it possible $\cos \phi (s)$ and $\sin \phi (s)$ to be nth order differentiable and not $\phi (s)$?

Question

Pressley's Differential Geometry book supposes that the angle between the tangent vector to a curve and any fixed vector is smooth (i.e. all orders of differentiablity exits and are continuous) and in order to prove that I approached a few ways and two of them blocked by the following two questions I need to prove so completing the proof for the mentioned problem:

1- Suppose that $t=(t_1,t_2)$ is a smooth function and $\langle t,n\rangle=0$ where $n$ is the unit vector perpendicular to $t$ formed by rotation of $t$ counterclockwise with angle $\pi/2$; how to prove that $n$ is smooth? (by intuition when t varies slowly so does n which is not only a rigorous proof but also suggests for just continuity).

2- $t=(t_1,t_2)=(\cos \phi (s), \sin \phi (s))$. Since $t$ is smooth so is $t_1$ and $t_2$. Since composition of a continuous function (here : sin and cos) and a discontinuous function (here: $\phi(s)$) still maybe continuous so much not possible to say about $\phi$; also arcsin and arccos are not smooth everywhere like sin and cos are; so is it is possible to deduce from smoothness of $t=(t_1,t_2)=(\cos \phi (s), \sin \phi (s))$, the smoothness of $\phi (s)$ at all?

Added: 3- If non of the two questions is answerable, so how to prove that that the angle between the tangent vector to a curve and any fixed vector is smooth, if the tangent vector is smooth?

Answer 1

Some hints:

1. You have an explicit formula for $n$. The components of $n$ are $$ (n_1, n_2) = \left( \frac{-t_2}{\sqrt{t_1^2 + t_2^2}}, \frac{t_1}{\sqrt{t_1^2 + t_2^2}} \right) $$

2. If $\phi(s)$ is a continuous function and you know $\cos(\phi(s))$ and $\sin(\phi(s))$ is smooth. Then in an interval where $\cos(\phi(s)) \neq 0$ you have by implicit differentiation $\phi'(s) = \dfrac{1}{\cos(\phi(s))} \dfrac{d}{ds} \sin(\phi(s))$ which is a ratio of two smooth functions and hence smooth.

   Now, away from the points $\cos(\phi) = \pm 1$, you have $\cos$ is (smoothly) invertible. And away from the points $\sin(\phi) = \pm 1$, you also have $\sin$ is (smoothly) invertible. Those two sets of points don't overlap.