Prove that there exists a smooth map $g\colon R\to R$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$. Let $f\colon S^1\to S^1$ be any smooth map. Prove that there exists a smooth map $g\colon\mathbb{R}\to\mathbb{R}$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.
The book told me to show that $g(2\pi)=g(0)+2\pi q$, then extend $g(t+2\pi) =g(t)+2\pi q$.
But I don't understand what they mean. How can I get function $g$ from $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$, and what is $g$?
 A: Take a particular value of $t$; compute $(\cos(t), \sin(t))$. Evaluate $f$ at that point, to get a new point $(\cos u, \sin u)$, where $u$ is defined only up to multiples of $2 \pi$. 
Now tentatively define $g(t) = u$. This gives a definition, but it's not necessarily continuous, right? (Or smooth). But that's the main idea: $g(t)$ is the angle for $f(\cos t, \sin t)$. And $q$ is an integer that you might have to add to make things work out smoothly at the endpoints of an interval.
To really solve the problem properly, you need to the path lifting theorem. I can write out detail if you like, but since you seemed to be confused by the basics, I thought I'd start here. 
One small suggestion: write '\cos' rather than 'cos' in math to make formatting look nicer.
Details: Let $I$ denote $[0, 2\pi]$. 
Let $$
h: I \to S^1: t \mapsto f(\cos t, \sin t).
$$
Since $h: I \to S^1$ is continuous, there's a map $H:I \to \mathbb R$, the universal cover of $S^1$ such that $h = p \circ H$, where $p : \mathbb R \to S^1: x \to (\cos x, \sin x)$ is the universal covering map. 
(This is because of the path lifting theorem.) 
Now compose $H$ with the map $u \mapsto u \bmod 2\pi$ to get a map 
$$
K: I \to \mathbb R : u \mapsto H(u \bmod 2\pi)
$$
which is evidently continuous as well. $K$ is the map $g$ that you're looking for. 
You now need to show that everything works out nicely, but that's fairly straightforward. The number $2\pi q$ that you're looking for is just $H(2 \pi) - H(0)$, which is an integer multiple of $2 \pi$ because of the conclusions of the path-lifting theorem, one of which is that  $p(H(0)) = p(H(2\pi))$. 
For smoothness, you've got several maps that are compositions of each other, and two out of every 3 are known to be smooth; it's not too hard to conclude that the third is as well. 
A: $$f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))=(\cos(g(t)+2\pi q),\sin(g(t)+2\pi q))$$
On the other hand 
$$f(\cos(t),\sin(t))=f(\cos(t+2\pi q),\sin(t+2\pi q))=(\cos(g(t+2\pi q)),\sin(g(t+2\pi q)))$$
From these two results you get 
$$g(t)+2\pi q=g(t+2\pi q)$$
where $q\in\mathbb{Z}$. $f$ can be thought as a map taking points from the unit circle and mapping them to another unit circle through the rule 
$$f(\cos(t))=\cos(g(t))$$
and 
$$f(\sin(t))=\sin(g(t))$$
For example of $g(t)=t$ is such a function and $f$ is the identity map.
