Properties of $\sin$ and $\cos$ In my textbook I have the following:
If $\alpha:[a,b] \rightarrow \mathbb{R} \backslash \{0\}$ is continuous, we can set 
$$\frac{\alpha}{||\alpha||}(t) = (\cos\gamma(t),\sin\gamma(t))$$
for some $\color{red}{continuous}$ $\gamma(t)$ ,up to jumps of $2 \pi n$ (1)
Intuitively I can see that this should be correct. But I can not seem to show why this is? 
I mean, we  know that $(\cos[\gamma(t) + 2n\pi],\sin[\gamma(t) + 2n\pi]) = (\cos\gamma(t),\sin\gamma(t))$
...

Could someone explain line (1) a little more?

 A: If $\alpha$ is continuoustly differentiable, it suffices to chose a primitive of $\displaystyle\frac{\alpha'}{\alpha}$ for $\gamma$.
Otherwise, it is a consequence of the path lifting theorem applied to the covering space of $\mathbb{S}^1$ by $\mathbb{R}$.
A: What you post does not say that $\gamma$ must be continuous.  It says that $\gamma$ can be chosen to be continuous.
A: $\frac{\alpha(t)}{||\alpha(t)||}$ is a unit vector for every $t$, so it lies on the unit circle. Now since $\alpha$ is continuous, $\frac{\alpha(t)}{||\alpha(t)||}$ is a continuous function on the unit circle, so we can parametrize it by $(\cos(\gamma(t)), \sin(\gamma(t)))$ for some continuous $\gamma(t)$.
A: So $\|\alpha(t)\|$ is continuous and never zero, so $F(t) = \frac{\alpha(t)}{\|\alpha(t)\|}$ is continuous (vector-valued) function of $t$.  Now $F(t)$ is always of norm 1, because
$$
\left\| \frac{\alpha(t)}{\|\alpha(t)\|} \right\|
=
\frac{\|\alpha(t)\|}{\| \alpha (t)\|} = 1
$$
We know that $(\cos \theta, \sin \theta)$ are points on the unit circle, so from that
$$
\frac{\alpha(t)}{\|\alpha(t)\|}
=
(\cos \gamma(t),\sin \gamma(t))
$$
Now why is $\gamma$ continuous?  Because $\cos \gamma(t)$ must be continuous and $\sin \gamma(t)$ must be continuous.  That doesn't necessarily imply that $\gamma(t)$ is continuous.  In fact, you could choose a $\gamma$ that is not continuous.  The $\gamma$ that you pick is not unique!
A simple intuitive argument that you can pick a continuous $\gamma$ is that if you change the coordinates of a point on the unit circle a little bit, you only need to change the angle a little bit.  As $\gamma(t)$ is the angle, this says that you should be able to pick a continuous $\gamma(t)$.
Be careful about simply applying $\arccos$ and $\arcsin$.  The $\gamma(t)$ that we chose can quite possibly span over several multiples of $2\pi$.  For example take $\alpha(t) = (\cos(10t),\sin(10t))$ for $0 \leq t \leq 2\pi$.  Taking $\arccos(\cos(10t))$ using the standard branch of arccos
does not end up with $10t$, it ends up with a spiky (although continuous) function.  Taking $\arccos$ will always end up with a continuous function, but most likely not even differentiable.  Also it must also work for $\arcsin(\sin(10t))$ which is a different function, so simply taking $\arccos$ and $\arcsin$ using the standard branches does not get you your $\gamma$.
