Prove that $f(x) = (\sin x, \cos x)$ defined from $\mathbb R$ to the unit circle is onto How do you prove that the function $f: [0,2\pi$) $\rightarrow S^1$ such that $f(x) = (\sin x, \cos x)$, where $S^1$ is the unit circle, is onto?
I was thinking considering two variables a and b in the range of f(x) and stating that f(arcsin(a)) and f(arccos(b)) produce x, but that only works for the x and y coordinates one at a time, not simultaneously. Help?
 A: There really isn't much more to this then you expect. Consider some arbitrary point in the codomain: $(\sin(a),\cos(a))$. By the definition of onto, all you have to do is show that there is some point in the domain $[0,2\pi)$ such that $f(x)=(\sin(a),\cos(a))$. Can you see how to take it from here?
A: Hint.
The function $$
\begin{array}{l|rcl}
f : & [0,\pi] & \longrightarrow & [-1,1]\\
    & x & \longmapsto & \cos x\end{array}$$ is onto.
A: Yes. Let $(y_{1},y_{2}) \in S^{1}$; let $r:= \sqrt{y_{1}^{2} + y_{2}^{2}}$. Then there is some $x \in [0, 2\pi[$ such that $x$ is the angle made by the vector $(y_{1},y_{2})$ with the horizontal axis and $y_{1} = r\cos x$ and $y_{2} = r\sin x$; but this shows that $f$ is surjective (onto).
A: This is one of the most basic facts of analysis. If you assume the high school definition of $\cos$ and $\sin$, together with their continuity, ranges, etc., it is of course obvious, as declared in the accepted answer. 
But if you start with an analytical definition of the trigonometric functions, say in terms of the exponential function,  of power series, or of ODE's, a full proof requires about two pages of work. You can find such a proof in the section The trigonometric functions, Chapter 8 of Baby Rudin.
