Show that a trig function is bijective. Let $g:(1,\infty)\rightarrow S^1\setminus\{(1,0)\}$ such that $g(x)=(cos(2\pi/x),sin(2\pi/x))$. I want to show that this map is a bijection.
For injectivity, I know that I need to show $x_1 = x_2$ whenever $g(x_1) = g(x_2)$. I was trying to show that it forces $2\pi/x_1 = 2\pi/x_2$ in the given domain. I tried to use trig relations here but I have no idea how should I do this.
For surjectivity, suppose $(a,b) \in S^1\setminus(1,0)$, I can set $x = 2\pi/cos^{-1}(a)$. Am I correct?
Thanks!
 A: If you define$$\begin{array}{rccc}f\colon&(1,\infty)&\longrightarrow&(0,1)\\&x&\mapsto&\dfrac1x\end{array}$$and$$\begin{array}{rccc}h\colon&(0,1)&\longrightarrow&S^1\setminus\{(1,0)\}\\&x&\mapsto&\bigl(\cos(2\pi x),\sin(2\pi x)\bigr),\end{array}$$then $g=h\circ f$. Clearly, $f$ is a bijection. So, all you need to prove is that $h$ is a bijection too.
If $h(x)=h(y)$, then $\cos(2\pi x)=\cos(2\pi y)$ and $\sin(2\pi x)=\sin(2\pi y)$. So,\begin{align}1&=\cos^2(2\pi x)+\sin^2(2\pi x)\\&=\cos(2\pi x)\cos(2\pi y)+\sin(2\pi x)\sin(2\pi y)\\&=\cos(2\pi(x-y))\end{align}and therefore $x-y\in\Bbb Z$. But $x-y\in(-1,1)$ too, and so $x-y=0$. That is, $x=y$.
And if $(a,b)\in S^1\setminus\{(1,0)\}$, take$$x=\begin{cases}\frac1{2\pi}\arccos(a)&\text{ if }b\geqslant0\\1-\frac1{2\pi}\arccos(a)&\text{ otherwise.}\end{cases}$$
A: Suppose $g(x)=g(y)$, then
\begin{align}
\cos\left(\frac{2\pi}{x}\right)=\cos\left(\frac{2\pi}{y}\right), \ \ \ \ \text{and} \ \ \ \ \ \ \sin\left(\frac{2\pi}{x}\right)=\sin\left(\frac{2\pi}{y}\right).
\end{align}
This imply
$$
\frac{2\pi}{x}=\frac{2\pi}{y}\ \ \ \ \text{or}\ \ \ \ \frac{2\pi}{x}=2k\pi-\frac{2\pi}{y},
$$
with $k\in\mathbb{N}$, for the first equality, and for the second equality
$$
\frac{2\pi}{x}=\frac{2\pi}{y}\ \ \ \ \text{or}\ \ \ \ \frac{2\pi}{x}=(2\ell+1)\pi-\frac{2\pi}{y},
$$
with $\ell\in\mathbb{N}$.
Now, if $\frac{2\pi}{x}=\frac{2\pi}{y}$ you are done.
Otherwise, if $\frac{2\pi}{x}=2k\pi-\frac{2\pi}{y}$ and $\frac{2\pi}{x}=(2\ell+1)\pi-\frac{2\pi}{y}$, you get $2k=2\ell+1$, absurd.
