How would I prove that $1-\sin(x)$ is one-to-one or not? I'm currently at the very beginning of my calculus course and I've run into an immediate issue in that I'm not entirely sure how to verify when a function is one-to-one. I know that they're considered one-to-one if $f(x)=f(y)$ when $y=x$, but when I apply this to $f(x)=1-\sin(x)$, the function appears as one-to-one even though I know it isn't unless the domain is restricted.
My work thus far is as follows:
Verify if $f(x)=1-\sin(x)$ is one-to-one
$f(a)=1-\sin(a)$ ;
$f(b)=1-\sin(b)$
$\Rightarrow1-\sin(a)=1-\sin(b)$
$\Rightarrow-\sin(a)=-\sin(b) \rightarrow \sin(a)=\sin(b)$
$\Rightarrow \arcsin(\sin(a))=\arcsin(\sin(b) \rightarrow a=b $
I know I'm doing something wrong, likely with the use of arcsin, but I'm not entirely sure.
 A: $f$ is not one-to one:
$f(x)=f(x +2 \pi)$.
A: Without any restriction on the domain, the function $f(x) = 1- \sin x$ is not one-to-one.
Your proof is good only when you restrict your domain for example to $(-\pi /2, \pi /2)$ 
A: Your definition of one-to-one isn't totally correct.  A better definition is "For all x,y if f(x) = f(y) then x = y".
Simply finding an example of a case where this is true isn't enough you have to take the next step of showing that there can be no other possible case.
A: From trigonometry,
$$1-\sin x=1-\sin y\iff x=y+2k\pi\lor x=\pi-y+2k\pi$$
which is not the relation we are looking for.
But if we restrict $x,y$ to the fourth and first quadrants, the term $2k\pi$ vanishes, as does $\pi-y$ (otherwise the angles would lie in other quadrants).
Then for
$$x,y\in\left[-\frac\pi2,\frac\pi2\right],x=y\iff1-\sin x=1-\sin y.$$

A: $$sin(x)=sin(y)$$
gives 
$$x=\sin^{-1}(\sin y)+2n\pi,\hspace{10pt} n\in Z$$ 
$$x=(2n+1)\pi-\sin^{-1}(\sin y),\hspace{10pt} n\in Z$$ 
thus $x\ne y$
$f(x)$ is not one-to-one
A: To show that $f$ is one-to-one, we need to show that if $f(a)=f(b)$, then $a=b$.
Suppose we have a $g$ so that $g(f(x))=x$ for all $x$; then
$$
f(a)=f(b)\implies g(f(a))=g(f(b))\implies a=b
$$
Thus, assuming that we have an $\arcsin$ so that $\overbrace{\arcsin}^g(\overbrace{\sin}^f(x))=x$ for all $x$ assumes what is to be proven.
However, the fact that $\sin(x)=\sin(x+2\pi)$ shows that $\sin$ is not one-to-one.
A: In more easy and simple way to show that the function  y=1-sin(x) is not one to one  Consider that
1-sin(0)=1=1-sin(pi) while 0<>pi
that is different Abscissas gives the ordinate.
