How to prove the function $\sin(x)$ is not onto \begin{align}
f:\mathbb{R}\rightarrow\mathbb{R},\:f\left(x\right)=\sin\left(x\right)\tag{1}
\end{align}
I know that it is not onto because for all values of $y$ past $[-1,1]$ there is no $x$. 
Graphically it makes sense but I'm finding it hard to do an actual proof for this function. 
 A: Since $\sin^2x+\cos^2x=1$, we see that $\sin^2x=1-\cos^2x\le 1$. Therefore
$$
-1\le |\sin x|\le 1
$$
for all $x\in\mathbb{R}$. In particular, there's no $x\in\mathbb{R}$ such that $\sin x=42$.
Why $42$? Well, that's obvious! ;-)
A: To show it is not one-to-one:
You only need to find two values $x_1, x_2$ with $x_1 \neq x_2$ such that $\sin(x_1) = \sin(x_2)$.
When you draw your "horizontal line test" the $x$-coordinates of the points of intersection are the $x_1$ and $x_2$ you want.

 Explicitly, $x_1 = 0, x_2 = \pi$ will work nicely.

To show it is not onto:
Just assert that there is no $x$ such that $\sin(x) = 2$, this is a reasonably "common" fact that I would think you are allowed to draw on.
A: Recall the definition for a surjection, $\forall y \in \mathbb R \exists  x\in \mathbb R  (f(x)=y)$ but since $2\not \in Im f$...
A: The sine is not onto because there is no real number $x$ such that $\sin x = 2$.
A: The sine function on the entire real line cannot be one to one because the values of the map repeat every integral multiple of $\pi/2$. By definition, such a function cannot be one to one. 
It cannot be onto $\mathbb{R}$ either because $\max\sin = 1$ and $\min\sin = -1$. Therefore, for $x$ in $\mathbb{R}$ whose absolute value is greater than $1$, $x$ is not in the range of the sine function. 
However, you can restrict the domain of sine in such a manner that it is one to one, such as from $0$ to $\pi/2$. On this restricted domain,the function is both one to one and onto. Which is why we can define the inverse sine function on this domain. 
A: A function is one to one may have different meanings. (1) one to one from x to f(x). (2) one to one from f(x) to x, and (3) both ways.
When you say, when y is out of [-1, +1], there is no x, that is the case-2 being not true. Further, since there exists multiple x values to one y in sin(x), that is the case-1 being not true. 
A: Since $\sin x$ is  periodic with period $2\pi$, it is enough to consider $\sin x$ over the interval $[0,2\pi]$. Also $\sin x$ being a continuous function defined on a compact subset $[0,2\pi]$, range of $\sin x$ must be a compact subset i.e. a closed and bounded subset of $\mathbb R$. So the range of  $\sin x$ cannot be whole of $\mathbb R$.
A: If you are letting $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x) = \sin(x)$, then, as others have pointed out, $f(x) = \sin(x)$ will not be onto because $\sin(x)\neq\gamma\in\mathbb{R}\setminus[-1,1]$; that is, not every value in $\mathbb{R}$ will be obtained by plugging in values from $\mathbb{R}$. 
However, it may help you to consider the function $f:\mathbb{R}\to[-1,1]$, where $f(x)=\sin(x)$. Can you see why this function would be onto as opposed to when $f:\mathbb{R}\to\mathbb{R}$? 
