Find a restricted domain so that the function is injective The sinus function is not injective when the domain is the whole $\mathbb{R}$. 
To find a restricted domain where the function is injective, can we do that only using the graph or is there also an other way? 
Such a resticted domain is for example $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right ]$. How can we find it? 
 A: Let $f: A \to B$ be any function.
The restriction of $f$ to the empty set is injective.

Let $f: A \to B$ be any function defined on a nonempty set $A$.
If $a \in A$ then $f$ restricted to the singleton set $\{a\}$ is injective.

Let $f: A \to B$ be any function defined on a nonempty set $A$. Let the range of $f$ be denoted by $D$.
Using the axiom of choice, there exist a subset $C \subset A$ such that the restriction of $f$ to $C$ defines a bijective correspondence between $C$ and $D$.

The function $f(x) = sin(x)$ has range $[-1,+1]$ and the function $g(x) = \text{arcsin}(x)$ is a right inverse of $f$. The range of $g(x)$ is the interval
$\quad [-\frac{\pi}{2},\frac{\pi}{2}]$
The function $f$ restricted to this set in injective.
If $h: [-1,+1] \to \Bbb R$ is any right inverse of $f$, then $f$ restricted to $h\big( [-1,+1] \big)$ will be injective.
A: Find the interval on the graph is fine.
As an alternative we can consider the trigonometric circle and choose any angular interval such that any value for $\sin \theta $ is reached exactly one time.
The more natural choice is $\theta\in \left[-\frac \pi 2, \frac \pi 2\right]$ but there are many other possibilities of course as for example


*

*$\theta\in \left[\frac \pi 2, 3\frac \pi 2\right]$

*$\theta\in \left[3\frac \pi 2, 5\frac \pi 2\right]$
and more in general all the intervals
$$[\theta_1,\theta_2]\subseteq  \left[-\frac \pi 2+k\pi, \frac \pi 2+k\pi\right]$$
works fine.
A: We know continuous, strictly increasing (decreasing) functions are always injective. Thus, we can just choose an interval where sin(x) is always increasing. 
