Conceptualization for proof of continuity of $sinx$ First noting that $|\sin(x)| &lt; |x|$ and that $-1 \leq\sin(x) \leq 1$ and $-1 \leq \cos(x) \leq 1$
For all real $x$ and $a$:
$$\sin(x) - \sin(a) = 2 \sin(\frac{x-a}{2}) \cos(\frac{x+a}{2})$$
using the fact that $|\cos{\frac{x+a}{2}}| \leq 1$ we get:
$$|\sin(x)-\sin(a)| \leq 2 |\sin(\frac{x-a}{2})| \leq |x-a|$$
This apparently shows that sin is continuous everywhere. I don't see why. Is it trying to use squeeze theorem? If so, would this be the logical continuation (for any $a \in R$):
$$\lim\limits_{x \to a} |\sin(x)-\sin(a)| \leq  \lim\limits_{x \to a} 2 |\sin(\frac{x-a}{2})| \leq \lim\limits_{x \to a} |x-a|$$
$$ 0 \leq  \lim\limits_{x \to a} 2 |\sin(\frac{x-a}{2})| \leq 0$$
Hence we get $\lim\limits_{x \to a} |\sin(\frac{x-a}{2})| = 0$ showing that it is continous ? Could someone confirm this idea?
The squeeze theorem is really on
$$0 \le |\sin(x) - \sin(a)| \le |x-a|.$$
Since $\lim_{x \to a} |x-a| = 0$, we obtain
$$\lim_{x \to a} |\sin(x) - \sin(a)| = 0,$$
which is equivalent to
$$\lim_{x \to a} \sin(x) = \sin(a),$$
which is the definition of continuity for the function $f(x)=\sin(x)$.
The $\epsilon-\delta$ definition of continuity is:

A function $f$ is continuous at $x_0$ in its domain if for every $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that whenever x is in the domain of $f$ and $|x − x_0| &lt; \delta$, we have $|f(x) − f(x_0)| &lt; \epsilon$. Again, we say $f$ is continuous if it is continuous at every point in its domain.

Basically can we always find a range of $\delta&gt;0$ around $x_0 = a$ where at every point the sin($x$) is within (as small as we choose) $\epsilon$-close to the sin($a$)? $\quad$ Yes. 

If you pick the $\delta$ to be $\epsilon$ itself, then any $x$ such that $|x- a|&lt;\epsilon$, will indeed satisfy
$|$sin($x$) - sin($a$)$| \leq |x - a| &lt; \epsilon$
This follows from definition really.