Prove $\cos(x)$ is continuous I'm stuck at a particular step and could use some help.
By definition, a function is continuous at $x=a$ iff $\lim_{x \to a} f(x) = f(a)$.
So I assume to prove $\cos(x)$ is continuous we must use the definition of a limit to show that: 
$$\lim_{x \to a} \cos(x) = \cos(a) \iff \forall \epsilon > 0, \exists \delta>0 : 0 < |x - a| < \delta \implies |\cos(x)-\cos(a)| < \epsilon$$
From some triangle proofs it can be shown that:
$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$
$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha) \sin(\beta)$
Subtract the two equations:
$$\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2\sin(\alpha) \sin(\beta)$$
Let $x = \alpha + \beta$ and $a = \alpha - \beta$. 
Then $\alpha = \frac{x+a}{2}$ and $\beta = \frac{x-a}{2}$, implying:
$$\cos(x) - \cos(a) = -2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)$$
So
$$\left|\cos(x)-\cos(a)\right| = 2\left|\sin\left(\frac{x+a}{2}\right)\right| \cdot \left|\sin\left(\frac{x-a}{2}\right)\right|$$
I feel like I am close because I was able to change it so at least we have an $x-a$ term but now I'm not sure where to go from here.

Note that
$$
\cos(x+h)=\cos(x)\cos(h) - \sin(x)\sin(h)
$$
and so
\begin{align}
\lim_{h \to 0}\cos(x+h) &= \lim_{h \to 0}\cos(x)\cos(h)-\lim_{h \to 0}\sin(x)\sin(h) \\
&= \cos x\lim_{h \to 0}\cos(h)-\sin(x)\lim_{h \to 0}\sin(h) \, .
\end{align}
The problem reduces to showing that $\sin$ and $\cos$ are continuous at $0$. For $\sin$, note that when $h>0$,
$$
0 < \sin(h) < h \, ,
$$
and since $\lim_{h \to 0^{+}}h=0$, by the squeeze theorem we find that $\lim_{h \to 0^+}\sin(h)=0$. The same argument can be made for $x<0$, but with the inequalities reversed. Hence, $\lim_{h \to 0}\sin(h)=0$. We can prove $\lim_{h \to 0}\cos(h)=1$ directly by the epsilon-delta definition of a limit. First, note that because $\cos$ is an even function, this problem reduces to proving that $\lim_{h \to 0^+}\cos(h)=1$. Now, if we set $\delta=\arccos(1-\varepsilon)$*, then
\begin{align}
& h < \arccos(1-\varepsilon) \\
\implies & \cos(h) > 1-\varepsilon \\
\implies & \cos(h) - 1 > -\varepsilon \\
\implies & 1 - \cos(h) < \varepsilon
\end{align}
and the result follows.

If $\varepsilon > 2$, then any value of $\delta$ suffices.
