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.

Use that $|\sin u|\leq\min\{|u|,1\}$, then $2\left|\sin\left(\dfrac{x+a}{2}\right)\right|\cdot\left|\sin\left(\dfrac{x-a}{2}\right)\right|\leq 2\left|\dfrac{x-a}{2}\right|=|x-a|<\epsilon$ if we take $\delta=\epsilon$.
