This is what I've already done. Can't think of how to proceed further
$|\cos(x)-\cos(y)|=\left|-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\leq\left|\frac{x+y}{2}\right||x-y|$
What should I do next?
This is what I've already done. Can't think of how to proceed further
$|\cos(x)-\cos(y)|=\left|-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\leq\left|\frac{x+y}{2}\right||x-y|$
What should I do next?
Hint: Any continuous function is uniformly continuous on a closed, bounded interval, so $\cos$ is uniformly continuous on $[-2\pi,0]$ and $[0,2\pi]$.
The cosine function is Lipschitz-continuous because its derivative is bounded. That follows from the mean value theorem. Every Lipschitz-continuous function is uniformly continuous.
postscript prompted by Tom Oldfield's comment: A function $f$ is Lipschitz-continuous iff there is some non-negative number $m$ such that for all $x,y$ in the domain of $f$ we have $|f(x)-f(y)|\le m|x-y|$. This is stronger than uniform continuity, in that every Lipschitz-continuous function is uniformly continuous, but not every uniformly continuous function is Lipschitz continuous. An example of a uniformly continuous function that is not Lipschitz-continuous is $x\mapsto\sqrt{1-x^2}$ on the interval $[-1,1]$. It's really easy to prove that Lipschitz continuity entails uniform continuity.