Prove that $\sin(x+\frac{\pi}{n})$ converges uniformly to $\sin(x)$. I've just starting learning uniform convergence and understand the formal definition. What I've got so far is: 
$|\sin(x+ \frac{\pi}{n}) - \sin(x)| < \epsilon \ \ \ \ \forall x \in \mathbb{R} \ \ \ \ $ for $n \geq N, \epsilon>0$
LHS = $|2\cos(x+\frac{\pi}{2n})\cdot \sin(\frac{\pi}{2n})| < \epsilon $
Am I going down the right route here? I've done some examples fine, but when trig is involved on all space, I get confused as to what I should be doing...
Any help at all would be VERY much appreciated, I have an analysis exam tomorrow and need to be able to practice this.
Thanks.

What you did so far is fine. If you want to continue in the direction you started, it remains to notice that
$$\left|2\cos\left(x+\frac{\pi}{2n}\right)\cdot \sin\frac{\pi}{2n}\right| \le 2\left|\sin\frac{\pi}{2n}\right|$$
and that for any given $\varepsilon>0$ you can choose $N$ such that you have inequality
$$2\left|\sin\frac{\pi}{2n}\right|<\varepsilon$$
for $n>N$.
Can you see why this is true?
