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.

Can you prove that  $\,\,\sin \pi/n\to 0\,\,,\,\,\cos \pi/n\to 1\,\,$? Then do some $\epsilon$-stuff, and use trigonometric identites of double angle to get Cauchy condition for uniform convergence:$$\left|\sin\left(x+\frac{\pi}{n}\right)-\sin\left(x+\frac{\pi}{m}\right)\right|\leq \left|\sin x\left(\cos\frac{\pi}{n}-\cos\frac{\pi}{m}\right)+\cos x\left(\sin\frac{\pi}{n}-\sin\frac{\pi}{m}\right)\right|$$
and remember that $\,\,\forall x\in\mathbb{R}\,,\,\,|\sin x+\cos x|<2$
