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.
 A: Use the fact that the sine function's derivative has absolute value of at most one to see that 
$$|\sin(x) - \sin(y)| \le |x - y|.$$
A: 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?
A: 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 $\,\,\forall x\in\mathbb{R}\,,\,\,|\sin x+\cos x|<2$
A: First find the pointwise limit, that is where $\displaystyle \lim_{n \to \infty} f_{n}(x)=\displaystyle \lim_{n \to \infty} \sin\Bigl(x + \frac{\pi}{n}\Bigr)$ goes to. You can easily see that this limit is same as $\displaystyle \lim_{y \to 0} \sin\bigl(x +\pi \cdot y\bigr)= \sin{x}$. So your function converges to $f(x)=\sin(x)$. Now to prove uniform convergence we have to look at the sequence $$d_{n} = \sup_{x \in \mathbb{R}} \: \Bigl\{\:\Bigl|\sin\Bigl(x + \frac{\pi}{n}\Bigr) - \sin(x)\:\Bigr|  : x \in \mathbb{R}\Bigr\}$$ and prove that $d_{n} \to 0$ for $n \in \mathbb{N}$. 
We do the derivative test. Look at $f(x) = \sin\Bigl(x + \frac{\pi}{n}\Bigr) - \sin(x)$, $f'(x) = \cos\Bigl(x+\frac{\pi}{n}\Bigr) - \cos(x)$. Equating this to $0$ we see that
\begin{align*}
\cos\Bigl(x + \frac{\pi}{n}\Bigr) - \cos(x) &= 2 \cdot \sin\Bigl(x + \frac{\pi}{2n}\Bigr) \cdot \sin\Bigl(-\frac{\pi}{n}\Bigr) = 0
\end{align*}
and so $x = -\frac{\pi}{2n}$.  From this we see that $d_{n} \to 0$ and hence the convergence is uniform.
This question might be worth looking:

  
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*Common Problems while showing Uniform Convergence of functions.
  

