I was asked to check whether $f_n(x)= n(\sin(x+\frac{1}{n}) -\sin x )$ convergence uniformly in $\mathbb{R}$. So I found the limit function $f(x) = \cos x $ and I tired to prove that $$ \underset{x \in \mathbb{R}}{\text{sup}}\{|f_n(x)-f(x)|\}\xrightarrow[]{n \rightarrow \infty} 0 $$
with some trigonometric identities but with no success. I would to get a hint or some help. Thanks!
$f_n(x)=2n\sin\frac{1}{2n}\cos(x+\frac{1}{2n})$.Therefore $|f(x)-f_n(x)|=|2n\sin\frac{1}{2n}\cos(x+\frac{1}{2n})-\cos x|\leq\\|2n\sin\frac{1}{2n}(\cos(x+\frac{1}{2n})-\cos x)|+|\cos x(2n\sin\frac{1}{2n}-1)|$.
Only need to show right hand side $\to 0$.This can be easily done from $\lim\limits_{n\to\infty}2n\sin\frac{1}{2n}=1$.