Using the sequential definition of uniform continuity to show $\sin(x)$ is uniformly continuous on $\mathbb{R}$

Question

I want to show $\sin(x)$ is uniformly continuous on $\mathbb{R}$. Let $\{a_{n}\}$ and $\{b_{n}\}$ be sequences such that $\lim_{n\to\infty}[b_{n} - a_{n}] = 0$. Then, we need to show $\lim_{n\to\infty} |\sin(b_{n}) - \sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?

Answer 1

By the mean value theorem:

$\sin(b_{n}) - \sin(a_{n})= \cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.

Can you proceed ?

Answer 2

By simple trigonometric manipulation:

$\sin(b_n) - \sin(a_n) = 2\sin(\frac{b_n-a_n}{2})\cos(\frac{b_n+a_n}{2})\le 2\sin(\frac{b_n-a_n}{2})$.