# Uniform continuity of $\sin(x)$

I know $\sin(x)$ is uniform continuous on the non negative real line i.e on $[0,\infty)$ i tried and get it is uniformly continuous on the non positive real line i.e on $(-\infty,0]$ is then $\sin(x)$ will be uniform continuous on $\mathbb R$ please help me thanks in advance

is similarly the function $\cos(x)$ is uniformly continuous on $\mathbb{R}$ . what about other trignometric functions?

• Other (than sine and cosine) trigonometric functions are not even pointwise continuous. – njguliyev Aug 18 '13 at 22:38

Since the function $$\sin$$ is differentiable on $\mathbb R$ and its derivative the function $$\cos$$ is bounded then the function $\sin$ is lipschitzian and then uniformly continuous on $\mathbb R$.
You go with the definition. You need to show that for every $\epsilon > 0$ there is $\delta>0$ such that if $|x-y| \leq \delta$, then $|\sin(x) - \sin(y)| \leq \epsilon$.
Since the derivative of $\sin(x)$ in absolute value is bounded by 1, it follows that $$\frac{|\sin(x) - \sin(y)|}{|x-y|} \leq 1$$ for all $x,y \in \mathbb{R}$. Therefore, for every $\epsilon >0$ we can take $\delta=\epsilon$ to satisfy the definition of uniform continuity.