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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?

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

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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.

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