# Thread: prove sin(x) is uniformly continuous.

1. ## prove sin(x) is uniformly continuous.

Prove directly from the definition of uniform continuity that $\sin(x)$ is uniformly continuous on $(-\infty,\infty).$

(Hint: a unit circle may help to make an estimation.)

2. prove that $|\sin x-\sin y|\le|x-y|$ holds for $x,y\in\mathbb R$ and you're done.

3. Anything wrong with this argument?

given $\epsilon,\delta > 0$
$|\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta$

4. Double post.

5. $|m-n|\leq |m|-|n|$ is wrong. Take m=0, n=1 for example.

6. Originally Posted by Bruno J.
$|m-n|\leq |m|-|n|$ is wrong. Take m=0, n=1 for example.
Typo; I was trying to apply the triangle inequality. It is not even necessary anyway. Any thoughts on the edited post?

7. Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.

8. Originally Posted by cgiulz
Prove directly from the definition of uniform continuity that $\sin(x)$ is uniformly continuous on $(-\infty,\infty).$

(Hint: a unit circle may help to make an estimation.)

$|\sin x-\sin y|=\left|2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|$

$|\cos\theta|\leq1$ and you can use the unit circle to show that $|\sin\theta|\leq|\theta|$.

Let $\delta=\epsilon$ and don't look back.

9. Originally Posted by Jose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
Yes, but how can it be done directly from the definition?

10. Originally Posted by Jose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
The only thing is that this requires him to assume $\sin x$ is differentiable and furthermore that it's derivative is $\cos x$. (If they haven't proved that $\sin x$ is continuous yet, it's unlikely they've proved its derivative is $\cos x$.)

11. Originally Posted by redsoxfan325
The only thing is that this requires him to assume $\sin x$ is differentiable and furthermore that it's derivative is $\cos x$. (If they haven't proved that $\sin x$ is continuous yet, it's unlikely they've proved its derivative is $\cos x$.)
This is true.

Go yanks

12. Originally Posted by cgiulz
Go yanks

I hope the Phillies can pull it off two years in a row.

13. You could say that $\sin x$ is unif. cont. in $[0,2 \pi]$ and since it's periodic it's unif. cont. in $\mathbb{R}$

Edit. Apparently I keep avoiding the definition of uniform cont.

14. Originally Posted by cgiulz
Anything wrong with this argument?

given $\epsilon,\delta > 0$
$|\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta$
I think this is correct, is it not?

15. That is true, yes, but I think you need to show that $|\sin x-\sin y|\leq|x-y|$, not just state it.

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