# prove sin(x) is uniformly continuous.

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• October 25th 2009, 06:36 PM
cgiulz
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.)
• October 25th 2009, 07:01 PM
Krizalid
prove that $|\sin x-\sin y|\le|x-y|$ holds for $x,y\in\mathbb R$ and you're done.
• October 25th 2009, 08:45 PM
cgiulz
Anything wrong with this argument?

given $\epsilon,\delta > 0$
$|\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta$
• October 25th 2009, 08:49 PM
Bruno J.
Double post.
• October 25th 2009, 08:51 PM
Bruno J.

$|m-n|\leq |m|-|n|$ is wrong. Take m=0, n=1 for example.
• October 25th 2009, 08:53 PM
cgiulz
Quote:

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?
• October 25th 2009, 08:55 PM
Jose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
• October 25th 2009, 09:04 PM
redsoxfan325
Quote:

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.
• October 25th 2009, 09:06 PM
cgiulz
Quote:

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?
• October 25th 2009, 09:07 PM
redsoxfan325
Quote:

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 $\cos x$.)
• October 25th 2009, 09:11 PM
cgiulz
Quote:

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 (Wink)
• October 25th 2009, 09:16 PM
redsoxfan325
Quote:

Originally Posted by cgiulz
Go yanks (Wink)

(Angry)

I hope the Phillies can pull it off two years in a row.
• October 25th 2009, 09:22 PM
Jose27
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.
• October 25th 2009, 09:32 PM
cgiulz
Quote:

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?
• October 25th 2009, 09:34 PM
redsoxfan325
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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