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

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

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- Oct 25th 2009, 06:36 PMcgiulzprove sin(x) is uniformly continuous.
Prove directly from the definition of uniform continuity that $\displaystyle \sin(x)$ is uniformly continuous on $\displaystyle (-\infty,\infty).$

(Hint: a unit circle may help to make an estimation.) - Oct 25th 2009, 07:01 PMKrizalid
prove that $\displaystyle |\sin x-\sin y|\le|x-y|$ holds for $\displaystyle x,y\in\mathbb R$ and you're done.

- Oct 25th 2009, 08:45 PMcgiulz
Anything wrong with this argument?

given $\displaystyle \epsilon,\delta > 0$

$\displaystyle |\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta$ - Oct 25th 2009, 08:49 PMBruno J.
Double post.

- Oct 25th 2009, 08:51 PMBruno J.

$\displaystyle |m-n|\leq |m|-|n|$ is wrong. Take m=0, n=1 for example. - Oct 25th 2009, 08:53 PMcgiulz
- Oct 25th 2009, 08:55 PMJose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.

- Oct 25th 2009, 09:04 PMredsoxfan325
Here are some tools that will help you.

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

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

Let $\displaystyle \delta=\epsilon$ and don't look back. - Oct 25th 2009, 09:06 PMcgiulz
- Oct 25th 2009, 09:07 PMredsoxfan325
The only thing is that this requires him to assume $\displaystyle \sin x$ is differentiable and furthermore that it's derivative is $\displaystyle \cos x$. (If they haven't proved that $\displaystyle \sin x$ is continuous yet, it's unlikely they've proved its derivative is $\displaystyle \cos x$.)

- Oct 25th 2009, 09:11 PMcgiulz
- Oct 25th 2009, 09:16 PMredsoxfan325
- Oct 25th 2009, 09:22 PMJose27
You could say that $\displaystyle \sin x$ is unif. cont. in $\displaystyle [0,2 \pi]$ and since it's periodic it's unif. cont. in $\displaystyle \mathbb{R}$

Edit. Apparently I keep avoiding the definition of uniform cont. - Oct 25th 2009, 09:32 PMcgiulz
- Oct 25th 2009, 09:34 PMredsoxfan325
That is true, yes, but I think you need to show that $\displaystyle |\sin x-\sin y|\leq|x-y|$, not just state it.