trigonemtric continuity proof

**lllll** · January 31st 2009, 06:54 PM

using the $\delta-\varepsilon$ definition prove that $\sin(x)$ is continuous an a given interval $[a,b]$ .

by the definition $\forall \ \varepsilon >0, \ \exists \ \delta >0$ s.t. $|x-c|<\delta \Rightarrow |f(x)-f(c)|<\varepsilon$

therefore

$|x-c|<\delta \Rightarrow |\sin(x)-\sin(c)|<\varepsilon$

where $|\sin(x)-\sin(c)|$ can be rewritten as $\bigg{|}2\cos\left(\frac{x+c}{2}\right) \cdot 2\sin\left(\frac{x-c}{2}\right)\bigg{|}= \bigg{|}2\cos\left(\frac{x+c}{2}\right)\bigg{|} \cdot \bigg{|}2\sin\left(\frac{x-c}{2}\right)\bigg{|}$

now since we are on a closed interval we can let the supremum equal $M$ , such that $M\geq0$

thus: $\bigg{|}2\cos\left(\frac{x+c}{2}\right)\bigg{|} \cdot \bigg{|}2\sin\left(\frac{x-c}{2}\right)\bigg{|} \leq M \cdot \sin\left(\frac{\delta}{2}\right)$

I'm not 100% sure but since $\delta>0$ the wouldn't $\sin\left(\frac{\delta}{2}\right) = 0$ ?

giving me $\bigg{|}2\cos\left(\frac{x+c}{2}\right)\bigg{|} \cdot \bigg{|}2\sin\left(\frac{x-c}{2}\right)\bigg{|} \leq M \cdot \sin\left(\frac{\delta}{2}\right)= M\cdot 0 = 0<\varepsilon$

I'm also wondering if we can apply the fact that $M$ is equal to a supremum if we weren't in a closed interval? If so then wouldn't:

$\bigg{|}2\cos\left(\frac{x+c}{2}\right)\bigg{|} \leq 1$ ?

**red_dog** · February 1st 2009, 12:20 AM

$\left|2\sin\frac{x-c}{2}\cos\frac{x+c}{2}\right|=2\left|\sin\frac{x-c}{2}\right|\cdot \left|\cos\frac{x+c}{2}\right|$

But $\left|\cos\frac{x+c}{2}\right|\leq 1$ and $\left|\sin\frac{x-c}{2}\right|\leq\frac{|x-c|}{2}$

So $\left|2\sin\frac{x-c}{2}\cos\frac{x+c}{2}\right|\leq |x-c|<\delta$

So we can take $\delta=\epsilon$

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