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Math Help - trigonemtric continuity proof

  1. #1
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    trigonemtric continuity proof

    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 ?
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  2. #2
    MHF Contributor red_dog's Avatar
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    \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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