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    this question could fall into any module tbh, but i'm stuck on it anyway:
    Show that for all x,  2 - \frac{4}{3}x^2 < \frac{sin 2x}{x}
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    Anyone please? It happened to appear on a past exam on Calculus... (but could also be an algebra question, either way I'm still stuck on it)
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    Amm.. is it like you need to simplify both equations?
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    No, I just need to prove that for all x,  2 - \frac{4}{3}x^2 is less than  \frac{sin 2x}{x} . And I have no idea how to do it! (No wait I think I do have an idea, do I express sin(2x) as its Taylor polynomial? And how many terms would I need in that polynomial to be sure?)
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    You'll need to use Taylor's theorem with remainder. It should be fairly obvious how many terms to take.
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    Find the maclaurin expansion of sin2x, group like terms then show that for all x RHS is greater than 0.


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    Find maclaurin expansion for sin2x then group like terms, take up to x^3 or more if you wish, show that RHS is greater than 0 for all x, or less than 0 if you go for the LHS.


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