Prove that sin x < x for x > 0

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The discussion focuses on proving that sin x < x for x > 0 using Taylor's theorem. Participants suggest calculating the Taylor series expansion of sin x, which is sin x = x - x^3/3! + x^5/5! - ... . By analyzing the alternating terms of the series, it is established that each subsequent term decreases in magnitude, leading to the conclusion that sin x is always less than x for positive values of x.

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Anyone knows how to prove it using Taylor's theorem?
 
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Calculate the Taylor series of sin x and then take a close look at the alternating terms.
 

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