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For 0 < x lt=pi/2, show that x-(x^3)/6< ...

For `0 < x lt=pi/2, `show that `x-(x^3)/6< sin(x) < x dot`

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To prove the inequality \( x - \frac{x^3}{6} < \sin(x) < x \) for \( 0 < x \leq \frac{\pi}{2} \), we will break it down into two parts: 1. Show that \( \sin(x) < x \). 2. Show that \( x - \frac{x^3}{6} < \sin(x) \). ### Step 1: Proving \( \sin(x) < x \) Let \( f(x) = \sin(x) - x \). ...
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