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Prove an inequality with a $\sin$ function

prove that $$\sin\theta\geq \frac{2}{\pi}\theta$$ for $0 \leq \theta \leq \dfrac{\pi}{2}$ My idea was to divide by $\theta$ take a $\lim$ when theta goes to $0$. But it only works for small $\theta$, not for the entire interval...

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marked as duplicate by amWhy, user7530, Thomas, Henry T. Horton, Erick Wong Jan 25 '13 at 18:02

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use $f(x)=\sin x$ is concave on $[0,\pi]$. –  tetori Jan 25 '13 at 17:30
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2 Answers 2

up vote 3 down vote accepted

Consider $$f(x)=\sin x-\frac{2}{\pi}x$$
Take it's derivative and find regions of increase/decrease in the region to find its minimum value and use this to prove your result.

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It's true that $$\frac{\pi}{2}\frac{\sin\theta}{\theta}\geq 1$$ since $\frac{\sin\theta}{\theta}$ is decreasing on $\theta \in (0,\pi/2]$.

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