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- Feb 14, 2012

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Let $0\le x \le \dfrac{\pi}{2}$. Prove that $\sin x \ge \dfrac{2x}{\pi}$.

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

-----

Let $0\le x \le \dfrac{\pi}{2}$. Prove that $\sin x \ge \dfrac{2x}{\pi}$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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Per definition any point on any secant to the graph lies below the graph (and on the graph at the end points).

But $y = \frac{2}{\pi}x$ is a secant to $\sin x$ on the given interval. Therefore $\sin x \geq \frac{2x}{\pi}$ for $x \in [0; \frac{\pi}{2}]$.

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