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Problem Of The Week #442 November 9th, 2020

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anemone

MHB POTW Director
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Feb 14, 2012
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to lfdahl for his correct solution (Cool) , which you can find below:
Since $\sin’’(x) = -\sin x \leq 0$ for $x \in [0; \frac{\pi}{2}]$, the function $\sin x$ is concave (downward).

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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