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I want to prove that $\left(\forall x \in \left( 0; \frac{\pi}{2} \right)\right) \left[ \sin(x) > \frac{2}{\pi} \cdot x \right]$. This is quite easy to see when drawing the functions, but I wonder if my proof here is correct:

From $(\sin(x))' = \cos(x)$ and $\left( \frac{2}{\pi} \cdot x\right)' = \frac{2}{\pi}$ and $\cos(0) > \frac{2}{\pi}$ follows that $\sin(x)$ grows quicker than $\frac{2}{\pi}$ for $x > 0$ and thus is larger right from the beginning.

Furthermore, let's assume there is an $x \in \left( 0; \frac{\pi}{2}\right)$ with $\sin(x) < \frac{2}{\pi} \cdot x$. As we proved that $\sin(0)' > \frac{2}{\pi}$, there needs to exist an $x \in \left( 0; \frac{\pi}{2}\right)$ with $\sin(x)'' = 0$.

$\sin(x)'' = (\cos(x))' = -\sin(x)$

$-\sin(x) = 0 \iff \arcsin(0) = x \iff x = 0$

As $0 \notin \left( 0; \frac{\pi}{2}\right)$, $\left(\forall x \in \left( 0; \frac{\pi}{2} \right)\right) \left[ \sin(x) > \frac{2}{\pi} \cdot x \right]$.

Basically I've found geometrically that the second derivative had to be zero if the sin were larger in the beginning but smaller in the beginning. Is there a theorem for that?

Is there a way to make the proof easier?

Thanks!

Hellstorm
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1 Answers1

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You can use the fact that $f(x) = - \sin x$ is convex on $[0, \pi/2]$ (for example, because its second derivative is positive on $[0,\pi/2]$). The convexity property (the graph is below any arc) yields $$ - \sin \left( (1-t) \times 0 + t \times \frac{\pi}{2} \right) \leq (1-t) \times f(0) + t \times f\left(\frac{\pi}{2} \right) $$ and $$ - \sin \left( \frac{\pi}{2} t \right) \leq - t \quad \mbox{and} \quad t \leq \sin \left( \frac{\pi}{2} t \right) $$ for any $t \in [0, 1]$.

user7440
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