# Prove an inequality with a $\sin$ function: $\sin(x) > \frac2\pi x$ for $0<x<\frac\pi2$

$$\forall{x\in(0,\frac{\pi}{2})}\ \sin(x) > \frac{2}{\pi}x$$ I suppose that solving $\sin x = \frac{2}{\pi}x$ is the top difficulty of this exercise, but I don't know how to think out such cases in which there is an argument on the right side of a trigonometric equation.

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Try to draw graphs of lhs and rhs – Norbert Oct 14 '12 at 2:16
$\sin x$ is concave on this interval, so if you draw a line joining two points of the graph, the graph of this functions will lie above this line. What line do you get if you try the points $(0,\sin 0)$ and $(\pi/2, \sin \pi/2)$? – Martin Sleziak Oct 14 '12 at 7:51
– Martin Sleziak Nov 22 '15 at 7:22

As one of the comments suggested, the easiest way is to draw a graph of sine and the line through $(0,0)$ and $(\frac{\pi}{2},1)$, and notice that one is above the other.

There's another way though; expanding on the hints above, consider the functions $f$ and $g$ defined by $$f(x) = \frac{\sin{x}}{x} \quad \text{and} \quad g(x) = x\cos{x} -\sin{x}$$ Then we have $$f'(x) = \frac{x\cos{x}-\sin{x}}{x^2} \quad \text{and} \quad g'(x) = -x\sin{x}$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, so $g$ is decreasing. But we also have $g(0) = 0$, so it follows that $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, so $f$ is decreasing. As $x$ goes from (close to) $0$ to $\pi/2$, $f$ decreases from $1$ to $2/\pi$, and your result follows.

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How did you get the g(x) function? – 0x6B6F77616C74 Oct 15 '12 at 18:43
Perhaps it would have been better for me to have defined $g$ after finding $f'$. The motivation for choosing $g$ to be what I did is that then $f'(x) = \frac{g(x)}{x^2}$. Basically we want to conclude that $f$ is decreasing, but we don't see that immediately from its derivative, so we introduce $g$ to figure things out. – sourisse Oct 15 '12 at 19:18

Hint: Consider the monotone property of $f(x)=\frac{\sin(x)}{x}$ on interval $[0, \pi/2]$.

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Consider the properties of $\max$ and $\min$ of some function $f$,

$f = \dfrac{\sin x}{x}$ in this case.

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