$$\forall x\in\left(0,\frac\pi2\right),\ \sin(x) >\frac2\pi x $$ I suppose that solving $ \sin x = \frac2\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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3Try to draw graphs of lhs and rhs – Norbert Oct 14 '12 at 02:16
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3$\sin x$ 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 07:51
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See also: http://math.stackexchange.com/questions/596634/mean-value-theorem-frac2-pi-frac-sin-xx1 – Martin Sleziak Nov 22 '15 at 07:22
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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.
sourisse
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1Perhaps 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
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Here is a simple solution.
Let $0<x<\frac\pi2$ be fixed. By Mean value theorem, there exists $y\in (0, x)$ such that $$\sin(x)-\sin(0)= \cos(y)(x-0).$$ Thus, $$\frac{\sin(x)}{x}= \cos(y).$$ Similarly there exists $z\in\left(x, \frac\pi2\right)$ such that $$\sin\left(\frac\pi2\right)-\sin(x)= \cos(z)\left(\frac\pi2-x\right).$$ Thus, $$\frac{1-\sin(x)}{\pi/2- x}= \cos(z).$$ As $0<y<z<\frac\pi2$ and $\cos$ is a strictly decreasing function in $\left[0,\frac\pi2\right]$ we see that $$\cos(z)<\cos (y).$$ Thus $$\begin{aligned}\frac{1-\sin(x)}{\pi/2- x}&< \frac{\sin(x)}x\\\Rightarrow \frac{1-\sin(x)}{\sin(x)}&< \frac{\pi/2- x}x\\\Rightarrow\frac1{\sin(x)}-1&<\frac\pi{2x}- 1\\\Rightarrow\frac2\pi&<\frac{\sin (x)}x.\end{aligned}$$ Since $y\in\left(0,\frac\pi2\right),$ therefore $\cos(y) <1$. Hence it is also clear that $$\frac{\sin(x)}x= \cos(y) <1.$$ Hence, for any $0<x<\frac\pi2,$ we get $$\frac2\pi<\frac{\sin(x)}x<1.$$
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Hint: Consider the monotone property of $f(x)=\frac{\sin(x)}{x}$ on interval $[0, \pi/2]$.
JACKY88
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Consider the properties of $\max$ and $\min$ of some function $f$,
$f = \dfrac{\sin x}{x}$ in this case.
Asaf Karagila
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ILoveMath
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