# How to show $\sin x \geq \frac{2x}{\pi}, x \in [0, \frac{\pi}{2}]$? [duplicate]

I have tried the following: $$f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0$$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.

$\sin(x)$ is concave on that range and the line pass through $(0,0)$ and $(\pi/2,1)$ is $y=\frac{2}{\pi}x$

Let $$f(x) = \frac{\sin{x}}{x} \quad\implies f'(x) = \frac{g(x)}{x^2}~~ \text{with} \quad g(x) = x\cos{x} -\sin{x}$$ and $$g'(x) = -x\sin{x}\le 0$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, then $g$ is decreasing whereas $g(0) = 0$.

Thus $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, hence $f$ is decreasing on $[0,\frac{\pi}{2}]$. That is $$\frac{\sin{x}}{x} =f(x) \ge f(\pi/2) =\frac{2}{\pi}~~for ~~~0\le x\le \pi/2$$

Note that $f$ first increases from $0$ to some value and then decreases again to $0$, so it must be greater than $0$ all this time. The precise value is where $f'(x)=0$ and the graph of $f(x)$ will roughly look like a parabola.BTW, i.e. $f(\arccos(2/\pi))=\sin(\arccos(2/\pi))-(2/\pi)\arccos(2/pi)=\sqrt{\pi^2-4}/\pi-(2/\pi)\arccos(2/pi)$.

I would say, you have sort of completed this question. What you want to show is that f(x) >= 0 . You can say f' has critical point at around 0.881 (solve f'(x) =0 ). Then you plug 0, 0.881, pi/2 into f(x). Then you show the statement sin(x) >= 2x/pi is true. It would be easier if you plot the graph and think about it. For plotting, I recommend, https://www.desmos.com/calculator

you can type, f(x) = sin(x) -2pi/x g(x) =d/dx f(x) and so on, to create 2nd, 3rd,,, derivative.

• Let $f(x) = \left\{\begin{array}{ll}\frac{\sin(x)}{x} & \mbox{if$x \in \left(0,\frac{\pi}{2}\right)$.} \\ 1 & \mbox{ if$x=0$.} \end{array}\right.$

• By Mean value theorem applied to the function $g(x)=\sin(x)$ we get a $t\in (0,x)$ such that $\cos(t) =\frac{\sin(x)}{x}.$

• Note for $t \in (0,x)$ we have $$f'(x)=\frac{x\cos(x)-\sin(x)}{x^{2}} =\frac{\cos(x)-\frac{\sin(x)}{x}}{x}=\frac{\cos(x)-\cos(t)}{x} < 0$$ for all $x \in (0,\frac{\pi}{2}]$.

• I don't understand what you want to prove... Mar 27 '17 at 14:29