Proving \frac{2}{Pi}x<sin x for 0<x<\frac{Pi}{2}

[Penultimate]

$$\frac{2}{Pi}x<sin x$$ for 0<x<$$\frac{Pi}{2}$$

I dint figure out how to write Pi in symbol. I don't have any idea what to do with this one.

 

[HallsofIvy]

For $\pi$ use [ itex ]\pi[ /itex ]. In general clicking on LaTex in any post will show the code for it.

To show that $\left(\frac{2}{\pi}\right)x\le sin(x)$ for $x< \frac{\pi}{2}$, look at the function $f(x)= sin(x)- 2x/\pi$. It is 0 at both x= 0 and $x= \pi/2$. It's only critical point is where $f'(x)= cos(x)- 2/\pi= 0$ or $cos(x)= 2/\pi$. Since $2\pi$ is less than one, that occurs at some point between 0 and $\pi/2$. Further, f"(x)= -sin(x) which is negative for all x between 0 and $\pi/2$

 

[Penultimate]

OK thanks a lot.