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

[Penultimate]

\frac{2}{Pi}x&lt;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&lt; \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&#039;(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.