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Prove this inequality.

  1. Jun 18, 2009 #1
    [tex]\frac{2}{Pi}x<sin x[/tex] for 0<x<[tex]\frac{Pi}{2}[/tex]

    I dint figure out how to write Pi in symbol. I dont have any idea what to do with this one.
     
  2. jcsd
  3. Jun 18, 2009 #2

    HallsofIvy

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

    To show that [itex]\left(\frac{2}{\pi}\right)x\le sin(x)[/itex] for [itex]x< \frac{\pi}{2}[/itex], look at the function [itex]f(x)= sin(x)- 2x/\pi[/itex]. It is 0 at both x= 0 and [itex]x= \pi/2[/itex]. It's only critical point is where [itex]f'(x)= cos(x)- 2/\pi= 0[/itex] or [itex]cos(x)= 2/\pi[/itex]. Since [itex]2\pi[/itex] is less than one, that occurs at some point between 0 and [itex]\pi/2[/itex]. Further, f"(x)= -sin(x) which is negative for all x between 0 and [itex]\pi/2[/itex]
     
  4. Jun 18, 2009 #3
    OK thanks a lot.
     
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