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Sin inequality proof , ##0 \leq 2x/\pi \leq sin x##

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution
    Hi

    How do I go about showing ##0 \leq \frac{2x}{\pi} \leq sin x ##?

    for ## 0 \leq x \leq \pi /2 ##

    I am completely stuck where to start.

    Many thanks.

    (I see it is a step in the proof of Jordan's lemma, but I'm not interested in this, and the proofs I find do not explain this actual step, ta).
     
    Last edited: Jan 21, 2017
  2. jcsd
  3. Jan 21, 2017 #2

    Mark44

    Staff: Mentor

    This isn't true in general. If ##x > \pi/2##, the expression in the middle is larger than 1.

    BTW, surround your TeX expressions with either ## (inline) or $$ (standalone) at beginning and end. A single $ character doesn't do anything.
    I edited your earlier post to fix this.
     
  4. Jan 21, 2017 #3
    edited to include range of x

    sorry I am aware of that for latex, running low on caffeine ! ta
     
  5. Jan 21, 2017 #4

    Mark44

    Staff: Mentor

    Try showing that ##\sin x - \frac 2 {\pi} x \ge 0## on that interval, possibly using the Maclaurin expansion of ##\sin x##.
     
  6. Jan 21, 2017 #5

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Draw a graph! That's a good place to start.
     
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