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Proof sin x < x for all x>0.

  1. Oct 11, 2015 #1


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    Hello all,

    I want to prove the following inequality.
    sin(x)<x for all x>0.

    Now I figured that I put a function f(x)=x-sin(x), and show that it is increasing for all x>0. But this alone doesn't prove it. I need to show we have inequality from the start. I can't show that lim f(x) as x->0 is positive cause this limit equals 0. I can show that we have equality at x=0, and only at x=0. Therefore, sin(x) <=x for all x>=0, and we only have equality at x=0. So sin(x)<x for all x>0. This doesn't seem the right way to do it though.

  2. jcsd
  3. Oct 11, 2015 #2


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    Homework Helper

    No, you are correct: [itex]x - \sin x[/itex] is zero at x = 0 and is thereafter strictly increasing, so [itex]x - \sin x[/itex] can't be zero or negative for [itex]x > 0[/itex].
  4. Oct 11, 2015 #3


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    Gold Member

    You can visualize it comparing the graphs of ## y(x)=x## and ##y(x)=\sin{(x)}##
  5. Oct 11, 2015 #4
    Hint: sin(0) = 0, and 0 < d(sin(x))/dx < 1 for 0 < x <= π/2 (why?).
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