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Inequality involving theta and sin(theta)

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to prove the following inequality for [tex]0\leq \theta\leq\frac{\pi}{2}[/tex]

    [tex] \frac{2}{\pi} \theta \leq\sin\theta[/tex]


    2. Relevant equations

    [tex] 0 \leq \theta\leq\frac{\pi}{2} [/tex]

    [tex] 0\leq \frac{2}{\pi}\theta\leq 1 [/tex]

    3. The attempt at a solution
    I've looked at the limit

    [tex] \lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1 [/tex]

    I've also looked at other inequalities involving [tex]\sin\theta[/tex]

    [tex]\sin\theta<\theta[/tex]

    I've also tried to take derivatives to see how fast [tex]\sin\theta[/tex] grows in comparison to [tex]\theta[/tex], but I have not managed to prove the inequality.
     
  2. jcsd
  3. Jun 3, 2009 #2

    Mark44

    Staff: Mentor

    That limit won't do you much good, since it is for x near zero.

    The graphs of y = 2x/[itex]\pi[/itex] and y = sin(x), you'll see that the sine curve is above the line, and that the two intersect at the origin and at ([itex]\pi[/itex], 1), and at no other points on the interval you're interested in. This isn't a proof, though, but you should be able to convey what the graph shows through calculus.

    You should be able to establish your inequality by showing that the graph of y = sin(x) is concave down on the interval (0, [itex]\pi[/itex]), meaning that the graph of y = sin(x) will be above the graph of y = 2x/[itex]\pi[/itex] except at the endpoints of your interval.
     
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