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Math Help - please help, analysis problem..

  1. #1
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    please help, analysis problem..

    topic is 3<π<2√3
    Hint: apply mean value theorem on function Sin on the interval [0, π/6]
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kevek View Post
    topic is 3<π<2√3
    Hint: apply mean value theorem on function Sin on the interval [0, π/6]
    MVT: If a function f is continuous on [a,b] and differentiable on (a,b), then there exists c \in (a,b) such that f'(c) = \frac{f(b)-f(a)}{b-a}

    so, \sin is continuous on [0,\pi/6] and differentiable on (0,\pi/6). thus, there exist a c in (0,\pi/6) such that \cos c = \frac{\sin \pi/6 - \sin 0}{\pi/6}

    \frac{3}{\pi} = \cos c

    since c in (0,\pi/6), \frac{\sqrt{3}}{2} < \cos c < 1.. thus, 1 < \frac{1}{\cos c} < \frac{2}{\sqrt{3}}

    therefore, we get 1 < \frac{\pi}{3} < \frac{2}{\sqrt{3}} which gives you the desired result..
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  3. #3
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    Quote Originally Posted by kalagota View Post
    MVT: If a function f is continuous on [a,b] and differentiable on (a,b), then there exists c \in (a,b) such that f'(c) = \frac{f(b)-f(a)}{b-a}

    so, \sin is continuous on [0,\pi/6] and differentiable on (0,\pi/6). thus, there exist a c in (0,\pi/6) such that \cos c = \frac{\sin \pi/6 - \sin 0}{\pi/6}

    \frac{3}{\pi} = \cos c

    since c in (0,\pi/6), \frac{\sqrt{3}}{2} < \cos c < 1.. thus, 1 < \frac{1}{\cos c} < \frac{2}{\sqrt{3}}

    therefore, we get 1 < \frac{\pi}{3} < \frac{2}{\sqrt{3}} which gives you the desired result..
    you are brilliant Thank you very much !!!
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