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Math Help - Prove x cosx < sinx on 0 < x < pi/2

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    Prove x cosx < sinx on 0 < x < pi/2

    I am asked to prove that \frac{2}{\pi} < \frac{sin x}{x} < 1 if 0 < x < \frac{\pi}{2}, where the definition of sin is given by sin(x) = \frac{1}{2}(e^{ix} + e^{-ix}). I think I have a bit of the problem figured out. I reason that I can solve the whole problem if I know the derivative of \frac{sin x}{x} is always negative, which it is if I am to trust a graphing calculator.

    Once that's obtained, I think the rest follows from: \displaystyle \lim_{x \rightarrow 0^{+}} \frac{\sin x}{x} = \lim_{x \rightarrow 0^{+}} cos(x) = 1, and \frac{sin(\frac{\pi}{2}}{\frac{\pi}{2}} = \frac{2}{\pi}.

    So I need to show that \frac{x cos x - sin x}{x^{2}} is negative within these bounds. But I don't see how to do so, or if I need to re-consider how I'm approaching the task.
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    I suppose I should add that we have defined \pi as the smallest positive number such that cos(\frac{\pi}{2}) = 0.
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  3. #3
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    You should know that

    -1 \leq \sin{x} \leq 1 for all x.

    So for  x > 0, you that means

    -\frac{1}{x} \leq \frac{\sin{x}}{x} \leq \frac{1}{x}.


    Now see what happens as x \to 0 and x \to \frac{\pi}{2}...
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    MHF Contributor Bruno J.'s Avatar
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    I would use that \frac{\sin x}{x} = \cos(x/2)\cos(x/2^2)\cos(x/2^3)\dots. If 0 \leq x < y \leq \pi/2, it is easy to see that \cos(x/2) > \cos(y/2),\: \: \cos(x/2^2) >\cos(y/2^2), etc., so that \frac{\sin x}{x} > \frac{\sin y}{y}.
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    MHF Contributor Bruno J.'s Avatar
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    Another way using the same idea : taking the derivative, we have \frac{d}{dx}\frac{\sin x}{x} = -\frac{\sin x}{x}\sum_{j=1}^\infty \frac{\tan x/2^j}{2^j}, which is \leq 0 at least for 0 \leq x \leq \pi.

    In fact, the derivative of \frac{\sin x}{x} is not always negative.
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    Quote Originally Posted by Prove It View Post
    You should know that

    -1 \leq \sin{x} \leq 1 for all x.

    So for  x > 0, you that means

    -\frac{1}{x} \leq \frac{\sin{x}}{x} \leq \frac{1}{x}.


    Now see what happens as x \to 0 and x \to \frac{\pi}{2}...
    I can't say that I follow. You're seeing what the function does at infinity and at +/- pi/2. But if I don't know about the monotonicity of the function in the interval I'm considering, it seems like the function could be absolutely anything in-between these points.
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  7. #7
    Member mohammadfawaz's Avatar
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    Use MVT on the interval [0,x] where 0 < x < \frac{\pi}{2}
    let f(x)=sin(x) then:  sin(x)-sin(0) = f'(c)(x-0) where 0 < c < x.
    So, sin(x)=xf'(c)=xcos(c).
    Now, note that the function cos(x) is decreasing on [0,\frac{\pi}{2}] hence, cos(c)>cos(x).
    Therefore: sin(x)=xcos(c)>xcos(x) if 0 < x < \frac{\pi}{2}
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