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Math Help - Very hard inequality

  1. #1
    Member roshanhero's Avatar
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    Very hard inequality

    Use the inequalities \sin \theta < \theta < \tan \theta for a suitable value of \theta to show that \pi lies between 3 and 2\sqrt{3}.
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    Quote Originally Posted by roshanhero View Post
    Use the inequalities \sin \theta < \theta < \tan \theta for a suitable value of \theta to show that \pi lies between 3 and 2\sqrt{3}.
    Let \theta = \sin^{-1}x then x < \sin^{-1} x < \frac{x}{\sqrt{1-x^2}} and then choose x = \frac{1}{2} .
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    Hello, roshanhero!

    A fascinating problem . . .


    Use the inequalities: \sin \theta \,<\, \theta \,<\, \tan \theta for a suitable value of \theta

    to show that \pi lies between 3 and 2\sqrt{3}.

    We have: . \theta \:>\:\sin\theta
    \sin\theta is an increasing function on the interval \left[0,\:\frac{\pi}{2}\right]
    . . Then: . \arcsin\theta \:>\:\theta

    Let \theta \,=\,\frac{1}{2}
    We have: . \arcsin\left(\frac{1}{2}\right) \:>\:\frac{1}{2}\quad\Rightarrow\quad \frac{\pi}{6} \:>\:\frac{1}{2}

    . . Hence: . {\color{blue}\pi \:>\:3}



    We have: . \theta \:<\:\tan\theta
    \tan\theta is an increasing function on the interval \left[0,\:\frac{\pi}{2}\right]
    . . Then: . \arctan\theta \:<\:\theta

    Let \theta = \frac{1}{\sqrt{3}}
    We have: . \arctan\left(\frac{1}{\sqrt{3}}\right) \:<\:\frac{1}{\sqrt{3}} \quad\Rightarrow\quad \frac{\pi}{6} \:<\:\frac{1}{\sqrt{3}}

    . . Hence: . \pi \:<\:\frac{6}{\sqrt{3}} \quad\Rightarrow\quad {\color{blue}\pi \:<\:2\sqrt{3}}


    Therefore: . \boxed{\;3 \;<\;\pi \;<\;2\sqrt{3}\;}

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    What about using the calculator ???
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  5. #5
    Member roshanhero's Avatar
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    How can we show that the inequality \sin \theta < \theta holds for all values of \theta greater than 0.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by roshanhero View Post
    How can we show that the inequality \sin \theta < \theta holds for all values of \theta greater than 0.
    There are many ways. For example, the assertion is clear if \theta\geqslant 1 so assume 0<\theta<1. We know that \sin(\theta)=\theta-\frac{\theta^3}{3!}+\cdots using some clever manipulation of the series, and using the fact that 0<\theta<1\implies \theta^{n\geqslant 1}\leqslant \theta we may conclude that \sin(\theta)=\theta+\text{negative terms}.

    Alternatively, the claim that \sin(\theta)<\theta is equivalent to 0<\theta-\sin(\theta)=\theta-\sin(\theta)-(0-\sin(0))=\int_0^{\theta}\left(1-\cos(x)\right)dx. Since an integral is positive precisely when it's integrand is positive over the region of integration all we need to show is that 0\leqslant x<\theta<1\implies 1-\cos(x)\geqslant 0 but, this follows readilly.


    Lastly, the easiest way is to just note that if f(\theta)=\theta-\sin(\theta) then f(0)=0,f'(\theta)=1-\cos(\theta)\geqslant 0 and so f(\theta)\geqslant 0,\text{ }\theta\geqslant 0.


    Or, actually. You could use the fact that f(x)=\sin(x) is Lipschitz (with Lipschitz constant one) to conclude that since \left|\sin(x)-\sin(y)\right|\leqslant |x-y|,\text{ }\forall x,y\in\mathbb{R}\implies |\sin(x)-\sin(0)|=|\sin(x)|\leqslant |x|. Which, is a stronger claim.
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