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How do I get:

$\cos{\theta} \lt \frac{\sin{\theta}}{\theta} \lt 1$

  • 2
    Maybe you should write $\le$ than <, since $\cos 2\pi n=1$.2012-02-21

2 Answers 2

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For $0< t<\pi/2$:

$\ \ \ \bullet$ Using similar triangles: $\color{darkgreen}{\tan t}={\color{maroon}{\sin t}\over\color{darkblue}{\cos t}} ={ {\text{length}( \color{darkgreen} {\overline{{IZ}})} }\over 1 }\quad \Longrightarrow \quad\color{darkgreen}{\tan t}=\text{length}(\color{darkgreen}{\overline{IZ}})$

$\ \ \ \bullet$ $t$ is the length of the arc $\color{orange}{IQ}$.

$\ \ \ \bullet$ Area of the circular sector $O\color{orange}{IQ}={t\over 2\pi}\cdot \pi\cdot 1^2={t\over2}$.

$\ \ \ \bullet$ Area of $\triangle OQI={1\over2}\cdot1\cdot\color{maroon}{\sin t}$.

$\ \ \ \bullet$ Area of $\triangle OIZ={1\over2}\cdot1\cdot\color{darkgreen}{\tan t}$.

From the diagram we have $ \text {area}(\triangle OQI) \le \text {area}(\text{circular sector} OQI) \le \text {area}(\triangle OZI) $ $ {1\over2}\cdot1\cdot\sin t\lt{1\over2} t\lt {1\over2}\cdot1\cdot\tan t $

$ \sin t\lt t\lt \cdot\tan t $

$ {1\over\sin t}\gt {1\over t}\gt {\cos t\over \sin t } $

$ \cos t\lt {\sin t\over t}\lt 1. $

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    @Gigili Using http://jsxgraph.org/2012-02-21
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Typically, this is used with the squeeze theorem to prove that $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$, so assuming you want your inequality to hold for $\theta$ near (but not equal to) $0$, consider the following picture, showing part of a unit circle for small $\theta>0$.

diagram

The length of the small (green) circular arc is $\theta$ and $\sin\theta<\theta$ since the perpendicular distance from the point on the circle to the $x$-axis is the shortest distance from that point to the axis, so $\frac{\sin\theta}{\theta}<1$. Similarly, $\theta<\tan\theta$ which is equivalent to $\theta\cos\theta<\sin\theta$ or $\cos\theta<\frac{\sin\theta}{\theta}$.

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    Not entirely sure why it is safe to conclude from the diagram that \theta < \tan \theta. Can you add a line arguing that?2012-02-21