How do I get $\cos{\theta} \lt \frac{\sin{\theta}}{\theta} \lt 1$?

Solution 1:

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

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

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

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

Solution 2:

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 $\theta>0$.

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 $\cos\theta<\frac{\sin\theta}{\theta}$.