How do I get $\cos{\theta} \lt \frac{\sin{\theta}}{\theta} \lt 1$? How do I get:
$$\cos{\theta} \lt \frac{\sin{\theta}}{\theta} \lt 1$$
 A: 
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
$$
\text {area}(\triangle OQI) \le 
\text {area}(\text{circular sector} OQI) \le
\text {area}(\triangle OZI) 
$$
$$
{1\over2}\cdot1\cdot\sin t\lt{1\over2} t\lt {1\over2}\cdot1\cdot\tan t
$$
$$
 \sin t\lt t\lt  \cdot\tan t
$$
$$
  {1\over\sin t}\gt {1\over t}\gt  {\cos t\over \sin t }
$$
$$
\cos t\lt {\sin t\over t}\lt 1.
$$
A: 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 $\cos\theta<\frac{\sin\theta}{\theta}$.
