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

How do I get:

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

Answer

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
$$

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

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