What is the supremum of $|\sin (t )/t|$? What is the supremum of $|\sin (t )/t|$?
There should be one, and it will be at least than 1, since $\lim_{t \to 0} \sin (t )/t = 1$.
Thoughts?
 A: Here's a nice proof. Consider the unit circle as pictured in this nice image from AOPS:

(source: artofproblemsolving.com)
The vertical line is $\sin(t)$ and the arc around the circle is $t$ since the radius is $1$. Thus, $|\sin(t)| < |t|$ for all $t$ and we get that the supremum of $\left| \frac{\sin(t)}{t}\right|$ is $1$.
In fact, we can extend this "visual" prove to give the inequality $t \le \tan(t)$ for the angle between $0$ and $\pi/2$ by considering the similar triangle with the adjacent leg and hypotenuse lying on the same ray but with the adjacent leg being the whole radius. Then the opposite leg is clearly bigger than the arc and has length $\tan(t)$.
A: Draw a unit circle through the origin. The positive x-axis intersects the circle at $(1,0)$. Suppose $0 < t < \pi/2$. The ray R obtained by rotating the positive x-axis counter-clockwise by $t$ radians intersects the circle at $(\cos t,\sin t)$. The arc of the circle from $(1,0)$ to $(\cos t,\sin t)$ is $t$ units in length, and is longer than the shortest distance from $(\cos t,\sin t)$ to the $x$-axis, which has length $\sin t$. That is
$$
       0 < \sin t < t, \;\;\; 0 < t < \pi/2.
$$
So,
$$
         0 < \frac{\sin t}{t} < 1,\;\;\; 0 < t < \pi/2.
$$
For $t \ge \pi/2$, one has $1/t \le \frac{2}{\pi}$ and $|\sin t| < 1$. So,
$$
             \left|\frac{\sin t}{t}\right| < \frac{2}{\pi} < 1, \;\;\; t \le \pi/2.
$$
Finally, $\sin t/t$ is an even function, which eliminates the need for analysis for $t < 0$. The supremum has to be 1 because the function is bounded by 1 for $t \ne 0$, and because $\lim_{t\rightarrow 0}\frac{\sin t}{t}=1$.
A: Note first that $\sin(t)/t$ is an even function, so it suffices to argue that $-1\le \sin(t)/t\le 1$ for $t\ge0$, with equality only for $t=0$ (where, of course, the quotient is understood as its limit). 
Let $f(t)=t-\sin(t)$, so $f(0)=0$ and $f'(t)=1-\cos(t)\ge0$ for all $t$, with $f'(t)>0$ except at isolated points. This means that $f$ is strictly increasing, so $f(t)>0$ for $t>0$ and, therefore, $\sin(t)/t<1$ for $t>0$. 
Also, the same analysis but now applied to $g(t)=t+\sin(t)$ gives us that $t+\sin(t)>0$ for $t>0$, so $\sin(t)/t>-1$. 
