How you prove that depends, as others have said, upon how you define sine and cosine. A common definition, in Calculus and PreCalculus texts is to define them in terms of the "unit circle"- the circle of radius 1 with center at the origin of a coordinate system. Given such a circle, to find cos(t) and sin(t), start at (1, 0) and measure around the circle (counter-clockwise if t> 0, clockwise if t< 0) a distance |t|. cos(t) is the x-coordinate of the final point, sin(t) is the y-coordinate of that point.
Now, draw a line from (cos(t), sin(t)) to the x-axis, a straight line from (cos(t), sin(t)) to (0, 0).
Those two lines, together with the x-axis itself, form a right triangle having legs of length y and x and so area of (1/2)sin(t)cos(t).
The "sector" having as boundaries the x-axis, the line from (cos(t), sin(t)) to (0, 0) and the arc of the circle, has area t/2 (the area of a circular sector with radius r and angle \theta has area \theta r/2. Here, r= 1 and \theta= t.)
Draw a vertical line at (1, 0) (tangent to the circle). That intersects the line from (cos(t) sin(t)) to (0, 0) at (1, tan(t)).
(The line from (cos(t), sin(t)) to (0, 0) has slope (sin(t)- 0)/(cos(t)- 0)= sin(t)/cos(t)= tan(t) and so has equation y= tan(t) x. Setting x= 1, y= tan(t).)
So this gives a right triangle with legs of length 1 and tan(t) and so area (1/2)(1)(tan(t)= (1/2)tan(t).
Now, clearly, the first right triangle is inside the circular sector which is inside the last right triangle. That is the area of each is less than the area of the next:
(1/2)sin(t)cos(t)\le (1/2)t\le (1/2)tan(t)
And obvious thing to do is multiply through by 2:
sin(t)cos(t)\le t\le tan(t)= \frac{sin(t)}{cos(t)}
Divide each part by sin(t). (We are taking the limit as t goes to 0 so t is NOT itself 0 and sin(t) is not 0)
cos(t)\le \frac{t}{sin(t)}\le \frac{1}{cos(t)}
Taking the reciprocal of each part reverses the inequalities:
\frac{1}{cos(t)}\ge \frac{sin(t)}{t}\ge cos(t)
or
cos(t)\le \frac{sin(t)}{t}\le \frac{1}{cos(t)}
That is the result you wanted. The application of that, of course, is to observe that cos(0) is 1 and cosine is a continuous function to taking the limit as t goes to 0, both left and right parts go to 0:
1\le \lim_{t\to 0}\frac{sin(t)}{t}\le 1
which proves that \lim_{t\to 0} sin(t)/t= 1.
