This method of approximation is a specific form of Buffon's needle and a better proof is given in proofs from the BOOK, which does not involve any calculus.

The number Pi (π) can be found through experiment by drawing parallel lines a constant distance apart and then randomly dropping matches half that length. Then

π ≈ (number of dropped matches) / (number of matches touching or crossing a line)
The more matches dropped, the better is the approximation.

This can be proved by relating it to the formula under a sine graph.

proof

For the first part of the proof we asume that this works and so the probability of a match crossing or touching a line is 1/π. It is also necessary to assume that the matches can land pointing any directrion with equal liklihood. Thus, each angle of a match to the direction of the parallel lines is also equally likely. From this we can see that when taking the length of the match as 1 unit the sin from the direction of parallel lines is the perpendicular distance between the two ends of the match. The probability of a match touching or crossing a line is perpendicular distance / 2. As the distance between the parallel lines is 2 and taking one end of the match for the other to cross the line it needs to be within the perpendicular distance to the next perpendicular line in the direction of the other end.

So if you draw a graph of probability of perpendicular distance against perpendicular distance / 2 the area under the graph is 1/π, probability of a match crosses or touches a line, and it is the same shape as a sin graph. So if you change the probability to radians by multiplying by half pi and the perpendicualar distance / 2 to sin x by multyplying by 2 you find that the area under a sine graph which uses radians is 1.

Now we work back the other way.

Construct a circle. With point A on the circumference which marks 0 radians. Then construct points E and F also on the circumference. F has a greater angle from A than e. Also add point g so that GF is perpendicular to OA and GE is parallel to 0A. As F and E get closer together FE is a better aproximation to the curve and OEF approaches a right angle. So, as OEF aproaches a right angle, become closer with OEG = GEF as both take OEG from a right angle. Also approach GFE = EOA.

So GF is the positive change in sin x from E to F, where x is in radians. So using trigonometry:

cos x = dsinx / dx and -sin x = dcosx/dx

Area under sinegraph until half pi under a sine graph is the same as that in a cosine graph as same as working back form half pi radians for sine.

so

area under y gradient graph

sin x cos x - sin x

So as sin π /2 = 1 the area under a radian sine graph until π /2 is 1, as required.