There is a story, apparently from Diogenes Laertius and Plutarch (and repeated here, and here, and here, and, well, here) that Thales was able to calculate the height of an Egyptian pyramid by the following method: he waited until the length of his own shadow was equal to his height, and then measured the shadow of the pyramid. He reasoned that at that moment the pyramid will (like him) cast a shadow as long as its own height.
This story seems to occur at a fairly early stage in the consideration of the problem. Using only middle-school mathematics, and at any moment when the sun is in the sky, the man could measure his shadow, his own height, and the pyramid’s shadow, and then calculate the height of the pyramid by proportional reasoning; he need not wait until the sun angle makes his shadow's length equal to his own height. Using trig tables instead, the man need never measure his own shadow or height at all.
But they say that Archimedes was able to calculate the value of $\pi$ by comparing the perimeters of polygons drawn such that the radius of a single circle is the radius of the smaller polygon and the apothem of the larger polygon. When I try this latter problem, it involves me in trig tables.
Am I right – Archimedes used trig tables because they were introduced during the 400-year window between the two lives? Or is there another explanation for the stories, if trig tables existed at both times or at neither time? Perhaps in the story about the pyramid the difficulty about using trig tables was just the practical one, measuring the sun’s angle?
Using the language of trigonometry it makes things easier to follow. But Archimedes himself did not know trig. But the extrapolation to classical geometry is not so difficult.
6 equilateral triangles can be put together to form a hexagon.
The sides of an inscribed hexagon are each of length $2\sin 30$ The sides of an excribed hexagon are each of length $2\tan 30$
$12\sin 30 \le 2\pi \le 12\tan 30$
$\tan 30 = \sqrt 3, \cos 30 = \frac {\sqrt 3}{2}, \sin 30 = \frac 12$
And if we had a formula to find $\sin \frac {\theta}{2}, \tan \frac \theta 2$ based on $\sin\theta, \cos\theta, \tan\theta$
then
$(2^n)6\sin \frac {30}{2^n} \le \pi \le (2^n)\tan \frac {30}{2^n}$
And we do:
$\tan \frac 12 \theta = \frac {\sin\theta}{1+\cos \theta}\\ \cos \frac 12 \theta = \frac {\sqrt {1+\cos \theta}}{2}\\ \sin \frac 12 \theta = \frac {\sqrt {1-\cos \theta}}{2}\\ $
We repeatedly apply the half angle theorems then we will get increasingly accurate bounds for $\pi$
But Archimedes did not have trig. What tools did he have at his disposal?
If $AD$ bisects $\angle BAC$ then $BD:DC = AB:AC$
From which we can derive:
$BD:DC = AB:AC\\ \frac {BD}{DC} = \frac {BC - DC}{DC} = \frac {BC}{DC} - 1\\ \frac {BC}{DC} = \frac {AB}{AC} + 1 = \frac {AB+AC}{AC}\\ \frac {DC}{AC} = \frac {BC}{AB+AC}$
Which is equivalent to the $\tan \frac 12 \theta$ formula above
It also seems that he used rational approximations of the trigonometric ratios at each step. Exactly how he calculated those ratios has been lost over time, but methods to approximate square roots were known to the Babylonians, and certainly would have been known to Archimedes.
A 30-60-90 triangle has rations approximately:
$BC:AC:AB \approx 153:265:306$ \frac {DC}{AC} = \frac {153}{571}$
Using the Pythagorean theorem: $\sqrt {153^2 + 571^2}\approx 591$ $DC:AC:AD \approx 153:571:591$
$6\cdot \frac {153}{571}<\pi<6\frac {571}{591}\\ 3<\pi<3.46$
Repeating the process gives us methods to approximate ratios for a sequence of right triangles each with an angle half the size of the one before.
If $E$ lies on the angle bisector of $DC$
$EC:AC:AD = 153:1162:1172$
$12\cdot \frac {153}{1162}<\pi<12\frac {153}{1172}\\ 3.10<\pi<3.22$
etc.