Over a million developers have joined DZone.
{{announcement.body}}
{{announcement.title}}

How medieval astronomers made trig tables

DZone's Guide to

How medieval astronomers made trig tables

· Big Data Zone
Free Resource

Need to build an application around your data? Learn more about dataflow programming for rapid development and greater creativity. 

How would you create a table of trig functions without calculators or calculus?

It’s not too hard to create a table of sines at multiples of 3°. You can use the sum-angle formula for sines

sin(α+β) = sin α cos β + sin β cos α.

to bootstrap your way from known values to other values. Elementary geometry gives you the sines of 45° and 30°, and the sum-angle formula will then give you the sine of 75°. From Euclid’s construction of a 5-pointed star you can find the sine of 72°. Then you can use the sum-angle formula to find the sine of 3° from the sines of 75° and 72°. Ptolemy figured this out in the 2nd century AD.

But if you want a table of trig values at every degree, you need to find the sine of 1°. If you had that, you could bootstrap your way to every other integer number of degrees. Ptolemy had an approximate solution to this problem, but it wasn’t very accurate or elegant.

The Persian astronomer Jamshīd al-Kāshī had a remarkably clever solution to the problem of finding the sine of 1°. Using the sum-angle formula you can find that

sin 3θ = 3 sin θ – 4 sin3 θ.

Setting θ = 1° gives you a cubic equation for the unknown value of sin 1° involving the known value of sin 3°. However, the cubic formula wasn’t discovered until over a century after al-Kāshī. Instead, he used a numerical algorithm more widely useful than the cubic formula: finding a fixed point of an iteration!

Define f(x) = (sin 3° + 4x3)/3. Then sin 1° is a fixed point of f. Start with an approximate value for sin 1° — a natural choice would be (sin 3°)/3 — and iterate. Al-Kāshī used this procedure to compute sin 1° to 16 decimal places.

Here’s a little Python code to play with this algorithm.

from numpy import sin, deg2rad
 
sin3deg = sin(deg2rad(3))
 
def f(x):
    return (sin3deg + 4*x**3)/3
 
x = sin3deg/3
for i in range(4):
    x = f(x)
    print(x)

This shows that after only three iterations the method has converged to floating point precision, which coincidentally is about 16 decimal places, the same as al-Kāshī’s calculation.

Source: Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry

Check out the Exaptive data application Studio. Technology agnostic. No glue code. Use what you know and rely on the community for what you don't. Try the community version.

Topics:

Published at DZone with permission of John Cook, DZone MVB. See the original article here.

Opinions expressed by DZone contributors are their own.

{{ parent.title || parent.header.title}}

{{ parent.tldr }}

{{ parent.urlSource.name }}