Mathematics: what India contributed to the ancient world

My previous post touched on the knowledge ancient Indian astronomers gleaned from the ancients.

Here, however, I'd like to highlight what they contributed to mathematicians at large. I exclude techniques they uncovered which were either independently used elsewhere or which generally never came into use, there being alternatives.

India's big contribution was in the development of techniques and tools.

A major practical trigonometric contribution the Indians made was the development of sines. This was to replace the tables of "chords" of angles first used for calculations by Hipparchus: the half-chord of double an angle corresponds to its sine, the chord of an angle being 120 times the sine of half the angle. Aryabhatta, in particular, seems to have realised that this value came up more frequently in trigonometric calculations than the more unwieldy chord. He even detailed the construction of the first sine table: for angles from 0 to 90 degrees in steps of 3.75 (i.e pi/48 radians).

Brahmagupta developed a numerical interpolation technique which we now know as the method of second-order differences.

And of course, the decimal place value number system as we know it, and the use of zero are Indian in origin. More on this in a later post.

Incidentally, the calculation of the sine of pi/48 is reasonably straigjtforward if you know basic trig identities. pi/24 = pi/6 - pi/8. pi/6 has standard sine and cosine values that can be read off an equilateral triangle. Those for pi/8 can be worked out using the half-angle formulae for pi/4, corresponding to an isosceles right triangle. Then use the difference formula to obtain the values for pi/24, and the half-angle formulae again to get the desired answer.

Aryabhatta might have missed a trick, though: calculating these values in steps of 3 or 1.5 degrees (pi/60 or pi/120) would have been a bit more effort but could still have been managed with the techniques of his day.

The trick is to obtain trig values for pi/5: this can be done from similar triangles obtained by drawing diagonals in a regular pentagon, making use of the golden ratio to obtain these values in terms of the square root of 5. One you have this, use pi/30 = pi/5 - pi/6, and the half-angle formulae then for pi/60 and again for pi/120.

Why not at intervals of 1 degree (pi/180) then? Well, the issue here is that you would need to solve a cubic equation to obtain this value from that of pi/60. The techniques for solving these analytically were not available until Cardano came along centuries later, they certainly cannot be solved geometrically. Moreover, attempting to solve the cubic for this case leads to an answer in terms of complex numbers, though the value is demonstrably real! The way out is to use approximations, but that theory would be more fully developed much later.