The Kerala School, as we saw in past articles, were big on options for jyā computations. First, they improved Aryabhata's jyā recurrence, then added on an improved interpolation method. Later, we saw the Mādhava jyā series, plus a partially pre-computed version of it. Finally, we have the श्रेष्ठं नाम वरिष्ठानां verse, which provides a precomputed jyā table with much better accuracy. What more could one ask for?
What if, they ask, we could find a recurrence relation without that pesky R – the radius of standard circle? We recollect that the improved recurrence relationship
विलिप्तदशकोना ज्या राश्यष्टांशधनुः कलाः ।
आद्यज्यार्धात्ततो भक्ते सार्धदेवाश्विभिस्ततः
etc.
K_n = K_{n-1} – {(\frac{a}{R})}^2.B_{n-1} \\
B_n = K_n + B_{n-1}
\)
where R is the radius, and a is the full-chord of the quarter circumference divided by 24, (ie 225′). To make this computable, they use the approximation \(a = 225′\). This is more accurate than the Āryabhaṭa approximation which sets the corresponding jyā to 225′, since the full-chord is between the jyā and the corresponding arc in length.
This leads to the approximations
\({(\frac{a}{R})}^2 \approx 233′ 30″\) (नीलोबालारिः or सार्धदेवाश्वि in bhūtasaṅkhyā).
This was later improved by Śankaravāriyar to \({(\frac{a}{R})}^2 \approx 233′ 32″\).
Can we remove the need for R here? It turns out we can, and this will lead us into an interesting diversion through triangles and cyclic quadrilaterals, which is a journey of interest in itself.
Ābādhas and Area of a Triangle
लम्बगुणं भूम्यर्धं स्पष्टं त्रिभुजे फलं भवति
Half of the base multiplied by the perpendicular to it is the area of a triangle
The first stop in our journey is the elementary expression for area of a triangle. Consider a विषमत्र्यश्रम् (scalene triangle) ABC, with the longest side BC called the भूमिः (base), and AB as the second longest side.
AD is the perpendicular from the भूमिः to the vertex A. We call it the लम्बः (perpendicular). D divides the भूमिः into two segments (आबाधः॒) BD and DC. We draw TR and US, of equal length to and parallel to AD such that they bisect BD and DC. It is evident from this figure that rectangle RSTU has the same area as triangle ABC. (Triangle ATP and PBR are congruent etc.). Therefore,
\(
\begin{align*}
A &= RS. AD \\
&= \frac{1}{2}. BC. AD \\
&= \frac{1}{2}. b. h
\end{align*}
\)
We also see the interesting relationships:
\(\begin{align*}
BD^2 – CD^2 &= (AB^2 – AD^2)- (AC^2-AD^2) \\
& = AB^2 – AC^2
\end{align*}
\)
Since BC = BD + CD, some elementary manipulation gets us to
BD &= \frac{1}{2} . (BC + (\frac{AB^2 – AC^2}{BC})) \\
DC &= \frac{1}{2} . (BC – (\frac{AB^2 – AC^2}{BC})) \\
\end{align*}
\)
Cyclic Quadrilaterals
The next thing we need to look at are cyclic quadrilaterals – quadrilaterals that are inscribed in circles – and their diagonals and areas. Bhāskarācārya had given the following formula for the area of a cyclic quadrilateral (approximate for non-cyclic quadrilaterals).
सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः ।
मूलमस्फुटफलं चतुर्भुजे स्पष्टमेवमुदितं त्रिबाहुके ॥
Take the sum of all sides, and divide it into half. Multiply the result by the four differences of itself with each side. The square root of the result is is the approximate area of a quadrilateral, and precise for a triangle. (This is exact for a cyclic quadrilateral)
\(
\begin{align*}
A &= \sqrt{s.(s-a)(s-b)(s-c)(s-d)} &\textit{cyclic quadrilateral} \\
A &= \sqrt{s.(s-a)(s-b)(s-c)} &\textit{triangle} \\
\end{align*}
\)
How do we prove these, and how does this help us come up with a jyā recurrence without using the standard radius R? That will be the subject of the next article, where we will look at the three diagonals of a cyclic quadrilateral. Yes, three. You read that right.
