Alongside calculus, the Kerala School also came up with some interesting results in Trigonometry, including the now commonly known method to calculate the sines (ज्या) of the sums are difference of two angles (arcs):
जीवे परस्परनिजेतरमौर्विकाभ्याम्
अभ्यस्य विस्तृतिदलेन विभज्यमाने ।।
अन्योन्ययोगविरहानुगुणे भवेतां
यद्वास्वलम्बकृतिभेदपदीकृते द्वे ।।
1) The two jyā (of the two arcs), multiplied by the kojyā of the other, and divided by half the diameter, by adding and subtracting, become the jyā of the similar operations (ie sum or difference of arcs).
2) Alternately, the sums and differences of the square roots of the differences between squares of the two jyā and the lamba can be used.
That is
\(
\begin{align*}
jyā(a \pm b) &= \frac{1}{R}.(jyā(a). kojyā(b) \pm jyā(b). kojyā(a)) \\
sin(A \pm B) &= sin(A).cos(B) \pm sin(B).cos(A)
\end{align*}
\)
The “extra” factor of 1/R , for those used to the modern form, comes from the fact that jyā and kojyā run from 0 to R, while modern sines and cosines run from 0 to 1.
What about the second part of the verse, what exactly is a lamba?
ज्यायोः परस्परं घातात् त्रिज्याप्तं लम्ब इष्यते
The product of the two jyā divided by the radius is the lamba
\(
\begin{align*}
jyā(a \pm b) &= \sqrt{jyā(a)^2 – lamba^2} \pm \sqrt{jyā(b)^2 – lamba^2} \\
lamba &= \frac{jyā(a) . jyā(b)}{R}
\end{align*}
\)
What the “lamba” is geometrically can be seen from the yukti below.
Yukti
How do we justify these results? Yuktibhāṣā provides two proofs, one for either part of the verse.
The first yukti
Consider arc \( EA_3 = EA_2 + A_2A_3\). We seek to find the jyā in terms of the jyā of the constituent arcs. We take the radius OA2, and draw A3L perpendicular to it, extending it to meet the circle at A1 such that A1A2 = A2A3. Jyā and kojyā of EA1, EA2, and EA3 are A1B1, A2B2, A3B3, and A1S1, A2S2, A3S3 respectively. We draw perpendiculars from L to the East and North axes as LC and LD respectively. LC meets A3S3 at Q and LD meets A3B3 at P. A1Q is the lamba, also equal to LP
A3PL and A2S2O are perpendicular triangles, while OLC and OA2B2, and OLD and OA2S2 are parallel. Hence, each set can be related with त्रैराशिकम्. This gets us:
\(\begin{align*}
PL &= \frac{S_2O . A_3L}{A_2O} \\
&= \frac{A_2B_2.A_3L}{A_2O} \\
& = \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3) \\
PA_3 &= \frac{A_2S_2 . A_3L}{A_2O} \\
&= \frac{1}{R}. kojyā(EA_2) . jyā(A_2A_3) \\
LC &= \frac{OL.A_2B_2}{A_2O} \\
&= \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3) \\
LD & = \frac{OL. A_2S_2}{ A_2O}\\
&= \frac{1}{R}. kojyā(EA_2) . kojyā(A_2A_3)
\end{align*}
\)
Using these, we can see that
\(
\begin{align*}
jyā(EA_3) &= A_3B_3 = LC + A_3P \\
jyā(EA_2+A_2A_3) &= \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3) + \frac{1}{R}. kojyā(EA_2) . jyā(A_2A_3) \\
jyā(EA_1) &= A_1B_1 = LC – A_3P \\
jyā(EA_2-A_2A_3) &= \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3) – \frac{1}{R}. kojyā(EA_2) . jyā(A_2A_3)
\end{align*}
\)
and also, as a bonus:
\begin{align*}
kojyā(EA_3) &= A_3S_3 = LD- PL \\
kojyā(EA_2+A_2A_3) &= \frac{1}{R}. kojyā(EA_2) . kojyā(A_2A_3) + \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3) \\
kojyā(EA_1) &= A_1S_1 = S_1Q + A_1Q = LD + PL \\
kojyā(EA_2-A_2A_3) &= \frac{1}{R}. kjyā(EA_2) . kojyā(A_2A_3) + \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3)
\end{align*}
\)
This demonstrates the first method of the verse. What about the second, which directly uses the lamba (PL) ?
The second yukti
We have seen that the lamba \( A_1Q = PL= \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3) \).
\(
\begin{align*}
A_3B_3 &= A_3P+PB_3 \\
A_3P &= \sqrt{A_3L^2 – LP^2}\\
PB_3 &= \sqrt{B_3L^2 – LP^2}
\end{align*}
\)
Now, A3L is evidently jyā(A2A3). What is B3L? Reasoning backwards, we can infer that it must be jyā(EA2). How do we demonstrate that? We already know that \( PB_3 = LC = \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3)\), so we can calculate
\(
\begin{align*}
B_3L^2 &= LP^2 + LC^2 \\
&= \frac{1}{R}. jyā(EA_2)^2. (jyā(A_2A_3)^2 + kojyā(A_2A_3)^2) \\
&= jyā(EA_2)^2
\end{align*}
\)
And therefore:
\(
\begin{align*}
jyā(a \pm b) &= A_3B_3 \\
&= \sqrt{jyā(a)^2 – lamba^2} \pm \sqrt{jyā(b)^2 – lamba^2}
\end{align*}
\)
In this case, we used the value of LC already inferred in the previous yukti. Can we make the two yuktis fully independent?
As it turns out, yes we can. Consider a circle centered at O', with triangle A3LB3 circumscribed. Once such circle is always possible. In this circle A3B3 and A3L would both be samastajyā (fullchords), and LB3 would be the samastajyā of their difference (by definition). In a circle of twice the radius of that construction, A3B3 and A3L would both be jyā (halfchords), and on the same circle, LB3 would have to be the jyā of their difference. Since only one such circumscribing circle can exist, therefore one such circle with A3B3 and A3L as jyā can exist. Therefore, it must be the current one, where A3B3 = jyā(EA3) and A3L = jyā(A2A3). Therefore LB3 = jyā(EA2).
Using जीवे परस्परन्यायः
This allows us to use both the Madhava jyā table (श्रेष्ठं नाम वरिष्ठानाम्) , and the Madhava jyā series (निहत्य चापवर्गेण …) to compute accurate jyā. We note that it is much easier to use the series for small arcs. Since we know the jyā at the tabular points accurately, and can calculate jyā and kojyā of small arcs accurately with the series, we use the following technique to calculate the jyā of any arbitrary arc:
- Find the nearest tabular arc, and read off its jyā and kojyā from the table
- Calculate jyā and kojyā of the difference of the desired arc and the nearest tabular arc using the series.
- These are combined using जीवे परस्पर to get an accurate jyā of the desired arc.
This gives us the superior accuracy of the निहत्य चापवर्गेण series, but with faster convergence that we see for smaller arcs. The श्रेष्ठं नाम वरिष्ठानाम् table is accurate to the thirds, and using this method we have a quick and accurate method of calculating jyā for arbitrary arcs that maintains similar accuracy.
