SNU Mathematics: Brahmagupta's Sine Interpolation

In a previous post we saw a very accurate approximation to the Sine function provided by the scientist Bhaskara who lived in the 7^th century AD. In this post, we present an alternative formula given by his contemporary Brahmagupta. This is based on starting with a table of Sine values. The standard tables presented by Aryabhata and others preceding Brahmagupta used intervals of 3^o45'. Intermediate values were then found by linear interpolation. Let's express this in our language:

We assume intervals of fixed width

$\delta$ . We are thus looking at the angles 0,

$\delta$ ,

$2\delta$ , ...,

$i\delta$ ,... Suppose a table is available for the Sines of these angles:

$\sin(0)$ ,

$\sin(\delta)$ ,

$\sin(2\delta)$ , ...,

$\sin(i\delta)$ , ... Our task is to estimate the value of

$\sin(x)$ when

$x$ is not a tabulated angle. Suppose

$x$ is between

$x_{i-1}=(i-1)\delta$ and

$x_i=i\delta$ . The total change of the Sine function over this interval is

$\sin(x_i)-\sin(x_{i-1})$ and we denote this by

$\triangle\sin_i$ . If this change were happening linearly then every unit change in the angle would produce a change of

$\frac{\triangle\sin_i}{\delta}$ in the Sine function. In moving from

$x_{i-1}$ to

$x$ we would therefore create a change of

$\frac{\triangle\sin_i}{\delta}(x-x_{i-1})$ This gives the following approximation, which is called linear interpolation:

$\sin(x)-\sin(x_{i-1}) \approx \frac{\triangle\sin_i}{\delta}(x-x_{i-1}) \quad\mbox{or}\quad \sin(x) \approx \sin(x_{i-1}) + \frac{\triangle\sin_i}{\delta}(x-x_{i-1})$
In linear interpolation we match the function which a straight line which meets it at 2 points:

$x_{i-1}$ and

$x_i$ . Brahmagupta's approach amounts to improving this by matching the function with a quadratic which meets it at 3 points:

$x_{i-2}$ ,

$x_{i-1}$ and

$x_i$ . To derive this formula (and we don't know the steps Brahmagupta took) we start with a form similar to the linear interpolation formula:

$\sin(x)\approx \sin(x_{i-1}) + (x-x_{i-1})p(x)$ where

$p(x)$ is a linear function

$Ax+B$ . This already does the right thing at

$x_{i-1}$ . We have to choose

$p(x)$ so that it also does the right thing at

$x_{i-2}$ and

$x_i$ . In other words, we need

$p(x)$ to satisfy the following:

$\sin(x_{i-2})=\sin(x_{i-1}) + (x_{i-2}-x_{i-1})p(x_{i-2})$

$\sin(x_{i})=\sin(x_{i-1}) + (x_{i}-x_{i-1})p(x_{i})$ These can be rearranged into

$p(x_{i-2}) = \frac{\triangle\sin_{i-1}}{\delta}$

$p(x_{i}) = \frac{\triangle\sin_{i}}{\delta}$
This is again a linear interpolation problem and we have already seen how to solve it:

$\begin{eqnarray*} p(x) &=&p(x_{i-2}) + \frac{x-x_{i-2}}{x_i-x_{i-2}}(p(x_i)-p(x_{i-2}))= \frac{\triangle\sin_{i-1}}{\delta}+\frac{x-x_{i-2}}{2\delta}\left(\frac{\triangle\sin_{i}-\triangle\sin_{i-1}}{\delta}\right)\\ &=& \frac{\triangle\sin_{i-1}}{\delta}+\frac{x-x_{i-1}+\delta}{2\delta}\left(\frac{\triangle\sin_{i}-\triangle\sin_{i-1}}{\delta}\right) = \frac{\triangle\sin_i+\triangle\sin_{i-1}}{2\delta}+\frac{x-x_{i-1}}{\delta^2}\left(\frac{\triangle\sin_{i}-\triangle\sin_{i-1}}{2}\right) \end{eqnarray*}$

We bring things to a close by putting this expression for

$p(x)$ back into our initial quadratic approximation. We get:

$\sin(x) \approx \sin(x_{i-1}) + \frac{x-x_{i-1}}{\delta}\left(\frac{\triangle\sin_i+\triangle\sin_{i-1}}{2}+\frac{x-x_{i-1}}{\delta}\left(\frac{\triangle\sin_{i}-\triangle\sin_{i-1}}{2}\right)\right)$ Brahmagupta used

$\delta=15^o =900'$ . So his base trigonometric table needed only five values (ignoring the trivial cases of 0 and 90 degrees). And here are graphs showing the accuracy of his formula between 60 and 75 degrees: