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: