Page:A History of Mathematics (1893).djvu/118

 divided the circle into quadrants, each quadrant into 90 degrees and 5400 minutes. The whole circle was therefore made up of 21,600 equal parts. From Bhaskara's 'accurate' value for $$\scriptstyle{\pi}$$ it was found that the radius contained 3438 of these circular parts. This last step was not Grecian. The Greeks might have had scruples about taking a part of a curve as the measure of a straight line. Each quadrant was divided into 24 equal parts, so that each part embraced 225 units of the whole circumference, and correspond to $$\scriptstyle{3\frac{3}{4}}$$ degrees. Notable is the fact that the Indians never reckoned, like the Greeks, with the whole chord of double the arc, but always with the sine (joa) and versed sine. Their mode of calculating tables was theoretically very simple. The sine of 90&deg; was equal to the radius, or 3438; the sine of 30&deg; was evidently half that, or 1719. Applying the formula $$\scriptstyle{\sin^2a+\cos^2a=r^2}$$, they obtained sin 45&deg;$$\scriptstyle{=\sqrt{\frac{r^2}{2}}=2431}$$. Substituting for cos a its equal $$\scriptstyle{\sin(90-a)}$$, and making $$\scriptstyle{a=60}$$&deg;, they obtained sin 60&deg;$$\scriptstyle{=\frac{\sqrt{3r^2}}{2}=2978}$$. With the sines of 90, 60, 45, and 30 as starting-points, they reckoned the sines of half the angles by the formula $$\scriptstyle{\text{ver }\sin 2 a = 2 \sin^2 a}$$, thus obtaining the sines of 22&deg; 30&prime;, 11&deg; 16&prime;, 7&deg; 30&prime;, 3&deg; 45&prime;. They now figured out the sines of the complements of these angles, namely, the sines of 86&deg; 15&prime;, 82&deg; 30&prime;, 78&deg; 45&prime;, 75&deg;, 67&deg; 30&prime'; then they calculated the sines of half these angles; then of their complements; then, again, of half their complements; and so on. By this very simple process they got the sines of angles at intervals of 3&deg; 45&prime;. In this table they discovered the unique law that if a, b, c be three successive arcs such that $$\scriptstyle{a-b=b-c=}$$3&deg; 45&prime;, then $$\scriptstyle{\sin a-\sin b = (\sin b-\sin c)-\frac{\sin b}{225}}$$. This formula was afterwards used whenever a re-calculation of tables had to be made. No Indian trigonometrical treatise on the triangle