सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

118 (2 H Cos 15 H sin 5) / R — (2 H Cos 25 H sin 5) / R = (2 H sin 5) / R (H Cos 15 — H Cos 25) = (2 H sin 5) / R × (2 H sin 20 H sin 5) / R Let now X be the point where we are to find the H sine (Here let us take it as x° after the previous interval for generalisation). Then Bhaskara's formula would give (H Cos 20 H Sin 10) / R — (2 H sin 5) / R² × (2 H sin 20 H sin 5) / 20 × x = (H Cos 20 H sin 10) / R — x / (10 R²) × 2 H sin 20 H sin² 5 = (2 H Cos 20 H sin 5 H Cos 5) / R² — 1 / R² × x / 10 × 2 H sin 20 H sin² 5 = (2 H sin 5) / R { (H Cos 20 H Cos 5 — x / 10 × H sin 20 H sin 5) / R } put now successively x = 0° and 10° to get the rectified differences at B and C respectively; then those rectified differences would be respectively H Cos 20 H sin 10 and (2 H sin 5) / R × H Cos 25. But (AB + BC) / 2 = (H Cos 20 H sin 10) / R (found above) and BC = H sin 30 — H sin 20 = (2 H Cos 25 H sin 5) / R In other words the rectified differences at B and C are respectively what exactly has been stated by Bhāskara. Hence Kamalākara's condemnation of Bhāskara is quite unjustified. Verse 17. To rectify the arcual difference to obtain the arc for a given H sine. Subtract as many H sine-differences as could be sub- tracted from the given H-sine. Half of the remainder multiplied by the difference of the preceding and succeed- ing H sine-differences and divided by the succeeding and