[Pramāṇa](https://indike.org/pramana/) › [Studies](https://indike.org/pramana/ins) › [Science, technology and education](https://indike.org/pramana/ins.science) › [Mathematics in Ancient and Medieval India](https://indike.org/pramana/ins.science.bag-1979) › c8.s3.85 # Mathematics in Ancient and Medieval India c8.s3.85 Mathematics in Ancient and Medieval India c8.s3.85 - **Language:** English ### [c8.s3.85](https://indike.org/pramana/ins.science.bag-1979.c8.s3.85) first quadrant be divided into n equal divisions each being dx. Let p₁, p₂, p₃\_n be the sine chords for parts of the arc x corresponding to parts of the arc represented by dx, 2dx, 3dx... . Let p'₁, p'₂, p'₃... be the corresponding sine differences. Then p'₁ = p₁ = R sin x $p'₁ - p'₂ = R sin x. ((dx)²)/(R²) = p₁ ((dx)²)/(R²)$ $or p'₂ = p'₁ - p₁ ((dx)²)/(R²)$ $= p₁ - p₁ ((dx)²)/(R²)$ $p'₃ = p'₂ - p₂ ((dx)²)/(R²) = p₁ - p₁ ((dx)²)/(R²) - p₂ ((dx)²)/(R²)$ $$ $$ Similarly, $p'\_n = p₁ - p₁ ((dx)²)/(R²) - p₂ ((dx)²)/(R²) - - p\_n-1 ((dx)²)/(R²)$ $p'₁ + p'₂ + p'\_n &= np₁ - \[(n-1)p₁ + (n-2)p₂ + / & + 2p\_n-2 + p\_n-1\] ((dx)²)/(R²)$ or p\_n = sum of all these sine differences $&= np₁ - \[(n-1)p₁ + (n-2)p₂ + ... / & + 2p\_n-2 + p\_n-1\] ((dx)²)/(R²)$ $&= whole arc - \[(n-1)p₁ + (n-2)p₂ + / & + p\_n-1\] ((dx)²)/(R²)$ - [Previous: c8.s3.84](https://indike.org/pramana/ins.science.bag-1979.c8.s3.84) - [Next: c8.s3.86](https://indike.org/pramana/ins.science.bag-1979.c8.s3.86) - [Up: c8.s3](https://indike.org/pramana/ins.science.bag-1979.c8.s3) --- Source: [https://indike.org/pramana/ins.science.bag-1979.c8.s3.85](https://indike.org/pramana/ins.science.bag-1979.c8.s3.85) · Pramāṇa