Bhinmal is a city in the northwest part of India. According to another online article about famous scientist he was the son of Jisnugupta, who was an astrologer (2014). Little is known about other members of his family, since this was also a long time ago. Brahmagupta could be considered to have many titles, even though he considered himself an astronomer, according to that famous scientist article he is “mainly remembered for his contributions to mathematics”.
He rose up in the ranks within the scientist community and became the director of the astronomical observatory of Ujjain. According to a biography, Brahmagupta wrote a couple outstanding books, one being Brahmasphutasiddhanta which covered topics in “mean longitudes of the planets; true longitudes of the planets; the three problems of diurnal rotation…” and much more (O'Connor & Robertson).The other book he wrote was called Khandakhadyaka, which covers similar things as the Brahmasphutasiddhanta did.Although he wrote books mainly about astronomical, he had some controversial and excellent mathematical ideas. In the article found on the famous scientist site, it is said that Brahmagupta was the “first person in history to see zero as a number with its own properties”. It is also said that he saw zero as a number you would get after subtracting a number by itself, and that “zero divided by any other number is zero”. Brahmagupta had also thought that zero divided by zero equals zero, which is wrong because this is actually undefined. Brahmagupta may have had some wrong thoughts about math, but one of the formulas he discovered was right on. It deals with “determining the area of a cyclic quadrilateral given only the four sides lengths” (“Brahmagupta’s Formula”). According to that article, a nice proof for this formula is, Given: A cyclic quadrilateral with sides a, b, c, d, and the area of K can be found as: \(K=\sqrt{(s-a)(s-b)(s-c)(s-d)}\) where \(s=\dfrac{a+b+c+d}{2}\) is the semi-perimeter of the quadrilateral. If we draw AC, we find that \([ABCD]=\dfrac{ab\sin{B}}{2}+\dfrac{cd\sin{D}}{2}=\dfrac{ab\sin{B}+cd\sin{D}}{2}\). Since \(B+D=180^{\circ}\), \(\sin{B}:\sin{D}\). Hence, \([ABCD]=\dfrac{(ab+cd)\sin{B}}{2}\). Multiplying by 2 and squaring, we get: \(4[ABCD]^{2}=\sin^{2}{B}(ab+cd)^{2}\). Substituting \(\sin^{2}{B}=1-\cos^{2}{B}\) results in \(4[ABCD]^{2}=(1-\cos^{2}{B})(ab+cd)^{2}=(ab+cd)^{2}-\cos^{2}{B}(ab+cd)^{2}\). By the Law of Cosines, \(a^{2}+b^{2}-2ab\cos{B}= c^{2}+d^{2}-2cd\cos{D}$. $\cos{B}=-\cos{D}\), so a little rearranging gives \(2\cos{B}(ab+cd)=a^{2}+b^{2}-c^{2}-d^{2}\). Now, to put everything together: \(4[ABCD]^{2}=(ab+cd)^{2}-\dfrac{1}{4}(a^{2}+b^{2}-c^{2}-d^{2})^{2}\) \(16[ABCD]^{2}=4(ab+cd)^{2}-(a^{2}+b^{2}-c^{2}-d^{2})^{2}\) \(16[ABCD]^{2}=(2(ab+cd)+(a^{2}+b^{2}-c^{2}-d^{2}))(2(ab+cd)-(a^{2}+b^{2}-c^{2}-d^{2}))\) \(16[ABCD]^{2}=(a^{2}+2ab+c^{2}+2cd-d^{2})(-a^{2}+2ab-b^{2}+c^{2}+2cd+d^{2})\) \(16[ABCD]^{2}=((a+b)^{2}-(c-d)^{2})((c+d)^{2}-)a-b)^{2}\) \(16[ABCD]^{2}=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-b+a)\) \(16[ABCD]^{2}=16(s-a)(s-b)(s-c)(s-d)\) \([ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}\) ("Brahmagupta's Formula").
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## AuthorHello, as you may know my name is Kendrick Hardison. This blog will be used to share my experiences, teach things I may have learned, advise others, and much more. If you have any questions or if you would like for me to write a post on something you are interested in, click here to get to my contact page. WARNING: Math may show up smaller than normal text on mobile devices.## Archives
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