How to show that the approximation sinx=x-(x^3/3!)+(x^5/5!) is accurate to four decimal places if 0<=x<=pi/4

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How to show that the approximation sinx=x-(x^3/3!)+(x^5/5!) is accurate to four decimal places if 0<=x<=pi/4

2007-10-04 13:15:00 · 2 answers · asked by kakalaky1 1 in Science & Mathematics ➔ Mathematics

2 answers

The error is bounded by the maximum possible value of x^7/7! on this interval, i.e. (π/4)^7 / 7! = 3.66×10^-5.

2007-10-04 13:19:05 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋

(5x-6)=e^3 with the help of multiplying with the help of e on the two facets subsequently x=((e^3)+6)/5 then variety this into calc and that's ur answer however the respond I even have given is in precise variety and subsequently greater precise Please vote for superb answer : )

2016-10-21 01:55:50 · answer #2 · answered by ? 4 · 0⤊ 0⤋