Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{x^{3}}{100} - 9$ using $LaTeX: \displaystyle x_0=9$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 9}{- \frac{3 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 9$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (9.0000000000) - \frac{- \frac{(9.0000000000)^{3}}{100} + \sin{\left((9.0000000000) \right)} + 9}{- \frac{3 (9.0000000000)^{2}}{100} + \cos{\left((9.0000000000) \right)}} = 9.6351498801$ $LaTeX: x_{2} = (9.6351498801) - \frac{- \frac{(9.6351498801)^{3}}{100} + \sin{\left((9.6351498801) \right)} + 9}{- \frac{3 (9.6351498801)^{2}}{100} + \cos{\left((9.6351498801) \right)}} = 9.5942992861$ $LaTeX: x_{3} = (9.5942992861) - \frac{- \frac{(9.5942992861)^{3}}{100} + \sin{\left((9.5942992861) \right)} + 9}{- \frac{3 (9.5942992861)^{2}}{100} + \cos{\left((9.5942992861) \right)}} = 9.5942142678$ $LaTeX: x_{4} = (9.5942142678) - \frac{- \frac{(9.5942142678)^{3}}{100} + \sin{\left((9.5942142678) \right)} + 9}{- \frac{3 (9.5942142678)^{2}}{100} + \cos{\left((9.5942142678) \right)}} = 9.5942142674$ $LaTeX: x_{5} = (9.5942142674) - \frac{- \frac{(9.5942142674)^{3}}{100} + \sin{\left((9.5942142674) \right)} + 9}{- \frac{3 (9.5942142674)^{2}}{100} + \cos{\left((9.5942142674) \right)}} = 9.5942142674$