Answer:
0.44290869
Step-by-step explanation:
The Maclaurin series for sin⁻¹(x) is given by
sin⁻¹(x) = x + [tex]x^{\alpha } _{n=1}[/tex] [tex]\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}[/tex]
Use the first five terms of the Maclaurin series above to approximate sin⁻¹ [tex]\frac{3}{7}[/tex]. (Round your answer to eight decimal places.)
Answer
sin⁻¹(x) = x + [tex]x^{\alpha } _{n=1}[/tex] [tex]\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}[/tex]
in the above equation [tex]x^{\alpha } _{n=1}[/tex] summation from n=1 to ∞
we are estimating this for the first 5 terms as follows
sin⁻¹(x) = x + [tex]\frac{1}{2} * \frac{x^{3} }{3}[/tex] + [tex]\frac{1*3}{2*4} * \frac{x^{5} }{5}[/tex] + [tex]\frac{1*3*5}{2*4*6} * \frac{x^{7} }{7}[/tex] + [tex]\frac{1*3*5*7}{2*4*6*8} * \frac{x^{9} }{9}[/tex]
sin⁻¹(x) = x + [tex]\frac{x^{3} }{6}[/tex] + [tex]\frac{3x^{5} }{40}[/tex] +[tex]\frac{15x^{7} }{336}[/tex] + [tex]\frac{105x^{9} }{3456}[/tex]
now to get
sin⁻¹([tex]\frac{3}{7}[/tex]) substitute
hence,
sin⁻¹([tex]\frac{3}{7}[/tex]) = [tex]\frac{3}{7} + \frac{\frac{3}{7} ^{3} }{6} + \frac{3 * \frac{3}{7} ^{5} }{40} + \frac{15* \frac{3}{7} ^{7} }{336} + \frac{105* \frac{3}{7} ^{9} }{3456}[/tex]
sin⁻¹([tex]\frac{3}{7}[/tex]) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482
= 0.44290869
