Answer:
0.44290869
Step-by-step explanation:
The Maclaurin series for sinā»Ā¹(x) is given by
sinā»Ā¹(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 sinā»Ā¹ [tex]\frac{3}{7}[/tex]. (Round your answer to eight decimal places.)
Answer
sinā»Ā¹(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] Ā summation from n=1 to ā
we are estimating this for the first 5 terms as follows
sinā»Ā¹(x) = x + Ā [tex]\frac{1}{2} * \frac{x^{3} }{3}[/tex] Ā + Ā [tex]\frac{1*3}{2*4} * \frac{x^{5} }{5}[/tex] Ā + Ā [tex]\frac{1*3*5}{2*4*6} * \frac{x^{7} }{7}[/tex] Ā + Ā [tex]\frac{1*3*5*7}{2*4*6*8} * \frac{x^{9} }{9}[/tex]
sinā»Ā¹(x) = x + Ā [tex]\frac{x^{3} }{6}[/tex] Ā + Ā [tex]\frac{3x^{5} }{40}[/tex] Ā +[tex]\frac{15x^{7} }{336}[/tex] Ā + Ā [tex]\frac{105x^{9} }{3456}[/tex] Ā
now to get
sinā»Ā¹([tex]\frac{3}{7}[/tex]) substitute
hence,
sinā»Ā¹([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] Ā
sinā»Ā¹([tex]\frac{3}{7}[/tex]) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482
Ā Ā Ā Ā Ā Ā = Ā 0.44290869
