Question

The Maclaurin series for sin−1(x) is given by sin−1(x) = x + [infinity] n = 1 1 · 3 · 5 (2n − 1) 2 · 4 · 6 (2n) x2n+1 2n + 1 . Use the first five terms of the Maclaurin series above to approximate sin−1 3 7 . (Round your answer to eight decimal places.)

Answer 1

Answer:

0.44290869

Step-by-step explanation:

The Maclaurin series for sin⁻¹(x) is given by

sin⁻¹(x) = x + $x^{\alpha } _{n=1}$ $\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}$

Use the first five terms of the Maclaurin series above to approximate sin⁻¹ $\frac{3}{7}$. (Round your answer to eight decimal places.)

Answer

sin⁻¹(x) = x + $x^{\alpha } _{n=1}$ $\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}$

in the above equation $x^{\alpha } _{n=1}$  summation from n=1 to ∞

we are estimating this for the first 5 terms as follows

sin⁻¹(x) = x +   $\frac{1}{2} * \frac{x^{3} }{3}$  +  $\frac{1*3}{2*4} * \frac{x^{5} }{5}$  +  $\frac{1*3*5}{2*4*6} * \frac{x^{7} }{7}$  +  $\frac{1*3*5*7}{2*4*6*8} * \frac{x^{9} }{9}$

sin⁻¹(x) = x +  $\frac{x^{3} }{6}$  +  $\frac{3x^{5} }{40}$  +$\frac{15x^{7} }{336}$  +  $\frac{105x^{9} }{3456}$  

now to get

sin⁻¹($\frac{3}{7}$) substitute

hence,

sin⁻¹($\frac{3}{7}$) = $\frac{3}{7} + \frac{\frac{3}{7} ^{3} }{6} + \frac{3 * \frac{3}{7} ^{5} }{40} + \frac{15* \frac{3}{7} ^{7} }{336} + \frac{105* \frac{3}{7} ^{9} }{3456}$  

sin⁻¹($\frac{3}{7}$) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482

           =  0.44290869