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

0.44290869

Explanation:

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

sin⁻¹(x) = x +
`x^(\alpha ) _(n=1)`
`(1.3.5...(2n - 1))/(2.4.6...(2n)) * (x^(2n + 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

sin⁻¹(x) = x +
`x^(\alpha ) _(n=1)`
`(1.3.5...(2n - 1))/(2.4.6...(2n)) * (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 +
`(1)/(2) * (x^(3) )/(3)` +
`(1*3)/(2*4) * (x^(5) )/(5)` +
`(1*3*5)/(2*4*6) * (x^(7) )/(7)` +
`(1*3*5*7)/(2*4*6*8) * (x^(9) )/(9)`

sin⁻¹(x) = x +
`(x^(3) )/(6)` +
`(3x^(5) )/(40)` +
`(15x^(7) )/(336)` +
`(105x^(9) )/(3456)`

now to get

sin⁻¹(
`(3)/(7)`) substitute

hence,

sin⁻¹(
`(3)/(7)`) =
`(3)/(7) + ((3)/(7) ^(3) )/(6) + (3 * (3)/(7) ^(5) )/(40) + (15* (3)/(7) ^(7) )/(336) + (105* (3)/(7) ^(9) )/(3456)`

sin⁻¹(
`(3)/(7)`) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482

= 0.44290869