Question

How do you find the maclaurin series expansion of `cos (x)^2`?

Answer 1

`cos^2x = 1 -x^2+1/3x^4-2/45x^6 + ...`

Explanation:

We seek a Maclaurin series expansion of `cos^2x`

Using the known series:

`cosx = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + ...`

We can write the product as:

`cos^2x = {1 - (x^2)/(2!) + (x^4)/(4!) - ...}{1 - (x^2)/(2!) + (x^4)/(4!) - ...}`

`\ \ \ \ \ \ \ \ \ = {1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ - (x^2)/(2){1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ + (x^4)/(24){1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ - (x^6)/(720){1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...} - ...`

`\ \ \ \ \ \ \ \ \ = {1 - (x^2)/(2) + (x^4)/(24) - (x^6)/(720) +...}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ - {(x^2)/(2) - (x^4)/(4) + (x^6)/(48) - ...}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ + {(x^4)/(24) - (x^6)/(48) + ...}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ - {(x^6)/(720) - ...} - ...`

`\ \ \ \ \ \ \ \ \ = (1) + (-1/2-1/12)x^2+(1/24+1/4+1/24)x^4`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ + (-1/720-1/48-1/48-1/720)x^6 + ...`

`\ \ \ \ \ \ \ \ \ = 1 -x^2+1/3x^4-2/45x^6 + ...`