Question

How to find a maclaurin series for `(cosx)^2`?

Answer 1

See below

Explanation:

MacLaurin series are given by

`sum_(n=0)^oof^(n)(0)/(n!)x^n`

Let `f(x)=cos^2x`, then `f(0)=1`

`f´(x)=-2cosxsinx=-sin2x` then `f´(0)=0`
`f´´(x)=-2cos2x` then `f´´(0)=-1`
`f´´´(x)=4sen2x` then `f´´´(0)=0`
`f^(IV)(x)=8cos2x` then `f^(iv)(0)=8`
`f^(v)(x)=-16sen2x` then `f^(v)(0)=0`
`f^(vi)(x)=-32cos2x` then `f^(vi)(0)=-32`

`1-x^2/(2!)+8/(4!)x^4-32/(6!)x^6+....`