Question

How do you find a `6^(th)` degree taylor series `of cos(2x)` centered at `pi/6`?

6th degree taylor series of cos(2x) centered at `pi`/6

Answer 1

`cos2x=1-2x^2+0.667x^4-0.0889x^6+0.006349x^8-0.000282x^10`

Explanation:

`cosx=1-x^2/(2!)-x^4/(4!)-x^6/(6!)-x^8/(8!)-x^10/(10!)`

Replacing x by 2x, we have

`cos(2x)=1-(2x)^2/(2!)+(2x)^4/(4!)-(2x)^6/(6!)+(2x)^8/(8!)-(2x)^10/(10!)`
`=1-4/2x^2+16/24x^4-64/720x^6-256/40320x^8-1024/3628800x^10`

`=1-2x^2+0.667x^4-0.0889x^6+0.006349x^8-0.000282x^10`