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

Question

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

2495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-02T15:01:25

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

#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#