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

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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

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

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