# 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

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

#### 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!)
$= 1 - \frac{4}{2} {x}^{2} + \frac{16}{24} {x}^{4} - \frac{64}{720} {x}^{6} - \frac{256}{40320} {x}^{8} - \frac{1024}{3628800} {x}^{10}$

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