1. Maclaurin series problem

Hi I need help with this problem...

Construct the first three nonzero terms and the general term of the Maclaurin series generated by the function and give the interval of convergence:

$\displaystyle cos(x+2)$

In a page of my textbook it gives the general Maclaurin series for just cos(x) but when I ask my teacher whether it's just simply plugging in the (x+2), she says it's not. The question also gives the hint that $\displaystyle cos(x+2) = (cos2)(cosx)-(sin2)(sinx)$ but I am not sure what you can do with this hint?

2. For the function...

$\displaystyle f(x)= \cos (x+2)$

... is...

$\displaystyle f(0)= \cos 2$

$\displaystyle f^{'}(0)= -\sin 2$

$\displaystyle f^{(2} (0) = - \cos 2$

$\displaystyle f^{(3)} (0) = \sin 2$

$\displaystyle \dots$

... so that the McLaurin expansion is...

$\displaystyle \cos (x+2) = \cos 2\cdot (1-\frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots) + \sin 2\cdot (-x + \frac{x^{3}}{3!} - \frac{x^{5}}{5!} + \dots)$

... and the series converges for any real or even complex x...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. cos(2)cos(x) = cos(2)(1 - x^2/2+......

sin(2)sin(x) = sin(2)(x +........

this should help now add them

or you could generate from scratch:

f(0) = cos(2)

f ' (0) = -sin(2)

f " (0) = -cos(2)

f(x) -> cos(2) -sin(2)x -cos(2)x^2/2

For the general term see attachment--for each k there are 2 terms