Help with Maclaurins Theorum

[nil101]

I need to "Express \(\displaystyle 2\cos ^2 \left( {\frac{x}{2}} \right)\) as a power series up to five terms using Maclaurins Theorum".

I know that \(\displaystyle \L \cos ^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x\)


so I figured that \(\displaystyle \L \cos ^2 \left( {\frac{x}{2}} \right) = \frac{1}{2} + \frac{1}{2}\cos x\)


and that \(\displaystyle \L 2\cos ^2 \frac{x}{2} = 1 + \cos x\)

Is this correct?

Also, are the first five terms

f(x) = 1 + cos(x) ................. f(0)=2
f'(x) = -sin(x) ..................... f(0)=0
f''(x) =-cos(x) .................... f(0)=-1
f'''(x) =sin(x) ..................... f(0)=0
f''''(x)=cos(x) ..................... f(0)=1

?

 

[pka]

\(\displaystyle \L
\cos (x) = \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k x^{2k} }}{{\left( {2k} \right)!}}}\)

 

[nil101]

... I figured that \(\displaystyle \L \cos ^2 \left( {\frac{x}{2}} \right) = \frac{1}{2} + \frac{1}{2}\cos x\)


and that \(\displaystyle \L 2\cos ^2 \frac{x}{2} = 1 + \cos x\) ... not so sure about this

After thinking about it more and struggling with my double angles and algebra I worked out:
\(\displaystyle \L
2\cos ^2 \frac{x}{2} = \frac{1}{4} + \frac{1}{4}\cos x\)

Is this right?

I really need to know that I have used the right corollary.

The Maclaurins bit I can work out myself ( I think)

Thanks

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\cos (2A) = 2\cos ^2 (A) - 1\quad \Rightarrow \quad \cos (x) = 2\cos ^2 \left( {\frac{x}{2}} \right) - 1 \\
2\cos ^2 \left( {\frac{x}{2}} \right) = \cos (x) + 1 \\
\cos (x) = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k} }}{{\left( {2k} \right)!}}} \\
\end{array}\)

 

[nil101]

Thank you, thank you, thank you.