First Four Nonzero Terms of Taylor Series I'm not acing Calculus II and while this problem might be easy to some, it's not to me. Any help will do.

Problem: Find the first nonzero terms of the Taylor Series for $f(x) = \cos x$ where
  $$a = {\pi\over 6}$$

My work:
\begin{array}{rlrll} 
f(x) =& \cos x & f(a) =& \cos {\pi\over 6} &= {\sqrt 3\over 2} \\
f'(x) =& - \sin x & f'(a) =& - \sin {\pi\over 6} &= -{1\over 2} \\
f ''(x) =&  - \cos x & f ''(a) =&  - \cos {\pi\over 6} &= -{\sqrt 3\over 2} \\
f ''' (x) =& - (-\sin x) = \sin x & f ''' (a) =& \sin {\pi\over 6} &= {1\over 2} \\
f^4(x) =& \cos x & f^4(a) =& \cos {\pi\over 6} &= {\sqrt 3\over 2} 
\end{array}
Taylor Series: 
$$a_n={f^n(a)\over n!}(x-a)^n$$
$$\eqalign{
  & \frac{{{f^0}(a)}}{{0!}}{(x - a)^0} + \frac{{f'(a)}}{{1!}}{(x - a)^1} + \frac{{f''(a)}}{{2!}}{(x - a)^2} + \frac{{f'''(a)}}{{3!}}{(x - a)^3} + \frac{{{f^4}(a)}}{{4!}}{(x - a)^4}  \cr 
  &  = \frac{1}{{0!}}\;\frac{{\sqrt 3 }}{2}(1) + \left( { - \frac{1}{2}} \right)\frac{1}{{1!}}\left( {x - \frac{\pi }{6}} \right) + \left( { - \frac{{\sqrt 3 }}{2}} \right)\frac{1}{{2!}}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{{3!}}\frac{1}{2}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{1}{{4!}}\frac{{\sqrt 3 }}{2}{\left( {x - \frac{\pi }{6}} \right)^4}  \cr 
  &  = \;\frac{{\sqrt 3 }}{2} - \frac{1}{2}\left( {x - \frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{2}\frac{1}{2}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{2}\frac{1}{{3 \cdot 2}}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{{\sqrt 3 }}{2}\frac{1}{{4 \cdot 3 \cdot 2}}{\left( {x - \frac{\pi }{6}} \right)^4}  \cr 
  &  = \;\frac{{\sqrt 3 }}{2} - \frac{1}{2}\left( {x - \frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{4}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{{12}}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{{\sqrt 3 }}{{48}}{\left( {x - \frac{\pi }{6}} \right)^4} \cr} $$
Is this problem solved correctly? If not, where did I go wrong? If you have any tips for the Taylor Series I'll take it also. Thank you.

Recall that $f(h+a)=\cos(h+(\pi/6))=\cos(h)\cos(\pi/6)-\sin(h)\sin(\pi/6)$. Hence, $$2f(h+a)=\sqrt3\cos(h)-\sin(h).
$$
Now, $\cos(h)=1-\frac12h^2+O(h^4)$ and $\sin(h)=h-\frac16h^3+O(h^4)$, hence
$$
2f(h+a)=\sqrt3(1-\tfrac12h^2)-(h-\tfrac16h^3)+O(h^4).
$$
The first four terms of the expansion of $f(h+(\pi/6))$ are the $1$, $h$, $h^2$ and $h^3$ terms, namely,
$$
f\left(h+\frac\pi6\right)=\frac{\sqrt3}2-\frac12h-\frac{\sqrt3}4h^2+\frac1{12}h^3+O(h^4).
$$
More generally, the $h^{2n}$ term of the Taylor expansion is $$\frac{(-1)^n\sqrt3}{2(2n)!}$$ and the $h^{2n+1}$ term of the Taylor expansion is $$\frac{(-1)^{n+1}}{2(2n+1)!}.$$
