Express $\cos 6\theta $ in terms of $\cos \theta$ I think I'm supposed to use the chebyshev polynomials, as in $$ \cos n \theta = T_n(x) = \cos(n \arccos x)$$
But no idea what now?

Since $$\cos 2a =2\cos ^{2}a -1, \qquad\sin 2a
=2\sin a\cos a,$$ $$\cos (a+b)=\cos a\cos b-\sin a\sin b,$$
 and
$$
\begin{eqnarray*}
\cos 3\theta  &=&\cos (2\theta +\theta ) \\
&=&\cos 2\theta \cos \theta -\sin 2\theta \sin \theta   \\
&=&( 2\cos ^{2}\theta -1) \cos \theta -2\sin ^{2}\theta \cos
\theta  \\
&=&( 2\cos ^{2}\theta -1) \cos \theta -2( 1-\cos ^{2}\theta
) \cos \theta  \\
&=&4\cos ^{3}\theta -3\cos \theta ,
\end{eqnarray*}
$$
we have
$$
\begin{eqnarray*}
\cos 6\theta  &=&\cos \left( 2\times 3\theta \right)  \\
&=&2\cos ^{2}3\theta -1 \\
&=&2( 4\cos ^{3}\theta -3\cos \theta ) ^{2}-1 \\
&=&32\cos ^{6}\theta -48\cos ^{4}\theta +18\cos ^{2}\theta -1.
\end{eqnarray*}
$$

ADDED: From the definition of the Chebyshev polynomials
$$
\begin{equation*}
T_{n}(x)=\cos (n\arccos x)\Leftrightarrow T_{n}(\cos \theta )=\cos n\theta
,\quad \theta =\arccos x,
\end{equation*}
$$
we get 
$$
\begin{eqnarray*}
T_{1}(x) &=&\cos (\arccos x)=x \\
T_{2}(x) &=&\cos (2\arccos x)=2x^{2}-1.
\end{eqnarray*}
$$
Since they satisfy the recurrence
$$
\begin{equation*}
T_{n+1}(x)=2xT_{n}(x)-T_{n-1}(x),
\end{equation*}
$$
we have
$$
\begin{eqnarray*}
T_{3}(x) &=&2xT_{2}(x)-T_{1}(x) \\
&=&2x( 2x^{2}-1) -x \\
&=&4x^{3}-3x \\
&& \\
T_{4}(x) &=&2xT_{3}(x)-T_{2}(x) \\
&=&2x( 4x^{3}-3x) -( 2x^{2}-1)  \\
&=&8x^{4}-8x^{2}+1 \\
&& \\
T_{5}(x) &=&2xT_{4}(x)-T_{3}(x) \\
&=&2x( 8x^{4}-8x^{2}+1) -( 4x^{3}-3x)  \\
&=&16x^{5}-20x^{3}+5x \\
&& \\
T_{6}(x) &=&2xT_{5}(x)-T_{4}(x) \\
&=&2x( 16x^{5}-20x^{3}+5x) -( 8x^{4}-8x^{2}+1)  \\
&=&32x^{6}-48x^{4}+18x^{2}-1.
\end{eqnarray*}
$$
Therefore
$$
\begin{eqnarray*}
\cos 6\theta  &=&T_{6}(\cos \theta ) \\
&=&32\left( \cos \theta \right) ^{6}-48\left( \cos \theta \right)
^{4}+18\left( \cos \theta \right) ^{2}-1 \\
&=&32\cos ^{6}\theta -48\cos ^{4}\theta +18\cos ^{2}\theta -1,
\end{eqnarray*}
$$
as above.
