Produce a power series for $\cos ^{2} 2 x$ as far as the term in $x^{6}$
From double angle formulae, $\cos 2 A=2 \cos ^{2} A-1$ (see Chapter 19).
from which, $\quad \cos ^{2} A=\frac{1}{2}(1+\cos 2 A)$ and $\quad \cos ^{2} 2 x=\frac{1}{2}(1+\cos 4 x)$
From Problem 1 ,
$$
\begin{aligned}
&\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\cdots \\
&\text { hence } \quad \cos 4 x=1-\frac{(4 x)^{2}}{2 !}+\frac{(4 x)^{4}}{4 !}-\frac{(4 x)^{6}}{6 !}+\cdots \\
&=1-8 x^{2}+\frac{32}{3} x^{4}-\frac{256}{45} x^{6}+\cdots
\end{aligned}
$$
$$
=\frac{1}{2}\left(1+1-8 x^{2}+\frac{32}{3} x^{4}-\frac{256}{45} x^{6}+\cdots\right)
$$
i.e. $\cos ^{2} 2 x=1-4 x^{2}+\frac{16}{3} x^{4}-\frac{128}{45} x^{6}+\cdots$