Find the Wronskian of $cos^2\theta$ and $1 + cos2\theta$
Let $f = cos^2\theta$ and $g = 1 + cos2\theta$, then
\begin{equation}
f' = -2sin\theta cos\theta = -sin2\theta \quad\quad\quad g' = -2sin2\theta
\end{equation}
since $W = fg' - f'g$, we get
\begin{align*}
W &= cos^2\theta(-2sin2\theta) - (-sin2\theta)(1 + cos2\theta)\\
&= -2cos^2\theta sin2\theta + sin2\theta(1 + cos2\theta)\\
&= sin2\theta(-2cos^2\theta + 1 + cos2\theta)\quad\quad\quad\quad by \quad(2cos^2\theta - 1 = cos2\theta)\\
&= sin2\theta(-cos2\theta + cos2\theta)\\
&= 0
\end{align*}
Therefore the Wronskian of $cos^2\theta$ and $1 + cos2\theta$ is $0$
