I1.1 If $a$ is the maximum value of $\frac{1}{2} \sin ^{2} 3 \theta-\frac{1}{2} \cos 2 \theta$, find the value of $a$.
$$
\begin{array}{l}
-1 \leq \sin 3 \theta \leq 1 \text { and }-1 \leq \cos 2 \theta \leq 1 \\
\frac{1}{2} \sin ^{2} 3 \theta \leq \frac{1}{2} \text { and }-\frac{1}{2} \cos 2 \theta \leq \frac{1}{2} \\
\frac{1}{2} \sin ^{2} 3 \theta-\frac{1}{2} \cos 2 \theta \leq \frac{1}{2}+\frac{1}{2}=1=a,
\end{array}
$$

Maximum occur when $\sin ^{2} 3 \theta=1$ and $-\cos 2 \theta=1$
i.e. $30=90^{\circ}+180^{\circ} n$ and $2 \theta=360^{\circ} m+180^{\circ}$, where $m, n$ are integers.
$$
\theta=30^{\circ}+60^{\circ} n=180^{\circ} \mathrm{m}+90^{\circ} \Rightarrow 60^{\circ} n=180^{\circ} \mathrm{m}+60^{\circ} \Rightarrow n=3 \mathrm{~m}+1 \text {; let } \mathrm{m}=1, n=4, \theta=270^{\circ}
$$

Remark: the original question is
If $a$ is the maximum value of $\frac{1}{2} \sin ^{2} \theta+\frac{1}{2} \cos 3 \theta$, find the value of $a$.
Maximum occur when $\sin ^{2} \theta=1$ and $\cos 3 \theta=1$
i.e. $\theta=90^{\circ}+180^{\circ} n$ and $3 \theta=360^{\circ} \mathrm{m}$, where $m, n$ are integers.
$\theta=90^{\circ}+180^{\circ} n=120^{\circ} \mathrm{m} \Rightarrow 3+6 n=4 \mathrm{~m}$, LHS is odd and RHS is even, contradiction.
The question was wrong because we cannot find any $\theta$ to make the expression a maximum.