Why the variance of Maximum Likelihood Estimator(MLE) will be less than Cramer-Rao Lower Bound(CRLB)? Consider this example. Suppose we have three events to happen with probability $p_1=p_2=\frac{1}{2}\sin ^2\theta ,p_3=\cos ^2\theta $ respectively. And we suppose the true value $\theta _0=\frac{\pi}{2}$. Now if we do $n$ times experiments, we will only see event 1 and event 2 happen. Hence the MLE should be
$$\mathrm{aug} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2}$$
where $m_1$ is how many times event 1 happened and $m_2$ is how many times event 2 happened.
Obviously, the solution to the above optimization problem(also it's our MLE) is $\pi/2$ which is a constant and has nothing to do with $m_1,m_2$.
But let's see what is our fisher information, whose inverse should give us a lower bound of the variance of any unbiased estimator:
$$\begin{align}
&\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\cos ^2\theta \right] ^2}{\cos ^2\theta}
\\
&=2\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\cos ^2\theta \right] ^2}{\cos ^2\theta}
\\
&=4\cos ^2\theta +4\sin ^2\theta 
\\
&=4
\end{align}$$
Hence there's a conflict, where do I understand wrong?
Thanks in advance!
Edit
If I calculate MLE not based on the observed data, but instead, based on all the possible outcomes before we do experiments, it should be:
$$\mathrm{aug}\underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2}\left[ \cos ^2\theta \right] ^{m_3}$$
taking ln we will have
$$\left( m_1+m_2 \right) \left( 2\ln\sin \theta -\ln 2 \right) +2m_3\ln\cos \theta $$
taking derivative w.r.t. $\theta$ and set derivative to be zero we will get
$$2\left( m_1+m_2 \right) \frac{\cos \theta}{\sin \theta}=2m_3\frac{\sin \theta}{\cos \theta}$$
Hence if true value $\theta_0=\pi/2$, $m_3$ will always be $0$, and we will always have a conclusion that $\hat\theta=\pi/2$ which will have no variance at all. Hence the variance of $\hat\theta$ is $0$ which against the rule of Cramer-Rao bound.

The first issue you have here is that your likelihood function does not appear to match the description of your sampling mechanism.  You say that you only observe events 1 and 2 happen, but if the sample size $n$ is known then this still fixes the number of times that event 3 happens (since $m_1 + m_2 + m_3 = n$).  Taking an observation $\mathbf{m} \sim \text{Mu}(n, \mathbf{p})$ gives the likelihood function:
$$L_\mathbf{m}(\theta)
= \bigg( \frac{1}{2} \sin^2 (\theta) \bigg)^{m_1 + m_2} \bigg( \cos^2 (\theta) \bigg)^{n - m_1 - m_2},$$
which gives the log-likelihood:
$$\ell_\mathbf{m}(\theta)
= \text{const} + (m_1 + m_2) \log (\sin^2 (\theta)) + (n - m_1 - m_2) \log (\cos^2 (\theta)).$$
As you can see, the likelihood function has an extra term that you have not included in your question.  We can see that $M_* = M_1+M_2 \sim \text{Bin}(n, \sin^2 (\theta))$ is a sufficient statistic in this problem so the problem is essentially one of binomial inference with a transformed probability parameter.  With a bit of calculus it can be shown that the MLE solves:
$$\sin^2 (\hat{\theta}) = \frac{m_1 + m_2}{n}
\quad \quad \quad \implies \quad \quad \quad 
\hat{\theta} = \text{arcsin} \bigg( \sqrt{\frac{m_1 + m_2}{n}} \bigg).$$
This estimator for the parameter is generally biased (it is unbiased in the case where $\sin^2 (\theta) = \tfrac{1}{2}$) so the applicable version of the Cramér–Rao lower bound in this case is the generalisation for biased estimators:
$$\mathbb{V}(\hat{\theta}) \geqslant \frac{|\psi'(\theta)|}{I(\theta)}
\quad \quad \quad \quad \quad
\psi(\theta) \equiv \mathbb{E}(\hat{\theta}).$$
The expectation function is:
$$\begin{align}
\psi(\theta)
&= \sum_{m=0}^n \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \cdot \text{Bin} (m|n,\sin^2 (\theta)) \\[6pt]
&= \sum_{m=0}^n {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \cdot \sin^{2m} (\theta) \cdot \cos^{2(n-m)} (\theta). \\[6pt]
&= \frac{\pi}{2} \cdot \sin^{2n} (\theta) - \frac{\pi}{2} \cdot \cos^{2n} (\theta) + \sum_{m=1}^{n-1} {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \cdot \sin^{2m} (\theta) \cdot \cos^{2(n-m)} (\theta), \\[6pt]
\end{align}$$
and its derivative (which appears in the bound) is:
$$\begin{align}
\psi'(\theta) 
&= \sum_{m=0}^n {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \frac{d}{d\theta} \bigg[ \sin^{2m} (\theta) \cdot \cos^{2(n-m)} (\theta) \bigg] \\[6pt]
&= n\pi \cdot \sin(\theta) \cos(\theta) [\sin^{2(n-1)} (\theta) +  \cos^{2(n-1)} (\theta)] \\[6pt]
&\quad + 
\sum_{m=1}^{n-1} {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \sin^{2m-1} (\theta) \cos^{2(n-m)-1} (\theta) \\[6pt]
&\quad \quad \quad \quad \quad \times \bigg[ 2m \cos^2 (\theta) - 2(n-m) \sin^2 (\theta) \bigg]. \\[6pt]
&= n\pi \cdot \sin(\theta) \cos(\theta) [\sin^{2(n-1)} (\theta) +  \cos^{2(n-1)} (\theta)] \\[6pt]
&\quad + 
2 \sum_{m=1}^{n-1} {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \sin^{2m-1} (\theta) \cos^{2(n-m)-1} (\theta) (m - n \sin^2 (\theta)). \\[6pt]
\end{align}$$
As you can see, this is a more complicated form for the Cramér–Rao lower bound.  Nevertheless, it should hold in this problem.
