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.

To make the equations easier I am gonna use only two states instead of three. $p_1 = f(\theta)$ and $p_2 = 1-f(\theta)$
You apply the formula  $$\mathcal{I}(\theta) = \sum_x{{ \overbrace{\left[ \frac{\frac{\partial}{\partial \theta} p\left( x;\theta \right) }{p\left( x;\theta \right) }\right] ^2}^{{logL^\prime}^2}}{p\left( x;\theta \right)}} \qquad \tag{1} $$
This derivative of the log likelihood is:
$$logL^\prime = \frac{ m_1 f^\prime(\theta) - m_2 f^\prime(\theta)}{m_1 f(\theta) + m_2 (1-f(\theta))}$$
The probability is
$$p(x;\theta) = f(\theta)^{m_1} (1-f(\theta))^{m_2}$$
And if we restrict to a single observation $m_1 = 1, m_2 = 0$ or $m_1 = 0, m_2 = 1$ then the sum in formula (1) becomes
$$\mathcal{I}(\theta) = \frac{f^\prime (\theta)^2}{f(\theta)} + \frac{f^\prime (\theta)^2}{1-f(\theta)} = \frac{f^\prime (\theta)^2}{f(\theta) (1-f(\theta))}  $$
then
$$\mathcal{I}(\theta)^{-1} = \frac{f(\theta) (1-f(\theta))}{f^\prime (\theta)^2}$$
and with $f(\theta) = \sin^2\theta$ this becomes the same as your result
$$\mathcal{I}(\theta)^{-1} =  \frac{ \sin^2\theta \cos^2\theta}{(2 \cos\theta \sin\theta )^2} = \frac{1}{4}$$
But with $f(\theta) = \theta$ then you get
$$\mathcal{I}(\theta)^{-1} =  \theta(1-\theta)$$
like in this question https://math.stackexchange.com/questions/396982/fisher-information-of-a-binomial-distribution
Maybe there is some regularity condition that is not satisfied? Doesn't the Cramer Rao bound only apply to continuous distributions and unbiased variables?
