4. If $\cos ^{4} \theta+\sin ^{4} \theta+(\cos \theta \cdot \sin \theta)^{4}+$ $\frac{1}{\cos ^{4} \theta+\sin ^{4} \theta}=\frac{41}{16}$, then $\sin ^{2} \theta=$
4. $\frac{1}{2}$.

Let $x=\cos ^{2} \theta \cdot \sin ^{2} \theta$.
From the given, we have
$$
\begin{array}{l}
1-2 x+x^{2}+\frac{1}{1-2 x}=\frac{41}{16} \\
\Rightarrow\left((1-x)^{2}-\frac{9}{16}\right)+\left(\frac{1}{1-2 x}-2\right)=0 \\
\Rightarrow\left(\frac{1}{4}-x\right)\left(\frac{7}{4}-x\right)+\frac{4 x-1}{1-2 x}=0 \\
\Rightarrow\left(x-\frac{1}{4}\right)\left(\frac{4}{1-2 x}+x-\frac{7}{4}\right)=0 .
\end{array}
$$

Notice that, $0 \leqslant x \leqslant\left(\frac{\cos ^{2} \theta+\sin ^{2} \theta}{2}\right)^{2}=\frac{1}{4}$.
Then $\frac{4}{1-2 x}+x-\frac{7}{4} \geqslant 4+0-\frac{7}{4}>0$.
Thus, $x-\frac{1}{4}=0 \Rightarrow \cos ^{2} \theta=\sin ^{2} \theta=\frac{1}{2}$.