1. If $\theta$ is an acute angle, and
$$
\cos \theta \cdot \cos \left(\theta-\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{10},
$$

then the value of $\sin \theta$ is $\qquad$ .
$=1 . \frac{2 \sqrt{5}}{5}$.
From the problem, we have
$$
\cos \theta\left(\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta\right)=\frac{3 \sqrt{2}}{10},
$$

which simplifies to $\cos ^{2} \theta+\cos \theta \cdot \sin \theta=\frac{3}{5}$.
Thus, $5 \cos ^{2} \theta+5 \cos \theta \cdot \sin \theta=3 \cos ^{2} \theta+3 \sin ^{2} \theta$, which means $2 \cos ^{2} \theta+5 \cos \theta \cdot \sin \theta-3 \sin ^{2} \theta=0$.
Therefore, $(2 \cos \theta-\sin \theta)(\cos \theta+3 \sin \theta)=0$.
Since $\theta$ is an acute angle, we know $\cos \theta+3 \sin \theta>0$, hence
$$
\sin \theta=2 \cos \theta \text {. }
$$

Thus, $\sin \theta=\frac{2}{\sqrt{2^{2}+1}}=\frac{2 \sqrt{5}}{5}$.