4. Given that the arithmetic mean of $\sin \theta$ and $\cos \theta$ is $\sin \alpha$, and the geometric mean is $\sin \beta$. Then $\cos 2 \alpha-\frac{1}{2} \cos 2 \beta=$ $\qquad$ .
4. 0 .

From the given, we have
$$
\begin{array}{l}
\sin \alpha=\frac{\sin \theta+\cos \theta}{2}, \sin ^{2} \beta=\sin \theta \cdot \cos \theta. \\
\text { Therefore, } \cos 2 \alpha-\frac{1}{2} \cos 2 \beta \\
=1-2 \sin ^{2} \alpha-\frac{1}{2}\left(1-2 \sin ^{2} \beta\right) \\
=\frac{1}{2}-2 \sin ^{2} \alpha+\sin ^{2} \beta \\
=\frac{1}{2}-2\left(\frac{\sin \theta+\cos \theta}{2}\right)^{2}+\sin \theta \cdot \cos \theta \\
=\frac{1}{2}-\frac{1+2 \sin \theta \cdot \cos \theta}{2}+\sin \theta \cdot \cos \theta=0 .
\end{array}
$$