4.70 $1-\sin ^{2} \alpha-\sin ^{2} \beta+2 \sin \alpha \sin \beta \cos (\alpha-\beta)$.
4.70 We have

\[
\begin{aligned}
& \sin ^{2} \alpha+\sin ^{2} \beta-(\cos (\alpha-\beta)-\cos (\alpha+\beta)) \cos (\alpha-\beta)= \\
& =\frac{1}{2}-\frac{1}{2} \cos 2 \alpha+\frac{1}{2}-\frac{1}{2} \cos 2 \beta-\cos ^{2}(\alpha-\beta)+\frac{1}{2} \cos 2 \alpha+\frac{1}{2} \cos 2 \beta= \\
& =1-\cos ^{2}(\alpha-\beta)=\sin ^{2}(\alpha-\beta)
\end{aligned}
\]

Now we find \(1-\sin ^{2}(\alpha-\beta)=\cos ^{2}(\alpha-\beta)\).