90 Given $\sin x+\sin y=\log _{2} 2, \cos x+\cos y=\log _{7} 4$, then the value of $\cos 2(x+y)$ is

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$90-\frac{7}{25}$. From
$$
\left\{\begin{array}{l}
\sin x+\sin y=\log _{7} 2 \\
\cos x+\cos y=\log _{7} 4
\end{array},\right.
$$

transform to
$$
\left\{\begin{array}{l}
2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}=\log _{7} 2 \\
2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\log _{7} 4,
\end{array}\right.
$$

by dividing the two equations, we get $\tan \frac{x+y}{2}=\frac{1}{2}$, thus
$$
\cos (x+y)=\frac{1-\tan ^{2} \frac{x+y}{2}}{1+\tan ^{2} \frac{x+y}{2}}=\frac{3}{5},
$$

therefore
$$
\cos (2 x+2 y)=2 \cos ^{2}(x+y)-1=-\frac{7}{25} .
$$