= Solution
\b[The assertion is false for the algorithm as printed.] Its second output depends only on $V$. For $0<V\leq1/2$ its sine factor is nonnegative. For $1/2<V<1$, its radius is at most $\sqrt{2\log2}$, so in every case $Y\geq-\sqrt{2\log2}$. Therefore $\Pr(Y<-2)=0$, whereas a standard normal variable has positive <probability> below $-2$. The printed output cannot be a pair of independent standard normals.
The minimal correction is to use $U$ in both logarithms. Let $R=\sqrt{-2\log U}$, $\Theta=2\pi V$, and set $X=R\cos\Theta$, $Y=R\sin\Theta$. Independence of $U,V$ gives independent $R,\Theta$ with densities $re^{-r^2/2}$ for $r>0$ and $1/(2\pi)$ for the angle. The polar-coordinate Jacobian is $r$, so the joint Cartesian density is
$$
\boxed{f_{X,Y}(x,y)=\frac1{2\pi}e^{-(x^2+y^2)/2}
=\frac{e^{-x^2/2}}{\sqrt{2\pi}}\frac{e^{-y^2/2}}{\sqrt{2\pi}}.}
$$
This proves independent standard normal outputs for the corrected <Box-Muller transform>. The private transcription retains the source's original $\log V$ and records this correction separately.
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