= Solution
Here $\Delta(t)=\sin^2B(t)$. For $f(b)=\sin^2b$, $f'(b)=\sin2b$ and $f''(b)=2\cos2b$. Since standard <Brownian motion> has <quadratic variation> $dt$, <Itô's formula> gives
$$
\boxed{d\Delta(t)=\sin(2B(t))\,dB(t)+\cos(2B(t))\,dt.}
$$
Therefore
$$
\boxed{V(t)=\sin(2B(t)),\qquad
D(t)=\cos(2B(t))=1-2\Delta(t).}
$$
The squared diffusion coefficient is $V(t)^2=4\Delta(t)(1-\Delta(t))$. Its sign depends on $B(t)$: one cannot replace $\sin(2B(t))$ by the nonnegative square root while retaining the same driving <Brownian motion>.
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