 Solution

Here $\Delta (t)={\sin }^{2}B(t)$. For $f(b)={\sin }^{2}b$, $f^{\prime }(b)=\sin 2b$ and $f^{\prime \prime }(b)=2\cos 2b$. Since standard Brownian motion has quadratic variation $dt$, Itô's formula givesThereforeThe 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.