I am trying to solve the following problem:
Give the Stochastic Differential Equations for $X_t = \cos B_t$ and $Y_t = \sin B_t$, where $B_t$ is the standard Brownian Motion.
I have used the following Ito formula:
$$dg(t,B_t) = \frac{\partial g}{\partial t}(t, B_t)\,dt + \frac{\partial g}{\partial x}(t, B_t)\,dB_t + \frac12\frac{\partial^2 g}{\partial x^2}(t, B_t)\,dt$$ where $g(t, B_t) = \cos B_t$ and $g(t, B_t) = \sin(t, B_t)$ for $X_t$ and $Y_t$ respectively.
This results in:
$$dX_t = -\frac12\cos B_t\,dt - \sin B_t\,dB_t\,,$$
$$dY_t = -\frac12\sin B_\,dt + \cos B_t\,dB_t\,.$$
But I wasn't sure if I was missing anything.
From \begin{align} X_t&=\cos B_t\,,&Y_t&=\sin B_t \end{align} you got by Ito's formula \begin{align} dX_t&=-\sin B_t\,dB_t-\frac12\cos B_t\,dt\,,&dY_t&=\cos B_t\,dB_t-\frac12\sin B_t\,dt \end{align} which can be written as a coupled SDE for $X$ and $Y\,,$ \begin{align} dX_t&=-Y_t\,dB_t-\frac12X_t\,dt\,,&dY_t=X_t\,dB_t-\frac12Y_t\,dt\,, \end{align} or as two uncoupled SDEs \begin{align} dX_t&=-\sin(\arccos X_t)\,dB_t-\frac12X_t\,dt\,,&dY_t=\cos(\arcsin Y_t)\,dB_t-\frac12Y_t\,dt\,. \end{align}