Let $B=(B^1,B^2)$ be a two-dimensional Brownian motion w.r.t. the Filtration $\mathcal{F}^B$. Show that $(M_t^2)_{t\in \mathbb{R}_{+}}:=(e^{B_t^1} \cos(B_t^2))_{t\in\mathbb{R}_{+}}$
I've tried it with Ito's formula but i failed to find an argument that it is a martingale.
I did Ito formula for $f(x,y)=e^x\cos(y)$, with $f_x(x,y)=e^x\cos(y)$, $f_y(x,y)=e^x\sin(y)$ and $f_{yy}(x,y)=-e^x\cos(y)=-f_x(x,y)$
$d(f(B_t^1,B_t^2))=f_y(B_t^1,B_t^2)dB_t^2+(f_x(B_t^1,B_t^2)+\frac{1}{2}f_{yy}(B_t^1,B_t^2))dB_t^1\\ = e^{B_t^1}\sin(B_{t}^2)dB_t^2-\frac{1}{2}e^{B_t^1}\cos(B_t^2)dB_t^1$
I know that $\cos$ and $\sin$ is bounded, but i don't know anything about $e^{B_t^1}$. So i don't know wich argument i need, the i can say that this function is an martingale.
Can anybody help me?