Use Ito's formula to prove that the following stochastic process is a $\{\mathcal{F_t}\}$- martingale.
a) $X_t = e^{\frac{1}{2}t}cosB_t \ \ \ \ (B_t \in \mathbb{R})$
So $d(e^{\frac{1}{2}t}sinB_t)=e^{\frac{1}{2}t}cosB_t dB_t$
now since $f(t):= e^{\frac{1}{2}t}cosB_t \in \mathcal{V}(0,T) for \ all \ T$
How does this mean that $X_t = e^{\frac{1}{2}t}sinB_t$ is a martingale?
Let $f(t,x) = e^{\frac{1}{2}t} \cos(x)$. What you want to show is that $f(t, B_t)$ is a $(\mathcal{F}_t)$-martingale. Using Ito's formula $$ df(t,B_t) = \frac{\partial f}{\partial t}(t, B_t) \, dt + \frac{\partial f}{\partial x} (t, B_t) \, dB_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(t, B_t)\,dt $$ Now $$ \frac{\partial f}{\partial t} = - \frac{1}{2}\frac{\partial^2 f}{\partial x^2} $$ so $$ df(t,B_t) = \frac{\partial f}{\partial x}(t,B_t) \, dB_t. $$ Can you deduce from here that $f(t,B_t)$ is a martingale?