I want to show $M(t)=e^{\frac{t}{2}}\sin(B(t))$ is a martingale by using Ito s formula Show that $M(t)=e^{\frac{t}{2}}\sin(B(t))$ is a martingale by using Ito s formula.
$B(t)$ is brownian-motion.
i must show $s \le t  $ then $E(M(t)|\mathcal F_{s})=M(s)$ but i dont know how use ito formula
 thanks for help
 A: Your exact question is answered on Quant Stackexchange here. Essentially, if you can express your stochastic process $M(t)$ as $M(t) = \int A(t) dB(t)$ where $B(t)$ is a martingale, then $M(t)$ is also a martingale. Since Brownian motion is a martingale, it suffices to express $M(t)$ as $\int A(t) dB(t)$ where $A(t)$ is something and $B(t)$ is Brownian motion. In the notation of stochastic calculus, you have to show
$$dM(t) = A(t) dB(t)$$
for some $A(t)$. 
Ito's formula allows you to calculate $dM(t)$ because $M(t) = f(t, B(t))$ where $f(x,y) = e^{x/2}\sin(y)$ and $B(t)$ is Brownian motion. Ito's formula gives you
$$dM(t) = \left(\frac{\partial f}{\partial x} + \frac{1}{2} \frac{\partial^2 f}{\partial y^2}\right) dt + \frac{\partial f}{\partial y}(t, B(t)) dB(t)$$
It doesn't matter what $\frac{\partial f}{\partial y}$ is. You just need to show that
$$\frac{\partial f}{\partial x} + \frac{1}{2} \frac{\partial^2 f}{\partial y^2} = 0$$
and it follows that $M(t) = (something) \times dB(t)$ is a martingale.
Edit: @Lost1 points out in the comments that this only proves that $M(t)$ is a local martingale. It also needs to satisfy some grwoth conditions in order to be a martingale. According to the Wikipedia article on "local martingale" it is sufficient that for every $\varepsilon > 0$ and every $t$ there exits a constant $C(\varepsilon, t)$ such that $|f(s,x)| \le Ce^{\varepsilon x^2}$ for all $x \in \mathbb{R}$ and all $0 \le s \le t$. In this case, $|f(s,x)| \le e^{s/2} \le e^{t/2}e^0$ seems to be sufficient. Feel free to edit if you don't think this is a good approach.
