Question: I'm asked to show if there exists $z$ in $\mathbb{C}$ such that, the two following conditions are simultaneously satisfied

$$|\sin(z)|>1, |\cos(z)|>1$$

For $|\sin(z)|^2$ I find $\displaystyle\frac{1-\cos(2z)}{2}$ wich is not a real number in general because $\cos(z)$ is not a real number. I don't know where is the mistake.

Answer: 
It is true that

$$\sin(z)^2 = \frac{1-\cos(2z)}{2}$$

but then you have to take the absolute value also on the right side. So:

$$|\sin(z)^2| = \left|\frac{1-\cos(2z)}{2}\right|$$

As for your exercise, write 

$$\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}, \cos(z) = \frac{e^{iz}+e^{-iz}}{2}$$

and try finding a value for $z$ such that the absolute value of both expressions is $>1$. Putting in purely imaginary numbers helps.
Wouldn't it be enough just to pick an $x$ such that $e^{ix}$ is pretty big? Then since you have $\sin x\approx {\frac1{2i}e^{ix}}$ and $\cos x\approx \frac12 e^{ix}$ you would have what you wanted.

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So for example, taking $x=-1000000000i$ ought to suffice.