What is $\overline{\sin({z})} $ equal to? What is
$\overline{\sin(z)}$
equal to?
 A: I will write $\exp(x)$ instead of $e^{x}$, they are synonyms. (Just a notational warning!)
Well, recall Euler's formula
$$ \exp(i\theta)=\cos(\theta)+i\cdot\sin(\theta).$$
Then we see
$$\exp(i\theta)-\exp(-i\theta)=2i\cdot\sin(\theta)$$ 
allows us to write
$$\frac{\exp(i\theta)-\exp(-i\theta)}{2i}=\sin(\theta).$$
So replacing $\theta$ with $z=x+iy$ in this case would produce $\exp(iz)-\exp(-iz)=2i\cdot\sin(z)$, where $z=x+iy$.
Addendum: Now consider the following:
$$\overline{\sin(z)} = \overline{\frac{\exp(iz)-\exp(-iz)}{2i}}$$
But look, this is just
$$\overline{\sin(z)}=\overline{\sin(x+iy)}$$
and the only place the imaginary part plays any role is the $iy$, we have
$$\overline{\sin(z)}=\sin(x-iy)$$
and this is precisely $\sin(\bar{z})$.
Looking on the right hand side, how can we say this? Well, we just change all the signs for $i$ and replace $z$ with $\bar{z}$, writing
$$\overline{\frac{\exp(iz)-\exp(-iz)}{2i}}=\frac{\exp(-i\bar{z})-\exp(i\bar{z})}{-2i}$$
But look, we may multiply the top and bottom by $-1$ producing
$$\overline{\frac{\exp(iz)-\exp(-iz)}{2i}}=\frac{\exp(-i\bar{z})-\exp(i\bar{z})}{-2i}=\frac{-\exp(-i\bar{z})+\exp(i\bar{z})}{2i}$$
which is precisely $\sin(\bar{z})$.
