Prove that $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$ using complex conjugation The Definition of $\sin$ and $\cos$ is given by the complex function,
$$f:\mathbb{C}\rightarrow\mathbb{C},\qquad
f(x)=e^{\text{i}x}=\cos(x)+\text{i}\sin(x)$$
I.e. cos is the real part of $e^{ix}$ and sin denotes the immaginary part of $e^{ix}$.
Now we know that $\forall z_i\in\mathbb{C}$
$$\overline{\sum_{i\in I}z_i}=\sum_{i\in I}\bar{z_i}$$
therefore since $e^x:=\sum_{k=0}^{\infty}\frac{x^k}{k!}$
$$\overline{e^{ix}}=e^{\overline{ix}}$$
for the left side I would get
$\cos(x)-\text{i}\sin(x)$
Can someone explain why we have for the right side
$\cos(-x)+ \text{i}\sin(x)$
And also how do we derive from the Formula 
$$\cos(-x)+ \text{i}\sin(x)=\cos(x)-\text{i}\sin(x)$$ 
that we got from above that
$\sin(-x)=\sin(x)$
The equation
$\cos(-x)=\cos(x)$ 
is clear
 A: It is not true that $\sin x$ is the imaginary part of $e^{ix}$, if $x$ is complex nonreal.
If $x$ is real, then
$$
\overline{ix}=\bar{i}\bar{x}=-ix
$$
Therefore, after noting that $\overline{e^z}=e^{\bar{z}}$ as you did,
$$
\overline{\cos x+i\sin x}=\overline{e^{ix}}=e^{\overline{ix}}=e^{-ix}=\cos(-x)+i\sin(-x)
$$
Hence
$$
\cos x-i\sin x=\cos(-x)+i\sin(-x)
$$
and the equality is proved.
Note that when $x$ is complex the derivation is slightly different, mainly because
$$
\overline{\cos x+i\sin x}\ne\cos x-i\sin x
$$
in general.
On the other hand, the definition of $\sin z$ for complex $z$ is
$$
\sin z=\frac{e^{iz}-e^{-iz}}{2i}
$$
and $\sin(-z)=-\sin z$ is obvious.

For completeness, if $z=a+ib$, then $e^{iz}=e^{-b+ia}=e^{-b}(\cos a+i\sin a)$, so the real part of $e^{iz}$ is $e^{-b}\cos a$, which is not $\cos z$ in general, for the simple reason that $\cos z$ is not itself real in general and you can observe that the real part is $\cos z$ if and only if $b=0$, that is, if $z$ is real.
A: Usually that first equation is a theorem (Euler's formula) rather than a definition.  But if you start from there, notice:
$$
    e^{-ix} = e^{i(-x)} = \cos(-x) + i \sin(-x)
$$
From the power series formula $e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$, we see that
$$
    e^{-ix} = \sum_{k=0}^\infty \frac{(-ix)^k}{k!} = \sum_{k=0}^\infty \frac{(\overline{ix})^k}{k!} = \overline{e^{ix}}
$$
Therefore
$$
    e^{-ix} = \overline{e^{ix}} = \overline{\cos(x) + i \sin(x)} = \cos(x) - i \sin(x)
$$
Now you just equate real and imaginary parts.
