Does Euler's formula give $e^{-ix}=\cos(x) -i\sin(x)$? Does Eulers formula give $$e^{-ix}=\cos(x) -i\sin(x)$$
I know that $$e^{ix}=\cos(x)+i\sin(x)$$ But how does it work when we have a $-$ in front
 A: Remember that $$\begin{align}-i &=-\sqrt{-1} \\ &=(-1)\sqrt{-1} \\ &=i^2\cdot i \\ &= i^3\end{align}$$ Thus, if
$e^{ix} = \cos(x) + i\sin(x),$ then $e^{-ix} = (e^{ix})^{-1}.$
Therefore,
$$\boxed{ \ e^{-ix} = \frac{1}{\cos(x) + i\sin(x)}. \ }$$

Edit:
As @Botond suggested, by using a conjugate method (used to rationalise denominators) as mentioned in his comment below, we get  the much nicer result, $$e^{-ix} = \frac{\cos(x) - i\sin(x)}{\cos(x)^2 + \sin(x)^2}.$$ Since $\cos(x)^2 + \sin(x)^2 = 1$ then yes,  correct indeed.
A: From the property that for any $z_1,z_2\in\mathbb{C}$, $$e^{z_1}\cdot e^{z_2}=e^{z_1+z_2}$$ we have that $$e^{ix}\cdot e^{-ix}=e^{ix-ix}=e^0=1$$ so $$e^{-ix}=\frac1{e^{ix}}=\frac{\overline{e^{ix}}}{e^{ix}\cdot\overline{e^{ix}}}=\frac{\overline{e^{ix}}}{|e^{ix}|^2}=\overline{e^{ix}}$$ using that $\cos^2x+\sin^2x=1$.
A: $$e^{ix}=\cos(x)+i\sin(x)\tag{1}$$
$$e^{-ix}=\cos(x)-i\sin(x)\tag{2}$$
To get from $(1)$ to $(2)$, you just replace $x$ with $-x$. You get the $-$ in front because $\sin(-x)=-\sin(x)$, but the identity is still the same.
A: The $\sin$ function is odd, so $\sin(-x)=-\sin(x)$.
The $\cos$ function is even, so $\cos(-x)=\cos(x)$.
Using these and Euler's formula, we can get that
$$e^{-ix}=e^{i(-x)}=i\sin(-x)+\cos(-x)=-i\sin(x)+\cos(x)$$
If you are not comfortable with it:
Let $u=-x$, then $$e^{-ix}=e^{iu}=i\sin(u)+\cos(u)=i\sin(-x)+\cos(-x)=-i\sin(x)+\cos(x)$$
And you can express the $\sin$ and $\cos$ in the following way:
$$e^{ix}+e^{-ix}=(i\sin(x)+\cos(x))+(-i\sin(x)+\cos(x))$$
$$e^{ix}+e^{-ix}=i\sin(x)+\cos(x)+-i\sin(x)+\cos(x)$$
$$e^{ix}+e^{-ix}=2\cos(x)$$
$$\frac{e^{ix}+e^{-ix}}{2}=\cos(x)$$
And
$$e^{ix}-e^{-ix}=(i\sin(x)+\cos(x))-(-i\sin(x)+\cos(x))$$
$$e^{ix}-e^{-ix}=i\sin(x)+\cos(x)+i\sin(x)-\cos(x)$$
$$e^{ix}-e^{-ix}=2i\sin(x)$$
$$\frac{e^{ix}-e^{-ix}}{2i}=\sin(x)$$
