Does Euler's formula give $e^{-ix}=\cos(x) -i\sin(x)$?

$$e^{ix}=\cos(x)+i\sin(x)\tag{1}$$ $$e^{-ix}=\cos(x)-i\sin(x)\tag{2}$$

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.

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 $$e^{ix}-e^{-ix}=(i\sin(x)+\cos(x))-(-i\sin(x)+\cos(x))$$ $$e^{ix}-e^{-ix}=i\sin(x)+\cos(x)+i\sin(x)-\cos(x)$$ $$e^{ix}-e^{-ix}=2i\sin(x)$$ $$\frac{e^{ix}-e^{-ix}}{2i}=\sin(x)$$

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,

$$\boxed{ \ e^{-ix} = \frac{1}{\cos(x) + i\sin(x)}. \ }$$

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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.