Real sinusoids with the same frequency are closed under addition. If
$$f(\omega) = A_1 \cos(\omega + \phi_1) + A_2 \cos(\omega + \phi_2)$$
Then there is some $A_3$ and $\phi_3$ so that: $$f(\omega) = A_3 \cos(\omega + \phi_3)$$
for real $A$ and $\phi$. The closed sinusoid can be found using polar notation $A_3 \angle \phi_3 = A_2 \angle \phi_2 + A_1 \angle \phi_1$.
Euler's definition is $e^{ix} = \cos(x) + i \sin(x) = \cos(x) + i\cos\left(x - \frac{\pi}{2}\right)$. I'm wondering if there is a way to use closure of sinusoidal addition to find complex numbers $Z_1$ and $Z_2$ such that $e^{ix} = Z_1 \cos(x + Z_2)$. Alternatively, is there a way to show that no such $Z_1$ and $Z_2$ exist?
My initial approach was to try to directly add $1 \angle 0$ and $i \angle -\frac{\pi}{2}$ using a right triangle of side lengths $1$ and $i$. The resulting magnitude $Z_1 = \sqrt{1^2 + i^2} = 0$ leads me to think that there may be no closed form for the sinusoid but I'm not sure.
This is just for my own random thoughts, but any insight (especially rigorous) would be greatly appreciated.
Assume a non-zero pair of such numbers exists. Then $$e^{ix}=Z_1 \cos(x + Z_2)=Z_1(\cos Z_2 \cos x-\sin Z_2 \sin x)$$ so $$1=Z_1\cos Z_2$$ $$i=-Z_1\sin Z_2$$ and $$Z_1^2=1^2+i^2=0$$ a contradiction. Therefore no such pair $(Z_1,Z_2)$ exists.
EDITS: Also note that this is in a sense the only function of the form $$a\cos x+b\sin x$$ for which no equivalent form $c\,\cos(x+d)$ exists. The problem results when $a^2+b^2=0$ - interesting.
In other words, the family of functions which don't admit such an expression are of the form $$a\cos x\pm ai\sin x=ae^{\pm ix}$$ Hmm... can anyone see why this is the case? Now you've got me interested ;)