trigonometry - Generalization of Euler's Formula

Euler's formula states that, for any real number x:

$$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$

Can it be generalized in that way?

$$ae^{ix}+be^{-ix}=c\cos(x+d)$$

where $a,b\in \mathbb{C}$ and $c,d\in \mathbb{R}$.
Of course if $a=b=1$ and $c=2$, $d=0$ this is the common Euler's fomula, but it is true that for every $a,b$ I can rewrite a sum of complex exponentials as a single cosine? If it is, what is the relationship between these constants?

Answer

Suppose your formula is true for any $x \in \mathbb R$. You can write it as

$$ae^{ix}+be^{-ix}- \frac{c}{2}(e^{id} e^{ix}+ e^{-id}e^{-ix}) = 0,$$

that is

$$\left(a - \frac{c}{2}e^{id}\right) e^{ix}+ \left( b - \frac{c}{2}e^{-id}\right) e^{-ix} =0.$$

Now we can use the fact that $e^{ix}$ and $e^{-ix}$ are linearly independent to get

$$a = \frac{c}{2}e^{id}, \quad b = \frac{c}{2}e^{-id}.$$

This implies

$$c = 2 (ab)^{1/2}, \quad \cos d = \frac{a+b}{2 (ab)^{1/2}}.$$

Since we require $c$ and $d$ to be real, then $ab$ has to be a positive real number and $a+b$ has to be real. This is possible only if $a$ and $b$ are complex conjugates: $b = \bar a$.