calculus - Why does Euler's formula have to be $e^{ix} = cos(x) + isin(x)$

In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.

The Formula
$$e^{ix} = \cos(x) + i\sin(x)$$
Differentiate
$$ie^{ix} = f'(x) + i g'(x)$$
Multiply original formula by $i$
$$ie^{ix} = if(x) - g(x)$$
Equate the differentiation and the multiplied version
$$f'(x) + ig'(x) = if(x) - g(x)$$
Equate real and imaginary (and cancel the i)

$$f'(x) = -g(x) \qquad g'(x) = f(x)$$

Then he goes on to explain $f(x) = \cos(x)$ and $g(x) = \sin(x)$. My question is why can't $f(x) = \sin(x)$ and $g(x) = -\cos(x)$? Can further proof be added to this proof to eliminate $f(x) = \sin(x)$ and $g(x) = -\cos(x)$?

Answer

Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.