Why does Euler's formula have to be $e^{ix} = \cos(x) + i\sin(x)$ 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 $g(x) = -\cos(x)$?
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.
Let z=cos(x)+isin(x)
When you 8derive it (in terms of x), 
 z'=-sin(x)+icos(x) 
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides  and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.