Trigonometry Identity: Prove that $\sin(a-b)=\sin a \cos b - \cos a \sin b$ First, I do not want a proof using $\sin(a+b)=\sin a \cos b + \cos a \sin b$. Second, I suspect that it has something to do with Euler's formula; $e^{ix}=\cos x + i\sin x$, but I am not sure. Can anyone give me some direction? Thanks in advance.
P.S. I apologize if this question is a duplicate. I did not find any when scrolling down the "Questions that may already have your answer" list.

I searched for an alternate way to deriving sum of angles relations, and by  congruence came up with this alternate derivation.  Perhaps it is acceptable.   
Take a coordinate pair say $(x,y)$ with angle $A$.  We have the definitions $\sin(A)= y/(x^2+y^2)^.5 $, and $\cos(A)= x/(x^2+y^2)^.5$. 
Now rotate, counterclockwise, the coordinate pair $(x,y)$ by an angle $B$ such that
$\angle A + \angle B = \angle C$
After rotation $(x,y)$ becomes $(x\cos B-y\sin B, x\sin B+y\cos B)$. (I'm leaving out a derivation for rotation since it is everywhere, See Rotate the graph of a function?). Since the angle of the rotated pair $(x,y)$ is $\angle C$ the following equations follow:
$$\cos(C)=\frac{x\cos B-y\sin B}{((x\cos B-y\sin B)^2+(x\sin B+y\cos B)^2)^.5}$$ 
Notice that:
$$(x\cos B-y\sin B)^2+(x\sin B+y\cos B)^2)= (x^2+y^2)$$
Hence,
$$\cos(C)=\frac{x\cos B-y\sin B}{(x^2+y^2)^.5}$$ 
And 
$$\cos(A+B)= {\cos(A)\cos(B)}-{\sin(A)\sin(B)}$$
Other Relations--------------
$$\sin(C)=\frac{x\sin B+y\cos B}{(x^2+y^2)^.5} = {\cos(A)\sin(B)}+{\sin(A)\cos(B)}$$
$$\sin(A+B)= {\cos(A)\sin(B)}+{\sin(A)\cos(B)}$$
We also have 
$$A\tan(y/x)= \angle A$$
 and 
$$A\tan\frac{x\sin B+y\cos B}{x\cos B-y\sin B}= \angle C$$
$$ \angle B +A\tan(y/x)= A\tan\frac{x\sin B+y\cos B}{x\cos B-y\sin B}$$
$$ \angle B +A\tan(y/x)= A\tan\frac{\tan B+y/x}{1-(y/x)\tan B}$$
Or, for example, by setting $Tan B = u$ and $ y/x = v$ 
$$ A\tan(u) +A\tan(v)= A\tan\frac{u+v}{1-uv}$$
For the tangent is straight forward.
$$tan(A + B) = \frac{x\sin B+y\cos B}{x\cos B-y\sin B}$$
$$tan(A + B) = (tan B + tan A) / (1 − tan A *tan B)$$
