How can I understand and prove the "sum and difference formulas" in trigonometry? The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true.
\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
\end{align}
So what I want to know is, 


*

*How can I prove that these formulas are correct?

*More importantly, how can I understand these formulas intuitively?


Ideally, I'm looking for answers that make no reference to Calculus, or to Euler's formula, although such answers are still encouraged, for completeness.

The following is (essentially) a graphical version of
an earlier answer,
relying on some simple vector operations in two dimensions
but without explicitly invoking matrices.
The argument is summarized in the following diagram.

This is meant to be a proof without words, but for clarity, I'll build up to the conclusion step by step below.
First, let's just consider the unit circle in the usual Cartesian plane.
Here it is, along with the $x$ and $y$ axes and unit vectors from the origin to the two points $(1,0)$ and $(0,1).$
We use $A=(1,0)$ and $B=(0,1)$ to denote the position vectors of these two points.

Next we add a point at an angle $\alpha$ counterclockwise around the circle from the point $A=(1,0).$
According to the unit circle interpretation of the sine and cosine, the coordinates of this point, and hence the coordinates of its position vector,
are $C = (\cos\alpha, \sin\alpha).$

But as illustrated in by the vector sum figure, this is equivalent to saying that $C$ is a linear combination of the two vectors $A=(1,0)$ and $B=(0,1),$
$$ C = (\cos\alpha) A + (\sin\alpha) B. $$
The equation above is a coordinate-free way of constructing a unit position vector rotated an angle $\theta$ counterclockwise from a given unit vector: find a unit vector at a right angle counterclockwise from the given vector, and the rotated vector is $\cos\theta$ of the given vector combined with $\sin\theta$ of the right-angle vector.
Now that we have the unit vector $C$, we construct a unit vector $D$ at a right angle clockwise from $C$:

The vector sum figure, rotated $90$ degrees from the figure in the previous diagram, shows that the $x$ coordinate of $D$ is the opposite of the $y$ coordinates of $C$
while the $y$ coordinate of $D$ is simply the $x$ coordinates of $C.$
If this is not obvious enough, you can verify it for a vector $C$ in each of the four quadrants.
Now we construct a unit vector $E$ at an angle $\beta$ counterclockwise from $C.$

We construct this vector as a linear combination of the given vector $C$
and the vector $D$ at a right angle counterclockwise from $C,$
that is,
$$ E = (\cos\beta) C + (\sin\beta) D. $$
Finally, we multiply the coordinates of $C$ by the scalar $\cos\beta$,
multiply the coordinates of $D$ by the scalar $\sin\beta$,
and add the resulting vectors coordinatewise to get the coordinates of $E$:
$$ E = (\cos\beta\cos\alpha - \sin\beta\sin\alpha,
        \cos\beta\sin\alpha + \sin\beta\cos\alpha). $$
But $E$ is a vector at an angle $\alpha + \beta$ counterclockwise from $A=(1,0),$
and therefore its coordinates give two trigonometric functions of $\alpha + \beta,$
\begin{align}
\cos(\alpha + \beta) &= \cos\beta\cos\alpha - \sin\beta\sin\alpha, \\
\sin(\alpha + \beta) &= \cos\beta\sin\alpha + \sin\beta\cos\alpha.
\end{align}
Simple rearrangement within the terms gets us the usual formulas for these functions.
