Proof of cos(A-B) and geometric intuition 
In the above figure, $a,b$ are unit vectors. The angle between them is $A-B$.
It is easy to see the green highlighted part is:
$$\cos A \cos B$$
This means the magenta part must be
$$\sin A \sin B$$
Is there a way to prove this elegantly?

My work:
I tried various similar triangle ratios, then a lot of trig simplification and arrived at the answer. I don't even want to attempt it again. It's a lot of mess.
I was hoping to get a more geometric intuition for the magenta segment. Any help?
Why I'm looking for geometric intuition:
The product $\cos A \cos B$ changes when the plane rotates
The product $\sin A \sin B$ changes when the plane rotates
But their sum $\cos A \cos B + \sin A \sin B $ doesn't change!
I  fully understand the algebra - rotation matrix is orthonormal and preserves the dot product: $$(Ra)' Rb = a'R'Rb = a'Ib = a\cdot b$$
Seeking geometric intuition if possible...
 A: Divide the vertical line from $a$ into two parts at the magenta line, say $x$ and $y$, such that $x+y=\sin \alpha$.
The angles formed at the endpoints of the vertical from $a$ are corresponding angles, and are equal to $\beta$.
The vertical from $a$ also splits the magenta line into $x\sin\beta$ and $y\sin\beta$, and $(x+y)\sin\beta = \sin\alpha\sin\beta$.
A: 
We know that equal chords of a make equal angles at the centre.
The chords $P_0 P_3$ and $P_1 P_2$ subtend equal angles at the centre.
Thus, $P_0 P_3 = P_1 P_2$.
Using the distance formula,
$$
\sqrt{[\cos(A - B) - 1]^2 + [\sin(A - B) - 0]^2} =
\sqrt{[(\cos B - \cos A)^2 + (\sin B - \sin A)^2 } \tag{1}
$$
Squaring both sides of (1), we get
$$
[\cos(A - B) - 1]^2 + \sin^2(A - B) = (\cos B - \cos A)^2 + (\sin B - \sin A)^2 \tag{2}
$$
Simplifying (2)  and using the formula $\cos^2 \theta + \sin^2 \theta = 1$, we get
$$
2 - 2 \cos(A - B) = 2 - 2 \cos A \cos B - 2 \sin A \sin B \tag{3}
$$
Simplifying (4), we get the trigonometric identity
$$
\cos(A - B) = \cos A \cos B + \sin A \sin B
$$
A: You should add something to your picture.

Let $\angle AOC=x$, $\angle BOC=y$, $OA=1$. Then $OB=\cos(x-y)$.
$\angle FAD=\angle BOC=y$.
Also $OB=OE+EB=OE+DF=OD\cos y+AD\sin y=\cos x\cos y+\sin x \sin y$. Then $\cos(x-y)=\cos x\cos y+\sin x \sin y$
