Proof of cos(A-B) and geometric intuition

Question

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?

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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...

Answer 1

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 $(x+y)\sin\beta = \sin\alpha\sin\beta$.

Answer 2

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 $$ \cos(A - B) = \cos A \cos B + \sin A \sin B $$

Answer 3

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 $\cos(x-y)=\cos x\cos y+\sin x \sin y$