# 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...

• You may be interested in checking answers here, especially my favorite by @Blue: math.stackexchange.com/questions/1292/… Commented Apr 1, 2022 at 16:25
• @Vasili that is a great resource! Thank you so much ! Commented Apr 1, 2022 at 16:32

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

• Ah! so the magenta line is just a projection of y coordinate of A over B ? If so, then the projection of A over B can be obtained by adding up individual projections of x,y coordinates separately... interesting... Commented Apr 2, 2022 at 1:37

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

• so the $P_3$ remains fixed as the plane rotates(because $A-B$ doesn't change)? Never seen this before! Brilliant proof!! Thank you! Commented Apr 1, 2022 at 16:19

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 $$\cos(x-y)=\cos x\cos y+\sin x \sin y$$
• Slick! Earlier I missed seeing $DF = AD\sin(y)$ and so went in a completely different dir. Thank you so much XD Commented Apr 1, 2022 at 16:12