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

Question

How to prove the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B $ by using cross product of two vectors?

Answer 1

Here's an outline. Let $\vec v = (\cos A, \sin A, 0)$ and $\vec w = (\sin B, \cos B, 0)$. These are two points on the unit circle in the $xy$-plane, and the angle between them is $\frac{\pi}{2} - (A+B)$. Prove this fact, and then use formula for the cross product of two vectors in terms of the angle between them and the fact that $\sin (\frac{\pi}{2} - \theta) = \cos \theta$.

Answer 2

Alternatively, it can be proved using:

$$\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$$ $$\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$

Answer 3

We begin in Figure 1 with the unit circle and the angles $a,b$, and $a-b$. Then we rotate $\triangle\rm OPQ$ about the origin until the point $\rm P$ coincides with the point $({1,0})$, as indicated in Figure 2. This rotation has no effect on the length of $\rm PQ$, thus whether we calculate it using the coordinates in Fig.1 or in Fig.2 will result in the same length.

Applying the distance formula gives: $$\eqalign{{\rm PQ}&= \sqrt{(\cos a-\cos b)^2+(\sin a-\sin b)^2}\\ &=\sqrt{\color{white}{\overline{\color{black}{ \cos^2a-2\cos a \cos b+\cos^2b+\sin^2a-2\sin a\sin b+\sin^2b }}}}\\ &=\sqrt{\color{white}{\overset{}{\overline{\overset{\color{white}{\overset..}}{}}}}} \underset{\displaystyle =2}{\overline{\underbrace{\overset{}{\color{black}{\cos^2a+\sin^2a+\cos^2b+\sin^2b}}}}}\overline{\overset{\color{white}{\overset{}.}}{\color{black}{\,-2\cos a \cos b-2\sin a\sin b}}}\\ &=\sqrt{\color{white}{\overline{\color{black}{ 2-2\cos a \cos b-2\sin a\sin b }}}}. }$$

Now using the coordinates in Fig.2 we get: $$\eqalign{{\rm PQ}&=\sqrt{(\cos(a-b)-1)^2+(\sin(a-b)-0)^2}\\ &=\sqrt{\cos^2(a-b)-2\cos(a-b)+1+\sin^2(a-b)}\\ &=\sqrt{2-2\cos(a-b)}. }$$

Equating the two yields: $$\eqalign{ \sqrt{2-2\cos(a-b)}&=\sqrt{\color{white}{\overline{\color{black}{ 2-2\cos a \cos b-2\sin a\sin b }}}}\\ 2-2\cos(a-b)&=2-2\cos a \cos b-2\sin a\sin b\\ \cos(a-b)&=\cos a\cos b-\sin a\sin b. }$$

Using the substitutions $b\leadsto-b$, and using the identities $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$, we get: $$\eqalign{\cos\big(a-(-b)\big)&=\cos a\cos(- b)-\sin a\sin(- b)\\ &=\cos a\cos b+\sin a\sin b.}$$

This proof is due to Cauchy I think.