why $\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\beta - \alpha)$? I'm studying linear algebra and there is a chapter in a book that says about unit vector and it says this $$ \cos\alpha \cos\beta + \sin\alpha \sin\beta = \cos(\beta - \alpha) $$ Why?? I'm newbie and need very detail answer. Does the alpha and beta just represent angle a and angle b and has no special meaning???
you can have a look at the attached picture from the book:

 A: Let $u = (\cos \theta, \sin \theta )$, and $v = (\cos \beta, \sin \beta)$, then $u\cdot v = ||u||\cdot ||v||\cdot \cos (\beta - \theta)$, and from this your identity follows.
A: Recall that the (counter-clockwise) rotation matrix by angle $\theta$ is 
$$
\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}.
$$
Consider rotating ccw by an angle of $\beta$ and then clockwise by an angle of $\alpha$.  This corresponds to the matrix:
$$
\begin{bmatrix}\cos(\beta-\alpha)&-\sin(\beta-\alpha)\\\sin(\beta-\alpha)&\cos(\beta-\alpha)\end{bmatrix}.
$$
On the other hand, this can be computed by applying the rotations one at a time and multiplying the matrices.  In other words, this rotation is the same as
$$
\begin{bmatrix}\cos(-\alpha)&-\sin(-\alpha)\\\sin(-\alpha)&\cos(-\alpha)\end{bmatrix}
\begin{bmatrix}\cos(\beta)&-\sin(\beta)\\\sin(\beta)&\cos(\beta)\end{bmatrix}=
\begin{bmatrix}\cos(\alpha)&\sin(\alpha)\\-\sin(\alpha)&\cos(\alpha)\end{bmatrix}
\begin{bmatrix}\cos(\beta)&-\sin(\beta)\\\sin(\beta)&\cos(\beta)\end{bmatrix}.
$$
Since these are two different ways of writing the same thing, the corresponding entries are the same, and hence $\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)=\cos(\beta-\alpha)$.
A: I think that some of the above answers use the property more so than prove it or state where it comes from. For a basic geometric proof, see http://www.themathpage.com/atrig/sum-proof.htm
