cos(alpha + beta)

I remember quite a bit of trig including some of the derivates and the Taylor expansions, but not the formulas for the sums of angles or their derivations.  Simple ones exist based upon rotations of the complex plane, but I wanted a straight geometric one.  I found one described by M. Bourne based on the unit circle abbreviated as follows:

We draw a circle with radius 1 unit, with point P on the circumference at (1,0).
We draw an angle alpha from the centre [sic] with terminal point Q  at (cos alpha, sin alpha).
We extend this idea by drawing:
   The angle beta with terminal points at Q (cos alpha, sin alpha) and R (cos(alpha + beta), sin(alpha + beta))
   The angle -beta with terminal point at S (cos -beta, sin -beta)
   The lines PR and QS which are equivalent in length.

Now we have
   PR squared = cos(alpha + beta) squared + sin(alpha + beta) squared
         = 2 - 2cos(alpha + beta)
    QS squared = (cos(alpha) - cos(-beta)) squared + (sin(alpha) - sin(-beta)) squared
         = 2 - 2 cos alpha cos beta + 2sin alpha sin beta

So cos(alpha + beta) = cos alpha cos beta - sin alpha sin beta.

If you remember some trig and draw your own diagram you will see that it works.