How do you prove that #cos(x-y) = cosxcosy + sinxsiny#?

#cos (a - b) = cos(a)*cos(b) + sin(a)*sin(b)#
can be demonstrated by first showing that
#cos (a + b) = cos(a)*cos(b) - sin(a)*sin(b)#
and then doing the conversion using the CAST principle as indicated.

1. I'm sure there are other ways to do this; but this is what I came up with. (it is pretty long).
2. My apologies for using #a# and #b# instead of #x# and #y#; I drew the diagrams below before checking what variables had been used in the request.

Part 1 : Show #cos (a + b) = cos(a)*cos(b) - sin(a)*sin(b)#

A triangle XQP has been constructed along the hypotenuse of triangle XYQ with angle #a# above angle #b# as in the diagram.

The line segment XP is identified as the unit length for all measurements in this system.

A rectangle is constructed with base XY by extending the line from Y through Q until a point Z is reached where PZ is parallel to the bottom (XY) (completion of the rectangle establishes point W)

Within triangle XQP is is clear that (since #|XP| = 1#)
#|XQ| = cos(a)#
and
#|PQ| = sin(a)#

Therefore, in triangle XYQ
#|XY| = cos(b)*cos(a)# (#cos(b)# scaled up by the #cos(a)#)

Similarly in triangle QZP
#|PZ| = sin(a)*sin(b)#

Since WZ is parallel to XY (by construction)
angle XPW = angle PXY = #a+b#
and
#|WP| = cos(a+b)#

From the diagram
#cos(a+b) + sin(a)*sin(b) = cos(a)*cos(b)#
or
#cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)#

Part 2 : Show that if #cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)#
then #cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b)#

#cos(a-b) = cos(a + (-b))#
so we can substitute to get
#cos(a-b) = cos(a)*cos(-b) - sin(a)*sin(-b)#

By the CAST quadrant diagram for trig. signs (below) we can see that
#cos(-b) = cos(b)#
and
#sin(-b) = -sin(b)#

Therefore, we can write:

#cos (a - b) = (cos(a) * cos(b)) - (sin(a) * (-sin(b) ))#

or
#cos (a - b) = cos(a)*cos(b) + sin(a)*sin(b)#